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You are here: Home / SMPS and Converters / How to Build a Boost Converter Circuit: Explained with Calculations

How to Build a Boost Converter Circuit: Explained with Calculations

Last Updated on July 5, 2025 by Swagatam 186 Comments

I have explained comprehensively how to build a boost converter circuit for converting a low level DC voltage inputs to a higher level DC voltage outputs. I have furnished all the required calculations s that you can design a customized boost converter circuit.

Table of Contents
  • A Practical Boost Converter Circuit Design using IC 555
  • Important Parameters:
    • Calculating the Part Values
    • Finalized Part values for the above 555 boost Converter Circuit
  • Construction Steps (How to Assemble)
  • Adding a Feedback
    • Video Demonstration for IC 555 Boost Converter circuit
  • How a Boost Converter Works (Theoretical Analysis)
    • Operating Principle
  • Boost Converter Configuration
  • Modes of Operation
  • Calculating, Inductance, Current, Voltage and Duty Cycle in a Boost Converter
    • Inductor Current Build-Up (Transistor ON Phase)
    • Inductor Discharge (Transistor OFF Phase)
    • Steady-State Operation
    • Power Stage Calculation
    • Determining Maximum Switching Current
    • Calculating Ripple Current
    • Maximum Output Current Check
    • Maximum Switch Current
    • Inductor Selection
    • Inductor Ripple Current Approximation
    • Rectifier Diode Selection
    • Output Voltage Setting
    • Input Capacitor Selection
    • Output Capacitor Selection
    • Important Equations for Boost Converter Power Stage

A Practical Boost Converter Circuit Design using IC 555

This simple circuit is built using an IC 555 circuit for boosting USB 5V to 24V, or any other desired level. The same design can be used for boosting a 3.7 V to 24 V from a Li-Ion cell.

555 boost converter circuit practical design

Let's assume we have the following parameters required for the above circuit:

Important Parameters:

  • Input Voltage (Vin): 5V
  • Output Voltage (Vout): 24V
  • Output Current (Iout): 1A
  • Switching Frequency (f): 50 kHz

Calculating the Part Values

Step 1: Output Power

Pout = Vout * Iout = 24V * 1A = 24W

Efficiency (η) = 85%

Pin = Pout / η = 24 / 0.85 ≈ 28.2W
Input Current (Iin) = Pin / Vin = 28.2 / 5 ≈ 5.64A

Step 2: Inductor Selection

Inductance (L) = (Vin * D) / (f * ΔI)
D = 1 - (Vin / Vout) = 1 - (5 / 24) ≈ 0.79
Ripple Current (ΔI) = 0.4 * Iin = 0.4 * 5.64 ≈ 2.26A
L = (5 * 0.79) / (50000 * 2.26) ≈ 35µH

Choose an inductor with L = 35µH, capable of handling at least 6A peak current.

Step 3: Diode Selection

Reverse Voltage: ≥ Vout = 24V
Forward Current: ≥ Iout = 1A

Choose a Schottky diode, such as MBR360 (60V, 3A).

Step 4: Capacitor Selection

Output Capacitor (C) = (Iout * D) / (f * ΔV)

Assume Ripple Voltage (ΔV) = 1V:

C = (1 * 0.79) / (50000 * 1) ≈ 16µF

Choose a low-ESR electrolytic capacitor, such as 100µF, 35V.

Step 5: Resistor Selection for 555 Timer

Frequency (f) = 1.44 / ((R1 + 2R2) * C1)

Assume C1 = 0.01µF:

R1 + 2R2 = 1.44 / (50 * 103 * 0.01 * 10-6) = 2.88kΩ

Set R1 = 1kΩ, then 2R2 = 1.88kΩ, so R2 = 940Ω.

Use standard values: R1 = 1kΩ, R2 = 1kΩ.

Finalized Part values for the above 555 boost Converter Circuit

L1: 35µH, 6A

D1: MBR360 (60V, 3A Schottky)

Cout: 100µF, 35V (low ESR)

R1 & R2: 1kΩ each

C1 (timing): 0.01µF

T1 (switch): BD31 or better NPN transistor capable of 6A or a MOSFET (e.g., IRF540N)

Construction Steps (How to Assemble)

Step 1: Build the Oscillator Circuit

Assemble the 555 timer circuit on a breadboard or PCB.

Use the formula for the 555 timer frequency:

f = 1.44 / ((R1 + 2R2) * C1)

Select the appropriate values for R1, R2, and C1 to achieve the desired switching frequency (e.g50 kHz).

Connect the output of the 555 timer (pin 3) to the base or the gate of the switching transistor/MOSFET through the resistor (e.g 100Ω).

Step 2: Connect the Inductor and Switching Transistor

Solder the inductor (e.g.35µH) in series with the input voltage and the collector (or drain) of the switching transistor.

Connect the emitter (or source) of the transistor to ground.

Step 3: Add the Diode

Solder the Schottky diode (e.g MBR360) between the inductor and the output capacitor.

Connect the cathode of the diode to the positive terminal of the output capacitor.

Step 4: Connect the Output Capacitor

Add an electrolytic capacitor (e.g., 100µF 35V) across the output terminals for to smooth out the boosted DC voltage.

Step 5: Feedback/Control (Optional)

For adjustable output voltage you can include a feedback circuit with a variable resistor (potentiometer) to adjust the duty cycle of the PWM signal.

Step 6: Add Input and Output Terminals

Solder the connectors or terminals for the input voltage (Vin) and output voltage (Vout).

Step 7: Verify Connections

Double-check all the connections to ensure that there are no shorts or loose connections.

Testing the Circuit

Connect the 5V DC power supply to the input terminals.

Measure the output voltage using the multimeter to ensure that it reaches the desired value (e.g.,, 24V).

Adjust the duty cycle (if applicable,) to fine-tune the output voltage.

Connect the load (e.g, a resistor or a small motor) to the output and measure the current.

Assembly Tips

Use the heat sink for the transistor or MOSFET to dissipate the heat during the operation.

Keep the connections between the inductor, switch, diode, and the capacitor as short as possible to minimize the losses.

Use the printed circuit board (PCB) for getting a robust and reliable assembly.

Adding a Feedback

The above circuit can be regulated with a feedback as shown below:

555 boost converter circuit with feedback control

The idea looks quite straightforward. IC 555 is configured as an astable multivibrator whose frequency is decided by the values of resistors and capacitor at pin#7 and pin#6/2.

This frequency is applied to the base of a driver transistor TIP31 (incorrectly shown as BD31).

The transistor oscillates at the same frequency and forces supply current to oscillate within the connected inductor with the same frequency.

The selected frequency saturates the coil and boosts the voltage across it to a greater amplitude which is measured to be around 24V.

This value can be tweaked to even higher levels by modifying the turns of the inductor and the frequency of the IC .

Video Demonstration for IC 555 Boost Converter circuit

How a Boost Converter Works (Theoretical Analysis)

A boost converter is a kind of SMPS or switch mode power supply which fundamentally works with two active semiconductors (transistor and diode) and with a minimum of one passive component in the form of a capacitor or an inductor or both for greater efficiency.

The inductor here basically is used for stepping up the voltage and the capacitor is introduced for filtering the switching fluctuations and for reducing current ripples at the output of the converter.

The input power supply which may be required to be boosted or stepped up could be acquired from any suitable DC source such as batteries, solar panels, motor based generators etc.

Operating Principle

The inductor in a boost converter plays the important of stepping up the input voltage.

The crucial aspect which becomes responsible for activating the boost voltage from an inductor is due to its inherent property of resisting or opposing a suddenly induced current across it, and due to its response to this with a creation of magnetic field and subsequently destroying of the magnetic field. The destroying leads to the releasing of the stored energy.

This above process results in the storing of the current in the inductor and kicking back this stored current across the output in the form of back EMF.

A relay transistor driver circuit can be considered a great example of a boost converter circuit. The flyback diode connected across the relay is introduced to short circuit the reverse back EMFs from the relay coil and to protect the transistor whenever it switches OFF.

If this diode is removed and a diode capacitor rectifier is connected across the transistor's collector/emitter, the boosted voltage from the relay coil can be collected across this capacitor.

Boost converter block diagram

The process in a boost converter design results in an output voltage that’s always higher than the input voltage.

Boost Converter Configuration

Referring to the following figure, we can see a standard boost converter configuration, the working pattern may be understood as given under:

When the shown device (which could be any standard power BJT or a mosfet) is switched ON, current from the input supply enters the inductor and flows clockwise through the transistor to complete the cycle at the negative end of the input supply.

Boost converter switching device working

During the above process the inductor experiences a sudden introduction of current across itself and tries to resist the influx, which results in the storing of some amount of the current in it through the generation of a magnetic field.

At the next subsequent sequence, when the transistor is switched OFF, the conduction of current breaks, yet again forcing a sudden change in the current level across the inductor.

The inductor responds to this by kicking back or releasing the stored current. Since the transistor is in the OFF position, this energy finds its path through the diode D and across the shown output terminals in the form of a back EMF voltage.

Function of diode in a boost converter

The inductor performs this by destroying the magnetic field which was earlier created in it while the transistor was in the switch ON mode.

However, the above process of releasing energy is implemented with an opposite polarity, such that the input supply voltage now becomes in series with the inductor back emf voltage. And as we all know that when supply sources join in series their net voltage adds up to produce a bigger combined outcome.

The same happens in a boost converter during the inductor discharge mode, producing an output which may be the combined result of the inductor back EMF voltage and the existing supply voltage, as shown the diagram above

This combined voltage results in a boosted output or a stepped up output which finds its path through the diode D and the across capacitor C to ultimately reach the connected load.

The capacitor C plays quite an important role here, during the inductor discharge mode the capacitor C stores the released combined energy in it, and during the next phase when the transistor switches OFF again and the inductor is in the storing mode, the capacitor C tries to maintain the equilibrium by supplying its own stored energy to the load. See the figure below.

Function of PWM and load in boost converter

This ensures a relatively steady voltage for the connected load which is able to acquire power during both the ON, and OFF periods of the transistor.

If C is not included then this feature is cancelled resulting in a lower power for the load and lower efficiency rate.

The above explained process continues as the transistor is switched ON/OFF at a given frequency, sustaining the boost conversion effect.

Modes of Operation

A boost converter may be primarily operated in two modes: the continuous mode, and the discontinuous mode.

In continuous mode, the inductor current is never allowed to reach zero during its discharging process (while the transistor is switched OFF).

This happens when the ON/OFF time of the transistor is dimensioned in such a way that the inductor is always connected back quickly with the input supply through the switched ON transistor, before it’s able to get completely discharged across the load and the capacitor C.

This allows the inductor to consistently produce the boost voltage at an efficient rate.

In the discontinuous mode, the transistor switch ON timing may be so wide apart that the inductor may be allowed to get discharged fully and stay inactive in between the switch ON periods of the transistor, creating huge ripple voltages across the load and the capacitor C.

This could make the output less efficient and with more fluctuations.

The best approach is to calculate the ON/OFF time of the transistor which yields maximum stable voltage across the output, meaning we need to make sure that the inductor is optimally switched such that it’s neither switched ON too quickly which might not allow it to discharge optimally, and nor switch it ON very late which might drain it an inefficient point.

Calculating, Inductance, Current, Voltage and Duty Cycle in a Boost Converter

Here we’ll discuss only the continuous mode, which is the preferable way to operate a boost converter. Let’s evaluate the calculations involved with a boost converter in continuous mode:

Inductor Current Build-Up (Transistor ON Phase)

While the transistor is in the switched ON phase, the input source voltage (Vi) is applied across the inductor, inducing a current (IL) build-up through the inductor for a time period (t). This can be expressed as:

dIL/dt = Vi/L

By the time the ON state of the transistor ends and it switches OFF, the current built up in the inductor can be calculated using:

ΔIL(on) = Vi × D × T / L

Where:

  • D is the duty cycle.
  • L is the inductance value of the inductor (Henry).
  • T is the switching period.

Inductor Discharge (Transistor OFF Phase)

When the transistor is OFF, assuming the diode offers minimal voltage drop and the capacitor (C) is large enough to maintain a constant output voltage, the output current (IL) can be deduced as:

Vi - Vo = L × dIL/dt

The current variation (ΔIL) across the inductor during the discharge phase can be expressed as:

ΔIL(off) = (Vi - Vo) × (1 - D) × T / L

Steady-State Operation

Under steady conditions, the energy stored in the inductor during the ON phase equals the energy released during the OFF phase. This implies:

ΔIL(on) + ΔIL(off) = 0

Substituting the expressions for ΔIL(on) and ΔIL(off):

Vi × D × T / L - (Vi - Vo) × (1 - D) × T / L = 0

Simplifying:

Vo / Vi = 1 / (1 - D)

Or equivalently:

Vo = Vi / (1 - D)

This shows that the output voltage in a boost converter is always higher than the input voltage across the full range of duty cycles (0 < D < 1).

The duty cycle can be calculated as:

D = 1 - Vi / Vo

Power Stage Calculation

To calculate the boost converter’s power stage, the following guidelines are necessary:

  1. Input Voltage Range: Vin(min) and Vin(max)
  2. Minimum Output Voltage: Vout
  3. Maximum Output Current: Iout(max)
  4. IC specifications: Choose based on the IC’s datasheet.

Determining Maximum Switching Current

The first step is to calculate the duty cycle at the minimum input voltage:

D = 1 - (Vin(min) × η) / Vout

Where:

  • Vin(min) = minimum input voltage
  • Vout = output voltage
  • η = efficiency (e.g., 80%)

Calculating Ripple Current

The ripple current can be calculated as:

ΔIL = Vin(min) × D / (f × L)

Where:

  • f = minimum switching frequency
  • L = inductor value

Maximum Output Current Check

Verify the IC’s maximum output current:

Iout(max) = [Ilim(min) - ΔIL / 2] × (1 - D)

Where:

  • Ilim(min) = minimum current limit from the IC datasheet
  • ΔIL = ripple current

If the calculated Iout(max) is insufficient, choose a new IC with a higher current limit or increase the inductor value to reduce ΔIL.

Maximum Switch Current

The maximum switch current can be calculated as:

Isw(max) = ΔIL / 2 + Iout(max) / (1 - D)

Inductor Selection

For inductors:

L = Vin × (Vout - Vin) / (ΔIL × f × Vout)

Choose an inductor with a current rating higher than Isw(max).

Inductor Ripple Current Approximation

Approximate the ripple current as 20% to 40% of Iout(max):

ΔIL = (0.2 to 0.4) × Iout(max) × Vout / Vin

Rectifier Diode Selection

For the rectifier diode, use:

If = Iout(max)

The diode’s power dissipation can be calculated as:

Pd = If × Vf

Where:

  • If = average forward current of the diode
  • Vf = forward voltage of the diode

Schottky diodes are recommended due to their low forward voltage drop and high peak current rating.

Output Voltage Setting

Most converters set the output voltage using a resistive divider network (this may be built-in for fixed output voltage converters).

With the given feedback voltage Vf and feedback bias current Ifb, the voltage divider can be calculated. The current through the resistive divider should ideally be about 100 times the feedback bias current:

Ir1_2 >= 100 * Ifb  ---------- (9)

Where:

  • Ir1_2 is the current through the resistive divider to GND.
  • Ifb is the feedback bias current from the datasheet.

This ensures less than 1% inaccuracy in the voltage measurement. However, lower resistor values can increase power loss in the resistive divider, which may be unsuitable for high-efficiency designs.

With the above condition, the resistors can be calculated as follows:

R2 = Vf / Ir1_2  ---------- (10)

R1 = R2 * [(Vout / Vf) - 1]  ---------- (11)

Where:

  • R1 and R2 are the resistive divider resistors.
  • Vf is the feedback voltage from the datasheet.
  • Ir1_2 is the current through the resistive divider to GND, as calculated in Equation 9.
  • Vout is the desired output voltage.

Input Capacitor Selection

The minimum value for the input capacitor is typically specified in the datasheet. This value is crucial for stabilizing the input voltage under the peak current requirements of a switching power supply.

Low ESR ceramic capacitors are preferred, and the dielectric material should be X5R or better. This avoids significant capacitance reduction due to DC bias or temperature effects.

If the input voltage is noisy, the capacitance value may need to be increased.

Output Capacitor Selection

To reduce output voltage ripple, small ESR capacitors are ideal. Ceramic capacitors with an X5R dielectric or better are recommended.

For converters with external compensation, any capacitor value above the minimum specified in the datasheet can be used.

However, the compensation must be adjusted accordingly. For internally compensated converters, the recommended inductor and capacitor values should be followed, or the datasheet guidelines should be used.

For secondary compensation, the following equations can help adjust output capacitor values for a desired output voltage ripple:

Cout_min = Iout_max * D / (fs * ΔVout)  ---------- (12)

Where:

  • Cout_min is the minimum output capacitance.
  • Iout_max is the maximum output current of the application.
  • D is the duty cycle calculated using Equation 1.
  • fs is the minimum switching frequency of the converter.
  • ΔVout is the desired output voltage ripple.

The ESR of the output capacitor introduces additional ripple, calculated as:

ΔVout_ESR = ESR * [(Iout_max / (1 - D)) + (ΔIL / 2)]  ---------- (13)

Where:

  • ΔVout_ESR is the additional output voltage ripple caused by the capacitor's ESR.
  • ESR is the equivalent series resistance of the output capacitor.
  • Iout_max is the maximum output current of the application.
  • D is the duty cycle from Equation 1.
  • ΔIL is the inductor ripple current from Equation 2 or Equation 6.

Important Equations for Boost Converter Power Stage

Maximum Duty Cycle:

D = 1 - (Vin_min * n / Vout)  ---------- (14)

Where:

  • Vin_min is the minimum input voltage.
  • Vout is the desired output voltage.
  • n is the efficiency of the converter (e.g., 85%).

Inductor Ripple Current:

ΔIL = Vin_min * D / (fs * L)  ---------- (15)

Where:

  • Vin_min is the minimum input voltage.
  • D is the duty cycle from Equation 14.
  • fs is the nominal switching frequency of the converter.
  • L is the chosen inductor value.

Maximum Output Current of the Selected IC:

Iout_max = [Ilim_min - ΔIL] * (1 - D)  ---------- (16)

Where:

  • Ilim_min is the minimum current limit of the integrated switch (from the datasheet).
  • ΔIL is the inductor ripple current from Equation 15.
  • D is the duty cycle from Equation 14.

Application-Specific Maximum Switch Current:

Isw_max = (ΔIL / 2) + (Iout_max / (1 - D))  ---------- (17)

Where:

  • ΔIL is the inductor ripple current from Equation 15.
  • Iout_max is the maximum output current of the application.
  • D is the duty cycle from Equation 14.

Inductor Calculation:

L = Vin * (Vout - Vin) / (ΔIL * fs * Vout)  ---------- (18)

Where:

  • Vin is the nominal input voltage.
  • Vout is the desired output voltage.
  • fs is the minimum switching frequency of the converter.
  • ΔIL is the estimated inductor ripple current (see Equation 19).

Inductor Ripple Current Estimation:

ΔIL = (0.2 to 0.4) * Iout_max * (Vout / Vin)  ---------- (19)

Where:

  • ΔIL is the estimated inductor ripple current.
  • Iout_max is the maximum output current required.

Rectifier Diode Parameters:

  • Forward Current:
If = Iout_max  ---------- (20)
  • Power Dissipation:
Pd = If * Vf  ---------- (21)

Where:

  • If is the average forward current of the rectifier diode.
  • Vf is the forward voltage of the rectifier diode.

Resistive Divider Network:

  • Current:
Ir1_2 >= 100 * Ifb  ---------- (22)
  • Resistor R2:
R2 = Vf / Ir1_2  ---------- (23)
  • Resistor R1:
R1 = R2 * [(Vout / Vf) - 1]  ---------- (24)

Output Capacitance:

Cout_min = Iout_max * D / (fs * ΔVout)  ---------- (25)

Ripple Due to ESR:

ΔVout_ESR = ESR * [(Iout_max / (1 - D)) + (ΔIL / 2)]  ---------- (26)

These equation provide a comprehensive guide for designing and analyzing the power stage of a boost converter.

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Filed Under: SMPS and Converters Tagged With: Boost, Build, Calculations, Converter, Explained

About Swagatam

I am an electronics engineer and doing practical hands-on work from more than 15 years now. Building real circuits, testing them and also making PCB layouts by myself. I really love doing all these things like inventing something new, designing electronics and also helping other people like hobby guys who want to make their own cool circuits at home.

And that is the main reason why I started this website homemade-circuits.com, to share different types of circuit ideas..

If you are having any kind of doubt or question related to circuits then just write down your question in the comment box below, I am like always checking, so I guarantee I will reply you for sure!



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Reader Interactions

Questions & Answers

Total Posts: 186 (Older Threads Archive)
Newest Oldest
g
May 26, 2018 • 8 years ago #60688

hello, how can i make the Simple Boost Converter using a single BJT circuit work with a solar panel to charge the battery?

Reply
SwagatamAdmin
May 26, 2018 • 8 years ago #60689

Hi, please see the diagram at the bottom of this article, it might be just what you are looking for

https://www.homemade-circuits.com/simplest-automatic-led-solar-light/

Reply
ge
June 2, 2018 • 8 years ago #60759

thanks for the reply.
however i would like to use 1 battery instead of 3 to keep the circuit as small as possible.

Reply
SwagatamAdmin
June 2, 2018 • 8 years ago #60762

All the circuits referred here use a single battery. So perhaps you can try the first design from the above article

Reply
Corne
December 15, 2021 • 5 years ago #107345

Good day Swag

I am looking for a diagram to build that can do the following. Dc booster
I got solar panels that is 53 volt and an element rated 48v resistance 7 ohm. If i connect directly it pull the panels down to 3volt.i dont want to buy a geyserwise mppt just for this.

Reply
SwagatamAdmin
December 15, 2021 • 5 years ago #107353

Hi Corne, a boost converter will not solve your problem, neither is it required.

A 48V 7 ohm element would require a current of 48/7 = 6 amps. So is your 53V panel rated to provide this much current? You can check this by momentarily connecting an ammeter across the solar panel during peak sunlight.

Your element is rated at 48 x 6 = 288 watts, so you will need a 300 watt panel to ensure that the voltage does not drop too much.

Reply
Shigida
May 31, 2018 • 8 years ago #60733

Good afternone sir!
I would like to use this boost converter (the first picture) to lit the LED at night and
as a charger (at day) of 4.5 v cell (3×1.5 cell) .The LEd position will be also the position of cells .And on the position of the cell (on the above picture) will come a small solar panel.can you show me how to do a current limiting that can be used for both task (maybe by useing poti?).
Thankyou in advance!

Reply
SwagatamAdmin
June 1, 2018 • 8 years ago #60744

Hello Shigida, I think for this low current application you could probably limit the current with a resistor in series with the battery positive or negative terminal. The resistor value could be calculated using Ohms law:
R = V/I
where V will be output voltage minus battery voltage and I will be the safe charging current of the battery. Make sure the max output is slightly lower than the battery’s full charge level.

Reply
Godson
July 13, 2018 • 8 years ago #61678

Hello sir Swagatam,
Thanks a lot for this post.
I would like to use the IC555 version to boost 3.7V Li ion battery to 12V. What modifications do I need to make to achieve this?
Will the output be suitable for powering circuits that have ICs in them?

Reply
SwagatamAdmin
July 14, 2018 • 8 years ago #61681

Thanks Godson, you can use the same design as shown in the above article

Reply
Raghavendra
September 23, 2018 • 8 years ago #64119

Thank you sir for replying

Reply
Norman Kelley
October 3, 2018 • 8 years ago #64450

Hi Swagatam! I have tested the above circuit which uses a 555 timer and the output is 38.6 volts using a 100uH through hole fixed inductor. I tried changing the pin 6 to pin 7 resistor to 10K with no change. I have 22uH, 47uH, 68uH, 100uH, 220uH, 330uH 470uH, 1MuH inductors. I tried 47uH inductor and the output voltage was 38.9v. I tried 220uH and the output voltage was about 37.8v. I tried 3v, 6v, 9v, 12v input with almost the same results. It just takes longer to reach max voltage as the supply voltage is reduced.What should I modify to get 20-24v output? I only have 20volt 1/2 watt Zener available. I looked at other posts which use a Zener to control the output voltage but they require a 1-watt Zener. Thanks!

Reply
SwagatamAdmin
October 3, 2018 • 8 years ago #64452

Hi Norman, you can adjust the voltage either by changing the number of turns of the inductor or simply by adjusting the 1K value between the pin#6 and 7.

Reply
Grace
December 23, 2018 • 8 years ago #64691

Thanks Swag, but can 3.7v lithium battery power the 555ic. You said +5v to 12v input.

Reply
SwagatamAdmin
December 24, 2018 • 8 years ago #64699

Hi Grace, yes you are right, normal 555 ICs will not work with 3.7V but the CMOS version (7)555 can be used which is rated to work with minimum 3V

Reply
Grace
December 23, 2018 • 8 years ago #64692

Please for the bjt booster , how can I reduce the voltage to 7v instead of 30v as designed to charge a smart phone

Reply
SwagatamAdmin
December 24, 2018 • 8 years ago #64702

sorry, you cannot use a 1.5V AAA cell to charge a smart phone.

Reply
BIJU K M
January 1, 2019 • 8 years ago #64785

What is the advantage to get add on a transistor BC547 feedback circuit? Please give me reply….

Reply
SwagatamAdmin
January 2, 2019 • 8 years ago #64786

it is for keeping the output voltage restricted to a specified limit, and preventing any rise beyond that limit.

Reply
BIJU MICHAEL
January 2, 2019 • 8 years ago #64790

Thanks for yours valuable reply.Yours support in electronic science field really dedicated…god bless
.

Reply
SwagatamAdmin
January 2, 2019 • 8 years ago #64791

It is my pleasure Biju!

Reply
Paul
March 2, 2019 • 7 years ago #65712

Hello,

Sir can i use N-channel mosfet instead of transistor? for T1 & BC547?
is there any modifications for this?
Thanks a lot..

Reply
SwagatamAdmin
March 2, 2019 • 7 years ago #65713

Hello Paul, mosfets will require minimum 9V to turn ON fully, so it cannot be in these applications.

Reply
Paul
March 2, 2019 • 7 years ago #65714

Thanks for the reply then can i use 13003 transistor for T1?

Reply
SwagatamAdmin
March 2, 2019 • 7 years ago #65717

I am sorry Paul, for BC547 you can use a mosfet since the gate would be receiving a boosted 12V so no problem.. MJE13003 won’t be suitable here!

Reply
Paul
March 2, 2019 • 7 years ago #65724

I cant find any TIP31 Its hard to find here in our place but i found TIP41 & TIP42, can it be used here in the circuit ?
Thanks…

Reply
SwagatamAdmin
March 3, 2019 • 7 years ago #65727

TIP41 is fine, it will work…

Reply
Paul
March 8, 2019 • 7 years ago #65800

Sir Can I use 100nf cap instead of 680 pf? i cant find that capacitor.. and 1n4007 diode instead of FR107.
thanks again..

Reply
SwagatamAdmin
March 8, 2019 • 7 years ago #65801

Hi Paul, There’s a huge difference between 100nF and 680pF, so it won’t work. You can try the first circuit instead.

Reply
John Smith
June 17, 2019 • 7 years ago #67730

Hmm, if Paul been asking about replacing T1 BJT with 13003 it would probably work, since 1300x family is just (high-voltage) NPN BJTs, not anyhow worse than any others. They are pretty typical in CCFLs and dumbest “electronic transformers”, esp these for “halogen lamps” and even some cheap SMPS designs.

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SwagatamAdmin
June 17, 2019 • 7 years ago #67731

It will work but with poor efficiency, that’s the reason we have such a huge range of devices designed for different voltage or current specs, otherwise the manufactures could have designed a single all purpose 1kv transistor : )

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John Smith
June 18, 2019 • 7 years ago #67735

Yes, but if one goes for joule thief or 555-driven circuit, especially with BJT as switch and FR107 as diode, they don’t have to expect superb efficiency or awesome output power in compact package, right?

This said, joule thief is funny thing to power small led or so out of 1.5 volt battery. Only few relatively exotic and expensive DC-DC ICs would start up from voltages that make “joule thief” happy.

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SwagatamAdmin
June 18, 2019 • 7 years ago #67744

Actually a joule thief circuit is supposed to be extremely efficient, that’s why it’s called a “joule thief”. Yes, it’s been one of the most interesting inventions so far, considering the fact that it can work even with voltages lower then 0.5 V.

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John Smith
June 18, 2019 • 7 years ago #67757

Joule thief key property that makes it interesting is that it runs down to like 0.3V or so, draining 1.5V batteries way below what most of other electronics could afford. Say, most CMOS-based ICs have a problem that 0.3V is way below of threshold voltage. Even “special” boost ICs like NCP1400 would start up from like 0.8V – then it would hold down to like 0.3V as well – because they start up, and then supply self out of own boosted output. However it implies startup problems if battery got below 0.8V and “cold” start happens.

This said, joule thief is kind of “blocking oscillator” – so it got no reasons to be terribly efficient. Waveforms aren’t really perfect, BJT drops some tenths of volt on C-E junction even if current is small, etc (FETs get edge in this regard generally, but most MOSFETs got threshold over 1V, so running less than 1V is a problem, lowest I know have Vg_th=0.8V). However joule thief gets its edge over many other circuits when it comes to using 1.5V batteries, draining them way more completely compared to most other circuits. So it would start and work out of batteries most equipment considers long “dead”. That’s where it gets extra power margin – partially negating its imperfections, so overall performance looks rather good.

p.s. I’ve finally got right one of these 10-year-led-blink, without getting it working I would have felt ashamed way too much – failing such a simple thing is a LOL. Grossly simplifying one (as suggested on youtube comments – just thief, cap and resistor). I’ve used 5.6uF 1206 SMD cercap, 5.1M 0805 resistor, PMBS3904 in sot 23 and hi-eff blue led – overall I’ve got shy 10uA average current while still getting very persuading bright blue flash about once in couple of seconds. I’ve also found funny enclosure for this little cheat, making it look like part of office alarm system, haha. A perfect joke for few vandal-unsafe places, granted most expensive part is battery and enclosure. And according to my computations battery would last … for hell knows how long, at 10uA average it would rather self discharge I guess. That’s probably best use of joulr thief I’ve faced to the date XD

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adofo
April 4, 2019 • 7 years ago #66293

adofo I have 5v 500ma solar panel. can I use your booster convert to run car tape

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SwagatamAdmin
April 4, 2019 • 7 years ago #66295

5 x 0.5 = 2.5 watts that’s too less for a car tape, it won’t work

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damilare
April 26, 2019 • 7 years ago #66677

sir I made this circuit but its not working I can’t find 680pf but am using 2A682k then I cant get tip31 and am using tip41 while testing it its just sparking as if I shot the circuit of the battery then the transistor is getting hot please help me

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SwagatamAdmin
April 26, 2019 • 7 years ago #66678

damilare, 2A682k = 6800pf, it’s not 680pF

In your present set up increasing the number of turns to 5 times more than the shown value, and check the results. Wind it on a ferrite rod.

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faith jumbo
June 19, 2019 • 7 years ago #67759

Goodday sir I really want to appreciate you for good work thank you very much please keep up

Sir please I build a boost converter using 555 timer to drive my mosfet, and I use a mosfet in place of the transistor but each time I power it up the mosfet get really hot please what could be the cause I have checked but can find any fault in the circuit

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SwagatamAdmin
June 19, 2019 • 7 years ago #67763

Hello faith, keep the frequency at around 50 kHz, and test by gradually increasing the turns until the MOSFET stops heating, this is perhaps the easiest trial and error way to optimize a boost converter.

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Faith
June 19, 2019 • 7 years ago #67766

Which turns is it the number of turns i.e the inductor

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SwagatamAdmin
June 19, 2019 • 7 years ago #67767

There’s only one component in 555 circuit which has turns, it’s L1

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Faith
June 23, 2019 • 7 years ago #67865

Sir I have seen my mistake all this while the diode was connected wrongly but now is working very well thanks for your assistance

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SwagatamAdmin
June 23, 2019 • 7 years ago #67870

Glad you could solve it!

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Faith
June 20, 2019 • 7 years ago #67791

Sir I did exactly what you said it still the same I noticed that when ever I connect the power supply it like both positive and negative are connected together it al hiways spark and if I try leave it for some second the wire Wil burnt out but if i remove the inductor it will not happen, and the output power is not boost please what could be the cause. The inductor I’m using is wound on a toroid core

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SwagatamAdmin
June 20, 2019 • 7 years ago #67793

How many turns did you use? and what at what frequency did you set your IC 555?

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Faith
June 23, 2019 • 7 years ago #67863

Sir I’m really confused here, I even try building the boost without the 555, i.e the simple circuit were I have just inductor, diode pull button switch and capacitor, when ever I pushed the switch the inductor Wil connect it to ground i.e shorting the circuit why this happening

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SwagatamAdmin
June 23, 2019 • 7 years ago #67868

The switching should be in millisconds, if you hold it even for 0.25 seconds it will create a short circuit. You can read the following article for more info:

https://www.homemade-circuits.com/how-boost-converters-works/

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Parsiva
July 4, 2019 • 7 years ago #68165

Hello sir
For giving the gate pulse to the BJt can i give it through Arduino?
like pwm technique

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SwagatamAdmin
July 4, 2019 • 7 years ago #68178

Hi Parsiva, you can do that, just make sure that the duty cycle and the frequency are optimally adjusted.

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marcelo
July 5, 2019 • 7 years ago #68212

Good evening sir,
very difficult to leave your pages. Your information is always precious.
Transistors in Darligton mode would lift me the current in these joule thief if powered by solar cell?
If not, is there any other way to get a current gain?
If it were a buck converter would there be a way to take advantage of the current of the dissipated voltage?
Thank you very much in advance.
A lot of Swag light

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SwagatamAdmin
July 5, 2019 • 7 years ago #68216

Thank you Marcelo, appreciate your interest very much!

A Darlington may help to increase the current transfer response. However, using many thin wires together for the winding may also help to gain more power from the design. A joule will normally work like a boost converter so not sure how it may be turned into a buck converter…but I don’t think this concept can be used effectively used for charging bigger batteries.

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BERNARD TENDENGU
October 31, 2019 • 7 years ago #71582

Dear Swag,
I do have a 12volt @75 amp dc battery and do intend to use to power about five led corn light bulbs and rating for each is 359 milliamps and wattage for each is 26 watts. I will be using the constant current method to drive the led driver to power the lead lights. The led driver can be a buck boost converter, sepic dc dc converter or cuk dc dc converter. The problem I am having is that of stepping down the high dc current from 75 amps to manageable one like say 13 amps. If it were ac it would be easier. What I am requesting from you is a general method which I can apply.

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SwagatamAdmin
October 31, 2019 • 7 years ago #71589

Dear Bernard, what is the voltage spec of the LED, there’s no need of a buck boost circuit, you can directly use the supply for the LEDs after configuring the LEDs appropriately.

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Hassan Khan
November 29, 2019 • 7 years ago #72592

can we use this circuit(555) to boost voltage from solar panel (12v to 15v)?

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SwagatamAdmin
November 30, 2019 • 7 years ago #72613

yes, it is possible.

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Miracletech
December 16, 2019 • 7 years ago #73113

Dear sir Swag, truly you have swag! I have been looking for this information for years but thanks to you I found it! Also, can the resistor be added after the toroidal core? Like directly at the base of the transistor?

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SwagatamAdmin
December 16, 2019 • 7 years ago #73128

Thank you Miracletech, yes you can try that for the first diagram. For more options you can also refer to this article:

https://www.homemade-circuits.com/1-watt-led-driver-using-joule-thief/

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miracletech
December 29, 2019 • 7 years ago #73685

Ok. Thank you so much.

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Andrew
May 31, 2020 • 6 years ago #79153

Please is there any other pwm oscillator ic I can use instead of 555 or can I equally use the popular viper22A or Rm6203 used for flyback converters, will it work.

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SwagatamAdmin
June 1, 2020 • 6 years ago #79163

You can use TL494 IC or similar, but VIPER can be difficult since it is configured to work with high voltage rectified AC.

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Andrew
June 20, 2020 • 6 years ago #79672

Ok I’ve just made the circuit, and am continuously having this problem of magnetostriction. According to my research, I find out that it is due to low frequency switching. So please can you tell me how I increase the switching frequency, and also the factors and components value that affect the frequency of operation.

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SwagatamAdmin
June 20, 2020 • 6 years ago #79679

If you are referring to TL494 circuit, you can change its Rt, or Ct values to change its frequency.

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Andrew
June 20, 2020 • 6 years ago #79681

No it’s the first version of the joule thief circuit.

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SwagatamAdmin
June 21, 2020 • 6 years ago #79691

Did you confirm the frequency with an oscilloscope? Since it’s a self oscillatory type circuit frequency is pretty much fixed. For higher frequency you can probably try the following version:

8X Overunity from Joule Thief – Proven Design

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Ordu Prudence
August 12, 2020 • 6 years ago #81274

Good day sir please for the first diagram I can’t really find a 1000uf capacitor and also the 0.0045uf
So can I try using 220uf and 22uf

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SwagatamAdmin
August 13, 2020 • 6 years ago #81296

Ordu, 1000uF can be replaced with 220uF, but the 0.0047uF value cannot be altered, you can adjust the value through series parallel capacitors if required.

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rudi
September 30, 2020 • 6 years ago #82731

Mr Swagatam, I was still wrong in wrapping the enameled wire on the toroid, could you give an example in the form of a picture, but not a symbol

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SwagatamAdmin
September 30, 2020 • 6 years ago #82742

Mr.Rudi, the winding is not critical at all. Just wind the two sides anyway you want, and after connecting it to the circuit if the circuit doesn’t work, just interchange the wire ends of any one of the winding, and your circuit will start working instantly.

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Eroka
October 2, 2020 • 6 years ago #82833

Good evening sir,pls I we like to learn more on dis DC 1.5 converter to DC 12 voit

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Vernon Tritchka
January 7, 2021 • 6 years ago #85954

Good morning
Working on a project that includes “boosting” 9v DC (battery) to 50 to 100v DC.
We’re not sure of the circuit design or components. We want it to be able to recharge for follow on use. (all going in a small package).
Have you done such a project that you could permit us to reference?
We appreciate the assistance.

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SwagatamAdmin
January 8, 2021 • 6 years ago #85968

Hello, you can try the first concept presented in the following article:

https://www.homemade-circuits.com/12v-car-laptop-charger-circuit-using/

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Vernon Tritchka
February 9, 2021 • 5 years ago #86849

Mr. Swagatam
I like your “simple boost circuit. I appreciate your response, the 12 v laptop charger circuit, I’ll study it. I find now that we need to get up to approx. 200 volts output from 9 volts input. I still need a bit of couching. Thanks again. Vernon

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SwagatamAdmin
February 9, 2021 • 5 years ago #86864

You are welcome Mr. Vernon!
For higher voltages you can using a small 9-0-9V 220V 100mA transformer, and use it in place of the inductor shown in the IC 555 circuit. This is hopefully help to accomplish the required results quickly!

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Seun
March 3, 2021 • 5 years ago #87514

Glad to find this article, Well done Swagatam.
Please I want a boost converter design in which I will use 100w/12v panel to charge 4*18ah/12v 48v system.
Thanks

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SwagatamAdmin
March 4, 2021 • 5 years ago #87521

Thank you Seun, I think the following design will be more suitable for your application:

https://www.homemade-circuits.com/high-power-dc-to-dc-converter-circuit-12-v-to-30-v-variable/

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ron
March 21, 2021 • 5 years ago #87846

hello im interested in making a earth battery im asking can i use a joule thief to take the volts form 0.5 to 12 volts and how to get more amps

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SwagatamAdmin
March 21, 2021 • 5 years ago #87850

Hi, Either you get amps, or volts, you can never get them together for free…

for an earth battery you may have to integrate huge number of plates in series parallel to get a decent power output

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Cris Paltenghe
March 24, 2021 • 5 years ago #87919

Swagatam, in your article on boost converters, specifically diagram “3.7 to 24V converter”, there is a 10K resistor on the output side. I can’t figure out what its purpose is. It appears that it would bleed energy from the output capacitor. What is its purpose?
Also, when does the diode conduct? It appears it conducts when T1 is off. When T1 is on the coil conducts, right? Please explain a little more where the “boost” comes from.
Thank you, regards, Cris

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SwagatamAdmin
March 24, 2021 • 5 years ago #87928

Cris, yes the 10k is the bleeder resistor, and will act like a load when no real load is connected.
When the transistor is ON, the coil keeps storing the charge until it has reached its peak voltage specification. Then when the transistor switches OFF, this stored peak voltage is fired back as a boosted voltage.
The peak voltage depends on the number of turns, higher turns will generate higher voltage, and proportionately lower current.

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Carlos
March 27, 2021 • 5 years ago #88030

Hello good afternoon, what is the output amperage of the circuit with Lm555 ..

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SwagatamAdmin
March 28, 2021 • 5 years ago #88050

It is 1 amp because TIP31 is used, you can try TIP35 for higher amps

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ningrat
October 14, 2021 • 5 years ago #101326

hello sir,
sir, if i use the first circuit “Simple Boost Converter using a single BJT” with 3.7volt input is there any component values ​​that need to be replaced?
is this circuit efficient for long time use?

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SwagatamAdmin
October 14, 2021 • 5 years ago #101373

Hello Ningrat, for 3.7V you don’t have to change any components in the first diagram, however the BJT specs actually depends on the load specs.

Yes the circuit is efficient for a long term use.

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Bob horseman
October 23, 2021 • 5 years ago #102315

Hello there can i use a peice of ferrite rod instead of a ferrite toroide.

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SwagatamAdmin
October 23, 2021 • 5 years ago #102344

yes you can do that!

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bob horseman
October 23, 2021 • 5 years ago #102357

Cheers mate that will save me from having to wait a week for delivery of toroid core.

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SwagatamAdmin
October 23, 2021 • 5 years ago #102359

Sure, no problem, and wish you all the best!

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Doctor B
June 11, 2022 • 4 years ago #125067

Please email me via dr.blackschleger@gmail.com so i can share a circuit photo with you, and,
happy Fathers’ Day, Professor Swagatam !

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SwagatamAdmin
June 11, 2022 • 4 years ago #125263

Thank you Doctor B, My email is homemadecircuits @ gmail . com

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Sharon
June 18, 2022 • 4 years ago #127843

These circuits look nice, and the principle is explained well.
Is there any circuits that can boost low voltages?
From as low as 30mV to 5V, other than using an LTC3108.
Or from 0.8V to 5V

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SwagatamAdmin
June 18, 2022 • 4 years ago #127921

Yes boosting 0.8V can be possible through a joule thief circuit. You can try the first circuit explained in the following article:

3 Best Joule Thief Circuits

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Sharon
June 20, 2022 • 4 years ago #128100

Thank you, I don’t really understand why, but I’m looking through the article.
Is it possible to boost from as low as 0.1V. without using a 1:100 current transformer (cause that’s not available in my region)

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SwagatamAdmin
June 20, 2022 • 4 years ago #128114

Without a transformer that can be impossible. Moreover I think 0.4V is the last limit which can be boosted using a joule thief circuit….0.1 is too low for boosting

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Alimon Yokoniya
August 30, 2022 • 4 years ago #132780

sir i want to know if it is possible to connect phone batteries together either in series or parallel. If it is possible i have a plan to connect four of them in parallel each of 2500mAH. But i want you to help me with a circuit that can be able to charge it automatically

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SwagatamAdmin
August 31, 2022 • 4 years ago #132788

Hi Alimon,
You can connect the batteries in parallel. However making an automatic charger can be difficult for a newcomer, therefore I would suggest using a high current LM317 based power supply and adjust its output to exactly 4.1 v so that over charging of the battery can be avoided.
4 batts in parallel would mean a total Ah rating of 10 Ah which will require a charging current of 5 amp. You can use a 9V 5 amp transformer to build the voltage/current source for the LM317 circuit.

You can try the following circuit for the mentioned purpose. Use only 2nos of 2N3055 in the circuit, that will be enough:

high current LM317 power supply circuit

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Moses
September 16, 2022 • 4 years ago #133263

Sir, I thought the minimum supply voltage for lm555 is 4.5v, how then is it able to boost 3.7 lion battery to 24volts. Is it really possible that way.

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SwagatamAdmin
September 16, 2022 • 4 years ago #133276

Moses, you can use the CMOS 7555 IC which are rated to work from 3 V onward.

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Abdul
April 5, 2023 • 3 years ago #141631

Yes I have recently made did experiment .by using mje 13003 transistor and 10k resistor and capacitor 35 volt 2200 uf. And the input is 3.7 while the output replect about 28 volt DC.but de battery did not last at all

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Philip
December 1, 2022 • 4 years ago #136762

Hi, what could you suggest I use to boost a dc 2.0 volt RSSI (relative signal strength indicator) voltage up to about 5 dc volts so I can run a bar graph module. This to show how much RF signal my radio is getting.

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SwagatamAdmin
December 1, 2022 • 4 years ago #136789

Hi, you can try the second or the third circuit from the above article, it should work for your application.

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Nedi Macalane
December 5, 2022 • 4 years ago #136916

Hi sir, thanx for wonderful circuits and your knowledge to, am my question is what if I use tip41c instead of BC 547 will the circuit function properly?

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SwagatamAdmin
December 5, 2022 • 4 years ago #136919

Thank you Nedi,
I don’t think TIP41C would work correctly instead of BC547, because BC547 has high gain and requires very little current to switch ON, while TIP41 might need a lot more current to switch ON. This might need modifications for the base resistors of the transistor.

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Vincent
March 29, 2023 • 3 years ago #141473

For the flyback circuit could I use 5v input instead of 1.5v?

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SwagatamAdmin
March 29, 2023 • 3 years ago #141476

You can use 5V, but then the output would be above 24V

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Vincent
March 31, 2023 • 3 years ago #141501

That is what I need, thank you

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SwagatamAdmin
March 31, 2023 • 3 years ago #141504

You are welcome! Glad you found the post helpful.

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Olusegun
April 14, 2023 • 3 years ago #141789

Please I need circuit diagram for 12vdc to 400vdc converter

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SwagatamAdmin
April 14, 2023 • 3 years ago #141790

You can try the following design:

https://www.homemade-circuits.com/how-to-convert-12v-dc-to-220v-ac-using/

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Breno1
May 24, 2023 • 3 years ago #142808

Hi @Swagatam. I really need your help with a boost converter circuit that I can use in my car to power up my laptop. The laptop takes in 18.5vdc at 3.5amps. thanks

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SwagatamAdmin
May 25, 2023 • 3 years ago #142834

Hi Breno1, You can try the following design. The inductor can be built by winding 25 or 30 turns using 1 mm super enameled copper wire over a ferrite core. The ferrite core can be a ferrite rod or a ferrite ring. The 24V zener can be replaced with a 20 V zener diode
switching solar boost converter circuit

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Breno1
May 26, 2023 • 3 years ago #142877

Thanks a lot for the circuit @Swagatam.
1. I have two inductors each has 15 turns, will it work if I connect them in series to get 30?
2. About the capacitors. Am I to us same type i.e polarized electrolytic capacitors?

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SwagatamAdmin
May 26, 2023 • 3 years ago #142883

@Breno1,
1) No two series coils will not work, you must build the coil over a single ferrite core
2) The nF capacitors can be disc ceramic type, uF capacitor can be an electrolytic.

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Chidonlite
June 28, 2023 • 3 years ago #143757

Good day sir,
Please can you help me on how to build 5V 2AMP, step up converter that works on voltage from 3v to 4.2v.
Using 50, or 100khz transistor oscillator.
I have a 5,000mah li-lon battery to power it

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SwagatamAdmin
June 28, 2023 • 3 years ago #143760

Hi Chidonlite,
I think you should try the first circuit from the above article. You can tweak and adjust the 12V zener diode to get the desired voltage at the output. For 2 amp current, the 2N2222 can be replaced with a TIP31 and the BC557 with a 2N2907…

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Chidonlite
June 29, 2023 • 3 years ago #143766

What will I change in the circuit if I replace the Bc557 with A1015 bjt

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SwagatamAdmin
June 29, 2023 • 3 years ago #143774

You can try it, no changes will be required.

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Magnalite
August 22, 2023 • 3 years ago #144838

can i put a coil of copper wire instead of an inductor

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SwagatamAdmin
August 22, 2023 • 3 years ago #144839

Yes copper coil will do but the inductance has to be matched correctly

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Sydney
June 13, 2024 • 2 years ago #152872

I’ve been following your educative posts and used most of them on my projects successfully. I thank you for that most sincerely. Currently, my question is; I want to boost 48v from a SMPS to 54v to charge two 48v 100AH connected in parallel ie these are EMS intelligent batteries. I need a schematic diagram.preferebly and possibly if this converter out put can be variable.Thanking you in advance.

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SwagatamAdmin
June 13, 2024 • 2 years ago #152880

Thank you for your kind feedback, glad I could be helpful.
I think you can try the first circuit from the following article, however this circuit will need to be experimented and modified appropriately as per the 48V input and 54V output.
The output can be made variable by removing the zener diode and the 10k, and replacing them with a potentiometer instead.
https://www.homemade-circuits.com/12v-car-laptop-charger-circuit-using/

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Maddy
June 17, 2024 • 2 years ago #152935

Sir, I have a 19V, 2.1A Laptop adapter! Please suggest me any boost converter circuit that can be made using locally available cheap components to up-convert above DC supply to 24V, atleast 1A DC! For one week or so, I have become a fan of your site. The more I dive into different articles, the more curious I become about the extent of knowledge still on your site but hidden from my eyes.

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SwagatamAdmin
June 17, 2024 • 2 years ago #152936

Thank you so much Maddy, I am glad you found the site helpful.

I think you should try the first circuit design from the following article:
https://www.homemade-circuits.com/12v-car-laptop-charger-circuit-using/
You will have to experiment a little with the coil to get the perfect desired results.
Let me know if you have any problems.

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Maddy
June 17, 2024 • 2 years ago #152956

Sir thank you for your quick response! I’ve a bit of problem here! As per your advise, the circuit should have a IRF640 but can I use IRF3205 there?

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SwagatamAdmin
June 18, 2024 • 2 years ago #152966

Hi Maddy, no issues, you can use any appropriate MOSFET whose voltage and current ratings are higher than the input/output ratings of the circuit.
So, an IRFF3205 would be perfectly fine here.

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Maddy
July 1, 2024 • 2 years ago #153347

All components gathered!! Except Ferrite core. I searched but ended with whooping prices!! Sir can I prepare a core using Powdered Iron as seen over the Internet!! Just using the term “DIY Ferrite core alternatives” gives you different sites that show you how one can make one using cheaper materials. What opinion do you have about this? Or the same cmt be made using Air core too?

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SwagatamAdmin
July 2, 2024 • 2 years ago #153352

Thanks Maddy,
Referring to the following diagram:
laptop charger 12 V to 19V converter circuit
Ferrite core will make the inductor a lot smaller and compact, whereas the same inductor can be quite bulky if an iron core is used and also for an iron core the frequency will need to be drastically reduced.
I don’t think there’s any cheaper alternative to ferrite cores, so it is not possible find any suitable alternative.
Air core is possible which will be even bigger in size than an iron core inductor.
If you are sure you can make a proper iron powder core, you can try it, but then the winding might need to be increased 5 times more and the frequency reduced proportionately.
The frequency can be reduced by increasing the base capacitors of the BC548 transistors.

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Maddy
July 2, 2024 • 2 years ago #153356

Sir just to get a ferrite core, I’ve extracted a torroidal ceramic magnet from small speaker. having 2″ external and 1″ internal dia. I could not find any other solution over the Internet for 3″ long, 1cm dia ferrite rod. The available ones were very small in size. Now I’m planning to demagnetize above torroid using AC current using shielded wires. I don’t know how long it wud take. But in the mean time, if you do know of any such cheap rod being sold online, please give me the link! Thanx again, Sirji !!

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SwagatamAdmin
July 2, 2024 • 2 years ago #153358

Hello Maddy,
the ferrite rod size is not critical at all, you can choose any standard ferrite rod, or even a standard ferrite ring can be used.
I don’t think a magnet can be converted into an inductive core by demagnetizing it.

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Maddy
July 13, 2024 • 2 years ago #154598

Sirji, I finally got my ferrite core at a decent price from a radio technician. Before putting everything to PCB, I’ve a final hurdle to remove! I wud be using this booster to power a small RF amplifier ckt. So the question is, should I shield above inductor inside some Mu-metal or steel type enclosure to prevent the noise from this booster from disturbing my RF amplifier ckt? Or the simplest method to do this is to keep both ckts far away from each other. Or another way is just to keep the inductor away! And if this is done, is there any effect over the output voltage? Please advise!! Thanx again!!

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SwagatamAdmin
July 14, 2024 • 2 years ago #154666

Maddy, Referring to the following booster converter circuit, this is suitable for high power loads, if your load is a small RF amplifier then this circuit may be an overkill:
laptop charger 12 V to 19V converter circuit
Regarding the RF noise, it can be eliminated by using an large filter capacitor across the supply terminals of your RF amplifier, in addition to the one which is already included at the boost converter circuit output.
Let me know if you have any further doubts.

Reply
Maddy
July 14, 2024 • 2 years ago #154671

Sirji! Thnx for the advise! But I’ve used some wrong terms here! Actually I was asking for the disturbance due to the magnetic field of the inductor to the Audio content being amplified in RF range! Presently the ckt is perfectly working with 19v adapter! But I was intending to raise the supply up to 24v. Searching the net I came across various methods said before. So I asked for your expert advise!! Thanx again!

Reply
SwagatamAdmin
July 14, 2024 • 2 years ago #154677

Thanks Maddy, glad to know the circuit is working for you.
Yes, in that case it is better to encapsulate the ferrite inductor inside a Mu-metal enclosure.
Let me know how it goes.

Reply
Maddy
July 20, 2024 • 2 years ago #155424

Sirji I have a small question! Most of the commercially available booster ckts have variable output facility with them. Can we have any modifications for our booster ckt have this freedom! Say by adding a pot at any stage to vary the output. This wud make it more versatile! Thanx in advance!!

Reply
SwagatamAdmin
July 20, 2024 • 2 years ago #155443

Maddy, the 555 IC boost converter is adjustable using the feedback pot.
The following designs are also adjustable:
https://www.homemade-circuits.com/high-power-dc-to-dc-converter-circuit-12-v-to-30-v-variable/
https://www.homemade-circuits.com/12v-car-laptop-charger-circuit-using/

Reply
Maddy
July 20, 2024 • 2 years ago #155451

Sirji, thanx for quick reply. Just for reminder, I’m already working on the car laptop charger ckt. I asked for the ways for making this adjustable! Please suggest some ways that can vary the output voltage without modifying the turns of the inductor!

Reply
SwagatamAdmin
July 20, 2024 • 2 years ago #155455

Maddy, you can modify it in the following manner to get an adjustable boosted output.
adjustable boost converter circuit

Reply
Daniel
August 31, 2024 • 2 years ago #159667

Hello sir, thanks for your informative response. I want to do a 555 boost converter with a feedback loop, someone recommended that for a 36v-48v I will need a zener diode in the feedback line, I don’t know the reason and where I will put it.

please kindly guide. thanks sir, Engineer Swagatam

Reply
SwagatamAdmin
August 31, 2024 • 2 years ago #159668

Thank you Daniel,
Yes you will need a feedback loop to be connected between the output and the shutdown pin of the control IC.
Please see the last diagram in which you can see the feedback is connected with the pin#5 of the IC555.
The feedback loop makes sure that the output voltage remains fixed as set by the feedback level.

Reply
Daniel
September 7, 2024 • 2 years ago #160181

please how will I set the feedback to appropriate level.’?

Reply
SwagatamAdmin
September 7, 2024 • 2 years ago #160201

Please build the basic 555 boost converter design first then I will tell you how to set the feedback…

Reply
Daniel
September 7, 2024 • 2 years ago #160228

I have built 555 ic converter as specified in the article but I am not getting the feedback, please guide, sir.

Reply
SwagatamAdmin
September 7, 2024 • 2 years ago #160232

Daniel, Try the following circuit, and use 100 turns for the coil over a ferrite rod.
Make sure to connect a 1n4007 diode across the collector/emitter terminals of the transistor.
boost
Check the output voltage, it must much higher than the supply input DC.
First confirm this, then we can go for the feedback.

Reply
Daniel
September 8, 2024 • 2 years ago #160337

Thanks sir, fine boosted the voltage from 12.5v to 22.7v, but not stable. please what next, sir?

Reply
SwagatamAdmin
September 9, 2024 • 2 years ago #160389

Daniel, now you can add the feedback as shown in the following circuit:
boost feedback
Please replace the 10k preset with a 100k preset, and replace the resistor between pin#7 and positive with a 10k, and remove the 10k connected parallel to the output capacitor, it is not required.

Reply
Daniel
September 10, 2024 • 2 years ago #160561

How do make the output stable

Reply
SwagatamAdmin
September 11, 2024 • 2 years ago #160567

If you put overload at the output, then the feedback will not work….and output voltage will drop..

Reply
Daniel
September 20, 2024 • 2 years ago #161010

please if I want to boost to 96v from 12v, how many turns to use and wire. guage. thanks sir.

Reply
SwagatamAdmin
September 21, 2024 • 2 years ago #161055

That will need some experimentation, but it is not difficult, just go on increasing the number of turns and also tweak the PWM/frequency of the 555 until you get the optimal outcome…

Reply
Daniel
September 7, 2024 • 2 years ago #160182

please I will know the appropriate frequency and inductor for a particular design.

Reply
SwagatamAdmin
September 7, 2024 • 2 years ago #160202

You will have to adjust it with some experimentation, and trial and error. I can help you with this…

Reply
Daniel
September 7, 2024 • 2 years ago #160229

I have tried but very tedious by trial and error. Thanks for always.

Reply
Daniel
September 19, 2024 • 2 years ago #160861

Hello sir, please I saw a 12v , 8A to 24v, 8A converter,

1. is this boost or buck converter because it increases the power?
2. please help Me with a circuit for it?
3. please how can I use such converter for infinity power supply?

Reply
SwagatamAdmin
September 19, 2024 • 2 years ago #160864

Hi Daniel,
That is a boost converter, but the output current will not be 8 amps, it will be around 3.5 amps.
Power can never be boosted, output power will be always lower than the input power.

Reply
Daniel
September 20, 2024 • 2 years ago #161007

Thanks sir, engineer Swagatam, you are correct, it was typo error on the converter presentation.

Reply
SwagatamAdmin
September 21, 2024 • 2 years ago #161054

No problem Daniel, glad it helped!

Reply
Akafachelen
January 3, 2025 • 2 years ago #167711

Greatings sir pls on your 1st diagram simple smps boost converter I tried to run the simulation but my proteous display 2 error sms
1 was stating no specification for D1 (FR107)
2nd pls is the zener operating ?

Reply
SwagatamAdmin
January 4, 2025 • 2 years ago #167718

Hi Akafac,
It is difficult to tell why the simulation is showing the error.
You can try the following design instead and let me know how it goes:
joule thief 1

Reply
Akafac helen
January 5, 2025 • 2 years ago #167745

Pls sir for the above circuit I’ll need the specifications of TR1 on the diagram

Reply
SwagatamAdmin
January 5, 2025 • 2 years ago #167746

For the exact specifications you can refer to the following diagram:
simple joule thief circuit
For the transistor, please use a TIP122, for handling 1 amp current.

Reply
Akafac helen
January 5, 2025 • 2 years ago #167747

Thanks very much sir pls I have seen and article online on this topic I want to share with you but your email isn’t reachable I have simulated it it’s almost on point but pls I need some guidance sir
Pls can you send me your email to my Gmail so that I forward to you the link to it sir ? Thanks

Reply
SwagatamAdmin
January 5, 2025 • 2 years ago #167748

Akafac, unfortunately I do not work with simulators, instead I use my knowledge based brain simulation and finally by testing through practical experimentation. I have tested the above circuit and it works wonderfully well.
If you any other questions regarding the circuit working, please let me know, i will try to help.

Reply
Akafac helen
January 10, 2025 • 2 years ago #167836

Greatings sir, pls my 5v to 12v boost converter I couldn’t calculate the inductor value and filtering capacitor value that will give me a constant 12v out I intend to take in 5v , send out 12v , out put current of 2A taking in to consideration a feedback system but my switch should be trigger by a pulse sign maybe generated by NE555 which is supply by a 5v dc in put which oscillate to generate the pwm to drive my switching regulator
Pls sir I need your total help again
My specifications are
1 input 5v , output 12v , output current 2A,
Thank

Reply
SwagatamAdmin
January 12, 2025 • 2 years ago #167846

Hello Akafac, here are the calculations for your application:

Specifications:

Input Voltage = 5V
Output Voltage = 12V
Load Current = 2 Amps

Duty Cycle Calculation:

D = 1 – (Vin / Vout)
D = 1 – (5 / 12)
D = 0.5833 (58.33%)

Output Power:

Pout = Vout * Iout
= 12 * 2
Pout = 24W

Input Current:

Assume efficiency (η) = 80% = 0.8
Iin = Pout / (Vin * η)
= 24 / (5 * 0.8)
= 24 / 4
Iin = 6A

Inductor Value:

L = (Vin * D) / (f * ΔIL)
Where ΔIL = Ripple Current = 20% of Iin = 0.2 * 6 = 1.2A
Assume switching frequency (f) = 50kHz = 50,000 Hz

L = (5 * 0.5833) / (50,000 * 1.2)
L = 2.9167 / 60,000
L = 48.61 µH
Choose standard value: 47 µH

Output Capacitor Value:

Cout = (Iout * D) / (f * ΔVout)
Where ΔVout = Output Ripple Voltage = 0.1V

Cout = (2 * 0.5833) / (50,000 * 0.1)
Cout = 1.1667 / 5,000
Cout = 233.33 µF
Choose standard value: 220 µF or use parallel capacitors.

Resistor Values for IC 555:

Frequency formula: f = 1.44 / ((R1 + 2R2) * Ct)
Assume Ct = 10 nF = 0.00000001 F

R1 + 2R2 = 1.44 / (f * Ct)
R1 + 2R2 = 1.44 / (50,000 * 0.00000001)
R1 + 2R2 = 1.44 / 0.0005
R1 + 2R2 = 2880 Ω

Choose R1 = 1 kΩ = 1000 Ω
2R2 = 2880 – 1000
2R2 = 1880 Ω
R2 = 1880 / 2
R2 = 940 Ω

Reply
Akafac helen
January 12, 2025 • 2 years ago #167861

Sir, thank very much God blessed you for your good works indeed I acknowledge your kind heart

Reply
SwagatamAdmin
January 12, 2025 • 2 years ago #167863

You are most welcome Akafac.

Reply
Akafac helen
January 12, 2025 • 2 years ago #167842

Greatings sir pls when calculating the capacitor value why did you assume ripple voltage of 1v, and for the inductor you assume o.4

Reply
SwagatamAdmin
January 12, 2025 • 2 years ago #167849

Hello Akafac,
we must assume Ripple Current as a Percentage of Load Current if Input Current is Not Clearly specified.
If input current Iin cannot be estimated and an “optimal” ripple current is assumed, use a rule-of-thumb based on the load current (Iout):

Typical Design: Assume Δ I = 20-30% of Iout.
High-Efficiency Design: Assume Δ I = 10-15% of Iout.

Example:

For Iout = 2A:

Δ I = 0.2 * 2 = 0.4A (20%).
Δ I = 0.1 * 2 = 0.2A (10%)

Reply
Akafac helen
January 13, 2025 • 2 years ago #167868

Fine boosted sir but the out put voltage is not stable it’s increasing from 12v to 15v

Reply
SwagatamAdmin
January 13, 2025 • 1 year ago #167871

Akafac, did you build it practically, or doing a simulation? because I cannot help with simulations, I can only help with practical working.
Try increasing the output capacitor value and check again.

Reply
Akafac helen
January 13, 2025 • 1 year ago #167885

Thanks very much sir, the PWM oscillator circuit with NE555 is working ok for 50khz, 58.33%
Pls I wish to add a feedback circuit to it pls how do I go about it?

Reply
SwagatamAdmin
January 14, 2025 • 1 year ago #167894

Thanks Akafac, the feedback circuit is given at the end of the post.

Reply
Akafac helen
January 14, 2025 • 1 year ago #167907

Good day sir hope you’re doing well, pls my teacher ask me to used ATMEGA 32 to generate to PWM
Instead of NE555 pls help sir
Thanks

Reply
SwagatamAdmin
January 14, 2025 • 1 year ago #167908

Sorry Akafac, my microcontroller programming knowledge is not good, so I may not be able to help you in this regard….

Reply
Akafac helen
January 15, 2025 • 1 year ago #167932

Thank you sir I applaud
You’re a kind person

SwagatamAdmin
January 16, 2025 • 1 year ago #167939

Thank you Akafac, for your kind words, I appreciate it. I Hope you find someone who can do it for you…

Akafac helen
January 16, 2025 • 1 year ago #167947

Yes sir it’s working well but the diagram you gave me and the calculations are perfect I’m fine now my circuit is ok thanks sir

SwagatamAdmin
January 16, 2025 • 1 year ago #167948

Thanks Akafac, Glad to know it is working. All the Best to you…

Akafac helen
February 21, 2025 • 1 year ago #168556

Hello sir, hope you’re doing well
Pls how can I trigger a MOSFET IRFZ44N with BJT since my pulse voltage is 5v and the vgs(gate voltage of the MOSFET is 10v)to trigger it appropriately ?

Reply
SwagatamAdmin
February 21, 2025 • 1 year ago #168560

Hello Akafac,
If you are having a separate 10V DC then you will need two BC547 to amplify and invert the 5V to 10V.
If you do not have a separate 10V then you may need a boost converter circuit.
1.5V to 3.3V boost converter circuit for LEDs

Reply
Jaden
April 15, 2025 • 1 year ago #173177

I am trying to make 2 similar circuits that can take 2 AA batteries and output 9, and 18V DC, would this process work for making those circuits or do you have other suggestions.
1 circuit needs to take 3V in and output 18V DC, the other needs to output 9V DC

Reply
SwagatamAdmin
April 16, 2025 • 1 year ago #173362

For your application a joule thief circuit might be more suitable:
https://www.homemade-circuits.com/1-watt-led-driver-using-joule-thief/

Reply
Ngang
May 3, 2025 • 1 year ago #175468

Good morning Sir.
Thank you for the great work you are doing.
Sir, I need a DC to DC converter circuit diagram to use in a mobile PA System that can deliver up to 100Watt or more from a 12V DC lithium battery

Reply
SwagatamAdmin
May 3, 2025 • 1 year ago #175495

Thank you so much Ngang, is it a boost converter or a buck converter, please provide the full specifications of the converter, I will try to figure it out….

Reply
Steele Braden
June 11, 2025 • 1 year ago #180653

Dear Sir, I have a situation where the input supply voltage is one volt.
This is the ONLY voltage available, so obviously a 555 cannot be used for the square wave signal to the transistor base.
What simple remedy is there for this.
I need about up to 8 volts output.
There is plenty of amperage available at the one volt input supply.
Thanks,
Steele Braden.

Reply
SwagatamAdmin
June 11, 2025 • 1 year ago #180667

Hello Steele, You can try a joule thief circuit concept, as explained in the following article:
https://www.homemade-circuits.com/1-watt-led-driver-using-joule-thief/
https://www.homemade-circuits.com/8x-overunity-circuit-using-joule-thief/

Reply
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