Four simple boost converter circuits are explained in this post, which can be built and used for converting a low level DC voltage input to a higher level DC voltage output.
What is a Boost Converter
A DC boost converter circuit is designed for stepping-up or boosting a small input voltage levels to a desired higher output voltage level, hence the name "boost" converter. Since these circuits basically step up a low voltage to a higher voltage levels, they are also know as step-up converters.
Although a boost converter circuit may involve many complex stages and calculations, here we will see how the same could be built using minimum number of components, and with effective results.
Basically a boost converter works by oscillating current though a coil or inductor, wherein the voltage induced in the inductor is transformed into a boosted voltage whose magnitude is dependent on the number of turns and PWM of the oscillation frequency.
Basic Working Principle of Boost Converter
Referring to the figure above, the basic working principle of a boost converter topology can be understood from the following points:
When the switch S is switched ON, the coil voltage UL becomes equal to the input voltage and the current through the coil begins increasing linearly.
Next, if the switch S is turned OFF, the coil ensures that current through it keeps flowing independent of how much the output voltage rises. The current at this moment flows via diode D.
In this situation, the coil voltage UL is negative and the output voltage is higher than the input voltage. As a result, the current flowing through the coil decreases linearly. During this period, the coil supplies the output with a stepped-up, boosted voltage.
After this, if the switch S is switched ON again, the process repeats. The above process repeats continuously as long as the switch S is switched ON/OFF repeatedly. This causes the output to get a continuous supply of a stepped-up voltage. The smoothing capacitor makes sure the boosted voltage is filtered correctly and is a pure DC.
You may also like: Make this 1.5 V to 3.6 V boost converter for LED flashlights
1) Simple 5 V to 12V Boost Converter Circuit
In the first concept as shown in the figure above, the input DC supply can be anywhere between 3 V and 5 V. We can adjust the output voltage to 12 V or some other desired voltage by tweaking the zener diode D2.
So, when output voltage tends to increase excessively, T2's operating point shifts, causing T1 to switch on for a shorter duration of time (or maybe not at all).
We can expect the stepped-up output voltage to be around 12.6V at a 20 mA output current. The input current at a 5 V input voltage will be around 64 mA.
This translates to an efficiency of 77%, which is not bad at all for such a straightforward circuit.
2) How to get +5V, -5V Dual Supply from 1.2 V
If you are looking for a circuit to boost 1.2 V NiCd supply to 5 V then you can use this second circuit below. Moreover, this circuit will allow you to get a dual +5V and -5V supply from a 1.2 V DC single supply input.
Parts List
- All resistors are 1/4 watt 5% unless specified
- R1, R5 = 1 K
- R2, R3 = 100 Ohms
- R4 = 2.2 K
- Capacitors
- C1 = 10 uF / 25 V Electrolytic
- C2 = 0.01 uF Ceramic Disc
- C3 = 1 uF / 25 V Electrolytic
- C4 = 0.1 uF Ceramic Disc
- C5 = 10 uF / 25 V Electrolytic
- Semiconductors
- D1, D2, D3 = FR107
- D4 = 5.1 V 1/2 watt zener diode
- T1 = 2N2222
- T2 = 2N2907
- Inductor
- L1 = 270 uH 500 ma
The step up converter is basically formed using T1, L1 and D1. Zener diode D4 functions as a feedback at the base of transistor T2 and provides the required stabilization for the circuit.
The maximum current output capacity of this circuit is around 10 mA at +/- 5V.
The circuit will provide a maximum efficiency of 60% which does not looks too impressive. However with an input DC of just 1.2 V you cannot except more than this.
3) Flyback Type 1.5 V to 30 V Boost Converter using a single BJT
The third boost converter circuit shown below uses a joule thief flyback topology.
Parts List
- R1 = 1K 1/4 watt
- D1 = 1N4148 or a Schottky diode such as FR107 or BA159
- T1 = any NPN power BJT such as TIP31, 2N2222, 8050 or BC139 (on heatsink)
- C1 = 0.0047uF
- C2 = 1000uF/25V
Inductor is built using 20 turns each of super enameled copper wire on a ferrite torroid T13. Wire thickness can be as per the output current requirement.
In the above design a single BJT and an inductor is all that's needed for visualizing an incredible 1.5V to upto 30V boost.
The circuit works using a joule thief concept and utilizes an inductor in the flyback mode for generating the specified high efficiency output .
Using a flyback concepts allows the two side of the transformer isolated and ensures better efficiency, since the load is able to operate during the OFF time of the BJT, which in turn prevents the BJT from overloading.
While experimenting I found that adding C1 drastically improved the performance of the circuit, without this capacitor the output current did not look too impressive.
4) 3.7 V to 24 V Converter
Now let's refer to our fourth step-up converter design which will boost a 3.7 V input DC to 24 V output DC. 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.
The above circuit can be regulated with a feedback as shown below:
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 Links for the above boost converter circuits are provided below:
can i put a coil of copper wire instead of an inductor
Yes copper coil will do but the inductance has to be matched correctly
What will I change in the circuit if I replace the Bc557 with A1015 bjt
You can try it, no changes will be required.
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
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…
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
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
https://www.homemade-circuits.com/wp-content/uploads/2023/05/switching-solar-boost-converter-circuit.jpg
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?
@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.
Please I need circuit diagram for 12vdc to 400vdc converter
You can try the following design:
https://www.homemade-circuits.com/how-to-convert-12v-dc-to-220v-ac-using/
For the flyback circuit could I use 5v input instead of 1.5v?
You can use 5V, but then the output would be above 24V
That is what I need, thank you
You are welcome! Glad you found the post helpful.
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?
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.
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.
Hi, you can try the second or the third circuit from the above article, it should work for your application.
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.
Moses, you can use the CMOS 7555 IC which are rated to work from 3 V onward.
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
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
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:
https://www.homemade-circuits.com/wp-content/uploads/2022/04/high-current-LM317-power-supply-circuit.jpg
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
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
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)
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
Hello there can i use a peice of ferrite rod instead of a ferrite toroide.
yes you can do that!
Cheers mate that will save me from having to wait a week for delivery of toroid core.
Sure, no problem, and wish you all the best!
Please email me via dr.blackschleger@gmail.com so i can share a circuit photo with you, and,
happy Fathers’ Day, Professor Swagatam !
Thank you Doctor B, My email is homemadecircuits @ gmail . com
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?
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.
Hello good afternoon, what is the output amperage of the circuit with Lm555 ..
It is 1 amp because TIP31 is used, you can try TIP35 for higher amps
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
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.
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
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
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
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/
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
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!
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.
Hello, you can try the first concept presented in the following article:
https://www.homemade-circuits.com/12v-car-laptop-charger-circuit-using/