The article provides us with all the formulas for enabling easy calculation of the power stage of a boost converter circuit having an integrated IC based operating system and in the continuous mode.

Contents

- 1 Standard Design of a Boost Converter
- 1.1 Evaluating the Highest Switching Current
- 1.2 Inductor Selection
- 1.3 Rectifier Diode Determination
- 1.4 Output Voltage Setting
- 1.5 Input Capacitor Selection
- 1.6 Output Capacitor Selection
- 1.7 Equations to Evaluate out the Power Stage of a Boost Converter
- 1.8 Maximum Duty Cycle:
- 1.9 Inductor Ripple Current:
- 1.10 Maximum output current of the nominated IC:
- 1.11 Application specific max switch current:
- 1.12 Inductor Approximation:
- 1.13 Inductor Ripple Current Valuation:
- 1.14 Current By using Resistive Divider Network for Output Voltage Positioning:
- 1.15 Recommended for You:

**Standard Design of a Boost Converter**

The following figure demonstrates the standard arrangement of a boost converter in which the switch is a part of the implemented IC.

Mostly, low power converters possess the diode substituted by an alternate switch, in-built into the converter. If this is the situation, pretty much all equations within this article apply except the power dissipation equation of the diode.

**Needed Guidelines to Calculate Boost Converter Power Stage**

The following 4 guidelines are necessary to Calculate Boost Converter Power Stage:

1. Input Voltage Range: Vin(min) and Vin(max)

2. Minimal Output Voltage: Vout

3. Highest Output Current: Iout(max)

4. IC Circuit employed to building the boost converter.

This is often mandatory, simply because certain outlines for the computations ought to be taken which may not be mentioned of the data sheet.

In the event that these limitations are familiar the approximation of the power stage normally

takes place.

### Evaluating the Highest Switching Current

The primary step to determine the switching current would be to figure out the duty cycle, D, for the minimum input voltage. A bare minimum input voltage is employed mainly because this results in the highest switch current.

*D = 1 - {Vin(min) x n} / Vout---------- (1)*

Vin(min) = minimum input voltage

Vout = required output voltage

n = efficiency of the converter, e.g. anticipated value may be 80%

The efficiency is put into the duty cycle calculation, simply because the converter is required to present the power dissipation also. This estimation offers a more sensible duty cycle compared to the formula without the efficiency factor.

We need to possibly allow an estimated 80% tolerance (that could be not impractical for a boost

converter worst case efficiency), should be considered or possibly refer to the Conventional Features portion of the picked converter's data sheet

**Calculating the Ripple Current **

The subsequent action for calculating the highest switching current would be to figure out the inductor ripple current.

In the converter datasheet usually a specific inductor or a variety of inductors are referred to as to work with with the IC. Therefore we must either use the suggested inductor value to calculate the ripple current, if nothing is presented in the datasheet, the one estimated in the Inductors list.

**S****election of this application note to Calculate Boost Converter Power Stage.**

*Delta I(l) = {Vin(min) x D} / f(s) x L ---------- (2)*

Vin(min) = smallest input voltage

D = duty cycle measured in Equation 1

f(s) = smallest switching frequency of the converter

L = preferred inductor value

Subsequently it has to be established should the preferred IC may be able to supply the optimum output

current.

*Iout(max) = [ I lim(min) - Delta I(l)/2 ] x (1 - D) **---------- (3)*

I lim(min) = minimal value of the

current restriction of the involved switch (highlighted in the data

sheet)

Delta I(l) = inductor ripple current measured in earlier equation

D =

duty cycle calculated in first equation

In case the estimated value for the optimum output current of the decided on IC, Iout(max), is below the systems expected greatest output current, an alternative IC with a slightly higher switch current control really needs to be employed.

On condition that the measured value for Iout(max) is probably a shade less than the expected one, you possibly can apply the recruited IC with an Inductor with larger inductance whenever it is still in the prescribed series. A larger inductance lessens the ripple current therefore enhances the maximum output current with the specific IC.

If the established value is above the the best output current of the program, the greatest switch current in the equipment is figured out:

*Isw(max) = Delta I(L) / 2 + Iout(max) / (1 - D) --------- (4)*

Delta I(L) = inductor ripple current measured in second equation

Iout(max), = optimum output current essential in the utility

D = duty cycle as measured earlier

It is actually the optimum current, the inductor, the enclosed switch(es) in addition to the external diode is required to stand up against.

### Inductor Selection

Sometimes data sheets furnish numerous advised inductor values. If this is the situation, you will want to prefer an inductor with this range. The larger the inductor value, the increased is the maximum output current mainly because of the decreased ripple current.

The cut down the inductor value, the scaled-down is the solution size. Be aware that the inductor really should invariably include a better current rating as opposed to the maximum current specified in Equation 4 due to the fact that the current speeds up with lowering inductance.

For elements wherein no inductor range ls handed out, the following picture ls a reliable calculation for the suitable inductor;

*L = Vin x ( Vout - Vin) / Delta I(L) x f(s) x Vout --------- (5)*

Vin = standard input voltage

Vout = preferred output voltage

f(s) = minimal switching frequency of the converter

Delta I(L) = projected inductor ripple current, observe below:

The inductor ripple current simply cannot be measured with the first equation, just because the inductor ls not recognized. A sound approximation for the inductor ripple current ls 20% to 40% of the output current.

*Delta I(L) = (0.2 to 0.4) x Iout(max) x Vout/Vin ---------- (6)*

Delta I(L) = projected inductor ripple current

Iout(max) = optimum output

current required for the application

### Rectifier Diode Determination

To bring down losses, Schottky diodes really needs to be considered a good choice.

The forward current rating considered necessary is on par with the maximum output current:

*I(f) = Iout(max) ---------- (7)*

I(f) = typical

forward current of the rectifier diode

Iout(max) = optimum output current important in the program

Schottky diodes include considerably more peak current rating in comparison with normal rating. That is why the increased peak current in the program is not a big concern.

The second parameter containing to be monitored is the power dissipation of the diode. It consists of to handle:

*P(d) = I(f) x V(f) ---------- (8)*

I(f) = average forward current of the rectifier diode

V(f) = forward voltage of the rectifier diode

### Output Voltage Setting

Most of the converters allocate the output voltage with a resistive divider network (that could be in-built

should they be stationary output voltage converters).

With the assigned feedback voltage, V(fb), and feedback bias current, I(fb), the voltage divider tend to be

computed.

The current with the aid of the resistive divider could perhaps be about one hundred times as massive as the feedback bias current:

*I(r1/2) > or = 100 x I(fb) ---------- (9)*

I(r1/2) = current in the course of the resistive divider to GND

I(fb) = feedback bias current from data sheet

This augments below 1% inaccuracy to the voltage evaluation. The current is additionally considerably larger.

The main problem with smaller resistor values is an increased power loss in the resistive divider, except the relevancy might be somewhat elevated.

With the above conviction, the resistors are worked out as listed below:

*R2 = V(fb) / I(r1/2) ---------- (10)*

*R1 = R2 x [Vout / V(fb) - 1] ---------- (11)*

R1, R2 = resistive divider.

V(fb) = feedback voltage from the data sheet

I(r1/2) = current due to the resistive divider to GND, established in Equation 9

Vout = planned output voltage

### Input Capacitor Selection

The least value for the input capacitor is typically handed out in the data sheet. This very least value is vital for steady the input voltage as a result of the peak current prerequisite of a switching power supply.

The most suitable method is to make use of reduced equivalent series resistance (ESR) ceramic capacitors.

The dielectric element needs to be X5R or higher. Otherwise, the capacitor could drop off most of its capacitance on account of DC bias or temperature (see references 7 and 8).

The value could in fact be raised if perhaps the input voltage is noisy.

### Output Capacitor Selection

The best method is to locate small ESR capacitors to lessen the ripple on the output voltage. Ceramic capacitors are the right types when the dielectric element is of X5R type or more efficient

In the event the converter bears external compensation, any kind of capacitor value above the advocated smallest in the datasheet can be applied, yet somehow the compensation must have to be altered for the selected output capacitance.

With internally compensated converters, the advisable inductor and capacitor values needs to be accustomed, or the information in the datasheet for adapting the output capacitors could be adopted with the ratio of L x C.

With secondary compensation, the following equations can be of help to regulate the output capacitor values for a planned output voltage ripple:

*Cout(min) = Iout(max) x D / f(s) x Delta Vout ---------- (12)*

Cout(min) = smallest output capacitance

Iout(max) = optimum output current of the usage

D = duty cycle worked out with Equation 1

f(s) = smallest switching frequency of the converter

Delta Vout = ideal output voltage ripple

The ESR of the output capacitor increases a dash more ripple, pre-assigned with the equation:

*Delta Vout(ESR) = ESR x [ Iout(max)/1 -D + Delta I(l)/2 ] ---------- (13)*

Delta Vout(ESR) = alternative output voltage ripple resulting from capacitors ESR

ESR = equivalent series resistance of the employed output capacitor

Iout(max)= greatest output current of the utilization

D = duty cycle figured out in first equation

Delta I(l) = inductor ripple current from Equation 2 or Equation 6

### Equations to Evaluate out the Power Stage of a Boost Converter

**Maximum Duty Cycle: **

*D = 1 - Vin(min) x n / Vout ---------- (14)*

Vin(min) = smallest input voltage

Vout = expected output voltage

n = efficiency of the converter, e.g. estimated 85%

### Inductor Ripple Current:

*Delta I(l) = Vin(min) x D / f(s) x L ---------- (15)*

Vin(min) = smallest input voltage

D = duty cycle established in Equation 14

f(s) = nominal switching frequency of the converter

L = specified inductor value

### Maximum output current of the nominated IC:

*Iout(max) = [ Ilim(min) - Delta I(l) ] x (1 - D) ---------- (16)*

Ilim(min) = smallest value of the current limit of the integral witch (offered in the data sheet)

Delta I(l) = Inductor ripple current established in Equation 15

D = duty cycle estimated in Equation 14

### Application specific max switch current:

*Isw(max) = Delta I(l) / 2 + Iout(max) / (1 - D) ---------- (17)*

Delta I(l) = inductor ripple current estimated in Equation 15

Iout(max), = highest possible output current required in the utility

D = duty cycle figured out in Equation 14

**Inductor Approximation: **

*L = Vin x ( Vout - Vin) / Delta I(l) x f(s) x Vout ---------- (18)*

Vin = common input voltage

Vout = planned output voltage

f(s) = smallest switching frequency of the converter

Delta I(l) = projected inductor ripple current, see Equation 19

### Inductor Ripple Current Valuation:

*Delta I(l) = (0.2 to 0.4) x Iout(max) x Vout/Vin ---------- (19)*

Delta I(l) = projected Inductor ripple current

Iout(max) = highest output current important in the usage

Typical Forward Current of Rectifier Diode:

*I(f) = Iout(max) ---------- (20)*

Iout(max) = optimal output current appropriate in the utility

Power Dissipation in Rectifier Diode:

*P(d) = I(f)
x V(f) ---------- (21)*

I(f) = typical forward current of the rectifier diode

V(f) = forward voltage of the rectifier diode

### Current By using Resistive Divider Network for Output Voltage Positioning:

*I(r1/2) > or = 100 x I(fb) ---------- (22)*

I(fb) = feedback bias current from data sheet

Value of Resistor Between FB Pin and GND:

*R2 = V(fb) / I(r1/2) ---------- (23)*

Value of Resistor Between FB pin and Vout:

*R1 = R2 x [Vout/V(fb) - 1 ] ---------- (24)*

V(fb) = feedback voltage from the data sheet

I(r1/2) = current

due to the resistive divider to GND, figured out in Equation 22

Vout = sought after output voltage

Smallest Output Capacitance, otherwise pre-assigned in the data sheet:

*Cout(min) = Iout(max) x D / f(s) x Delta I(l) ---------- (25)*

Iout(max) = highest possible output current of the program

D = duty cycle figured out in Equation 14

f(s) = smallest switching frequency of the converter

Delta Vout = expected output voltage ripple

Excess Output Voltage Ripple owing to ESR:

*Delta Vout(esr) = ESR x [ Iout(max) / (1 - D) + Delta I(l) / 2 ---------- (26)*

ESR = parallel series resistance of the employed output capacitor

Iout(max) = optimum output current of the usage

D = duty cycle determined in Equation 14

Delta I(l) = inductor ripple current from Equation 15 or Equation 19

Tun says

Sir, I need your help on a circuit capable to step up dc of between 5v and 15v to 30v

Swagatam says

Tun, you can use a IC 555 based boost converter circuit or a "joule thief" concept can also be tried

Tun says

Hi Sir, i need your help sir on how to obtain an inductance of 40microhenry from a coil with (0.02inc.) on ferrite core(d=0.2inch. and h=1.5inch.). thank u sir. All I need is number turns needed.

Swagatam says

Hi Tun, that can be extremely difficult using a formula, so you can better try it practically using an L meter

if you don't have an L meter then could use the following concept for measuring it

http://www.homemade-circuits.com/2014/08/15-v-inductance-meter-circuit.html

Tun says

thank you sir.

Kiran Guru says

Hi, i have designed a boost converter for 12 to 24v. input is given from solar panel (two 18v/1a) in parallel through lm338 regulator fixed for 12v.ive made a inductor for boost convrtr having 38 turns of .5mm enmld cu wire on torroid frm cfl and cpr of 4.7uf/250 and mosfet irf540 to drive a light load 10 watt. but this arrangment is consuming too much power causing heating of ic and mosfet and light does not glow well even at full sun hrs, suggest me some ideas to counter this problem.

Swagatam says

Hi, in any inductor based converter, the frequency must correctly match the inductance of the inductor, otherwise the inductor can get hot and the driver transistor can burn…make sure the frequency is correctly adjusted such that the inductor the inductor is forced to oscillate at its natural frequency.

Kiran Guru says

Thanks for your quick reply.I am using arduino with tlp 250 to drive mosfet. change in duty cycle and time period via arduino is not causing any variation in output.so plz give me some ideas to match the frequency.

Swagatam says

PWM adjustment will alter the output current of the converter, you will have to adjust the frequency to optimize the resonance, until this is done, the converter will keep consuming high amount of current….