In this post I will explain the various parameters required for designing a correct buck converter inductor, such that the required output is able to achieve maximum efficiency.

In our previous post I explained the basics of buck converters and realized the important aspect regarding the transistor's ON time with respect to the periodic time of the PWM which essentially determines the output voltage of the buck converter.

In this post we'll go a little deeper and try to evaluate the relationship between the input voltage, switching time of the transistor, output voltage and the current of the buck inductor, and regarding how to optimize these while designing a buck inductor.

## Buck Converter Specifications

Let's first understand the various parameters involved with a buck converter:

**Peak inductor current, (** **i _{pk}**

**)**= It's the maximum amount of current that an inductor can store before getting saturated. Here the term "saturated" means a situation where the transistor switching time is so long that it continues to be ON even after the inductor has crossed its maximum or peak current storing capacity. This is an undesirable situation and must be avoided.

**Minimum Inductor Current, (****i _{o}**

**)**= It's the minimum amount of current that may be allowed for the inductor to reach while the inductor is discharging by releasing its stored energy in the form of back EMF.

Meaning, in the process when the transistor is switched OFF, the inductor discharges its stored energy to the load and in the course its stored current drops exponentially towards zero, however before it reaches zero the transistor may be supposed to switch ON again, and this point where the transistor may switch ON again is termed as the minimum inductor current.

The above condition is also called the continuous mode for a buck converter design.

If the transistor does not switch ON back before the inductor current has dropped to zero, then the situation may be referred to as the discontinuous mode, which is an undesirable way to operate a buck converter and may lead to an inefficient working of the system.

**Ripple Current, (Δi =** **i _{pk}**

**-**

**i**

_{o}**)**= As may be seen from the adjoining formula, the ripple

**Δ**i is the difference between the peak current and minimum current induced in the buck inductor.

A filter capacitor at the output of the buck converter will normally stabilize this ripple current and help to make it relatively constant.

**Duty Cycle, (D = ****T _{on}**

**/T)**= The duty cycle is calculated by dividing the ON time of the transistor by the periodic time.

Periodic time is the total time taken by one PWM cycle to complete, that is the ON time + OFF time of one PWM fed to the transistor.

**ON time of the Transistor (** **T _{on}**

**= D/f)**= The ON time of the PWM or the "switch ON" time of the transistor may be achieved by dividing the duty cycle by the frequency.

**Average output current or the load current, (****i _{ave}**

**= Δi / 2 = i**

_{load}**) =**It's obtained by dividing ripple current by 2. This value is the average of the peak current and the minimum current that may be available across the load of a buck converter output.

**RMS value of Triangle wave irms = √{****i _{o}**

^{2}

**+**

**(Δi)**

^{2}

**/ 12} =**This expression provides us the RMS or the root mean square value of all or any triangle wave component that may be associated with a buck converter.

OK, so the above were the various parameters and expressions essentially involved with a buck converter which could be utilized while calculating a buck inductor.

Now I will explain how the voltage and current may be related with a buck inductor and how these may be determined correctly, from the following explained data:

Remember here we are assuming the switching of the transistor to be in the continuous mode, that is the transistor always switches ON before the inductor is able to discharge its stored EMF completely and become empty.

This is actually done by appropriately dimensioning the ON time of the transistor or the PWM duty cycle with regard to the inductor capacity (number of turns).

### V and I Relationship

The relationship between voltage and current within a buck inductor may be put down as:

**V = L di/dt**

or

**i = 1/L 0ʃtVdt + i _{o}**

The above formula may be used for calculating the buck output current and it holds good when the PWM is in the form of an exponentially rising and decaying wave, or may be a triangle wave.

However if the PWM is in the form of rectangular waveform or pulses, the above formula can be written as:

**i = (Vt/L) + i _{o}**

Here Vt is the voltage across the winding multiplied by the time for which it's sustained (in micro-secs)

This formula becomes important while calculating the inductance value L for a buck inductor.

The above expression reveals that the current output from a buck inductor is in the form of a linear ramp, or wide triangle waves, when the PWM is in the form of triangular waves.

Now let's see how one may determine the peak current within a buck inductor, the formula for this is:

**ipk = (Vin – Vtrans – Vout)Ton / L + i _{o}**

The above expression provides us the peak current while the transistor is switched ON and as the current inside the inductor builds up linearly (within its saturation range*)

### Calculating Peak Current

Therefore the above expression can be used for calculating the peak current build-up inside a buck inductor while the transistor is in the switch ON phase.

If the expression io is shifted to the LHS we get:

**i _{pk} - i_{o} = (Vin – Vtrans – Vout)Ton / L**

Here Vtrans refers to the voltage drop across the transistor's collector/emitter

Recall that the ripple current is also given by Δi = ipk - io, therefore substituting this in the above formula we get:

**Δi = (Vin – Vtrans – Vout)Ton / L ------------------------------------- Eq#1**

Now let's see the expression for acquiring the current within the inductor during the switch-OFF period of the transistor, it may be determined with the help of the following equation:

**i _{o} = **

**i**

_{pk}- (Vout – VD)Toff / LAgain, by substituting ipk - io by Δi in the above expression we get:

**Δi = (Vout – VD)Toff / L ------------------------------------- Eq#2**

The Eq#1 and Eq#2 can be used for determining the ripple current values while the transistor is supplying current to the inductor, that is during it's ON time..... and while the inductor is draining the stored current through the load during the transistor switch OFF periods.

In the above discussion we successfully derived the equation for determining the current (amp) factor in a buck inductor.

### Determining Voltage

Now let's try to find a expression which may help us to determine the voltage factor in a buck inductor.

Since the Δi is common in both Eq#1 and Eq#2, we can equate the terms with each other to get:

**(Vin – Vtrans – Vout)Ton / L = (Vout – VD)Toff / L**

**VinTon – Vtrans – Vout = VoutToff – VDToff**

**VinTon – Vtrans – VoutTon = VoutToff - VDToff****VoutTon + VoutToff = VDToff + VinTon – VtransTon**** Vout = (VDToff + VinTon – VtransTon) / T**

Replacing the Ton/T expressions by duty cycle D in the above expression, we get**Vout = (Vin – Vtrans)D + VD(1 – D)**

Processing the above equation further we get:

**Vout + VD = (Vin – Vtrans + VD)D**

or

**D = Vout - VD / (Vin – Vtrans – VD)**

Here VD refers to the voltage drop across the diode.

#### Calculating Step Down Voltage

If we ignore the voltage drops across the transistor and the diode (since these can be extremely trivial compared to the input voltage), we can trim down the above expression as given below:

**Vout = DVin**

The above final equation can be used for calculating the step down voltage that may be intended from a particular inductor while designing a buck converter circuit.

The above equation is the same as the one discussed in the solved example of our previous article "how buck converters work.

In the next article we'll learn how to estimate the number of turns in a buck inductor....please stay tuned.

Nil says

Hello

I am using a ready made module of Buck convertor based on ic XL 4016 . .My input is 36v dc & output is 24vdc @ 1.5amps RO pump as a load. But when i give load my output voltage reduce at level 23.5 volt.

I need minimum 23.9 voltage atlases. Can please tell me reason of reduce 0.5 volt ?

Swagatam says

Hi, did you try adjusting the FB resistive divider values so that the 0.5V is corrected? By the way a 0.5V error in 24V is too small to be considered.

Nil says

Dear

i am just set preset value for 24 v dc output , Instead of that some other resister (feedback ) also need to set for adjust low dropout. I mean its low drop output voltage becomes high. U are right 0.5 v is too low, but due to this error i get change in pressure as well as flow of water with my system.

if u suggest direct 36v to 24v dc having very very low drop out then good for me.

Swagatam says

In that case a reasonably good alternative is LM338 IC based power supply. You can adjust its output precisely to the desired value. However being a linear IC this IC may become significantly hot. Nevertheless you can add a large heatsink to it for controlling the heat. I could have suggested the 7824 IC but it can supply only upto 1 amp

bilal says

AOA.SIR.

I will share a circuit with plz help me how to find current in the circuit.

Swagatam says

Sorry Bilal, circuit diagram sharing is not possible here, you can explain it verbally, if possible I may try to help!

Jose Luis says

Hi Sir,

Very interesting article.

I have a doubt about the expression √{i2o + (Δi)2/12} =

What is i2o? Do you divide the addition by 12 or only the ripple?

could you give an example?

Tahk you very much

Jose Luis

Swagatam says

Thank you Jose for pointing the issue, please check it now I have corrected the subscripts in the formula so that they make sense.

Tolu says

This calculations seem complex, please can you suggest the optimum inductor I can use.

Swagatam says

for manual tweaking you can use the first circuit from this article:

https://www.homemade-circuits.com/designing-solar-inverter-tutorial/

initially use inductor having identical number turns equal to the supply input value, then tweak the frequency and PWM to get the maximized output

Abu hassan says

Thanks

victory says

Pls help me with how i can make a 500va transformer

Swagatam says

details are furnished here:

https://www.homemade-circuits.com/2012/02/how-to-design-your-own-inverter.html

victory says

Please sir help me with a circuit that will automatically turn on my inverter at night and off during daytime my battery is 12v 38amp

Swagatam says

you can use any one of the designs presented in the following article:

https://www.homemade-circuits.com/2012/01/how-to-make-light-activated-day-night.html

Unknown says

ok Mr Swagatm we stay tuned for the next section how to estimate turnns of coils sir. thank u Alex

Swagatam says

sure Alex, you'll see it soon, let me first post a few pending requests from the readers in my meantime….

VIJAY AJ says

Sir can. U post the calculation for boost converter

Swagatam says

Vijay, you'll see it soon in one of my future articles.

pro says

sir, i sent another video to your email,it is not like the other one

i just what know the name of the transistor that was use

Swagatam says

the video is not opening because you may have not toggled the "share" button.

tell me the power input (voltage and current), I'll tell you which transistor to use