# Buck Converter

We know that people use voltage divider circuits, voltage regulators and Zener diodes for stepping down the DC Power but there is the limit for attaining the required power. That’s why we need such electronic circuits from which we can attain required DC Power. Fig 1 shows the basic figure for buck converter. Figure 1: Basic buck converter

So, let’s analyze the output waveforms for it. When switch will be in position 1, output will be equal to Vg and output will be equal to zero when switch will be in position 2. It is the SPDT switch (the ideal one) which will be realized later. Due to this switching, a rectangular waveform is produced which is shown below. Figure 2: Rectangular waveform produced

So, as we apply Fourier analysis on the above waveform then for the DTs time, the area will be DTsVg and zero for the D’Ts time. So, the equations can be written as,

<vs> = (1/Ts)∫vs(t)dt                     limit from 0-Ts

Solving it yields,

<vs> = (1/Ts)(DTsVg) = DVg

So, the average value of vs(t) is equal to Duty cycle(D) times the input voltage. Where duty cycle is the fraction of time for which SPDT switch spends time at position 1 and it is a number between 0 and 1. So, in figure 1 only the filter insertion is required because we want pure DC power and 2nd order filter will reduce the switching harmonics to the extent and then the output will be approximately equal to input voltage rather than the losses included.

v = <vs> = DVg

The 2nd order filter will be designed using lossless elements like capacitor and inductor due to which current and voltage ripples will be reduced much down to the level which won’t affect our system. Figure 3: Filter Insertion for Harmonics Elimination

The buck converter in figure 3 allows total control over the input supplied. But the low pass 2nd order filter needs voltage and current analysis which includes inductor volt-second balance and capacitor-charge balance principles. And small-ripple approximation greatly helps in doing this analysis.

Let’s do the analysis of inductor and capacitor waveforms. It is not completely possible to build a perfect low pass filter which will remove the harmonics. So, we build such a low pass filter which must allow some small amount of high frequency harmonics to reach the required Power level. Therefore, the output equation can be written as

v(t) = V + vripple(t)

As we can see that the output equation contains our DC component V but also includes small AC component due to allowing the high frequency harmonics by low pass filter designing. The vripple(t) must always be small in a well-designed converter because our objective is to produce DC power and when we apply the small-ripple approximation, the output becomes approximately equal to input applied. The equations can be written as

||vripple(t)|| << V

And                                                                              v = V

And the waveform can be seen in the following figure. Figure 4: Output Waveform which includes vripple(t)

Now analyze the inductor current waveform over one complete interval of time. The inductor current can be found out by integrating the inductor voltage waveform. When the switch is in position 1, the inductor one side is connected to the input and the circuit becomes as shown below. Figure 5: Switch in position 1

So, the inductor voltage can be written in equation form as

vL = Vg – v(t)

When we apply small-ripple approximation, the equation reduces to

vL = Vg – V

Now, we can find the inductor current using the following equations.

vL(t) = LdiL(t)/dt

diL(t)/dt = vL(t)/L = (Vg – V)/L

Hence, it is a constant inductor current increasing slope because inductor voltage is constant.

Now, when the switch is in position 2, the inductor one side is connected to the ground and the circuit reduces as below. Figure 6: Switch in position 2

So, the inductor voltage can be written in equation form as

vL = – v(t)

When we apply small-ripple approximation, the equation reduces to

vL = – V

Now, we can find the inductor current using the following equations.

vL(t) = LdiL(t)/dt

diL(t)/dt = vL(t)/L = – V/L

Hence, it is a constant inductor current decreasing slope because inductor voltage is constant.

The inductor voltage and current waveform is as below: Figure 7: Inductor voltage and current waveforms

If we want to calculate the inductor current ripple ΔiL which is of real interest because we want to pass the minimal ripple current through our system. We know the slopes and length of the sub-interval therefore; the equation can be derived as follows:

(Change in iL) = (slope)(length of subinterval)

2ΔiL = ((Vg – V)/L)(DTs)

ΔiL = ((Vg – V)/2L)(DTs)

L = ((Vg – V)/ 2ΔiL)(DTs)

These equations determine the inductance and inductor current ripple for the buck converter.

Now, when come to inductor volt-second balance we can see from the figure 7; we simply take the average find out in the following equations:

0 = <vs> = (1/Ts)∫vs(t)dt                     limit from 0-Ts

Solving the equation, we get

V = DVg

Inductor volt-second balance enables us to derive the expression for the DC component of the output voltage of converter.

Similar arguments can also be applied on capacitor. The capacitor equation can be written as

iC(t) = CdvC(t)/dt

Integration over one period of time interval yields

vC(Ts) - vC(0) = (1/C)∫iC(t)dt                       limit from 0-Ts

So, capacitor charge-balance equation can be written as

0 = <iC> = (1/Ts)∫iC(t)dt                     limit from 0-Ts

Now, have a look at the switch realization of the buck converter. First, convert SPDT switch to two SPST switches and then define the voltage polarities and current direction on those switches. Then realize their I-V curve graphs and then place the respective switches in place of those SPST switches.

The final buck converter looks like below: Figure 8: Designed Buck Converter

When it comes to the PCB manufacturing, the inductor toroidal core is selected according to the international standard and the its winding wire is selected using American wire gauge system. The diode and MOSFET are selected according to the circuit designing that for how much rated power you want from the circuit. The PCB should be compact and Aluminum one because it will soak all the heat generated from the buck converter because it is of high-power circuit. The gate driving circuitry will be on-board to avoid the losses.

This PCB can be manufactured from the https://raypcb.com