Design and simulation of a voltage-controlled Buck converter feedback circuit

Aim: To design and simulate a buck converter with a voltage-control analog feedback circuit and obtain the required waveforms.

Procedure:

Given that:

Input Voltage (`V_i`) = 50 V

Duty ratio (D) = 0.4 or 40%

Switching Inductor (L) = 3 mH

Output Capacitor (C) = `100 muF`

Switching Frequency (`F_s`) = 20 kHz

Load Resistance (R) = `20 Omega`

Calculations:

The Voltage ratio of a buck converter is

`V_o/V_i = D`

`therefore V_o = D*V_i`

`therefore V_o = 0.4*50 = 20 V`

Therefore, the output voltage of the designed buck converter should be 20 V.

Now, for designing a feedback circuit we have to assume some parameters, like

Equivalent Series Resistance of the Capacitor (`R_(ESR)`) = `8 mOmega`

Ramp Voltage (`V_(ramp)`) = 2.5 V

Reference Voltage (`V_(ref)`) = 8 V

Oscillatory Voltage (`V_(osc)`) = 20 V

Now, calculating the poles and zeros of the converter

Frequency of the double poles (`F_(LC)`) = `1/(2pisqrt(LC))`

`therefore F_(LC) = 1/(2pisqrt(3*10^-3*100*10^-6)) = 290.57 Hz`

The frequency at zero (`F_(ESR)`) = `1/(2pi*R_(ESR)*C)`

`therefore F_(ESR) = 1/(2pi*8*10^-3*100*10^-6) = 199 kHz`

Assuming the crossover frequency to be 1/5 of the switching frequency

`therefore F_o = F_s/5`

`therefore F_o = 20000/5 = 4000 = 4 kHz`

Since, `F_(LC) < F_o < F_s/2 < F_(ESR)`, type III-B compensator is suitable for this application.

The compensator poles and zeros are chosen `70^o` phase lead at the crossover frequency.

Now, calculating the poles and zeros of the compensator

`F_(Z2) = F_osqrt((1-sintheta)/(1+sintheta))`

`therefore F_(Z2) = 4000sqrt((1-sin70^o)/(1+sin70^o)) = 705.3 Hz`

`F_(P2) = F_osqrt((1+sintheta)/(1-sintheta))`

`therefore F_(P2) = 4000sqrt((1+sin70^o)/(1-sin70^o)) = 22.685 kHz`

`F_(Z1) = 0.5*F_(Z2)`

`therefore F_(Z1) = 0.5*705.3 = 352.65 Hz`

And, `F_(P3) = F_s/2`

`therefore F_(P3) = 20000/2 = 10000 = 10 kHz`

Now, calculating the values of the parameters of the compensator

Choosing `C_(f3)` = 2.2 nF

`R_(f3) = 1/(2pi*C_(f3)*F_(P2))`

`therefore R_(f3) = 1/(2pi*2.2*10^-9*22.685*10^3) = 3189 Omega`

Choosing `R_(f3) = 3190 Omega`

`R_(f1) = 1/(2pi*C_(f3)*F_(Z2))`

`therefore R_(f1) = 1/(2pi*2.2*10^-9*705.3) = 99380.75 Omega`

Choosing `R_(f1) = 99.381 kOmega`

`R_(f2) = (R_(f1)*V_(ref))/(V_o-V_(ref))`

`R_(f2) = (99381*8)/(20-8) = 66254 Omega`

Choosing `R_(f2) = 66.254 kOmega`

`R_(c1) = (2pi*F_o*L*C*V_(osc))/(V_i*C_(f3))`

`therefore R_(c1) = (2pi*4000*3*10^-3*100*10^-6*20)/(50*2.2*10^-9) = 1370876.79 Omega`

Choosing `R_(c1) = 1370.877 kOmega`

`C_(c1) = 1/(2pi*R_(c1)*F_(Z1))`

`therefore C_(c1) = 1/(2pi*1370877*352.65) = 0.329 nF`

Choosing `C_(c1) = 0.35 nF`

`C_(c2) = 1/(2pi*R_(c1)*F_(P3))`

`therefore C_(c2) = 1/(2pi*1370877*10000) = 11.609 pF`

Choosing `C_(c2) = 11.6 pF`

The following figure shows the schematics of the designed buck converter with voltage-controlled analog feedback circuit using Spice based TINA-TI simulation software

The following table shows the components used and their parameters for designing the above buck converter

Sr. No.	Components	Parameters
1	DC Input Voltage Source	50 V
2	Switching Inductor	3 mH
3	Output Capacitor	100 uF
4	Equivalent Series Resistance	`8 mOmega`
5	Load Resistance	`20 Omega`
6	Resistor `R_(f1)`	`99.381 kOmega`
7	Resistor `R_(f2)`	`66.254 kOmega`
8	Resistor `R_(f3)`	`3190 Omega`
9	Capacitor `C_(f3)`	2.2 nF
10	Resistor `R_(c1)`	`1370.877 kOmega`
11	Capacitor `C_(c1)`	0.35 nF
12	Capacitor `C_(c2)`	11.6 pF
13	Reference Voltage	8 V
14	Ramp Voltage	2.5 V

Results:

Carrying out the transient analysis as per the above figure, we get the following results

Zoomed-in:

a. The output voltage transient and steady-state response:

Zoomed-in:

The output load voltage shoots up to a maximum of 44.45 V at the transient state and settles to a required steady-state value of 20 V in 6 msecs.

b. The Inductor Current:

Zoomed-in:

`I_(L(max)) = 1.1 A` and `I_(L(min)) = 0.899 A`

Therefore, the average inductor current = (1.1+0.899)/2 = 0.9995 A

And ripple in the inductor current = 1.1-0.899 = 0.201 A.

c. The Mosfet Current and Voltage:

Zoomed-in:

Mosfet Current = 1.09 A

Mosfet Peak Voltage = 50.71 V

d. The Diode Current and Voltage:

Zoomed-in:

Diode Current = 1.1 A

Diode Inverse Peak Voltage = -49.8 V

Conclusion:

Thus, a buck converter with a voltage-control analog feedback circuit using a type III-B compensator is designed and simulated and the required output waveforms are obtained.