I have made a buck converter with negative feedback on Falstad:




Here is the code of the circuit:

$ 1 3.125e-7 382.76258214399064 50 5 50 5e-11
R 192 128 160 128 0 0 40 5 0 0 0.5
f 208 176 208 128 40 1.5 0.02
w 224 128 272 128 0
d 288 240 288 128 2 default
w 288 128 272 128 0
l 288 128 368 128 0 1 0.0020841426863125808 0
w 368 240 288 240 0
g 288 240 288 256 0 0
r 432 240 432 128 0 1000
w 432 128 368 128 0
w 432 240 368 240 0
O 432 128 496 128 1 0
w 272 240 288 240 0
a 208 256 208 192 8 15 -15 1000000 2.079598055202733 8.323315996821838 100000
w 208 192 208 176 0
w 432 128 432 80 0
w 432 80 528 80 0
w 528 80 528 288 0
w 528 288 224 288 0
w 224 288 224 256 0
R 192 256 192 304 0 1 100000 15 0 0 0.5
c 368 128 368 240 4 0.000009999999999999999 2.079598055202733 0.001 0
o 11 4 0 20482 5 0.1 0 1
o 1 4 0 4099 10 0.003125 1 2 1 3

Suppose we had a buck converter with no feedback.The relationship between voltage output and voltage input would be Vout = Vin*D where D=duty cycle of a mosfet.However in my case the output voltage is fed back to the input of my mosfet so the expression becomes Vout = Vin*D(Vout) and because it is fed to the inverting pin of the opamp it is more precise to say Vout = Vin*D(-Vout).

When I run the simulation I change the AC amplitude of the AC source connected to the non-inverting amplifier.The output voltage peaked at 2.08V when I set the AC amplitude of the AC source at 15V , increasing the amplitude of the AC source doesnt change the DC output voltage of the buck converter while setting the AC source at less than 15V decreases the DC output voltage of my buck converter.Why?

 

I modified the circuit a bit https://tinyurl.com/28fqldsk ← Falstad

what we must realize here is that the divided output voltage is compared against the arbitrary amplitude and DC-offset input PWM
▲
such does not give us an unarily determined output voltage regulation
RATHER an output LC tank defines some time constant for di/dt input that has to compared against the Net current loss on the output capacitor (which is in average proportional to output voltage drop dv/dt) -- the point is the voltages that go into Op Amp must be both comparable either to di/dt or to dv/dt ←← however it sets the hold output voltage not the set output voltage regulation

 

https://www.digikey.se/sv/articles/...ching-dc-dc-regulator-efficiency-at-low-loads

https://www.ti.com/document-viewer/lit/html/SSZTC38

 

Look

 

I have made a buck converter with negative feedback on Falstad:

View attachment 360801


Here is the code of the circuit:

$ 1 3.125e-7 382.76258214399064 50 5 50 5e-11
R 192 128 160 128 0 0 40 5 0 0 0.5
f 208 176 208 128 40 1.5 0.02
w 224 128 272 128 0
d 288 240 288 128 2 default
w 288 128 272 128 0
l 288 128 368 128 0 1 0.0020841426863125808 0
w 368 240 288 240 0
g 288 240 288 256 0 0
r 432 240 432 128 0 1000
w 432 128 368 128 0
w 432 240 368 240 0
O 432 128 496 128 1 0
w 272 240 288 240 0
a 208 256 208 192 8 15 -15 1000000 2.079598055202733 8.323315996821838 100000
w 208 192 208 176 0
w 432 128 432 80 0
w 432 80 528 80 0
w 528 80 528 288 0
w 528 288 224 288 0
w 224 288 224 256 0
R 192 256 192 304 0 1 100000 15 0 0 0.5
c 368 128 368 240 4 0.000009999999999999999 2.079598055202733 0.001 0
o 11 4 0 20482 5 0.1 0 1
o 1 4 0 4099 10 0.003125 1 2 1 3

Suppose we had a buck converter with no feedback.The relationship between voltage output and voltage input would be Vout = Vin*D where D=duty cycle of a mosfet.However in my case the output voltage is fed back to the input of my mosfet so the expression becomes Vout = Vin*D(Vout) and because it is fed to the inverting pin of the opamp it is more precise to say Vout = Vin*D(-Vout).

When I run the simulation I change the AC amplitude of the AC source connected to the non-inverting amplifier.The output voltage peaked at 2.08V when I set the AC amplitude of the AC source at 15V , increasing the amplitude of the AC source doesnt change the DC output voltage of the buck converter while setting the AC source at less than 15V decreases the DC output voltage of my buck converter.Why?

Probably because the average DC voltage of the input is zero, and something odd happens like phase reversal when the input is driven negative without a negative supply.