1
\$\begingroup\$

I am trying to simulate boost converter using voltage-mode controller as the schematic below. However, the circuit doesn't work as expected.
Output voltage is supposed to be equal to reference voltage which is 8V but in this case it is only 3V. Also the output of comparator is always 0 which means the switch is permanent OFF. I tried to tune PI controller parameters but it doesn't help much.
Is there a systematic method to design this without having to tune a lot?

Schematic:

enter image description here

Output voltage and inductor current waveforms:

enter image description here

Vsaw, VPI3, Vsw signals:

enter image description here

VPI1, VPI2:

enter image description here

\$\endgroup\$
3
  • 1
    \$\begingroup\$ You need to think how this circuit is supposed to work and how it behaves in the simulator. The output voltage is too low so the VPI3 ends up as 1V, that 1V should result in an increase in the duty cycle of Vsw but it doesn't because Vsaw has a maximum of 1V as well so comparing VPI3 and Vsaw results in 0V and no switching. This is a boost converter, would it work if the DutyCycle of Vsw would be very large or very small? How can you guarantee that the DuCy of Vsw stays within these limits? \$\endgroup\$ Jun 5, 2018 at 7:02
  • \$\begingroup\$ You probably have a signal inversion - try swapping the input pins of the comparator. And also try to draw it without all those background dots on - they make it a headache to read. \$\endgroup\$
    – Andy aka
    Jun 5, 2018 at 8:37
  • \$\begingroup\$ @Bimpelrekkie: I tried to increase sawtooth amplitude to 2V while keeping upper limitter 1V and it works. However, if I increase sawtooth amplitude to 2V and increase upper limitter to 1.5V and this doesn't work. What is the reason for this? \$\endgroup\$
    – emnha
    Jun 5, 2018 at 11:11

2 Answers 2

1
\$\begingroup\$

Looking at the circuit I think it might start to work if you swap the inputs around: -

enter image description here

VPI3 is limited at +1 volt due to the signal clipper and this forcing the comparator inputs to not align. I reckon it's just a signal inversion thing.

Swapping (inverting) the signals at the voltage sensor would also do the job.

\$\endgroup\$
9
  • \$\begingroup\$ Thanks. I tried to swap tje comparator or voltage but it doesn't work. The output voltage is very small, about 20mV. \$\endgroup\$
    – emnha
    Jun 5, 2018 at 11:02
  • \$\begingroup\$ OK that means the boost switch is on permanently. Previously it was off permanently. Thinking..... Try simplifying and also try it with the comparator swapped back. The problem I have is that it is not 100% clear whether you have an unwanted inversion or whether your limiter is fouling things up. So simplify by removing the limiter and making the PI controller = 1. \$\endgroup\$
    – Andy aka
    Jun 5, 2018 at 11:28
  • \$\begingroup\$ I removed the limitter keeping PI controller and it works. However, if I remove PI then it doesn't work. \$\endgroup\$
    – emnha
    Jun 5, 2018 at 11:46
  • \$\begingroup\$ OK does it works without the limiter in the original placement of the comparator or with its inputs inverted as I suggested might be the problem? \$\endgroup\$
    – Andy aka
    Jun 5, 2018 at 11:52
  • \$\begingroup\$ What happens if you remove the limiter and PI and swap the comparator inputs? \$\endgroup\$
    – Andy aka
    Jun 5, 2018 at 11:54
0
\$\begingroup\$

I find the voltage sensor part a bit suspect. It should be a differential amplifier with gain of 1, with Vref connected to the inverting input and the output voltage connected to the positive voltage. In the original post, the output voltage is 5-ish, the reference input is 8-ish, so VPI1 should be -3V, but it's showing 5-ish. Check that.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.