I'm currently learning about negative feedback and voltage mode control for various DC-DC converters. I'm very confused about how the control voltage changes the pwm signal, along with how the reference voltage and sensor gain are chosen. The pwm generator works by comparing the input control voltage to a ramp voltage. If the control voltage is greater than the ramp voltage the output is high and if it is lower than the ramp voltage the output is low. This results in a series of square pulses as shown below: enter image description here

Let's say we have the following buck converter with an output voltage of 15V: enter image description here

Why is a 5V reference voltage selected? Can it be anything? The sensor gain for this problem is selected to be 1/3. Is it 1/3 because we want the nominal output voltage to be equal to the reference voltage? I am quite lost from the part after the error comparator. Let's say that the voltage being fed-back is exactly 5V so Ve = 5-5 = 0V. If Ve is 0, how is the pwm generator able to generate any pulse signal to drive the transistor since the control voltage will be 0 coming out of the compensator?

Also, how is the ramp voltage value chosen? I would imagine that it would depend on the gain of the compensator since if the gain is too large, then the control voltage would always be higher than the ramp voltage.


1 Answer 1


"If Ve is 0, how is the pwm generator able to generate any pulse signal to drive the transistor since the control voltage will be 0 coming out of the compensator?"

In a proportional negative feedback system, the controlled output will never track the target exactly. There will be some error, but if the overall loop gain is high, that error will be small.

When "integral" feedback is added to a system, the error is integrated (and possibly added to a proportional error and/or a differential error), so that even a tiny error will eventually cause the system to respond to correct that error. The error will then tend to become increasingly small over time.

Without knowing the nitty gritty of the compensator, I can nevertheless note that it is characterized by G(s). Those particular symbols (G and s) are typical of a what is called a transfer function (The ratio of the Laplace Transform of the output of a component to the Laplace Transform of the input). The mere use of a transfer function in this spot, suggests that there may be some "integral" feedback here. (Or not. It isn't specified, so I don't know for sure.)

  • \$\begingroup\$ So when we design the control loop, are we assuming that the error is never zero which is why there will always be some duty ratio coming out of the pwm generator? I guess I'm confused about how that specific duty ratio (in this case about 0.536) is maintained when the error is really low if the ramp voltage comparison is used to generate the pwm signal. \$\endgroup\$
    – ggg123
    Commented Dec 4, 2020 at 1:00
  • \$\begingroup\$ Yes, there will always be some duty ratio coming out of the pwm generator, if only because no system tracks perfectly. Suppose for a moment, that rapid response was not necessary. If the error is integrated, by the compensator, a small positive error signal will cause the compensator output to grow. That will cause the duty cycle to grow, which will cause the output voltage to grow. Then the error signal becomes 0. But when the error signal is 0, the output of the compensator (if it is a pure integrator) will just remain at whatever value it was before the error became 0. \$\endgroup\$ Commented Dec 4, 2020 at 1:05
  • \$\begingroup\$ Oh that makes sense. Then, should the ramp voltage peak be selected based on the control voltage that comes out of the compensator that has a zero error input? \$\endgroup\$
    – ggg123
    Commented Dec 4, 2020 at 1:55
  • \$\begingroup\$ A good question. I don't know the answer off the top of my head, but I think, it doesn't really matter, because it's Vc that controls the modulation, not Ve, and there could be amplification between Ve and Vc. However, I think it is certainly convenient, because you want a stable source for generating the ramp function, and you already have a stable source in the reference voltage, so why not give the reference voltage a dual purpose. But again, I think, you could have a separate source/ different voltage. \$\endgroup\$ Commented Dec 4, 2020 at 2:08
  • \$\begingroup\$ got it, thank you! \$\endgroup\$
    – ggg123
    Commented Dec 4, 2020 at 2:49

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