My previous question was about a hysteretic-style buck converter:

Buck Converter Control Scheme- Why isn't just a comparator not enough?

By what I gather from previous answers, a hysteretic-style buck converter would not work efficiently as the switching frequency is not constant, resulting in output ripple.

Here is the schematic of standard Voltage Control Mode scheme: enter image description here

The control loop has two main parts: error amplifier and voltage comparator. An error amplifier is a differential amplifier with a high gain.

I understand that to maintain switching frequency constant, the error voltage is compared with a sawtooth(of fixed frequency) and the output of comparator (VR) controls the duty cycle.

What is the function of an error amplifier here?

Even if the output was directly compared with a sawtooth wave, the switching frequency would have remained constant.

What I've read is that an error amplifier is essentially an integrator in this case. How does an integrator help here? Is the function of capacitor is just frequency compensation?

Help me out! This is a difficult topic for me and there aren't many resources available :(

** Update ** The output of integrator would be a ramp with negative slope at very low frequencies. How does that help? Why only an integrator?

enter image description here

  • \$\begingroup\$ who said hysteretic-style would result in increased ripple? the point of hysteretic control is to ensure a specific ripple... the frequency will vary w.r.t. load and yes this will increase switching losses. But lets keep the FUD out of this. As long as the bandwidth of the controller is greater than load response it is fine. SwitchReluctance machines are usually controlled with hysteretic, soft-switching schemes \$\endgroup\$
    – user16222
    Commented Jan 3, 2016 at 18:20
  • \$\begingroup\$ That's not the schematic of a voltage-mode converter. You're comparing the output of the error amp with the current in the inductor. A traditional voltage mode PWM has a sawtooth generator on that input to the comparator and the PWM output of the comparator driving the switch. Your schematic is showing current-mode control. \$\endgroup\$
    – John D
    Commented Jan 3, 2016 at 19:40

1 Answer 1


Your schematic is showing current-mode control.

In voltage-mode control the error amplifier compares the output voltage to a reference, then provides an output that is compensated and compared to a sawtooth wave that will vary the PWM duty cycle to drive the error lower.

At low frequencies (esp DC) we want the error to be as small as possible, so we want the error amp/compensation to look like an integrator. Any error will then build up and drive the PWM to force the error to zero. Only zero error will allow the output of the compensation to reach steady state at DC.

If we just used an integrator for the compensation we would have to close the loop at a very low frequency which would mean poor transient response (poor rejection to disturbances). That's because an integrator provides 90 degrees of phase shift and we would have to close the loop well below the L-C output filter resonance which provides another 180 degrees of phase shift. We also wouldn't be able to control transients causing the output filter to ring because our control bandwidth would be lower than the LC resonant frequency. Still, we want to put a pole at DC for good DC regulation.

The output filter L-C has 2 poles. So in order to cancel those two poles we can put 2 zeroes in the compensation. Now we still have that 90 degree phase shift until the switching frequency effects and error amp bandwidth start to come in.

With the two zeroes we form in the compensator we get a couple of poles as well (if we want them or not).

So we put one additional pole in the compensation somewhere around half the switching frequency to filter out noise and ripple, and a pole at the capacitor ESR zero frequency to cancel that. Now we can close the loop at a reasonable frequency with a reasonable phase margin and still get good DC accuracy.

  • \$\begingroup\$ Could you please check the update? \$\endgroup\$
    – JGalt
    Commented Jan 4, 2016 at 13:50
  • 1
    \$\begingroup\$ The schematic now shows a conventional voltage-mode fixed-frequency PWM buck converter. If what I wrote above is not clear you should start to study control theory, that will help you understand what the purpose of the compensation is. \$\endgroup\$
    – John D
    Commented Jan 4, 2016 at 17:57
  • 1
    \$\begingroup\$ Thank you I get the basic idea of why the compensation is required here. I need to study control theory properly to fully understand your answer. I'd thought that the only reason why an integrator is used is for frequency compensation. But according to your answer, "At low frequencies (esp DC) we want the error to be as small as possible, so we want the error amp/compensation to look like an integrator. Any error will then build up and drive the PWM to force the error to zero." How does using an integrator reduce the error? \$\endgroup\$
    – JGalt
    Commented Jan 4, 2016 at 18:22
  • 1
    \$\begingroup\$ The output of the compensation changes the PWM duty cycle in the direction to reduce the error, correct? So consider what the output of an integrator does given even the smallest amount of steady-state error. The step response of an integrator is a ramp, right? So any DC error at all will result in the integrator output ramping up or down and changing the PWM until the error is exactly zero. (Assuming an ideal amplifier, etc.) \$\endgroup\$
    – John D
    Commented Jan 4, 2016 at 19:08
  • \$\begingroup\$ Thank you! I'll have to study control theory properly to analyze compensation networks and understand the topic further :) \$\endgroup\$
    – JGalt
    Commented Jan 5, 2016 at 2:48

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