0
\$\begingroup\$

Looking at modern ICs, we prefer current control feedback over voltage control feedback, due to the speed of the feedback. Current control will react faster than voltage control. Here are 2 images, first of voltage control PWM feedback and secondly of current control PWM feedback

voltage control PWM feedback

current control PWM feedback

But why is current control fundamentaly faster?

Both images come from the RECOM DC/DC book of knowledge

\$\endgroup\$
  • 1
    \$\begingroup\$ Your example pictures are not relevant to the question. This is because both applications are relatively low speed and one is trying to regulate voltage whilst the other is trying to limit current. Both signals fed back are converted (or are) voltages hence they do not typify why some op-amps are voltage feedback and some are called current-feedback types. \$\endgroup\$ – Andy aka Jan 3 at 8:10
  • \$\begingroup\$ @J.Joly: Thinking aloud: is it not that for inductive loads the voltage feedback tells us very little whereas the current is more directly related to the output power? \$\endgroup\$ – Transistor Jan 3 at 10:01
  • \$\begingroup\$ @Transistor, uhm yeah, your voltage can change immediatly, while your current changes gradually. But it is not because it can change immediatly that it will change. I'm not quiet sure, but I don't think that there is a DC DC converter for example that changes immediatly and drasticly its voltage. \$\endgroup\$ – J. Joly Jan 3 at 10:31
2
\$\begingroup\$

All transistors have GBW limit whether they are BJT’s or FETS. Yet common emitter or source configurations are unity gain thus maximum BW. This is why ECL and CML and current mode comparators are fastest yet have lower input impedance.

But in this analog case the CM PWM does not have to integrate the voltage and simply compares the voltage error with the current * Rsense =Vs to match the scaled down voltage error in order to cut off the driver (reset). The voltage error ripple is reduced by detection on each pulse rather than the integration of many pulses from the filter time constant which must be many pulses in delay to reduce ripple voltage.

The Current Mode Feedback (CMF) reduces the inner loop to a much faster 1st order LPF for smaller Step load variations.

The Voltage Mode Feedback (VMF) with a 2nd order group delay response is twice as much even if the BW were the same, yet we know the pulse feedback has much higher loop bandwidth than the switching frequency so the small step delays are much less.

  • added

But given the Volt-second saturation limits of the inductor's charge current, there are limits to the benefit of the CML feedback when the PW rises to 100% for large step load or large input step voltage changes. These tradeoffs improve the small step response delay more than the large step response but in both cases improve stability and ringing.

\$\endgroup\$
  • \$\begingroup\$ So to conclude, due to the fact that with current mode you see at the input what changes and with voltage mode you only see what changes at the output. Which makes current mode faster, due to the delay your output has on the change of your input? \$\endgroup\$ – J. Joly Jan 3 at 10:26
  • \$\begingroup\$ The group delay of a 1st order LPF at same cutoff is already 1/2 of a 2nd order LPF yet they have different breakpoints \$\endgroup\$ – Sunnyskyguy EE75 Jan 3 at 17:51
1
\$\begingroup\$

I disagree with the statement that current-mode control is faster than voltage-mode control. The control-to-output transfer functions are certainly different between the techniques and probably easier to compensate with current-mode control. However, the controlled parameter in both cases remains the same: the duty ratio \$D\$. It is the way you construct it which differs between the adopted approaches. Remember that available volt-seconds are identical for a given structure whether you talk about voltage- or current-mode control.

In voltage-mode control or VM, you directly control \$D\$ via a comparator observing the error voltage and an artificial sawtooth. In current-mode control or CM, the artificial sawtooth is an image of the inductor current whose peak is controlled by the error voltage, hence the name also found in the literature of indirect duty ratio control. But if you take a boost converter for instance, the maximum crossover frequency is limited by the RHP zero and it is the same in CM and VM. You have a similar situation with a buck-boost or a flyback converter. If you take a buck converter operated in voltage-mode control, there is no theoretical upper crossover limit below half of the switching frequency. More complicated with current-mode control where subharmonic poles hamper the response at half the switching frequency.

Oui, the dynamic response of a CM converter is of 1st-order at low frequency (3rd-order converter in general for the 3 basic structures) and thus easier to control - in most cases - than the VM counterpart. Also, mode transition from CCM to DCM and vice versa is less troublesome in CM than in VM, making CM the most popular control scheme. But a lot of dc-dc bricks are still operated in VM not CM: better open-loop output impedance, no need for slope compensation which distorts current limit, feed-forward brings excellent input rejection in VM etc. As a preliminary conclusion, no straight answer!

\$\endgroup\$
  • \$\begingroup\$ okay so I read that if you look at the situation with a dynamic load, CM will react faster? \$\endgroup\$ – J. Joly Jan 3 at 10:24
  • \$\begingroup\$ It all depends on the compensation strategy, where the poles and zeros are placed and what crossover is selected. As I said in the text, the goal is to change current in an inductor and whether you deal with CM or VM, you are stuck by physical laws, volts over \$L\$. I have practical examples where two different CM and VM designs lead to the same response. \$\endgroup\$ – Verbal Kint Jan 3 at 10:30
  • \$\begingroup\$ I guess by speed they mean Step Response to load or input changes \$\endgroup\$ – Sunnyskyguy EE75 Jan 4 at 12:24
0
\$\begingroup\$

Maybe this answers your question:

A constant voltage regulator is responsive only to changes in the load voltage and adjusts the duty cycle accordingly. As it does not directly measure load current or input voltage, it must wait for a corresponding effect on the load voltage with any changes in the load current or input voltage. This delay affects the control characteristics of the switching regulator so that there are always several clock periods required for stabilization. The control loop must therefore be compensated to avoid overshoot or output voltage instability.

(Source: Section 1.2.5 PWM-Regulation Techniques from the RECOM DC/DC Book Of Knowledge)

So, the VM won't change the duty cycle until the output is already wrong, because it measures the output. While CM can change the duty cycle of the fly to make the output stable, according to the changes in current...

\$\endgroup\$
  • \$\begingroup\$ Why load current would be better than load voltage I have no idea... \$\endgroup\$ – Jona Van Puyvelde Jan 3 at 10:54
  • \$\begingroup\$ Sad the document is behind a gimme-all-your-personal-data-wall :/ \$\endgroup\$ – try-catch-finally Jan 3 at 13:48
  • \$\begingroup\$ The book bit.ly/2QmFWG8 The answer lies in group delay for small steps in my answer , yet by design choice e.g .1/3 max \$\endgroup\$ – Sunnyskyguy EE75 Jan 3 at 18:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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