Two Synchronous Buck Converters as Bridge Drive to Peltier

Question

I am looking to control a peltier without using PWM at the device leads (filtered PWM). The usual circuits I have seen is by having a regulated supply as the source to the top FETs on an H-bridge to flip the polarity to go back and forth for heating and cooling. The regulated supply can then be adjusted to control the applied voltage and current.

Of course with me going into circuit imagination land, I was thinking of how else I might want to do this. I got an idea from an application circuit I saw one time:

Instead of using two high power opamps, I am was thinking to instead use two synchronous buck converters. I can hold one at a steady voltage, and then adjust the other one from a microcontroller (fiddle with the feedback pin) to control the heating and cooling current. When I was looking through specialized peltier ICs, I came across exactly that type of setup: https://www.analog.com/media/en/technical-documentation/data-sheets/1923f.pdf

Though these types of ICs seem really expensive and are hard to find in stock nowadays so I went ahead and thought perhaps I can go ahead and just grab two buck converters and do the same. And as luck has it, I found an app circuit on it with the same idea:

So my idea is to do the same but with a cheaper and more readily available buck converter: http://www.aosmd.com/res/data_sheets/AOZ2261AQI-15.pdf

The way I plan on fiddling with the feedback pin will be like this: https://www.edaboard.com/threads/design-of-pwm-controlled-smps-output-voltage.377215/

Okay FINALLY my question, is this a legit way of controlling a peltier? I mean, it makes sense to me, and after researching synchronous buck converters for a few hours to get me more educated on the topology, it seems okay. The app circuits do bring me some confidence on legitimacy as well. My worry is how the current will be for either buck converter when it is sinking current. From what I understand, since the topology allows for continuous conduction mode, sinking current is not a big deal. Though I am unsure if there will be current sinking eventually to the source at the top FETS (usually called Vin for buck converters). I do not want any current going back to the voltage being bucked and I don't think it will, but I am having a little difficulty conceptualizing the flow of the return current. For more info, I plan to use this circuit with a microcontroller as part as a temperature controlled PID loop.

Answer 1

A synchronous buck converter contains a half bridge, and a half bridge can both source and sink current. A synchronous buck converter can force continuous conduction mode just by always having one or the other switch on -- and such a converter would act like one half of an H-bridge circuit, complete with regulation.

The LM2576 in your last picture uses a diode (D1) as the bottom-side switch of the converter. This means that the regulator circuit cannot actively sink current. While the regulator can work in continuous-conduction mode, it does so when the load is pulling enough current so that current in L1 never reverses.

In theory you could parallel D1 in that circuit with a FET, and turn it on after you disable the controller chip. To me this seems -- dangerous. As an alternative, you could roll your own H-bridge circuit complete with LC filters and just command PWM to it -- this gets pretty deep into switched amplifier design, though.