In a recent question it was asked whether a boost converter could be used to "recycle" power output by a buck converter being tested. There are quite a few commercial regenerative electronic loads available on the market. However, a) they tend to be of the power-line tie in type, sending their power to AC mains, rather than to a DC line, and b) I have been unable to find schematics for either commercial or DYI regenerative loads.

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In the circuit above, (from the question previously mentioned), \$V_{in}\$ is tied to the output of the buck converter. Whenever two power supplies are tied together, there may be contention for which power supply controls the voltage. For the circuit to work, and not settle at 0 volts as a stable point, the external power supply must "win" in any contention with the buck converter. The external power supply must provide power to make up all losses in the loop. One way to ensure that the external power supply is the "master" in this contention is to have the external power supply have a very low impedance, and the output of the boost converter a high impedance. Using an AC mains, or a lead acid battery for the external power supply satisfies the first condition. Using the boost converter in current regulation mode satisfies the second.

However, it would be nice to be able to use external power supplies which contain elements that block "back flow". Consider the following circuit:


simulate this circuit – Schematic created using CircuitLab

Provided the load can draw all of the current from the current mode boost converter (at the external supply voltage), the external supply will control the voltage. However, if the load cannot accept all that current at that voltage, the voltage across the load will rise.

Now, one may say "hold on a minute", the current into the load must be greater than the current delivered by the boost converter, else we would have a "free energy source". However, this is not so. It is only required that on average the energy supplied to the buck/boost loop equals the energy dissipated there. It is only required that on average the current from the boost converter must be less than the current drawn by the buck converter. However, both converters have elements that store energy, and simulations show that indeed, the buck may temporarily draw less current than the boost supplies, resulting in a significant rise in the tie-point voltage.

Ideas that immediately pop up as solutions are 1) use a bigger capacitor from \$V_{in}\$ to ground, or 2) use a shunt resistor from the buck \$V_{out}\$ to ground. A problem with 1) is that we don't know in advance how much energy might be stored in the buck converter, unless we know the details of the buck converter in advance. We might, but it would be nice not not have to rely upon such knowledge. A problem with 2 is that it reduces the amount of current recycled. Neither of these problems are a project killer. However, I wonder if there is a better way.

My question. Can anyone think of any other way to tie together an external supply which might restrict current back-flow, with a boost converter in current regulation mode, such that the voltage is determined by the external supply, even if the load does not accept all the current the boost converter is temporarily able to supply? Changes to the circuitry of the boost supply are absolutely acceptable. Adding a low-value sense resistor between the external supply and the tie-in point is also acceptable.

  • \$\begingroup\$ Somehow I am not getting the overall point. I wonder if there could be a simpler way to attain the overall goal. In any event, by putting an inductor between two H bridges you can create a universal bi-directional boost buck that can basically do whatever you want (as long as you don't need isolated ground). \$\endgroup\$ – mkeith Mar 26 at 5:03
  • \$\begingroup\$ Your link does not refer to the prior question. \$\endgroup\$ – Russell McMahon Mar 26 at 10:14
  • \$\begingroup\$ @RussellMcMahon thanks for catching that. fixed. \$\endgroup\$ – Math Keeps Me Busy Mar 26 at 12:26
  • \$\begingroup\$ I think the basic idea of the boost converter is fine. But instead of using a boost converter that is set up to regulate output voltage, you could use one that regulates input current, and is variable. So in essence, it is a boost converter programmable load. You don't need to worry about preventing backflow into the supply. Both converters are less than 100% efficient. This way the boost converter is not fighting with the supply either. Only the supply is regulating voltage. The boost just injects current. \$\endgroup\$ – mkeith Mar 26 at 16:49

This can regulate V1 or V2 regardless of which one is higher, depending on how the FET's are switched. It could also regulate current with appropriate feedback. And it can regulate irrespective of power flow direction. So, for example, V1 could be a battery. V2 could be the output from a buck or boost converter. The load current could be regulated.

I just kind of made up the inductor and capacitor values. Obviously it would depend on the application.


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ This is the sort of alternative thinking that I am looking for. However, I am not sure how your solution will solve the problem of one supply "overpowering" another supply with an embedded diode. Suppose V2 is the external supply, and V1 is the output of the buck converter under test. How does one regulate this circuit so that power flows from the V1 side to the V2 side without "overpowering" V2, (which I will assume is NOT an ideal voltage source, but contains a diode within it)? On the one hand you need to measure V2 for feedback. On the other hand, V1 can push the voltage of V2 up. \$\endgroup\$ – Math Keeps Me Busy Mar 26 at 5:45
  • \$\begingroup\$ V2 could be the regulator you are testing. V1 could be the input for V2. Basically if you put this H-bridge structure between two voltages it can move power whichever way you need to move it to regulate whichever thing you want to regulate (for example maybe you want to regulate the output current of V2, the buck you are testing). \$\endgroup\$ – mkeith Mar 26 at 5:49
  • \$\begingroup\$ If I understand you correctly, you are suggesting putting this circuit between the tie-point (between the external power supply and the boost converter) and the buck converter under test. Then if the boost converter raises the tie-point voltage too high, this converter will regulate it back down. Am I understanding you correctly? \$\endgroup\$ – Math Keeps Me Busy Mar 26 at 5:57
  • \$\begingroup\$ No. Use this as the boost converter is what I was thinking. But I actually also think a boost converter would work, as long as you set it up so it is regulating input current. Then it is like a constant current load on the Device Under Test (DUT). That is, the buck converter. \$\endgroup\$ – mkeith Mar 26 at 16:42
  • \$\begingroup\$ This probably does not address the underlying question. I may delete it if I muster up the determination to write a better answer. \$\endgroup\$ – mkeith Mar 26 at 16:50

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