1
\$\begingroup\$

enter image description here

I would like to test the efficiency of some buck converters. I am thinking of using two current sensors like INA219 (or similar) connected to an Arduino and using the serial monitor to gather data.

  • Vin: 12V
  • Vout: 5V, 3.3V for now.
  • Iout: 8-10A max. maybe 15-20A in future.

Instead of dissipating the output voltage of the buck converter into heat using a lamp or a power resistor, I would like to re-convert the buck output using a boost converter to be reused by the buck. It’s like a “power circle."

  • Can I use the output current limit of the boost converter to create gradual steps to measure efficiency over Iout?
  • How can I control Ilim resistor with my microcontroller or do I have to do that manually with a potetiometer?
  • Lastly, does Vin main act as a supplemental supply while the entire output of the boost ist being used by the buck?
\$\endgroup\$

4 Answers 4

-2
\$\begingroup\$

Instead of dissipating the output voltage of the buck converter into heat using a lamp or a power resistor, I would like to re-convert the Buck output using a boost converter to be reused by the Buck.

That won't work. Once the "feeding back" boost converter is producing sufficient output voltage to "supplement" the main power source, the load on the buck converter drops to that required to keep the boost converter operating in a no-load situation.

In other words your idea is flawed.

\$\endgroup\$
7
  • 2
    \$\begingroup\$ The RoadRunner tried something like that! \$\endgroup\$
    – Kartman
    Mar 22, 2021 at 11:23
  • \$\begingroup\$ Are you suffering from a misapprehension that this is over-unity or something? \$\endgroup\$
    – Neil_UK
    Mar 22, 2021 at 14:28
  • \$\begingroup\$ No Neil, I'm not and it won't work how you think it will. With no real load on the buck converter, the boost will only take what power it needs to supply current to the buck input and, given that the buck has no real load on it then it can't be used to mimic anything other than the lightest loading condition. \$\endgroup\$
    – Andy aka
    Mar 22, 2021 at 14:31
  • \$\begingroup\$ @Andyaka I suspect that you are (unusually) wrong here. Assume the Vin source is low impedance (eg lead acid battery as Neil suggests). If you apply a light nominal load to the buck converter at say 5V and then program the boost converter to provide say 100 Watts out at 12 V, as the boost rapidly loads down the buck, the buck responds by ramping up output power. While instability may occur it's not certain to and should be designable not to be. \$\endgroup\$
    – Russell McMahon
    Mar 23, 2021 at 12:15
  • \$\begingroup\$ @RussellMcMahon the op specifically said in his own words "to be reused by the Buck." - in other words, not to be fed back into the lead acid (aka low impedance) battery or source and, what you appear to be suggesting. \$\endgroup\$
    – Andy aka
    Mar 23, 2021 at 12:18
1
\$\begingroup\$

Yes, this will work, subject to stability and startup concerns.

Let's do some sums. Say the buck converter is supplying 10 watts. It will need to consume (say) 11 watts. The boost converter, with an input of 10 watts can only produce 9 watts. This leaves the external power source to inject an extra 2 watts, to cover the losses in both converters.

Stability is a real concern, as (assuming a bench power supply) you have three connected systems all with feedback loops controlling them. There is every possibility that they could fight each other, leading to wild and possibly damaging swings of voltage and current.

Ideally the power source should be something unimpeachable, like a lead acid battery, with a very low impedance, capable of sourcing and sinking large currents. Use a current-output battery charger to keep it topped up. An alternative would be a very large capacitor, providing a low impedance, and an ability to source and sink current, with tame behaviour.

The boost converter should be current output. Its output voltage will then be defined by the power source. The buck converter will then consume all the current the boost converter produces, with the extra current to cover losses being drawn from the power supply.

Startup may be an issue if the converters have problem into zero load, or have large input storage capacitors which allow them to momentarily draw less input power than they are providing. Ideally, you would start with the boost converter set to zero output current. The buck converter will then be drawing its quiescent current, plus an input for the boost converter's quiescent current. As you increase the programmed boost converter output current, it will throw more load onto the buck. It may help stability if you have a small real load on the buck output, just to increase the current required at 'no load'.

\$\endgroup\$
7
  • \$\begingroup\$ Where's the power going? The power has to go somewhere. The op wants it to go back into the buck converter. \$\endgroup\$
    – Andy aka
    Mar 22, 2021 at 14:48
  • \$\begingroup\$ I would like to re-convert the Buck output using a boost converter to be reused by the Buck. \$\endgroup\$
    – Andy aka
    Mar 22, 2021 at 15:29
  • 2
    \$\begingroup\$ @Andyaka Losses. This sort of thing is done all the time. I read a lovely article testing the right-angle drive gearboxes used for train traction, a few MW each IIRC. Four were arranged in a square, and driven by a small motor at rated speed. Then a variable angle coupling was used to 'torque up' the ring to rated torque. Result was up to full power circulating in the ring, with the motor supplying the losses. Same thing here. Even 90% efficient converters will need the voltage source input to be supplying 20% of the rated circulating power. \$\endgroup\$
    – Neil_UK
    Mar 22, 2021 at 15:51
  • \$\begingroup\$ @Neil_UK I have done the same for long term testing battery chargers, use an inverter as load to feed mains power back to the input and supply only the losses via external DC power supply to save on the electricity bill. The same is used in HVDC testing too in MW range but only pay for the kW of losses via DC current injection. \$\endgroup\$
    – winny
    Mar 23, 2021 at 7:52
  • 1
    \$\begingroup\$ @VladimirCravero Completely correct. The energy balance, the thermodynamics, is just fine. But if you want to try this, I would caution about the practical potential problem of stability. Power converters can be built to do funny things in edges cases like no or low load, or overload. This is why I recommend use a real low impedance dumb battery as the power input to break any vicious feedback. See winny's comment, he uses the mains as the low impedance power supply, which will work just as well for controlling any unwanted instability. \$\endgroup\$
    – Neil_UK
    Mar 23, 2021 at 9:17
0
\$\begingroup\$

Instead of dissipating the output voltage of the buck converter into heat using a lamp or a power resistor, I would like to re-convert the buck output using a boost converter to be reused by the buck. It’s like a “power circle."

This will work if the boost converter is run in current regulation mode (constant current mode), and the main power source supports current flowing back into itself (like a lead-acid battery, or a power supply with sufficient output capacitance). The feedback network should be designed so that you can set the current set-point of the boost converter if you want to test your buck converter at various loads. The maximum voltage for the boost converter should be set above the supply voltage, so that the boost works in current regulation mode, rather than voltage regulation mode.

Can I use the output current limit of the boost converter to create gradual steps to measure efficiency over Iout?

Yes, if your boost converter feedback network is designed correctly.

How can I control Ilim resistor with my microcontroller or do I have to do that manually with a potetiometer?

It would be easier with a potentiometer. With a microcontroller, you have several options.

  • You could drive a digitial potentiometer
  • You could drive a DAC
  • You could output a pwm signal, and filter it sufficiently so that it gives a stable voltage. I wouldn't recommend this, because ripple in your control voltage will cause errors in your output current.

You refer to an \$I_{lim}\$ resistor. Of course you will need to measure the current somehow, so you will need a resistor. But don't fall into the trap of thinking that the current control needs to be accomplished with a simple resistor as it may appear in a reference circuit in a chip's datasheet. You can also use an op amp to combine a voltage generated by a shunt resistor with a control voltage, used to provide current feedback. That is why an DAC may be used to vary the current.

Lastly, does Vin main act as a supplemental supply while the entire output of the boost ist being used by the buck?

Yes, all the losses in the two converters must be made up by the main power supply.

\$\endgroup\$
0
\$\begingroup\$

I agree with others here that this might work, but you probably need some load to "calm" everything down, especially during startup. But: is this setup actually suitable to test buck and boost converters? There are several issues here that will/may severely affect the performance of both the converts, so the measured performance might be quite different than what it would be in a more conventional circuit.

In addition to that there are several caveats that will require some specific conditions for the converters. Those conditions are unique for this circuit, so you might not be able to test the converters in any useful way. Sure, you can learn to build a rather interesting setup, but I can almost guarantee that if you change out a converter you need to adjust loads and capacitances so it will start up and run correctly.

If your goal is to test the converters, I would build a circuit for that: a setup where you can easily change parameters like capacitance and load and investigate the parameters that are important to your final circuit.

\$\endgroup\$

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.