Undersupplied LM2576 buck-boost converter: Prevention, and where is all the power going?

Question

I had an LM2576 (ADJ variant) lying around and tried building a buck, and then an inverting buck-boost converter, purely for academic purposes. I have next to no prior experience with converters (or power management in general).

For the inverting buck-boost, I tried to follow the data sheet, but didn't have the right inductors, nor the right diode, but I thought that even though it may not work particularly well, it should work to some degree, right? Instead of the suggested 68uH, I used a 330uH power inductor rated at 800mA, and instead of a Schottky diode I used a regular one rated at 1A (also tried standard 1N4148). The goal was to make -5V from 9V, give or take with a load of no more than 50mA or so.

^{simulate this circuit – Schematic created using CircuitLab}

Background: It didn't work initially (had slightly different parts here and there), and on my lab PSU it looked like it short circuited. Some searching suggested that I need different inductor or output capacitor values, and it's totally important that they're 100% correct, and a Schottky diode, so after messing around with the parts I have at hand, I was getting partial success. Turns out that was a red herring, after reading the datasheet again (in particular the topic about the undervoltage lockout circuit), I tried disconnecting VIn, then turning on the PSU and letting C1 charge, and then connecting VIn, and it worked despite my borked parts.

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It appears that the problem was that the lab PSU was ramping up its output voltage too slowly, in combination with the overcurrent protection kicking in at 500mA, so output voltage dropped below 2V while the circuit was drawing a constant 500mA, and the LM2576 never had a chance to boot up. Depending on the inductor and capacitor used, the output is between 0V and around -2V or so.

My question is, when it's drawing a steady 500mA in the broken state, where exactly is all that power going? If the output is 0V, then it can't be the diode, so it must be in the LM2576 itself, but where exactly? How come it's not enough to turn on the circuit properly? My only guess is that the LM2576, being advertised as a pure buck-converter, regulated its duty cycle close to 100% (my intuition says a buck-boost converter should operate around 50% to 60%), so at close to 100% the inductor essentially only sees DC and nothing really happens. I don't have an oscilloscope (yet), but my multimeter wasn't able to pick up any frequency at the VOut pin in the faulty state (but was able to pick up the switching frequency when the circuit works correctly), which would support this theory.

To be clear, I know that several things are going wrong here, but I would like to understand why I'm observing what I'm observing when it's in this wrong state.

If that is the case, then how do commercial or industrial applications reliably protect against entering in this faulty (and potentially dangerous) state in the first place? Even with an undervoltage lockout circuit present, the supply could be too weak, no? (imagine a battery powered device)

Answer 1

I'm suspecting that your 1N4007 is causing issues. The datasheet of your LM2576 specifically says not to use it. According to the 1N4007's datasheet, the forward voltage is 1V at 1A, which is causing a very large voltage drop. Not to mention, there's also a matter of the reverse recovery time of the diode, which is 2us. That is not ideal for high frequency switching, even if your duty cycle is nearly 100%. With the LM2576 switching frequency of 52kHz, or a ~19us period, having 2us of reverse recovery time is going to take up a rather large portion of time for your LM2576 to try to perform at a sufficient efficiency.

Answer 2

My question is, when it's drawing a steady 500mA in the broken state, where is all that power going?

Its going into heat, if you have a thermal camera or some way to measure the temperature you could see which component it's going into. Some of the power is going into the diode, the 1N4007 has a very large forward voltage (1V@1A) typically you want a very low forward voltage so less power is consumed by the diode. You also must use the right inductor. Also make sure that you have the adjustable part as some parts have the feedback divider built into the part.

How come it's not enough to turn on the circuit properly?

Also make sure that you view the input and output on an oscilloscope, sometimes really fast switching can confuse some cheap bench supplies current measuring.

Answer 3

it worked despite my borked parts

It "worked" for some low value of "work", with unloaded output or with a very light load :)

1. The inductor, unless it was designed to be a power inductor, would likely saturate when the output had heavier load. The converter will not work then and go into thermal shutdown rather quickly as it would get super-hot from having shorted output.

2. The diode will dissipate lots of power due to slow reverse recovery and forward voltage 2-3x that of a Schottky. This would also manifest with the output loaded.

The higher inductor value is not that big of a problem. The converter will work, slightly less efficiently perhaps. The important rating is the saturation current of the substitute inductor. If that is lower than the original inductor, the inductor will saturate once the load current is high enough. Once saturated, the inductor current will grow very quickly - limited only by the series impedance of the converter's switch and the input capacitor. The converter will detect that and open the switch to protect itself and the inductor. The problem then is that nothing further limits the current through the diode once the switch in the converter opens. This may do further damage, depending on circumstances.