Measuring average current through a boost circuit

Thread Starter

MrSoftware

Joined Oct 29, 2013
1,564
I've got a typical boost circuit and I need to measure the average current across it. I put a .24 ohm resistor in series with the diode on the output side and measured across it with the scope, see the screen shot below, but the negative voltages make me think my reading is not valid and I'm actually picking up emf from the inductor. I guess it could also mean the diode is too slow and current is going the wrong way, but that seems unlikely. Can anyone offer tips for measuring the average DC current here?

Oddly the 420mVrms shown in the scope screen shot would give me 1.75A, which is about what the circuit is drawing from my bench supply.

Output side of the circuit with the .24 ohm resistor:

upload_2018-10-10_19-38-27.png

Scope:

upload_2018-10-10_19-29-33.png
 

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ebp

Joined Feb 8, 2018
2,332
In a boost converter operating in continuous current mode (the inductor current never drops to zero) the inductor current is a triangle wave offset from zero by a varying amount. The switch current and the diode current are "chopped." The switch current is a ramp-on-pedestal with a rising ramp and the diode current is a falling ramp on pedestal. Again, the pedestal height varies. The peak to peak ramp amplitude also varies. If you assemble the switch current and the diode current into one waveform, they will be the same as the inductor current waveform.

The question, then, is which current's average do you want to measure?
What are you input and output voltages and what is the load?

Of course if you are measuring the diode current with the resistor, the scope "ground" must go to the right hand end of the resistor as you have drawn it (the positive output of the supply). Some of the noise will be "real", some of if probably isn't. To get a decent waveform you need to use a probe tip grounder, not a ground lead for the probe. This is especially true if you have to resort to using a probe at 1:1.
Of course no other probe can be connected to the circuit unless you use the same point for ground.

The diode you are using is a Schottky, so it will have "negligible" reverse recovery time.
 

Thread Starter

MrSoftware

Joined Oct 29, 2013
1,564
Thanks for the reply. The goal is really to get the average current leaving the boost circuit so that we can put an appropriate fuse on it, on the boosted side. It's powering a piezo driver circuit and the piezo is playing an ever changing set of tones, just to make it more interesting..

I ended up using 2 probes across the resistor and had the scope show me the difference. I think the scope does not calculate the difference as frequently as it acquires data points (or the resolution isn't high enough) because the difference data looks very binary. See below, the blue and yellow (under the blue) are the channels 1 and 2 output, and the purple is channel 1 minus channel 2, or the voltage across the resistor. It's really difficult to get an average voltage from that. So I ended up using the scope output to guestimate the peak current, then used an rms volt meter (fluke 87v) to get what is hopefully a reasonably accurate average DC voltage. We're going to order a few fuses around the value that we think is right and see what happens.


upload_2018-10-12_8-55-47.png
 

ebp

Joined Feb 8, 2018
2,332
Assuming the output of the boost is filtered with a capacitor, the sensible place to try to measure the current is after the capacitor.

Differential measurements with oscilloscopes by using two channels and math is often pretty unsatisfactory. It looks to me like it is essentially quantization difference between two channels with gains such that there really isn't any difference between them except for quantization limitations/error. This is a problem with producing a difference numerically post-digitization rather than as an analog difference before digitizing. It works OK-ish, sometimes.

Fuses are resistors, so they respond to the RMS current, not the average. The ratio of RMS to average depends very much on the waveform.

I assume the fuse is to protect the load, not the boost circuit.
 

Thread Starter

MrSoftware

Joined Oct 29, 2013
1,564
I think you're correct about the comparison post-digitization. There are capacitors in the circuit that didn't quite make the screenshot, just to the right of the resistor. And now that you mention it.. I bet the waveform would have been more friendly if I had put the resistor further down the line after the capacitors, that's a great point. The diode was convenient since it's in a relatively big package, which made scabbing on the resistor relatively easy, but I may have to go back and try that again with the resistor relocated to see if the results are any easier to read. The fuse is required for the device to pass intrinsic safety requirements. Intrinsic safety is not an area where I normally work so I'm just helping to collect data for the real decision makers on this one. :)
 

noweare

Joined Jun 30, 2017
101
The diode is reversed biased when the switch is on so that is why it looks like that. I don't know about the negative voltages and gaps but when your switching an inductor you will see spiking.
 
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