Estimating peak voltage with a cap-diode setup?

I decided to try to estimate peak voltage in my circuit, by connecting a 2 uF capacitor in series with a diode to my DC bus. The hope was that the capacitor would be only charged and never discharged.

I removed varistors from the system.

I ran my bridge for a little while and took DC voltage readings across the cap.

My estimate of peak voltage in the DC rail at 150 amps was 240-300V. I was getting 240V most of the time, but once got 300.

My question is, just how much can I trust this number.

i

Thanks. The cap was rated for that voltage. The diode was rated for

600V and after a while of testing, blew. I suspect that I exceeded its current rating, I should have put a resistor in series. i

If you put a resistor in series, then the voltage on the cap will take longer to catch up with the peak voltage (that is, the cap will charge more slowly).

I'm a little surprised the cap didn't blow... the cap and the diode both have effective series resistance (ESR), so they both dissipate power; very generally speaking diodes are a bit better at dissipating power than caps, and caps blow up rather more dramatically. But you still haven't said what kind of cap (nor what kind of diode) you were using, so I guess all bets are off.

Idunno about that. I tacked a few 0.1uF 400VDC Cornell DME's on my induction heater testing tank circuit, and when they feel like it, they just melt and ooze.

:o}

Tim

-- Deep Fryer: a very philosophical monk. Website:

The OP wasn't forthcoming about what type of caps he was using, and with a

2uF value, it could have been anything - for instance, five 10uF 50V electrolytics in series :-) I didn't want to assume.

I have some Wima caps and such. They are low ESR. I in any case, ideally, there should be as little energy dissipated, only accumulated in the cap once.

i

So, probably polypropylene film? They should be low leakage, so most of the leakage should be through the rest of the circuitry.

Discharge time is just like charge time: t(.7) = RC. That is, time to discharge by 70% (down to 30%) = RC. So, assuming your meter has 10MEG input resistance, and your capacitance is 2uF, you should see t(.7) = 20 seconds.

Are you seeing less than that?

It is challenging to make a peak detector that will hold the peak for much longer than a minute or so. It can certainly be done, but it takes rather more sophisticated techniques.

You can use something like a LabJACK (USB data acquisition device) to take readings every second for however long you like, and plot them in a spreadsheet or such. That way you don't need your peak hold to be very long. Of course, the LabJACK won't be able to handle hundreds of volts, so you'll need a resistive divider just like others have suggested.