Estimating peak voltage with a cap-diode setup?

[Ignoramus12834]

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

 

[Walter Harley]

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.

Meaning that you had a circuit like this?

-------o----------------->

|
V
-
|
SOURCE o-------. LOAD
| |
--- (METER)
--- |
| |

-------o-------o--------->

If so, then the meter would fairly accurately measure peak voltage, minus
the Vbe of the diode (about 0.7V), gradually falling due to leakage in the
capacitor. The peaks will charge the capacitor through its effective series
resistance, which (depending on what kind of cap you used) might be as high
as 10 ohms or might be a thousandth of that. To charge to 2/3 of a voltage
difference takes RC of time, so if it was 10 ohms at 2uF, that would mean
that if the V on the cap was 100V and a 200V peak came along for 20usec, the
cap would then show 170V. If the peak lasted 40usec, it would get to 2/3 of
the remaining difference, i.e., 190V. And so on. Leakage and effective
series resistance are orders of magnitude higher in electrolytic caps than
in other dielectrics such as polypropylene, so your results will depend on
what kind of cap you used. I trust you used a cap that was rated for that
much voltage...

 

[Ignoramus12834]

Meaning that you had a circuit like this?

|
V
-
|
SOURCE o-------. LOAD
| |
| |


If so, then the meter would fairly accurately measure peak voltage, minus
the Vbe of the diode (about 0.7V), gradually falling due to leakage in the
capacitor. The peaks will charge the capacitor through its effective series
resistance, which (depending on what kind of cap you used) might be as high
as 10 ohms or might be a thousandth of that. To charge to 2/3 of a voltage
difference takes RC of time, so if it was 10 ohms at 2uF, that would mean
that if the V on the cap was 100V and a 200V peak came along for 20usec, the
cap would then show 170V. If the peak lasted 40usec, it would get to 2/3 of
the remaining difference, i.e., 190V. And so on. Leakage and effective
series resistance are orders of magnitude higher in electrolytic caps than
in other dielectrics such as polypropylene, so your results will depend on
what kind of cap you used. I trust you used a cap that was rated for that
much voltage...

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

 

[Walter Harley]

[...]
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.

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.

 

[Tim Williams]

generally speaking diodes are a bit better at dissipating power than
caps, and caps blow up rather more dramatically.

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.

}

Tim

 

[Walter Harley]

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.

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.

 

[Ignoramus20878]

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

 

[Walter Harley]

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.

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.