Capacitor charging problem

"john" schrieb im Newsbeitrag news: snipped-for-privacy@i12g2000prf.googlegroups.com...

Hello John,

The output stage of your current source will run into saturation beyond the compliance range. This will result in distortion and less amplitude on your 100kOhm load.

Best regards, Helmut

On a sunny day (Tue, 27 Nov 2007 10:51:01 -0800 (PST)) it happened john wrote in :

No, it will initially show a short! That is why it starts at zero volt across the capacitor. .6mA in 100kOhm makes 60V, you only have 15V, you example makes no sense.

If you omit the resistor for a moment, then the voltage across the capacitor is The charge Q = C * U = I * t

So U = (T * t) / C C in Farad, I in ampere, and t in seconds. The voltage across the capacitor will then rise LINEAR with time, until it reaches the 15V (theoretically). However, with the resistor in parallel, the max voltage will be limited by I * Rp, or the 15V, whatever is lower. In this case 15V is lower then 60, so 15V And the charge current in the capacitor is no longer linear, as as teh voltatge rises, the current through the resistor is increasing, substract that from the current in the capacitor.

The capacitor will charge like this:

see under 'Time domain considerations'. There is no longer a linear rise in voltage.

Hello,

The leakage current will charge the capacitor upto its maximum voltage , because the leakage current does not have any other place to go it will keep charging the capacitor. Now, my questions are

1. will the current source shut down?

1. for longer times, the charge will break the dielectric and the leakage current will increase?

Please advice! John

On a sunny day (Wed, 28 Nov 2007 12:59:12 -0800 (PST)) it happened john wrote in :

1) provide current diagram 2) what leakage current 3) what ARE you trying to drive, so what IS your load. 4) what frequency are you talkinga bout when you said 'AC' in a previous posting. 5) why do you drive the load, or try to drive the load with a current osurce.

Answers:

1) 2) 3) 4) 5)

CUL

If you're looking for something strange to happen, you have only one real practical unknown - the leakage current of the capacitor. This will only be a factor if the leakage current develops a signifigant voltage across a 100K resistor.

When you're considering practical components, ratings and breakdown, you'll be dissapointed to know that there's no guarantee that anything will or will not happen outside paper ratings, unless published otherwise or practically demonstrated by somebody with the time money and need to do so.

Leakage in electrolytics can increase with shelf age. There's a typical method recommended for 'reforming' higher voltage parts safely.

RL

Late at night, by candle light, legg penned this immortal opus:

The whole thing is probably unanswerable from a theoretical point of view. He states that the CS has a compliance of 15 V but the capacitor has a 10 V rating. So once the cap's charged beyond 10 V it may die with a bang or a whimper and possibly just take it like a hero. Most likely just start leaking, had an odd one like that the other day. A leaky cap making very low frequency noise, it would charge up for a bit, then drop a few volts, charge up again and so on at an average period of a couple of seconds, very irregular both in period and peaks. It was below rated voltage so probably just old age that got it.

- YD.

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