Help! electrolytic capacitor 5000 MFD needed

[John Fields]

On Sun, 12 Mar 2006 20:50:41 +1100, Daniel
<dxmm@nospam.albury.net.au> wrote:

John Fields wrote:

---
Most of my multimeters, on the 200 ohm full scale range, supply
about 1A into a short, so if we consider the ohmmeter a constant
current source it'll take:

dv C 1V * 5E-3F
t = ------ = ------------ = 5 seconds
I 1e-3A

to charge the cap up to a volt through the ohmmeter. Not exactly
what I'd call "almost zero time"

All fine and dandy if the Ohm-meter were a constant current source and
the capacitor could just keep absorbing those electrons, but neither is
the case so your logic fails.

---
Not at all. On the higher resistance ranges the ohmmeter _does_
approximate a constant current source; at least for the purpose of
this discussion.
---

(As a side note, I'll give you 5E-3F is the same as 5000 microfarads,
but 1e-3A is not the one amp you say your multimeter can supply into a
short.

---
Yes, but that one amp '1A', earlier, was a trypo. It should have
read: "1mA", so the math is correct as it stands. I'm surprised you
didn't catch that, since it's not likely _any_ common ohmmeter is
going to push 1 amp through the resistance it's measuring.

Even my trusty old Simpson 260 on the 'R times 1' range only puts
about 100mA through a 100mA d'Arsonval movement with an internal
resistance of 1.08 ohms.
---

If you fix this in the workings, t = 5 milli-seconds, which I
would call "almost zero time" when talking about digital multi-meter
response time.)

---
It doesn't need to be fixed, though. All that needed to happen was
for the typo to be corrected, so the 5 seconds stands at 5 seconds,
which even for a relatively slow digital multimeter with an update
rate of 2 readings per second isn't "almost zero time".

--
John Fields
Professional Circuit Designer