# RC Reality Check

Discussion in 'Electronic Basics' started by Ben Schaeffer, Nov 27, 2003.

1. ### Ben SchaefferGuest

I have been running some simple series RC experiments and found something
funny, at least based on the math of RC time constants. After charging
fully, the voltage drop across the resistor is *not* 0 in typical series RC
Circuit using batteries.

Voltage Drop
Testing with a
470 ohm resistor: the drop is not measurable
10k ohm resistor: drops 1%
100k ohm resistor: drops 9%.

I have experimented with 5v, 9v, and 12v using .01 uf, .22uf, and 4.7uf
capacitors with the same results. At this rate 10Mohm of resistance would
make the capacitor 0.

I even attempted to "cheat" by getting the capacitor fully charged through 2
parallel resistors and then remove one, but the circuit quickly normalized
to the same value as if I started with one resistor.

Anyone have some better math on this relationship?

2. ### John PopelishGuest

Are you using two meters, one across the cap and one across the
resistor?

If so, they form a voltage divider in parallel with the RC.

3. ### Ben SchaefferGuest

Just using one meter, measuring ground and node between resistor and
capacitor.

4. ### John PopelishGuest

Sorry, but I have no idea where ground is in the circuit. The meter
should be connected across the resistor. The actual resistance is the
parallel combination of the resistor and the meter's resistance. If
the voltage does not go to zero, open the capacitor (effectively a
capacitor very near zero capacitance.) If you still read voltage,
something is wrong with your meter.

5. ### Ben SchaefferGuest

Let me explain what I am seeing with a 9v battery with 100k resistor and
..01uf Cap -
Voltages:
ground to Vcc = 8.47
ground to node between R and C = 0 (obviously, as no current flowing)
node between R and C to Vcc = 7.71

6. ### John PopelishGuest

There is no ground on a 9 volt battery, unless you run a wire from one
of its terminals and poke it into the ground. It just has a positive
and negative terminal. And you haven't told me which terminal
connects to the resistor, and which to the capacitor, and where your

One more time.

Are the two leads of your volt meter connected to the two ends of the
capacitor, or to the two ends of the resistor?

7. ### Ben SchaefferGuest

Didn't mean to confuse things - I meant negative when I said ground.

Circuit is:
9v positive---resistor---capacitor---9v negative

Measuring voltage with a 9v battery with 100k resistor and .01uf Cap

black meter probe to 9v negative and red meter probe to 9v positive = 8.47v
black meter probe to 9v negative to red meter probe to node between resistor
and capacitor = 7.71
black meter probe to node between resistor and capacitor and red meter probe
to 9v positive = 0

By the way, have you tried this?

8. ### Jacobe HazzardGuest

You have 0 drop across the resistor because there is almost no current
flowing through the cap once it has charged. This agrees with the math fine.

The reason you measure less than full battery voltage across the capacitor
is because some current is flowing through the meter itself, causing a
voltage drop to develop across the resistor. If your meter has an impedance
of 1M, then the math works out again:
8.47V * 1M / (1M + 100k) = 7.71V (approx)
This is because the 100k resistor and the meter form a voltage divider, and
the cap basically does nothing. Try taking the cap out of the circuit, and
measuring with 9V <-> resistor <-> meter <-> 0V
and you should get the same value.

Most newer, digital multimeters have a 10M impedance, so either
B) your 100k resistor is actually 1M or

Since you tried with different voltage sources and resistors, we can
probably rule out B and C, and say that A is the case. Try looking for the
specs in the manual.

If you did try the experiment again with a 1M resistor, I think your would
measure very close to 50% of the battery voltage across the cap. A 10M
resistor would give you about 1/11 the battery voltage, or a 91% drop across
the resistor.

9. ### John PopelishGuest

Many times. With the meter connected this way, it is forming a
voltage divider with the other resistor. The 100k resistor shows a
calculated (8.47-7.71) .76 volt drop. The meter has about ten times
as much drop (7.71), so it must have about ten times as much
resistance (10*100k=1meg).

So you have proven that your volt meter has a resistance of about 1
meg ohm. replace the resistor with 1 meg and you should read about
half of the battery voltage.

10. ### Ben SchaefferGuest

Thanks for the explanation. I forgot to say earlier that I tried measuring
the capacitor after pulling it out and it discharged through the meter so
quickly I never got a reading close to the 8.47v :-(.

11. ### Ben SchaefferGuest

Ah, I got it now! 100k is the highest resistor I have now, but I am getting
some more resistors and will try this again. Thanks.

12. ### John PopelishGuest

Try it with a 1000 uf electrolytic and you should have several seconds
of nearly 8.47 volts. A 1 meg ohm meter has a 1000 second time
constant with a 1000 uf capacitor. That means that the voltage will
fall to about 1/3 of the initial value in 1000 seconds (neglecting the
capacitor's own internal leakage resistance).

13. ### Ben SchaefferGuest

I tested the largest I had. I am getting some electrolytic capacitors and
will finally see it.

14. ### Dan AkersGuest

Ben Schaeffer wrote;
"I tested the largest I had. I am getting some electrolytic capacitors
and will finally see it."
____________________________________
Re;
Be aware Ben, that as the capacitance of your capacitors under test
increases, so does the steady state full charge leakage current. That
is to say, that you will not always be able to charge the cap to full
battery voltage and see the current drop to the theoretical "zero". As
a rule of thumb for electrolytics, the
"acceptable" leakage current is approximated by:

I=kC+0.3

Where I is the leakage current in mA, C is the capacitance in uF and k
is a constant based on the capacitors rated voltage.

For 3-100 volts k=0.010
For 101-250 volts k=0.020
For 251-350 volts k=0.025
For 351-500 volts k=0.040

Thus, for a 10V, 100 uF cap, you should expect about 1.3mA leakage
current at rated voltage. This could potentially result in a needed
130V drop across your 100k series charging resistor and a 140V supply to
get you there!! The point being, that you will probably need to reduce
your series resistance with high capacitance electrolytics.

Hope that helps...

-Dan Akers

15. ### Peter BennettGuest

When doing this sort of thing (or any measurement, really), you must
be aware of how your measuring instrument and technique might
influence the measurement.

If you hold the capacitor leads between your fingers and the meter