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RC Reality Check

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

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  1. 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. 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. Just using one meter, measuring ground and node between resistor and
    capacitor.
     
  4. 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. 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. 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
    two meter leads are connected.

    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. 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. 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
    A) your meter is unusual
    B) your 100k resistor is actually 1M or
    C) your battery is REALLY dead.

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

    Dan Akers Guest

    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. 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
    probes, your own body resistance will also affect your measurement.
    To see this effect, just hold the two meter probes in one hand, and
    notice the indicated resistance - I can get anything from a couple
    hundred K to a megohm, depending on how moist my fingers are, and how
    hard I squeeze them on the probes.
     
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