Equation for capacitor discharge

Discussion in 'Homework Help' started by pastaperson, Nov 1, 2009.

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I've got the following circuit here:

I want to find an equation to model the capacitor discharging. If I'm correct, this is the needed equation:

V = V0 * e^(-t/RC)

In my case, I am using a nine volt battery, 1K resistor and 5μF Capacitor. When I calculate the RC I get 0.064, which surely can't be right here?

I'm using Circuit Wizard, and this is the output of graph when it runs on the three probe points (as above):

Basically, when the equation for the capacitor discharges meets the equation of the voltage divider, the LED will turn off. I'm trying to find this equation so that I can include it in my write up.

- pasta
2. JoeJesterAAC Fanatic!

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Your image doesn't show a 9V discharge curve. The tau is only 5 milliseconds and the 9 V should have decreased to zero in about 25 milliseconds.

You might want to ensure you have a good discharge path.
3. t_n_kSenior Member

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Yes indeed - the discharge path into the LM741 negative input will be of fairly high resistance. I doubt the 10k plays any significant part in the discharge time.
4. t_n_kSenior Member

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I mean the 1k - R10.
5. t_n_kSenior Member

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In the interests of circuit predictability, the 1k should probably be in series with the switch to limit the current pulse magnitude drawn from the 9V supply when the switch is pressed - otherwise you have an effective momentary "short circuit" when C is even partially discharged.

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I've since fixed that. Its meant to act as a debounce, so I am pretty happy with the way it discharges.

What I'm confused with is 0.064 and why it isn't discharging correctly? I'm assuming the formula is correct here, but I'm missing something with my circuit?

When I graph 9 * e^(-1/0.064) over the domain [0, 1], I get the following image:

Which certainly fits what JoeJester was saying
7. Jony130Senior Member

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To proper "discharge" you need to add resistor parallel to C1 to get good discharge path.
Now capacitor is discharge by base current of LM741 and this current is very small typ. 20nA, max 200nA.
So to discharge 5uF capacitor to 0V you have to wait
C=Q/U=I*t/U

t=(C*U)/I=(5uF*9V)/200nA=225s

For example, if we have a capacitor charge to Vcc=9V and we wont to discharge capacitor to Vc=4.5V at time t=5s
We need parallel resistor R

$R=\frac{t}{C*In(\frac{Vcc}{Vc})}=1.5M$

R = t / ( C*In(Vcc/Vc) )
= 5s / ( 5uF * In(9/4.5) )= 5/(5uF*0.693)=5/3.465u=1.44MΩ=1.5MΩ
Last edited: Nov 2, 2009
8. JoeJesterAAC Fanatic!

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If you want the LED to energize when you press the switch and then de-energize after a delay when you release the switch, you might want to use the circuit illustrated in the 091101_B picture. 091101_A illustrates the waveforms.

If you want delay energizing the led, (less than one tau), then place the resistor in series with the switch.

The resistor placement in the OPs diagram is in series with the High-Z OPAMP input, making the discharge path the Input Z and the resitor.

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