# Analyzing an RC circuit that has a bias

Discussion in 'General Electronics Chat' started by LectricGuy, Apr 2, 2013.

1. ### LectricGuy Thread Starter New Member

Apr 2, 2013
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Folks,

I'm trying to come up with a formula for the attached circuit. It's a standard RC with a twist. A capacitor is in series with a resistor which goes to Vcc. A switch can then drop another resistor across the capacitor. Suppose the capacitor is fully charged, and then the switch is closed. What is the voltage across the capacitor over time?

Thanks for your thoughts.

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2. ### Jony130 AAC Fanatic!

Feb 17, 2009
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What is your guess about the voltage? Kept in mind that after transient response capacitor act just like a open circuit.

3. ### crutschow Expert

Mar 14, 2008
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Is this homework?

You calculate the DC voltage when the switch is closed which is the final voltage of the capacitor. Then you calculate the Thevenin equivalent of the two resistors and use the formula for the exponential decay of an RC circuit to find the voltage change of the capacitor with time.

4. ### LectricGuy Thread Starter New Member

Apr 2, 2013
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Thanks for the replies. No, this is not homework; it's to help understand how a cap can support a small MCU during the sleep to awake transition when the MCU suddenly needs a lot of power.

Perhaps my question was posed poorly. Obviously, the final voltage is just the resistor-divider result. But what is V(t) after the switch closes but before the cap reaches its steady-state voltage? It can't be the simple exponential time-constant thing since R1 is providing some charge into the cap as R2 is discharging it.

5. ### Jony130 AAC Fanatic!

Feb 17, 2009
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You will be surprised. But V(t) will be the simple exponential time-constant.
t = R*C = R1||R2 *C1
Vo = Vcc and V∞ = Vcc * R2/(R1 + R2) and after 5t circuit reach V∞.

6. ### crutschow Expert

Mar 14, 2008
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As Jony130 noted, the Thevenin equivalent of the two resistors takes all that into account and it just becomes a simple RC exponential decay.

7. ### LectricGuy Thread Starter New Member

Apr 2, 2013
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Thanks, guys. One more question though: if R1 (the resistor from the cap to the power supply) were 0 ohms, then the time constant is 0 (since the two resistors are in parallel), so the capacitor discharges instantly. But if R1 is 0 ohms, the capacitor is tied directly to the power supply so shouldn't discharge at all. What am I missing?

8. ### crutschow Expert

Mar 14, 2008
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If R1 is zero, then obviously the start and end voltages are the same in the RC exponential decay equation. There is no discharge. So that's what you are missing.

9. ### screen1988 Member

Mar 7, 2013
310
4
Just curious! How can you that time-constant so quickly?
I have to write KCL, KCV and then using Laplace,... to find Vc(t). Until get the result I know that t = R*C = R1||R2 *C.

10. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,901
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To find time-constant in this type of circuit. We need to find resistance seen from the capacitor plates pint of view. Resistance seen by C1 capacitor.
And we can use Thevenin equivalent to find resistance seen by a capacitor.

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11. ### screen1988 Member

Mar 7, 2013
310
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Aha, I see that with Thevenin equivalent I can find time-constant easily.
But I don't get find resistance seen from the capacitor plates point of view.
Why to find equivalent R Vcc has to be connected to ground?

12. ### Jony130 AAC Fanatic!

Feb 17, 2009
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See this example:
Let as close our circuit (10V battery and two 10K resistor ) in the black-box. We only have a access to A and B terminals.
And now let as try to find out what is inside the box, without opening the black-box.
What we can do?
Well we can measure voltage across AB terminals.
So we have Vth = 5V
We can also connect external load resistor and measure the corresponding voltage and current.
But we are brave enough and we shot-out A and B terminals.
And measure the short circuit current Isc.
Isc = 10V/10kΩ = 1mA
Based on only those two measurements we can draw the following conclusion.
Our black-box is seen by the outside world as a ideal 5V voltage source with internal resistance equal to
Rth = Vth/Isc = 5V/1mA = 5kΩ
So we can remove our black-box from the circuit. And we replace the black-box with his equivalent circuit. The 5V ideal voltage source with 5KΩ internal resistance.
And I hope that now you can see that our capacitor see this equivalent circuit.
So the time-constant is equal to t = R*C = 5kΩ*1uF.

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13. ### screen1988 Member

Mar 7, 2013
310
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Oh, thanks Jony! I like your pictures.
I knew this method.
When you said:
I thought that you have another way to find "resistance seen from the capacitor plates point of view" instead of Thevenin.