Thevenin Resistance and the Time Constant

[paulmdrdo]

Homework Statement

Homework Equations

Rth = R||R+R (not sure?), τ=Rth*C

The Attempt at a Solution

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So, what I ultimately wanted to determine is the time constant of this first order op-amp circuit.
I first tried to calculate the thevenin resistance seen by the capacitor by using
Rth = R||R+R (honestly I'm not quite sure if this is the right formula for getting the Rth) then the time constant would be τ = Rth*C.

I know there's another method using nodal analysis then solving for the resulting D.E. and observing the term containing the exponential function but I want to know the method without resorting to finding the D.E.

Thank you!

PS
Ideally condition is assumed for this problem.

 

[Tom.G]

Probably not the approach you are looking for, but here goes (with ideal components).

1. What is the op-amp gain as configured?
2. Assuming the R in series with the capacitor is open, what voltage is at the junction of the R's?
3. Reconnect the R in series with the capacitor.
4. What is the equivalent resistance that the capacitor sees? At what voltage?
5. If these impact the capacitor charge, how much and in which direction?

Can you deduce the end-point of the above scenario?

Cheers,
Tom

 

[paulmdrdo]

Probably not the approach you are looking for, but here goes (with ideal components).

1. What is the op-amp gain as configured?
2. Assuming the R in series with the capacitor is open, what voltage is at the junction of the R's?
3. Reconnect the R in series with the capacitor.
4. What is the equivalent resistance that the capacitor sees? At what voltage?
5. If these impact the capacitor charge, how much and in which direction?

Can you deduce the end-point of the above scenario?

Cheers,
Tom

I still see the resistance seen by the cap to be Rth = R||R+R.

 

[Tom.G]

So do I.
Do you account for the changing voltage at the junction of the three R's?

 

[paulmdrdo]

So do I.
Do you account for the changing voltage at the junction of the three R's?

How would that affect the calculation for Rth? Please bear with me.

 

[Tom.G]

It doesn't, your Rth calc is correct.

Sorry, I jumped ahead to the second part of your post where you are trying to find the Time Constant of the overall circuit. The circuit complicates the Time Constant calculation because the cap is not discharging simply thru a resistance directly to its other plate. Do you see that?

 

[paulmdrdo]

It doesn't, your Rth calc is correct.

Sorry, I jumped ahead to the second part of your post where you are trying to find the Time Constant of the overall circuit. The circuit complicates the Time Constant calculation because the cap is not discharging simply thru a resistance directly to its other plate. Do you see that?

Sorry, I'm trying figure out what you are saying but I still don't get it. Can you elaborate more on this part?
The time constant turned out to be 3RC but I don't understand the method used to calculate it.

 

[Tom.G]

The time constant of an RC circuit is defined as the time it takes to reach 1/e of its final value, about 63% of ultimate charge or discharge. This also assumes an Ideal Voltage Source for charging (resp. discharging) and strictly linear (WRT voltage and current) of the R and the C.

The circuit you are working with meets the strictly linear RC requirement but NOT the ideal voltage source requirement. You have correctly derived the Rth and the C is given. In a typical RC discharge circuit, the voltage source would be fixed. If the voltage source is not zero then the final voltage at the capacitor would equal the source voltage.

In your case however, the Rth connects to a voltage source that varies directly with the capacitor voltage, by virtue of the feedback from the op-amp output.

Unfortunately I'm not up to the math for this one. If you desire, I can request that others chime in with a more suitable presentation. What say?

Cheers,
Tom

 

[The Electrician]

Sorry, I'm trying figure out what you are saying but I still don't get it. Can you elaborate more on this part?
The time constant turned out to be 3RC but I don't understand the method used to calculate it.

What do you mean when you say "the time constant turned out to be 3RC"? Did the problem solve itself? Did your instructor hand out a solution sheet, or work it out on the board? Or did the textbook give that result? You say you don't understand the method; what method would that be?

If you want us to help you understand the method, you'll have to show us the method. Post a picture of the method as it was given to you.

 

[paulmdrdo]

What do you mean when you say "the time constant turned out to be 3RC"? Did the problem solve itself? Did your instructor hand out a solution sheet, or work it out on the board? Or did the textbook give that result? You say you don't understand the method; what method would that be?

If you want us to help you understand the method, you'll have to show us the method. Post a picture of the method as it was given to you.

Hi, this is the solution supplied by the book. And I understand this method actually. But what confuses me is when I tried to use another approach in solving for the time constant. My method went like this: first I determined the Rth seen by the capacitor and I came up with Rth = R||R+R from this the time constant is τ=Rth*C. Now, when I compared this with the book's answer the Rth seen by the cap is just 3R. Why is that? That is actually what I don't understand.

 

[The Electrician]

Refer to this schematic where I've added numbers to distinguish the 3 resistors; their resistances remain R ohms:

Your calculation of Rth assumes that the right end of R1 is essentially grounded--connected to zero volts, but the voltage there is not zero volts.

Imagine removing the capacitor and connecting a current source of 1 amp to the + input of the opamp. Set up the nodal equations and solve. You will find that the voltage at the + input is 3R volts. Since the current source is injecting 1 amp there, the resistance seen at that node is 3R ohms.

Or, without even solving equations you can reason like this. If you inject 1 amp at the + input, 1 amp will enter the V1 node from the left end of R2. The voltage across R1 is the same as the voltage across R2 because the opamp is connected as a unity gain amplifier, and the voltage Vo is the same as the voltage at the + input. Thus 1 amp flows out of the left end of R1. KCL says that then 2 amps flows downward through R3, so the voltage at V1 is 2R volts. The voltage across R2 is R volts (because it has 1 amp flowing to the left due to the injected 1 amp) which gives a voltage of 3R at the + input.

 

[paulmdrdo]

Refer to this schematic where I've added numbers to distinguish the 3 resistors; their resistances remain R ohms:

View attachment 240340

Your calculation of Rth assumes that the right end of R1 is essentially grounded--connected to zero volts, but the voltage there is not zero volts.

Aren't we suppose to short to ground the voltage source connected to R1 namely Vo? Can you tell me why it's not valid in this case?

 

[The Electrician]

Check the details of finding the Thevenin equivalent in your textbook.

Also, have a look at what Wikipedia says: https://en.wikipedia.org/wiki/Thévenin's_theorem

Go down to the section titled "Calculating the Thevenin equivalent", and you will find this:

"The Thévenin-equivalent resistance RTh is the resistance measured across points A and B "looking back" into the circuit. The resistance is measured after replacing all voltage- and current-sources with their internal resistances. That means an ideal voltage source is replaced with a short circuit, and an ideal current source is replaced with an open circuit. Resistance can then be calculated across the terminals using the formulae for series and parallel circuits. This method is valid only for circuits with independent sources. If there are dependent sources in the circuit, another method must be used such as connecting a test source across A and B and calculating the voltage across or current through the test source."

The opamp in your circuit is a dependent source. The voltage at Vo is controlled by (and is the same as) the voltage at the + input. You can't just set the output Vo to a short to ground.