1. Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.
    Dismiss Notice

How to find the time constant?

Discussion in 'Homework Help' started by Heidi, Sep 27, 2013.

  1. Heidi

    Heidi Member

    Joined:
    May 3, 2013
    Messages:
    213
    Likes:
    11
    Location:
    Taiwan
    Dear friends,

    In the circuit below, I would like to find its time constant by trying to convert that circuit into an equivalent one that contains a voltage source, one resistor and one capacitor. Could you please give me some clues as to how to proceed? thank you!
     

    Attached Files:

  2. MikeMl

    MikeMl Well-Known Member Most Helpful Member

    Joined:
    Mar 17, 2009
    Messages:
    11,139
    Likes:
    566
    Location:
    AZ 86334
    You cant.
     
    Last edited: Sep 27, 2013
  3. Heidi

    Heidi Member

    Joined:
    May 3, 2013
    Messages:
    213
    Likes:
    11
    Location:
    Taiwan
    Here is a solution which I downloaded from the Internet.

    What I can't understand is why that method works (if the method is correct).
    Is it possible to use Thevenin's theorem here? And how?
     

    Attached Files:

  4. dave

    Dave New Member

    Joined:
    Jan 12, 1997
    Messages:
    -
    Likes:
    0


     
  5. MikeMl

    MikeMl Well-Known Member Most Helpful Member

    Joined:
    Mar 17, 2009
    Messages:
    11,139
    Likes:
    566
    Location:
    AZ 86334

    DF84.jpg

    It doesn't even work if we reduce the amplitude of V2 to 0.6666V (which the author of the attachment overlooked).
     
    Last edited: Sep 28, 2013
  6. MikeMl

    MikeMl Well-Known Member Most Helpful Member

    Joined:
    Mar 17, 2009
    Messages:
    11,139
    Likes:
    566
    Location:
    AZ 86334
    Here is an equivalent circuit, but it is not the Thevinin Equiv. of Circuit A DF84a.jpg
     
  7. steveB

    steveB Well-Known Member Most Helpful Member

    Joined:
    Jan 16, 2009
    Messages:
    1,307
    Likes:
    639
    I think it might be possible in a sense. I'm in a restaurant so I can't be sure, but I sketched in on a napkin and it seemed to work out. However, I had to cheat with scaling and make an equivalent voltage source that didn't really scale in the right way. Also, I had to assume a perfect DC source for Vi. I'll check it later when I get home. Basically, I wrote on the differential equation for one capacitor and then considered that i2=-i1*C1/C2 if Vi does not change in time, where i1 and i2 are the capacitor currents. Following this through reveals a differential equation that looks a lot like a series RC circuit, but the voltage source is not quite right. But, if you cheat with units and change the value of the voltage source, you can make it work.

    Basically, I'll agree with MikeML and say this is not really the right way to do it, but it might be the intention of the problem. However, if the goal is just to find the equivalent time constant, then this is simple and it is just R1 in parallel with R2 and C1 in parallel with C2.
     
    Last edited: Sep 27, 2013
  8. Ratchit

    Ratchit Well-Known Member

    Joined:
    Mar 12, 2008
    Messages:
    1,955
    Likes:
    83
    Heidi,

    The time constant of the circuit only has meaning with respect to the energy storage elements in the circuit. In the circuit you submitted, there are two storage elements (capacitors), and therefore TWO possible time constants. I will calculate the time constant of C1. C2 will have a similar calculation. Using the potientiometer method, I will find the voltage across the C1 capacitor when a 1 volt step is applied.

    Heidi.JPG

    As you can see, the exponential term of "e" is a time constant of the resistors in parallel and the sum of the capacitors. In this symetrical circuit, each cap will have the same time constant. But, I believe the solution posted on the web is correct for the wrong reason.

    Ratch
     
  9. steveB

    steveB Well-Known Member Most Helpful Member

    Joined:
    Jan 16, 2009
    Messages:
    1,307
    Likes:
    639
    Finding the time constant is straightforward and the given soliution is correct for the right reason. Basically two storage components does not guarantee the system is second order with two time constants. One can write this system with one state equation, hence only one time constant.

    The only uncertainty is whether the first order system can be written as an equivalent simple series RC circuit driven by a simple voltage source. I think you can get close, but strictly I don't think it is rigorously correct. Hence, I agree with MikeMl.

    I'm reasonably certain of this but to be absolutely certain I need to derive the state space system formally. But I can pretty much see how it will go. You can write the state equation for one capacitor, but the the rate of change of voltage on the other cap is constrained by the rate of the voltage on the other cap and the rate on the input voltage. Hence, the second state equation is redundant and it's a first order system. Putting this in standard state space form is then tricky because the state equation depends on the derivative of the input, hence the state variable must be redefined to get the standard form without derivatives of the input signal. Presumably the new state will be a linear combination of the two capacitor voltages.
     
  10. Ratchit

    Ratchit Well-Known Member

    Joined:
    Mar 12, 2008
    Messages:
    1,955
    Likes:
    83
    I would like to see a linear network with two storage elements that can be represented by a first order DE equation. What is the correct reason for the solution? What does an "equivalent" circuit of a resistor and a capacitor mean with respect to the time constant? Does the first order representation hold good for all values? I look forward to seeing your derivation.

    Ratch
     
  11. Heidi

    Heidi Member

    Joined:
    May 3, 2013
    Messages:
    213
    Likes:
    11
    Location:
    Taiwan
    I want to thank you all for helping solve the problem and thank you for your precious time.

    I have read all the posts carefully. I'm not familiar with Laplace transformation yet, don't know what state equation for a capacitor means, but I see something interesting appear on the exponential in Ratchit's derivation for the capacitor voltage.
    So I think I need to focus on some simpler circuits first, think carefully about what a time constant means for an equivalent RC circuit.
    I'll go back to that problem later.

    Thank you again!
     
  12. MrAl

    MrAl Well-Known Member Most Helpful Member

    Joined:
    Sep 7, 2008
    Messages:
    11,049
    Likes:
    961
    Location:
    NJ
    Hi,

    I believe this is an aberration of some more simple question that was to be asked within a known framework of assumptions that were meant to be taken for granted as being part of the total problem.

    To make a long story short, you cant really answer this question because it is ill posed. There's no time constant for a voltage source in parallel with either one capacitor or any number of capacitors connected in series regardless of any resistors in parallel with them. That's because the time to charge one or two in series with a voltage source is zero, and also to charge one or two in zero time takes an infinite current, which is not practical. So theoretically you'd have to place some series resistance in series with something like the voltage source for example.

    So the circuit is not theoretically sound to begin with for a DC source, and for an AC source it would have to start up at exactly zero volts and with an AC source that starts at 0v the time constant would be meaningless anyway. The phase shift would then be of interest.

    Circuit simulators will either give incorrect results varied by how they handle voltages with capacitors and min series resistances, and some will actually reject this circuit outright and post an error about connecting voltage source and capacitors directly together without even attempting to provide any calculation points.

    So the first thing we would need here is clarification of what exactly they are looking for here. It is probably that they intended to solve the circuit for a voltage source that acts as the input voltage to the network and that voltage source has some internal impedance but knowing where this problem came from (seeing other problems just before it for example) might help clear this up.

    On the practical side, this kind of circuit can be used as a bias network where the resistor ratio sets the reference bias voltage, while the two capacitors add filtering but at the same time force the output voltage to reach the target value in a very short time. With only one capacitor as filter it would take much longer to reach the DC operating point. But the time to charge is then dependent on the power supply impedance plus the network, not the network alone. The short time charge voltage is then dependent mostly on the capacitor values, which would be set to oppositely match the two resistor values.
     
    Last edited: Sep 28, 2013
  13. steveB

    steveB Well-Known Member Most Helpful Member

    Joined:
    Jan 16, 2009
    Messages:
    1,307
    Likes:
    639
    Mr Al,

    You make a very good point about the way the circuit is driven with a voltage source. But, I still see one time constant for the circuit. If you drive one capacitor with a voltage source, then you don't have a time constant because the voltage source imposes the dV/dt on the capacitor and there is no differential equation which determines the capacitor response. But, with two caps, there can be an interplay between the capacitors and a time constant is relevant for that dynamics.

    A simple minded way to say it is that this improper voltage drive reduces the system order by one, from what you would normally expect. I think that is the cause of the confusion here. This is what I was trying to explain above. I'll write out the equations to be sure, and I'll try it again with a source resistance on the voltage source to be sure that the second time constant is restored.
     
  14. Ratchit

    Ratchit Well-Known Member

    Joined:
    Mar 12, 2008
    Messages:
    1,955
    Likes:
    83
    As far as I know, the simplest definition of a time constant (TC) relates to how fast a storage element (L or C) energizes from a step voltage or current, or de-energizes from an initial voltage or current with nothing applied. Each single capacitor or inductance in a circuit has its own TC. That is why I calculated the equation of C1's voltage and observed the exponent of the "e" term. Any voltage or current equation of a storage element is guaranteed to produce an "e" term if there is a resistance in the storage element circuit, but it might also contain other non "e" terms, too. This definition applies to any linear circuit with any number of storage elements.

    Ratch
     
  15. The Electrician

    The Electrician Active Member

    Joined:
    Jan 13, 2009
    Messages:
    551
    Likes:
    70
    A calculation of the impedance seen by the voltage source gives:



    This would lead one to believe that there would be two time constants in the response. But the voltage source ideally has zero internal impedance, and that changes things.

    I find it easier to intuit the behavior by considering a step in the output of the source from some constant voltage to zero. The caps will be charged to voltages determined by the ratio of R1 and R2 after a long time, and a sudden step of the source's voltage to zero is equivalent to replacing the source by a short. In that case, it's plain that the capacitors and resistors will suddenly be connected in parallel, and there will be only one time constant. The behavior for a step increase of the source voltage will result in the same sort of behavior. The zero internal impedance of the source effectively connects the resistors and capacitors in parallel.

    However, to remove the problem of infinite current needed to charge the caps in series when the source voltage makes step excursions in voltage, put a small resistor in series with the source. Even 1 microohm is sufficient to remove the theoretical difficulty.

    I wired up 2 equal 1k ohm resistors with a .2 uF and a .1 uF capacitor and drove them with square waves from a function generator having 50 ohms output impedance. There is a short time constant involving the 50 ohms of the generator plus the capacitors. During this relatively short (a few microseconds) time constant, the caps charge up to voltages in the ratio of their capacitances; then there is a longer time constant where the voltages across the caps become equal because of the equality of the resistors.

    Here are some scope captures showing first the short initial time constant of the charging (or discharging) of the caps at the leading edge of the rising (and falling) step of voltage due to the non-zero impedance of the generator, and then the subsequent equilibration of the cap voltages. The green trace is the applied step from the generator; the blue and purple traces are the capacitor voltages. One can see the initial step of the blue trace to half the voltage of the purple trace.

    Edit:
    One can infer from the result shown in post #16 that if the source resistance is not exactly zero, these two time apparent time constants won't be exactly equal; but, the 50 ohm impedance of the generator is low enough that the measured cap voltages appear visually to have the same time constant, illustrating the effect of the low resistance of the driving source.

    [​IMG]

    [​IMG]

    [​IMG]
     

    Attached Files:

    Last edited: Sep 28, 2013
    • Like Like x 1
  16. The Electrician

    The Electrician Active Member

    Joined:
    Jan 13, 2009
    Messages:
    551
    Likes:
    70
    Next, I replaced the .2 uF capacitor with a .01 uF capacitor so that the R1 C1 and R2 C2 time constants would be in the ratio 10:1. But, we still see that the apparent time constants of the voltage transients across the capacitors are equal. Other than the change in one capacitor, the setup is the same as in the previous post.

    [​IMG]

    [​IMG]
     

    Attached Files:

    • Like Like x 1
  17. The Electrician

    The Electrician Active Member

    Joined:
    Jan 13, 2009
    Messages:
    551
    Likes:
    70
    Harking back to the impedance expression in my first post, we would think that if the voltage source were not acting as a short, there would indeed be two time constants (if the caps aren't equal). I placed a 100k ohm resistor in series with the generator to make it look more like a current source rather than a voltage source. Here are scope captures of the voltages across the capacitors; the caps are still .01uF and .1uF and the resistors are both 1k. There are indeed two time constants.

    [​IMG]

    [​IMG]
     

    Attached Files:

    • Like Like x 1
  18. MikeMl

    MikeMl Well-Known Member Most Helpful Member

    Joined:
    Mar 17, 2009
    Messages:
    11,139
    Likes:
    566
    Location:
    AZ 86334
    All of this is just about an improperly-compensated 10X o'scope probe.
     
  19. steveB

    steveB Well-Known Member Most Helpful Member

    Joined:
    Jan 16, 2009
    Messages:
    1,307
    Likes:
    639
    The Electrician,

    I agree with your opinion. I sense we are all saying similar things, but you explained it very well. Ratch says there are two time constants, but he calculated them to be the same value, which is just another way of saying the circuit is first order. However, once that source resistance is added, it should indeed be two truly different time constants and the circuit is second order.
     
  20. Ratchit

    Ratchit Well-Known Member

    Joined:
    Mar 12, 2008
    Messages:
    1,955
    Likes:
    83
    I aver that there are two TC's that correspond to the two storage elements. Here's why. The fact that the two time constants are equal does not mean that each storage element should not be considered to have a separate TC. Notice in my equation for the voltage, that if the products R1,C1 and R2,C2 are equal, then the two time constants cancel out, and the voltage across each cap is dependent only on R1 and R2. Since the circuit is topologically symetrical, inserting a source resistance will change the TC's of each cap equally. An interesting situation can occur if the values of the components are not equal. One of the caps may be energized to a higher voltage than the other one. During de-energizing, one of the caps will bring the other cap to a higher voltage than it was when the voltage source was on.

    Ratch
     
  21. steveB

    steveB Well-Known Member Most Helpful Member

    Joined:
    Jan 16, 2009
    Messages:
    1,307
    Likes:
    639
    Well, saying there is one or two time constants when they seem to be equal is somewhat of a pointless debate. The important thing is that the system becomes first order and there in only one degree of freedom from a dynamic point of view.

    What about the case of two caps in series or two caps in parallel (or two inductors arranged as such)? Is there two time constants or one? Both cases become first order, and just because there are two elements does not mean the dynamics is second order and that there are two equal time constants.

    But this is somewhat of a strange case, so I can live with someone saying there are two equal time constants. I just dont' think it means much.
     

Share This Page