Calculate Voltage across resistor of rlc circuit

Discussion in 'Homework Help' started by darkknight, Feb 20, 2011.

  1. darkknight

    Thread Starter Active Member

    Oct 7, 2009
    41
    0
    knowing that the circuit is critically damped:
    l = 33mh
    c = .001µf
    r = 9.15 kohms

    i want to calculate vr at certain times frames but i am doing some wrong

    i am using the the formula

    i(t) = (V/L)*t*e^(-R/2L)

    R=2√(.033*.000000001) = 11489.1

    R/2L = 174077

    E^174077 gives me 0 so that throws the solution off, can anyone help me on what i am doing wrong?


    i(t) = (V/L)*t*e^(-R/2L)
     
    Last edited: Feb 20, 2011
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    The exponential term should include time 't' as well. You show a constant (-R/2L) as the exponent. That should be multiplied by 't'. At 1kHz the half cycle intervals are 500usec.

    Presumably you are trying to predict the resultant resistor voltage waveform with respect to time....?
     
  3. darkknight

    Thread Starter Active Member

    Oct 7, 2009
    41
    0
    yes that is what i trying to do, know the voltage across the resistor at several time points but

    even if i multiply the exponent by t, i get

    at time at 1 microsecond
    (R/2L)*t = 174077*.000001 = .011489

    e^ans = 1.012

    (4/.033)*.00001*1.012 = .0000123

    is this is way off what i got measuring it, i can figure out what is wrong
     
    Last edited: Feb 20, 2011
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    That's just telling you the value of the current at 1ms when you really want the current waveform over the entire interval from 0 to time t.

    At 1kHz the charge and discharge half cycles will only be 500us - not the full 1ms interval.

    So you would need to calculate & plot the current and hence resistor voltage values for sufficient points to give a true representation of the output waveform.

    You could do this in EXCEL series table say and do the plot as part of the same worksheet.
     
  5. darkknight

    Thread Starter Active Member

    Oct 7, 2009
    41
    0
    i have a table with values measured and now i have to calculate and compare the measured and calculated but i am having difficulty calculating it.

    i am struggling with the formula
     
  6. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    You'll probably need to show some working - what you've done so far.

    Also note that your assumed solution is not exactly correct. You have a repetitive source waveform whose period is much longer than the circuit time constant. This means the capacitor will be charged to ±4V (one or the other) at the start of a source half cycle. So the effective source voltage would be ±8V (again, one or the other) at the start of any half cycle.

    It would also be good if you showed your measured waveforms as a cross check.
     
  7. darkknight

    Thread Starter Active Member

    Oct 7, 2009
    41
    0
    values i got were for example

    at 1 microsec i measured .75V
    at 2 microsec i measured 1.6V
    at 3 microsec i measured 2.2V
     
  8. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    That looks about right.

    Attached my graph ...
     
  9. darkknight

    Thread Starter Active Member

    Oct 7, 2009
    41
    0
    but my difficulty is proving that through written form, i am having trouble with the formula
     
  10. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    O.K. - let's take the 2usec (2e-6) case

    i(t)=4/(.033)*(2e-6)*exp(-(2e-6)*11,489/(2*.033))=0.000171A

    So Vr(t)=11,489*.000171=1.96V

    Which is somewhere near what you read at 2us. Measurement errors could be significant and we don't know the tolerances on the values you gave - how did you determine the critically damped condition from your observations? R looks to be adjustable according to some experimental condition.

    I would have expected about twice that value for the circuit configuration you posted (and as indicated in my graph) but let's forget that for the moment.

    Was your actual source voltage 4V peak-to-peak? What was the source value at 2us on the other scope channel.
     
    darkknight likes this.
  11. darkknight

    Thread Starter Active Member

    Oct 7, 2009
    41
    0
    yes it was 4 p-p and we only used one scope and i see that your values are close to the measured, i have to relook at my calculations
     
  12. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Comparing with your 1us calculation I have i(t)=0.000102 A which would give a resistor voltage of 1.17V. Again this is somewhat different from your observed reading.

    Keep in mind the point that the inductor will also have some winding resistance which may or may not be significant in relation to the calculated critically damped resistance value of 11,489Ω.
     
  13. darkknight

    Thread Starter Active Member

    Oct 7, 2009
    41
    0
    something should be wrong because if its 4 p-p it should go over 2V correct?
     
  14. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    So what was your maximum reading for the resistor voltage & at what time?
     
  15. darkknight

    Thread Starter Active Member

    Oct 7, 2009
    41
    0
    at 6 now that i am checking my notes i measured 2.75
     
  16. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    If you have access to or can download a Spice based circuit simulation program you would be able to better appreciate what is happening in the circuit. For example you could download TINA-TI. There are several available for free - albeit with certain restrictions on simulation capability.
     
  17. darkknight

    Thread Starter Active Member

    Oct 7, 2009
    41
    0
    i dont have that program and not familiar with the software and time if not on my side :(

    additionally this part is incorrect

    i(t)=4/(.033)*(2e-6)*exp(-(2e-6)*11,489/(2*.033))=0.000171A

    So Vr(t)=11,489*.000171=1.96V

    its supposed to be

    Vr(t)=9150*.000171=1.56V

    correct?

    we want vr which is the voltage against the resistor
     
  18. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Yes - sorry about that. The R value is what you want. I was distracted by the critical damping value of 11,489Ω. Forgot to check R was different.

    That's a good outcome since the answers will be much closer to your observed values.

    Also, I withdraw my comments about the factor of 2 in the theoretical value. Since it is 4V peak-to-peak the capacitor only charges to ±2V. So the original formula using 4V is correct.
     
  19. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    So I finally obtain this waveform using Excel
     
  20. darkknight

    Thread Starter Active Member

    Oct 7, 2009
    41
    0
    i appreciate your help with this issue i had... thanks
     
Loading...