DC Chopper Circuit

Discussion in 'Homework Help' started by shespuzzling, Sep 28, 2010.

  1. shespuzzling

    Thread Starter Active Member

    Aug 13, 2009

    I am having trouble with a homework problem on a DC chopper circuit. I attached the circuit as a jpeg. The diode is a freewheeling diode to prevent arcing when the switch is opened or closed. Also, it is given that E1 is greater than E2.

    The question is 2-fold:

    a) Show that the maximum difference between Imax and Imin (i.e. Imax - Imin) occurs when the duty cycle alpha=0.5

    b) Obtain the value of the load inductance L so that Imax-Imin= 0.2*Iavg; where Iavg=200A, E1=400V, alpha=0.5 and T=10^(-3) sec

    My first problem is the fact that there is no resistance in the circuit...I am having trouble writing a differential equation. If I assume the switch has been opened and closed repeatedly for a while, and then it is opened again at t=0, there should be residual current in the loop with the inductor, and I think that should be the Imin, so the equation I wrote was:

    L(di/dt)-E2=0. I don't think this can be correct though, because then you would have i(t)=(E2/L)*t and that doesn't make any sense because if the switch was kept opened, the field in the inductor would dissipate and you'd be left with a short cirucited battery (E2) and ultimately no current. I don't really know how to proceed from here (I don't have much of a background in differential equations)...any help you could give would be greatly appreciated.

  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    Yes it's a curious problem.

    If you have any condition other than E2=D*E1 the mean current will either increase linearly unbounded or decrease linearly until Imin = 0A.

    In any case during the inductor charging phase (switch on) the current will increase from Imin to Imin+[(E1-E2)/L]*t where t is on the interval from 0 to D*T.

    During the discharge phase (switch off) the current will decrease from Imax to Imax-[E2/L]*t where t is on the interval from 0 to (1-D)*T

    T=1/f - where f is the switching frequency and D is the duty cycle.

    To prove the maximum ripple condition you will best make the assumption that at any given condition E2=D*E1. If you are not sure how to do this part I (or another member) will give you some clues later - when you have tried doing some of the analysis. Post what you can work out now.
  3. shespuzzling

    Thread Starter Active Member

    Aug 13, 2009
    Thanks for the ideas. So, I think I follow you for the first part (when the switch is closed) and I got the same equation for that period (assuming it's so fast it's can be assumed to be linear):

    i1(t)=((E1-E2)/L)*t + Imin for 0<t<D*T

    but for the second equation, I'm not sure how you got the negative in front of E2/L (even though it makes perfect sense, because it's going from max to min).

    Since the switch is suddenly opened, I thought the polarity of the inductor would reverse and you'd get:

    L(di/dt)=E2, and solving this for the interval D*T< t < T you would get: i2(t)=(E2/L)*(t-D*T) + Imax. I know this doesn't make sense, because this would mean that as t increases, so does i2(t), and that's not correct. I guess I can't figure out why this is the wrong way to go about it. If I use an altenrnate interval, like the one you suggested: 0<t<(1-D)*T, using i2(0)=Imax as the initial condition, I get the same thing.

    I've played around with this a lot, trying to find different expressions for Imax-Imin and I keep getting something along the lines of a constant multiplied by (T-DT) or (DT-T) and in both of those cases I can't see why D would have to equal .5 to get a max.

    The problem makes sense intuitively....in order to get the max difference you need to let Imax get as high as possible and also let Imin get as low as possible....thus the duty cycle should be a half...but I'm still struggling with proving that.

    Many thanks for the help.
  4. shespuzzling

    Thread Starter Active Member

    Aug 13, 2009
    I did some more work...

    I tried working the problem from this point:

    i1(t) = ((E1-E2)/L)*t + Imin
    i2(t) = Imax - (E2/L)*t

    I used i1(D*T)=Imax and i2(T)=Imin and then tried figuring out what Imax-Imin was and I'm still not getting anything. Even with the assumption that E2=D*E1 (not sure why I would assume that in the first place), the solution is circular or gives me D=1 or D=0.
  5. shespuzzling

    Thread Starter Active Member

    Aug 13, 2009
    I think I got it!!!

    Over-engineering is always my downfall (unfortunately). It just took me a while to recognize why E2=D*E1, but I had it right in front of me the whole time.

    Thanks so much for your help, I think I'll actually be able to turn this homework in on time!