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voltage peaks&time constant for inductor for square wave

Discussion in 'Electronics Homework Help' started by perchick, Jan 9, 2013.

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  1. perchick


    Nov 2, 2012
    i have a question about this circuit:
    Rb=470, R1=R2=1k (near the inductor) and L=2mH, Vcc=5V
    the input is 0 and 5 in a square wave. the inductor responds when a current is changing. the response when the input voltage goes from 0 to 5 is much smaller then the response for 5 to 0. here is the pulses on Vx:
    you can see in the graph that for 0->5 there is a low peak of about 2.4V and for input 5->0 the peak is 9.6. my first question is why is that? as I understand, when the transistor is in cutoff there is no current Ic so the VCC see the two resistors in the first moment so the voltage in Vx will be (Vcc-0.2)*R/2R. when the input goes from 5->0 there is a current through the inductor so the inductor acts like short circuit and we do not count the parallel resistor so the Vx will be (Vcc-I(L)*R)-0.2
    is there something wrong with this logic? I mean, i know i got the right values, I'm asking if my thought process is ok.
    the second question is how to calculate the time constant? for some reason im getting the values 4.4usec (0->5) and 2,2usec(5->0). is there any difference between calculating the time constant for the inductor (L/R) and the voltage peak in Vx?
    thank you for your help :)
  2. Laplace


    Apr 4, 2010
    The inductor only acts like a short circuit beginning about 5 LR time constants past a transient event. At the beginning of a transient event the inductor resists any change in the current flowing through it. That is why you can model the inductor as an open circuit when the transistor switch initially closes, it takes 5 LR time constants for the current to reach its steady state value. But when there is current flowing and the transistor switch opens, there is current flowing in the inductor. Where will this current flow for the ensuing 5 LR time constants? The voltage across the inductor will rise to whatever value is necessary to keep the current flow at the same value immediately following the transient event as it was before the event.

    And yes, the LR time constants are different in this case for the turn-on and turn-off events.
  3. perchick


    Nov 2, 2012
    thank you for your help
    let me see if i got the time constant right: when the transistor is in cutoff there is no current so the time constant is L/R but the current going out from the inductor is going to 2 resistor so the time constant will be for L/(R||R)=2m/0.5k=4usec.
    when the transistor goes from saturation to cutoff there is initially Ic current so the inductor acts like short circuit so all the current is passing through one resistor. the time constant will be L/R=2m/1k=2usec

    I hope you can help me with another question about a different circuit.
    this is the circuit:
    we keep B in 5V constant and A in pulse input from 0 to 5. the graph on the left shows the output when the voltage on A raises. i wanted to ask if someone knows the reason why the output looks the way it does. when A goes from 5->0 there is no waves, just normal delay time. i understand that the reason for this is parasitic capacitance in the diodes when we shift them from conducting to cutoff. i wanted to know what exactly happens there. when the diodes turns off, the charge build on them is discharging in some kind of resonance frequency but how does those charges get to the output??? and how do this charges going back to the circuit and discharging through the transistor and lower and lower amplitudes?
    let me clarify my question: the raise time is exponential and some how it got mix with sinusoidal function. why is that?
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