H-bridge BJT power dissipation.

Discussion in 'Homework Help' started by lam58, Mar 25, 2014.

  1. lam58

    Thread Starter Member

    Jan 3, 2014
    69
    0
    Hi, I'm a bit stuck. The attached image below shows an H-bridge DC motor controller with a supply of 12v and a motor current of 3 amps. Would I be right in saying the power dissipated in Q1 would simply be 36W i.e. 3amp*12v and the power dissipated in Q4 would be 3amp * 1.2v i.e. Ic*Vce(sat)?


    Note: when Q1 is on, Q4 is on and Q2,Q3 are off and vice versa.
     
  2. crutschow

    Expert

    Mar 14, 2008
    13,016
    3,235
    You are half right. The power dissipated in both transistors that are ON equals the motor current times their respective saturation voltage Vce(sat). You are dropping most of the voltage across the motor so it can't also be dropping across Q1.

    But your Vce(sat) is rather high because you are using a 3A max. transistor at 3A. You should typically operate a transistor at <75% of the max. ratings, with 50% being a good limit for reliability. That should drop the Vcd(sat) to below a half volt and reduce the dissipation in the transistor.
     
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