H-Bridge Configuration USING BJT, Power dissipated

Discussion in 'Homework Help' started by Iam72, Mar 23, 2016.

  1. Iam72

    Thread Starter New Member

    Mar 23, 2016
    1
    0
    Hello,

    I found a similar question on another Forum however it didn't help me much. I want to work out the power dissipated in in the transistors in the circuit attached and to be honest I am struggling a little bit, If anyone can give me any hints, or start me off in the right direction that would really be great.
    please note that when Q1 and Q4 are ON Q2 and Q3 are OFF and vice versa, when Q1 and Q4 are OFF Q2 and Q3 are ON.
    Power supply is 12v and the current through the motor is 3A.
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    Power dissipated in transistor in BJT is equal to Ptot = Vce *Ic. And from data sheet we can read Vce(sat) = 1.2V for Ib = 375mA. So the power is 1.2V*3A = 3.6W.
    But also notice that for TIP31 Ic_max = 3A so you must use a "bigger" BJT and also we do not know the base current, so we cannot be sure that the BJT is in saturation.
    https://www.fairchildsemi.com/datasheets/TI/TIP31C.pdf
     
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