Question on BJT transistor sizing (W)

Discussion in 'General Electronics Chat' started by mbohuntr, Jul 6, 2013.

  1. mbohuntr

    Thread Starter Active Member

    Apr 6, 2009
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    If I am using a piezo siren as a load, and it consumes 120mA, and I use it as a common emitter with a 1K base resistor, How do I determine the wattage for the transistor? I_{}b(12mA) + I_{}c(120mA) X V (12V) = Watts? = 1.3 watts? It doesn't seem like it should be that easy...
     
  2. #12

    Expert

    Nov 30, 2010
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    Let me take a stab at it...

    (12ma times Vbe) + (120ma times Vce) = P
    (.012 x .7) + (.12 x 12) = P
    .0084 + 1.44 = 1.4484W
     
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  3. €Hunter

    New Member

    Jul 6, 2013
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    Looks about right to me.
     
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  4. €Hunter

    New Member

    Jul 6, 2013
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    On a more serious note it looks terrible.
     
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  5. #12

    Expert

    Nov 30, 2010
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    I see. The piezo is in the collector side. That removes the 12V load from the transistor and the power rating goes way down...to about .0084W + Vce sat times .12A

    A schematic might have gotten through to me better.
    I'm not quite fully awake right now.
     
    Last edited: Jul 6, 2013
  6. €Hunter

    New Member

    Jul 6, 2013
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    Hard stuff. Anyone who disagrees is either a fool or a liar.

    Wattage for the transistor would be the current that it is switching multiplied by the voltage drop across the junction.
     
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  7. mbohuntr

    Thread Starter Active Member

    Apr 6, 2009
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    So. . 7* 132ma?
     
    Last edited: Jul 6, 2013
  8. #12

    Expert

    Nov 30, 2010
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    No. Post #5 has it right. 12 ma times .7 (Vbe) plus .12A times the saturation voltage from collector to emitter, which is usually less than .7 volts.

    You haven't told which transistor you are using so I can't look up the Vce (saturated) of the transistor.
     
  9. mbohuntr

    Thread Starter Active Member

    Apr 6, 2009
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    Sorry, forgot that... Right now I have a TO 92 3904 in the circuit.
     
    Last edited: Jul 6, 2013
  10. #12

    Expert

    Nov 30, 2010
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    Vce Sat at 120 ma seems to be .15 volts.
    Vbe at Ic = 100ma = .9V

    Power = (.012 x .9Vbe) + (.12 x .15Vce sat)
    P = .0108 + .018
    P = .0288 watts
     
    Last edited: Jul 6, 2013
  11. mbohuntr

    Thread Starter Active Member

    Apr 6, 2009
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    Thank-you very much! Both you and CHunter! That is why I like this site so much. Good people, great help! It seems like I'm within the safe zone for this circuit. Thanks again! :D
     
  12. #12

    Expert

    Nov 30, 2010
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    That's what the "Thanks" button is for. Bottom right corner of the posting box. Hey, it's the only pay I get here.:(
     
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