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Understanding biasing and ohms law

  1. Sep 13, 2011 #1
    Hello, I am trying to get my head around biasing.
    I do understand the principles.
    In the attached circuit the Collector current is 2.5mA and the collector resistor is 1K. Therefore my circuit here is biased at 2.5V. The power supply is 5V.
    So that would give a 2.5V drop over the collector resistor. The bit I dont understand is where does the other 2.5V drop come from to bring the VCC down to zero potential ?

    Thanks for any help. This is just elementary stuff but that is where I am at.
     

    Attached Files:

  2. jcsd
  3. Sep 13, 2011 #2
    You got it up side down. You need to have the resistor on the emitter to 0V. You need to put the collector at above 3.2V, then you put 2.5V+0.7V=3.2V at the base to set up 2.5mA of emitter current which give approx 2.5mA at the collector.

    Collector resistor like in your circuit does not effect current through the transistor in normal operating condition.

    OK, the normal way of looking at NPN transistor circuit is that it operate at Forward Reverse bias. The base emitter is forward biased and the base collector is reverse biased. So.

    1) For the transistor to turn on, you need to have about 0.7V from B to E and keep the C voltage above B.

    2) Set up the operating current by an emitter resistor which in my example is 1K. If I put 3.2V at the base, the emitter voltage would be about 2.5V( remember 0.7V from B to E?). This will put 2.5V across the 1K resistor and therefore the emitter current is 2.5mA.

    3) Collector is [itex] \beta\;[/itex] times base current. So collector current is emitter current minus base current.

    To summerize:

    [tex]I_E=\frac{V_B-0.7V}{R_E}[/tex]


    [tex] I_E= \beta I_B \;\Rightarrow I_C= I_E-I_B[/tex].
     
    Last edited: Sep 13, 2011
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