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Homework Help: NPN Transistor

  1. Jan 9, 2008 #1
    1. The problem statement, all variables and given/known data

    an npn transistor is used to drive a constant current I through three ‘white’ LEDs, connected in series. If the required current value is I = 25 mA, and the current-gain of the transistor is β = 300, find the value of the resistor R, giving your answer to 3 significant figures. You may assume for the operating conditions specified that the base-emitter voltage of the transistor is 600mV, and that its saturation voltage is 150mV.

    I should point out that the base of the transistor is connected to a voltage divider, with a resistor of 10k above a resistor R.

    2. Relevant equations

    Base Current = [tex]\frac{collector current}{gain}[/tex]

    3. The attempt at a solution

    Hi guys looking for a bit of help on a past paper for my instrumentation class.

    In this problem, I've tried to find the base current by dividing 25 by 300, and I'm a bit stumped as to the next step. Do i simply need to to divide the base current by the saturation voltage, or is that completely wrong?
  2. jcsd
  3. Jan 9, 2008 #2
    It's been a while since working with transistor circuits, and I think I completely forget what saturation currents and voltages were, whoo

    But anyways there should be some voltage going into that voltage divider, so you can figure out Vin(voltage at base) and the base current via that, and base current times gain is collector current. Since the base current depends on R, you can figure out R. There may be more nuances in there but I think that's the idea
  4. Jan 9, 2008 #3

    Here's a circuit diagram if that can give anyone a better idea of the circuit in question.

    Attached Files:

  5. Jan 9, 2008 #4


    User Avatar

    Staff: Mentor

    You set the 25mA by setting the emitter voltage, because that determines the current through the emitter resistor. Ve is Vb - 600mV, so that tells you what you need for Vb. But in setting Vb, you need to account for some base current that bypasses the base resistor R.

    BTW, the base resistor R and the emitter resistor "68R" cannot possibly be referring to the same "R". Does the diagram mean that the emitter resistor is 68 Ohms?

    Also, keep in mind that the current through the emitter resistor is really the LED current plus the base current. Granted, 1/300 is a small number, but to be accurate (and receive full points for the answer), you need to keep track of that current.

    That should be enough info to get you going again. Post up your calcs when you're done, and we'll take a look.
  6. Jan 9, 2008 #5

    Ok, thanks very much I'll get it finished tomorrow. Oh and yes the 68R is simply referring to Ohms.

    Edit; Still can't figure it out :(
    Last edited: Jan 10, 2008
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