Voltage across R2 in forward bias problem

Discussion in 'Homework Help' started by metelskiy, Jun 24, 2011.

  1. metelskiy

    Thread Starter Member

    Oct 22, 2010
    66
    3
    I'm a little confused about the following problem if I'm approaching it in the right way:
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    I need to find voltage across R2 and here is my approach.
    First I figured out that diode is forward bias and Vtotal=5V-0.7V=4.3V
    Than I calculated total current I=4.3V/23kΩ=0.187mA
    To calculate votage across R2=0.187mA*1kΩ=0.187V
    Did I do this correctly? This problem is a multiple choice and the number I got is not among the listed ones. I'm trying to figure out the approach rather than a straight answer. Thanks.
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Your initial assumption that the diode is forward biased with 0.7V is where you went astray.

    If you think about the circuit it should be clear that the diode voltage can never reach 0.7V.

    Try working out the voltage across R2 without the diode in place and see what that result would mean in terms of the forward bias of the diode.
     
  3. asullivan

    New Member

    Oct 23, 2010
    2
    0
    Kirchoff's law states that all voltage drops shall be the same in a // circuit..therefor calulate the V drop across the 22K Ohm res. and subtract that V from the source V... The difference willl be the answer.
    (Hint, To calc. the V drop across R1 you will need to find total current...)
     
  4. WellGrounded

    Member

    Jun 19, 2011
    32
    2
    The current of the circuit is:

    5.0 Volts / (22Kohm + 1Kohm) = 0.000217 Amperes

    The voltage across 1Kohm is 0.000217 Amp. x 1,000 Ohms = 0.217 V

    Because this is less than the 0.6 Volt trip point of the diode then the diode has no effect on the circuit. The diode would only start to have an effect if the 22Kohm was reduced to below ~7.33Kohm.

    The 1Kohm resistor needs 0.6 V /1,000 Ohms = .000600 current passing through it to create 0.6V before the diode becomes active.

    (5.0V - 0.6V) / 0.0006 Amp.[1 Kohm resistor] = 7.33Kohm

    The diode diverts the excess current away from its' parallel load while maintaining a (almost)constant voltage across the load.

    If you substitute the 22Kohm resistor with a 4.4Kohm resistor then 0.000600 Amperes goes through the 1 Kohm resistor and 0.000400 Amperes goes through the diode.

    Danny
     
  5. WellGrounded

    Member

    Jun 19, 2011
    32
    2
    Whoops, I double checked your post and didn't answer your question.

    Current is 5.0V / (22,000 + 1,000) = 0.000217 Amperes

    The voltage across the 22Kohm resistor is 22,000 Ohms X 0.000217 Amp. = 4.77 Volts

    Danny
     
  6. Kerim

    Member

    Mar 3, 2011
    35
    0
    I like to add that you were right to be confused :)

    Later, when you will work with bipolar junction transistors (BJT), this diode might represent the forward junction between the base and the emitter of an npn transistor. R1 and R2 would be the bias resistors. In the given circuit, since the voltage on R2 doesn't reach the diode forward voltage (which may start around 0.65 V), the transistor is said to be in the off state (no current at its output, collector).
    I mean that the idea of this circuit presents a good example of a real situation in some practical circuits.
     
  7. metelskiy

    Thread Starter Member

    Oct 22, 2010
    66
    3
    Thanks everything for explaining, it was helpful.
     
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