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Find current and PD across a circuit with diodes

  1. Sep 27, 2016 #1
    1. The problem statement, all variables and given/known data
    When answering the questions below, you may assume that the source resistance is negligible and that each diode has a potential difference of 0.6 V across it when conducting. For the circuit below calculate (a) the current in the 30 Ω resistor, (b) the current in the 60 Ω resistor, (c) the PD across the 40 Ω resistor.

    99ec75b6cc4f.jpg

    Answers: (a) zero, (b) 0.054 A, (c) 2.16 V

    2. The attempt at a solution
    (c) 1/R = 1/30 + 1/40 → R = 20 Ω → total R = 20 + 40 = 60 Ω. I = V / R = 6 / 60 = 0.1 A. V40 = I R40 = 0.1 * 40 = 4 V.

    (a) VParallel = 6 - 4 = 2 V. I30 = VParallel / R30 = 2 / 30 = 0.067 A.

    (b) V = 2 / 60 = 0.033 A.

    Doesn't fit the answer. I looked at the theory behind the diods. Still don't understand why the current in the 30 Ohm resistor is zero.

    5fa3ca3bb8bf.png

    Is it because current goes from the cathode part and therefore it is zero? Or because there are two diods and therefore the second one should have a current of zero?
     
    Last edited: Sep 27, 2016
  2. jcsd
  3. Sep 27, 2016 #2

    phinds

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    The diode in series with the 30ohm resistor is reverse biased. Do you understand what that means? The little picture you have included of a forward biased diode does not apply to a reverse biased diode.
     
  4. Sep 27, 2016 #3

    CWatters

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    I also recommend answering the questions in the order they were written. The answer to "a" will be a big help for "b" and "c".
     
  5. Sep 28, 2016 #4
    Hm, for (a) I would say it should be like this: since electrons move from the negative polarity of the 6 V, more specifically: downwards, right and then up to the 30 Ω and 60 Ω diodes. So, since the current flow is only possible when the diode arrow is faced towards the electron flow, like on this picture: 03246.png , therefore the current is equal to zero in the 30 Ω, since the position of the diode does not allow current flow.

    Should this be correct?

    (Theory from here.)
     
  6. Sep 28, 2016 #5

    CWatters

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    Correct.

    If there is no current flowing in the 30R branch perhaps you can see a way to simplify the circuit?
     
  7. Sep 28, 2016 #6
    6bf683c50db7.jpg

    Like this?

    For (b) we need to find I60 Ω. I = V / R or I = V / 60. How to find V? V = I R. Or V = V1 + V2 + V3. Or IR = IR1 + IR2 + IR3.

    40 + 60 = 100 Ω. So I = 6 / 100 = 0.06 A. V60 Ω = 0.06 * 60 = 3.6 V.
     
    Last edited: Sep 28, 2016
  8. Sep 28, 2016 #7

    CWatters

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    Exactly.
     
  9. Sep 28, 2016 #8
    Yes, but how do we proceed further? We don't know R for the diode or the V for the two resistors.

    RTotal = RDiode + 60 + 40

    6 = 0.6 + V60 Ω + V40 Ω

    Update: I can only calculate (c) algebraically like R of 60 and 40 Ohms = 100 Ohms. 40 / 100 = 0.4 of the voltage is taken by the 40 Ohm resistor. Both resistors take 6 - 0.6 = 5.4 V, so 5.4 * 0.4 = 2.16 V goes to the 40 Ohms one. But it doesn't look like physics. (b) is, in that case: 5.4 * 0.6 = 3.24 V. I = V / R = 3.24 / 60 = 0.054 A.
     
  10. Sep 28, 2016 #9
    You're right, you don't know the resistance of the diode. But let's consider what the problem statement says: the potential across the diode is 0.6V when conducting. This is regardless of the current across it. Hence, we cannot consider that Ohm's law, V=RI, is applicable to the diode, as the only possibility for this is that R adjusts itself according to the level of current. Too complicated, and not enough information in the problem statement for this case.

    What you could do is to use a very frequent approximation: consider the resistance of the diode negligible when compared to the other resistances in the circuit. This correspond to the model of an ideal diode: infinite resistance when reverse polarised, negligible resistance and constant V drop when directly polarised. In fact, I think the statement should have suggested this explicitly. Under this assumption, you can solve your problem.

    Obviously, this approximation is only good if the internal resistance of the diode is really negligible compared to the total resistance in your circuit, like in the present case.
     
  11. Sep 28, 2016 #10
    Thank you. What do you think of this "method":
    If we drop the resistance of the diode then we have I2 = V / R2 = 6 / 60 = 0.1 A. It is relatively close to the answer of 0.057 A, but still not the precise one.
     
  12. Sep 28, 2016 #11
    Both methods you propose are too messy for my liking, and they don't show the reasoning behind the calculations. Show structure and order in your solutions, and offer the lecturer(?) a line of reasoning. For example:

    We will consider that the diode is ideal, e.g. RD=0. Because of the drop across the resistor is VD=0.6V, we have that the resistances see a potential drop of:
    VR1+R2=VA-VD=...

    Hence, the current through the circuit is:
    IT = ... (complete)
    Therefore, the voltage drop across the 40Ω resistor is:
    VR1=...
    Bonus: We can verify that the voltage drop across the 60Ω resistor is:
    ...
    When we add together the voltage drops across the resistors we obtain VR1+R2
     
  13. Sep 28, 2016 #12
    1. 6 VTotal = 0.6 VDiode + V60 Ω resistor 1 + V40 Ω resistor 2 → both resistors have a reading of V60 Ω resistor 1 and 40 Ω resistor 2 = 6 V - 0.6 V = 5.4 V.

    2. Their both resistance R60 Ω resistor 1 and 40 Ω resistor 2 = 60 Ω + 40 Ω = 100 Ω.

    3. Since we need to find the current of the 60 Ω resistor 1, we assume that it consumes: 60 Ω / (40 Ω + 60 Ω) = 0.6 (or 60 %) of the V60 Ω resistor 1 and 40 Ω resistor 2 = 5.4 V. So the 60 Ω resistor 1 has: V60 Ω resistor 1 = V60 Ω resistor 1 and 40 Ω resistor 2 * 0.6 = 5.4 V * 0.6 = 3.24 V.

    4. We now find the current I60 Ω resistor 1 = V60 Ω resistor 1 / R60 Ω resistor 1 = 3.24 V / 60 Ω = 0.054 A.

    5. The second resistor consumes: 40 Ω / (40 Ω + 60 Ω) = 0.4 (or 40 %) of the V60 Ω resistor 1 and 40 Ω resistor 2 = 5.4 V. So the 40 Ω resistor 2 has: V40 Ω resistor 2 = V60 Ω resistor 1 and 40 Ω resistor 2 * 0.4 = 5.4 V * 0.4 = 2.16 V.
     
    Last edited: Sep 28, 2016
  14. Sep 28, 2016 #13

    gneill

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    Your results are correct, and your solution method is sound if a bit convoluted.

    An appropriate approach for these sorts of problems that will deemed both succinct and logical is to apply the basic circuit laws. In this case Ohm's Law and Kirchhoff's Voltage Law (KVL). State that you are applying these rules and then write KVL around the loop:

    upload_2016-9-28_10-20-40.png

    Ohm's Law applies for the resistors, while the battery has a fixed potential difference and diode potential change is a given. Solve for ##I##.
     
  15. Sep 28, 2016 #14
    Hm, indeed that's easier. V = V1 + V2 + VDiode → 6 = IR1 + IR2 + 0.6 → 6 = 60 * I + 40 * I + 0.6 → 100 * I = 5.4 → I = 0.054 A. V2 = IR2 = 0.054 * 40 = 2.16 V. Thank you!
     
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