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Diode voltage drop (connected in parallel with a resistor)

  1. Feb 1, 2012 #1

    I am a computer engineer that specialized more in software but I am trying to remember the electronics part.

    In the book I read there is an exercise where a diode (forward-biased) is connected in parallel with a resistor and then both of them are connected in series with another resistor.
    This simple circuit is connected to a dc power source.

    If I consider a voltage drop on the diode (either 0.7 or 0.3, etc) the value of the current that passes through the diode, I can calculate it correctly.

    The problem is when I can think of the diode as being ideal (with 0 drop voltage). In this case I can't seem to find the value of the current that passes through the diode (as in the answer).

    If the voltage drop on the diode is 0 can I consider the circuit as with no diode (2 series resistors connected to a dc voltage source)?

    If the voltage drop is 0, the voltage on the resistor connected in parallel with the diode will also be 0 ?
  2. jcsd
  3. Feb 1, 2012 #2


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    Yes, that is OK. If the diode was perfect, then it would be like a short circuit across the lower resistor if it was forward biased.

    So, there would be no current through the lower resistor and the entire supply voltage would appear across the top resistor.

    So, you could calculate the current in the diode just by using Ohm's Law on the top resistor.
  4. Feb 1, 2012 #3
    I think I can picture your circuit.
    If the diode is considered to be ideal then you are correct to say the volt drop =0.
    It also means that the diode has zero resistance so the current in the circuit will be given by the supply voltage divided by the series resistor.
    The parallel resistor is effectively 'shorted out' by the diode.
    Does this give a sensible answer?

    Sorry... just seen your response vk6kro
    Last edited: Feb 1, 2012
  5. Feb 1, 2012 #4
    Thank you for the quick answers.

    This means the answer in the book is wrong (when the ideal diode is considered).
    They obtained a current through the diode less then current through the series resistor.

    There is also a note in the book that says it's ok to consider an ideal diode when the voltage drop on it is less then 10% on the circuit voltage. But this is messing with the calculations. I will have 0 A at anything in parallel with that diode.
    Of course in the final circuit all the values would be taken into considerations but even so, is this ok to do (consider some components ideal in some specific situations) ?

    In faculty I only had 2 classes about analog electronics but I don't remember to ever consider a diode as an ideal one in computations.
  6. Feb 1, 2012 #5


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    Do they tell you the supply volts and the load resistance? The actual diode drop will depend on the current but you would normally assume some nominal drop like 0.6V - depending on the specific diode used.
  7. Feb 1, 2012 #6

    Perhaps there's a slight confusion of what "ideal diode" means? The real diode has the current as an exponential function of the voltage; as that's very complicated, there are 3 popular approximations (each one of them is an "ideal" diode thingie, but different ones)

    (a) One is Vd = 0V for any Id > 0; that's a pretty poor model
    (b) One is Vd = 0.6V for any Id > 0; that's a pretty popular model
    (c) One is Vd = 0.6V + Id.Rd for any Id > 0; that's a less popular model

    If one takes the ideal model (b), then you assume Vd = 0.6V, so the current on the second resistor (the one in series) is (Vo - 0.6) / R2. This is the current that passes between the diode and the first resistor. The current on the first resistor is 0.6 / R1. Then you take the current from the second resistor (Vo - 0.6) / R2 and subtract the one on the first resistor - that will be the current on the diode Id = (Vo - 0.6) / R2 - 0.6/R1

    So the book makes sense, if that's what the book did.
  8. Feb 1, 2012 #7
  9. Feb 2, 2012 #8
    The book is All New Electronics Self Teaching Guide.

    @fbs7 the aproximation for the ideal diode is (a).

    If I do the calculations considering a voltage drop of 0.7 on the diode (forward-biased), the result is as they say in the book.

    If I do the calculations considering a voltage drop of 0 on the diode (ideal and forward-biased) what I obtain is different from the book.

    In the book an ideal diode is one that has 0 voltage drop (forward-biased).

    The schematic is a very basic one:

    Vs ---- R1 ---- 1. diode (forward-biased) --- Vs
    2. R2 --------------------------- Vs

    1 and 2 are connected in parallel to R1.

    Vs = 10 V, R1 = 10kΩ, R2 = 1KΩ,
    I need to find the current through the diode (Id).

    Considering a voltage drop of 0.7 on the diode Id = 0,23mA (as in the book).
    If I consider an ideal diode (in the book this means 0 voltage drop <=> ignore the diode) Id = 0.3 mA but how b/c it shouldn't if I understand correctly.
    Last edited: Feb 2, 2012
  10. Feb 2, 2012 #9
    I think you're referring to problem 9 on page 67 of that book?


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  11. Feb 2, 2012 #10
    There's something strange in this question.
    If we assume "ideal diode" then no current will flow through R2 (1K) resistor.
    All R1 current (Vs/R1=1mA) will flow through the diode.
  12. Feb 2, 2012 #11


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    A perfect diode is a concept used where the voltage drop of a diode is regarded as trivial and the forward drop is taken as zero.

    You have a 10 Volt supply and a 10 K and a 1 K resistor in series across the power supply. Then you put a diode across the 1 K resistor.

    If the diode is taken as zero ohms and has zero forward volts, then the 1 K resistor is not doing anything because it is short circuited.
    So, there is 10 Volts across 10 K resistance so there is a current of 1 mA flowing. (10 V / 10000 ohms)

    If you did assume the diode had 0.7 volts across it, then the 1 K resistor also has 0.7 volts across it.
    So the 1 K has 0.7 mA flowing in it. (0.7 volts / 1000 ohms)
    The 10 K has 9.3 volts across it, so it has a current of 0.93 mA flowing in it. (9.3 V / 10000 ohms)
    By subtraction, then, the diode must have 0.93 mA - 0.70 mA or 0.23 mA. (Which is what you got.)

    You may be interested to note that a simulation of this gave the actual diode current as 0.4025 mA and the 1 K resistor's current as 0.54316 mA.
    This is because the diode's actual current is being restricted by a relatively large 10 K resistor and the voltage is less than 0.7 volts.
    The diode voltage in that case was 0.543 volts.

    If you find a genuine mistake in a book, write to the author or the publisher. Authors appreciate readers doing this.
  13. Feb 2, 2012 #12
    @fbs7 that is the problem.
    @vk6kro & @ Jony130 that is also my answer on 0v voltage drop (but not the author).
    @vk6kro regarding the simulation (i use multisim) I know it's different in the "real/virtual world".

    Thank you all.
    In conclusion the book result is wrong (if they consider 0V drop voltage) and I should write to the author to add it to errata.
  14. Feb 2, 2012 #13
    0.3 mA is not correct for an ideal diode: if Vd = 0 then the diode behaves like a short as long as it is in forward active mode. I think the answer would be 1 mA = 10V / 10 kΩ

    But, I don't think that's what the book's answer is referring to. If the voltage drop across the diode is Vd, then its current is

    Id = (Vs - Vd) / R1 - Vd / R2 = Vs / R1 - Vd (1/R1 + 1/R2)

    I think that the approximation he is doing in the book is this: if Vs >> Vd, then Vs - Vd ≈ Vs, and the formula above is approximately

    Id = Vs / R1 - Vd / R2 = 10 / 10 kΩ - 0.7 / 1 kΩ = 1 mA - 0.7 mA = 0.3 mA

    So he's just skipping the subtraction of 0.7 V from 10V. That is, he's not really considering a perfect diode, but is just ignoring the 0.7V when finding the current through the rest of the circuit.

    The writing in the answer is telling about that: he says "Ignore Vd in this case", which means ignore Vd when adding/subtracting with much larger numbers. It's just a little trick to make mental calculations - my father does that a lot. I just use the calculator :)
    Last edited: Feb 2, 2012
  15. Feb 2, 2012 #14
    I've noticed I can get the result he has, in that way but I think it is confusing and actually wrong (and thus i opened this topic), since up until this point in the book, if the voltage drop on the diode is less then 10% of the source voltage, he will consider it to be 0V.

    I am saying wrong because in one step of the same calculation the diode is considered ideal (as the author understands by ideal = 0v) and in the other step it has the ≈ value of a silicon diode (0.7).

    At least it is wrong from a math point of view because it breaks the law of identity.
  16. Feb 2, 2012 #15

    Yep, I don't like very much the way it is presented in the book, either.

    It's really a matter of approximating calculations, that is, Vs - Vd ≈ Vs if Vs >> Vd. That's a good approximation if Vs = 127V and Vd = 0.7V and I do that all the time, but to use that for 10V and 0.7V... that I'm not so sure.

    I think it's just old-style teaching. Nowadays with computers, symbolic manipulators, circuit simulators and all sorts of tools available to resolve these things with more precision, I don't think such approximations are all that needed.
    Last edited: Feb 2, 2012
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