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Homework Help: Find voltage in diode circuit

  1. Nov 9, 2012 #1
    1. The problem statement, all variables and given/known data
    Find voltage at Vo1 and Vo2
    The left part is the schematic used in the question. The right side is the engineering schematic I drew up.

    3. The attempt at a solution
    I'm not sure what to do here.
    because there is a Si and Ge diode that means the Ge will be actived first because the forward voltage is only 0.3V (vs 0.7V of the Si diode).

    So how should I look at this circuit then?
    ignore the Si diode and then Vo1 is the voltage across R2 and the Ge diode, and Vo2 is the voltage across only the Ge diode

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  3. Nov 9, 2012 #2


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    Yes. Try assuming the Si diode isn't conducting (eg remove it). Check the voltage that would appear at Vo1. If it's >0.7 then the Si diode is conducting so put it back.
  4. Nov 9, 2012 #3
    Ok so when ignoring the Si diode the voltage at Vo1 is 13.4V, >0.7V therefore it is conducting

    so looking at the whole circuit. I could be completely confused, but because Si diode is parallel to R2+Ge diode, is the total voltage drop across R2&Ge diode only 0.7V?

    edit: so Vo1=0.7V and Vo2=0.3V

    seems too easy thats why i asked,,:)
    Last edited: Nov 9, 2012
  5. Nov 9, 2012 #4

    rude man

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    The answer depends on the i-V characteristics of both diodes.

    You seem to assume that the voltage across the Si diode is 0.7V irrespective of the finite current thru it, also 0.3V for the Ge. Under this assumtion, both diodes will be on regardless of the resistor values.

    In reality, diodes like these have i ~ i0exp(V/26mV) with the Si i0 lower for the Si than for the Ge. Under this model, the outcome depends on the values of the resistors.

    If you are given a different model, tell us ...
  6. Nov 9, 2012 #5
    All the questions prior to this one assume the voltage drop across Si is 0.7V, and Ge is 0.3V.
    But I do know about the V-I characteristics. However, I know about Shockley's equation: I=I0(exp(qVa/kT)-1), haven't seen i ~ i0exp(V/26mV) before.

    In this case I guess I should write a range of simultaneous equations..
    The voltage drop across the Si Diode will still be equal (and thus limit) the voltage across R2+Ge diode.. right?

    But how do I apply the formula, what is I0?
    Last edited: Nov 10, 2012
  7. Nov 10, 2012 #6

    rude man

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    The shockley equation & mine are essentially the same except at very low currents. You'll note that kT/q = 26 mV near room temperature.

    Pick your i0 so that at the 1-10 mA level you get 0.3V for the Ge diode and 0.7V for the Si.

    The thing for you to notice is that the current thru the 1K is very much larger than the current thru the 0.47K.
  8. Nov 11, 2012 #7
    If you're concerned about whether the diodes are fully on or not you can't really assume the diodes are a specific size can you? :). The value will depend on junction area, which is selected by the manufacturer for different applications. If this question required you to find an exact voltage and current, it would have to supply the diode characteristics, either in a graph or as equation parameters.


    Because the diodes have exponential characteristics, their voltage drops will only change by tenths of a volt across ten decades of changes in current. This is easy to see:

    Ic = Is eVbe/VT
    Vbe = VT ln (Ic/Is)

    ΔVbe = VT ln (Ic2/Ic1)

    At room temperature, Vt = 26mV so a thousand fold increase in current only means a change of 0.18V in voltage drop. If your answer can be accurate to a few tenths of a volt, it is safe to assume the voltage drops across the silicon diode is 0.7 and 0.3 for the germanium. If that works in your circuit, you are good.

    That almost constant 0.7/0.3 volt drop depends on the diodes achieving a certain amount of minimum current. This is because the i/v characteristic has some rapidly changing i/v relationship below 0.7/0.3 before it finally settles into near constant voltage drop for currents greater than some minimum amount. You can find currents in the circuit assuming the 0.7/0.3 volt drops to see if that min current is achieved. This is what rudeman is suggesting but again, you cannot know what this min current is without diode characteristics.

    For the exact solution, you would need to solve Ic=Is eVbe/VT for Vbe and stick that into your KVL equations. You will find the resulting equations are not linear and probably won't have a closed form solution. The only way to solve these is to use a computer / calculator or by trial and error. The latter assumes some start voltage drop (like 0.7), working through the circuit equations, and then finding a better guess. Repeat until the guess is not much different from the assumed.

    You can also solve these graphically using load lines.

    Be careful of the accuracy of these 'exact' solutions too. Diode characteristics are statistically distributed. If they tell you your diode has a certain Is, they mean that particular batch of diodes has a mean Is as given with standard deviation of x. The temperature at the junction will only be at room temperature just after the circuit is switched on. The temperature will rise until a steady state is reached which depends on thermal resistance between the junction and environment and the temperature of the environment. There is resistance in the lead heading into the diode and there is resistance in the doped materials making up the diodes. There are many, many things that will contribute to the exact values of i,v that nature finds and many of those things are not knowable. We just want an answer that is accurate enough for the application.
  9. Nov 11, 2012 #8


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    It's pretty clear that the current through the Si is high enough for 0.7V to be a reasonable approximation for Vo1.

    Then you effectively have 0.7V feeding the Ge via a 470 Ohm...

    First stab for the current through the Ge diode would be (0.7-0.4)/470 = 0.66mA which is just about enough for most Ge diodes to be considered "On" eg for 0.4V to be a reasonable approximation for Vo2.

    Would be operating in the top right quarter of this chart...


    As aralbrec said, if it needs to be more accurate you need more data on the diodes.

    EDIT: Actually looking at that chart 0.25-0.3V would be a more reasonable value for Vo2 based on that diode data.
  10. Nov 11, 2012 #9

    rude man

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    There is no such thing as a diode being "fully on".

    The weak dependence of V on i is what legitimizes using values of is to give the specified V (0.3V and 0.7V) in the 1 - 20 mA range we are dealing with here.

    To avoid the transcendental equations you allude to, a simplified diode i-V model within the range of currents of interest could be invoked., such as: i = 0, V < VTH; i = kV, V > VTH, which is why I asked the OP if such a model was introduced in his studies.
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