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schematic

simulate this circuit – Schematic created using CircuitLab

I'm confused about whether zener is on or off due to not being able to know voltage on V0. I tried to find by applying KCL & KVL, but current on the diode (Id) and the resistor (Ir) is not equal to each other. I'm using the shockley diode equation.(Id=Is(e^(Vd/Vt))). This current source being connected to zener diode instead of a voltage source makes something complicated. Which way should I follow to find V0? Hope one can give me a trick.

Thanks in advance.

circuit

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    \$\begingroup\$ I don't see Id or Ir marked on your schematic so I can't follow your reasoning. Also, please show all of the work you have done so far. \$\endgroup\$
    – Elliot Alderson
    Commented Dec 2, 2018 at 21:08
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    \$\begingroup\$ What would Vo be if you assume that the Zener is not conducting? Is it higher than the Zener voltage? \$\endgroup\$
    – Phil G
    Commented Dec 2, 2018 at 21:12
  • \$\begingroup\$ What I'm trying to do is that if I find the current flowing through resistor (here I use Id current for each branch one by one because Ir and Id are the same, but to find Id , Vd has to be known. The given Vd values is for constant voltage drop model that requires relatively higher voltage than ,in this circuit, V0 , therefore I am not sure whether to use it) I would find voltage on resistor so then the V0. \$\endgroup\$
    – Klementayn
    Commented Dec 2, 2018 at 21:48
  • \$\begingroup\$ @Klementayn Assume the zener is off. This just means that you have two resistors in parallel, or \$\frac23\$k, times 3 mA. This works out to 2 V. Add 600 mV for your diodes and this would account for the entire current. But at 2.6 V, the zener isn't going to conduct current (leakage only.) So the assumption is supported. Do you need to be more detailed than this? \$\endgroup\$
    – jonk
    Commented Dec 2, 2018 at 21:49

2 Answers 2

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There are only two options here the zener is ether conducting or it isn't

If it is:

$$\dfrac{Vo-Vz}{R1}+\dfrac{Vo-Vd}{R2}+\dfrac{Vo-Vd}{R3} = I1$$

If it isn't:

$$\dfrac{Vo-Vd}{R2}+\dfrac{Vo-Vd}{R3} = I1$$

Where Vo is the voltage a junction of the diodes, Vz is the zener voltage and Vd is forward diode drop (0.6V).

What do these say? Does only one make sense?

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Treat the current source as the variable. At what current does the diode cut on? This is solved by assuming the diodes current is zero, and solving for the current needed to raise the zener diode voltage to 3V. Now that you know what current is going to turn on the diode, is 3mA greater or less than this current? With the state of the zerer solved, the rest of the problem is simple.

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  • \$\begingroup\$ Other way around. You don't know if there is sufficient voltage to turn allow current in the Zener diode. So, ignore the Zener initially and see what voltage is present at node V0. \$\endgroup\$
    – Dwayne Reid
    Commented Dec 3, 2018 at 5:07
  • \$\begingroup\$ Both methods work. I think the current first method is more deterministic as it involves no hypotheticals. \$\endgroup\$
    – skeptonomicon
    Commented Dec 3, 2018 at 5:28
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    \$\begingroup\$ That doesn't make sense to me. The value of current is fixed and specified. It is trivial to calculate the voltage at node V0 when the Zener diode is disconnected. It is then trivial to re-calculate the voltage at Node V0 should there be sufficient voltage to allow the Zener diode to conduct. \$\endgroup\$
    – Dwayne Reid
    Commented Dec 3, 2018 at 5:43
  • \$\begingroup\$ Did you try to solve it my way to see how hard it is? Both methods are trivial. This is a trivial problem that can be solved by inspection alone, however, if the student had a more complex problem in the future, with 5 legs with zeners, than the difference in the methods might be more obvious. I was trying to teach a method that will be useful in the future. \$\endgroup\$
    – skeptonomicon
    Commented Dec 6, 2018 at 2:48
  • \$\begingroup\$ why not post an answer where you solve the problem using the method that you suggest. \$\endgroup\$
    – Dwayne Reid
    Commented Dec 6, 2018 at 16:49

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