Diagram of Diode Circuit

I came across this problem while I was working through diode circuits in my textbook, doing problems where we have to graph the output voltage as a function of the input voltage or current. I approached the problem by looking at the end behavior of the function, when I_in is a really large positive or negative value. So when I_in is positive infinity, D2 is on and D1 is off and when I_in is negative infinity, D1 is on and D2 is off.

I thought that at the point where a diode turns on/off, the circuit is between two states and its behavior isn't really defined in a constant voltage model of a diode circuit. So I thought that the circuit has to be in both states and as a result, the two portions of the graph representing those two states must intersect at the point where the diode transitions. In other words, Vout must be a continuous function of I_in.

I found that when D1, D2 are off, Vout is 0V and when D1, D2 are on, Vout is the turn on voltage of D2. This meant in order to ensure continuity in between the two "edge states", D1 and D2 had to turn on and they never could have both been turned off (in general, I also think that a circuit with n diodes can only have n+1 "portions" in its graph).

With this, I was able to conclude that for -infinity < I_in < 2VD_on/R1, Vout = I_inR_1 - V_DON and for 2*VD_on/R1 < I_in < V_DON, Vout = V_DON.

However, to check my work, I simulated the circuit and found that I had the right equations for the graph but the transition between the two pieces happened at I_in = 0 so that at this point, V_out is both -V_DON and V_DON.

So now I'm really confused because this is the first circuit where my approach didn't work and I'm not sure how to get the actual answer if we don't have continuity of the graph.

  • \$\begingroup\$ Diodes don't just turn on and off. They have a defined relationship between voltage and current. \$\endgroup\$
    – Finbarr
    Commented Mar 26, 2022 at 18:39
  • \$\begingroup\$ Yeah but in the textbook I'm working through, we assumed that the diodes are ideal with a turn on voltage. In the simulation, I also used ideal diodes with a turn on voltage. \$\endgroup\$
    – snowball
    Commented Mar 26, 2022 at 18:44
  • \$\begingroup\$ I'm not sure what your actual question is. You might want to edit your post to make it clear. \$\endgroup\$
    – Finbarr
    Commented Mar 26, 2022 at 18:58
  • \$\begingroup\$ I can't fully follow your train of thought, but with this ideal constant voltage diode model and this circuit there is no way that both diodes are conducting at the same time. Positive current will pass through D2, negative current will run through D1 and R1. \$\endgroup\$ Commented Mar 26, 2022 at 19:01
  • \$\begingroup\$ @LarsHankeln Oh ok I see what you mean. But I think even with the graph going from D1 on, D2 off to D1 off, D2 on, it still ends up being the same graph since D1 on, D2 on gives the same output voltage as D1 off, D2 on. \$\endgroup\$
    – snowball
    Commented Mar 26, 2022 at 19:08

4 Answers 4


It depends on what level your circuit analysis is at. You have to know what diode model to use.

If your diode model is the elementary "0.7V drop if forward biased, ESR=0, otherwise open", then the analysis is:

  • \$ I_{in}=0 \to V_{out}={\rm undefined}\$, since we're trying to measure voltage of an open circuit, and the circuit analysis doesn't allow that
  • \$ I_{in}>0 \to V_{out}=0.7\ {\rm V} \$
  • \$ I_{in}< 0 \to V_{out}=-0.7\ {\rm V} + R_1\cdot I_{in} \$

Note that with either positive or negative current, the voltage "immediately" rises to the diode drop at least.


The Shockley diode equation, gives the relation between the voltage across and the current trough a forward biased diode:


Where \$\text{I}_\text{D}\$ is the diode current, \$\text{I}_\text{S}\$ is the reverse bias saturation current, \$\text{V}_\text{D}\$ is the voltage across the diode, \$\text{q}\$ is the electron charge, \$\text{k}\$ is the Boltzmann constant, \$\text{T}\$ is the temperature and \$\eta\$ is the ideality factor.

So, now for your circuit. When \$\text{I}_\text{in}>0\$ we know that \$\text{D}_1\$ prevents current to flow. So the output voltage is given by:


Now, when \$\text{I}_\text{in}<0\$ we know that \$\text{D}_2\$ prevents current to flow. So the output voltage is given by the solution to the following system of equations:

$$ \begin{cases} \begin{alignat*}{1} \text{V}_\text{out}&=-\left(\text{V}_{\text{D}_1}+\text{V}_\text{R}\right)\\ \\ \text{I}_\text{in}&=\text{I}_{\text{S}_{\text{D}_1}}\left(\exp\left(\frac{\displaystyle\text{q}\text{V}_{\text{D}_1}}{\displaystyle\eta_{\text{D}_1}\text{k}\text{T}_{\text{D}_1}}\right)-1\right)\\ \\ \text{V}_\text{R}&=\text{I}_\text{in}\text{R} \end{alignat*} \end{cases}\tag3 $$

So, we end up with:


  • \$\begingroup\$ This is of course an approximation. Schockley equation's beauty is that it's not discontinuous: you should use it no matter the polarity of the diode current/voltage :) Then a closed form solution needs a series expansion - and that's often what numerical solvers do, in fact. \$\endgroup\$ Commented Apr 1, 2022 at 20:22

The diode model you are using can have any voltage across the diode from a large negative voltage to a forward voltage < Vf without current flowing. It follows that (assuming the resistor is not open circuit) in the given circuit at zero current, there can be any voltage from -Vf to Vf across the circuit.

As the current increases from zero, the voltage will change differently in one direction than the other (because of the resistor), and you can calculate that behavior.

Real diodes do not behave like that (holding a static charge or whatever), of course. They conduct a current for any voltage applied, positive or negative, so at zero current there will be zero voltage.


When working with current sources connect to diodes you cannot use the ideal diode model. You have to use the Shockley diode equation to find the current and voltages of each diode.


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