Problem with finding output voltage of a diode circuit

Question

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.

Answer 1

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.

Answer 2

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

$$\text{I}_\text{D}=\text{I}_\text{S}\left(\exp\left(\frac{\text{q}\text{V}_\text{D}}{\eta\text{k}\text{T}}\right)-1\right)\tag1$$

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.

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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:

$$\text{V}_\text{out}=\frac{\eta_{\text{D}_2}\text{k}\text{T}_{\text{D}_2}}{\text{q}}\cdot\ln\left(1+\frac{\text{I}_\text{in}}{\text{I}_{\text{S}_{\text{D}_2}}}\right)\tag2$$

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:

$$\text{V}_\text{out}=-\left(\frac{\displaystyle\eta_{\text{D}_1}\text{k}\text{T}_{\text{D}_1}}{\displaystyle\text{q}}\cdot\ln\left(1+\frac{\displaystyle\text{I}_\text{in}}{\displaystyle\text{I}_{\text{S}_{\text{D}_1}}}\right)+\text{I}_\text{in}\text{R}\right)\tag4$$

Answer 3

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.

Answer 4

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.