# How to analyze this diode circuit?

I am trying to analyze this simple diode circuit by hand, but I can't seem to get too far.

simulate this circuit – Schematic created using CircuitLab

Using circuit lab, it is evident that current will flow through both diodes, which makes sense to me conceptually, however, trying to analyze using the Constant Voltage Drop model results in an unsolvable circuit.

simulate this circuit

I have tried to use superposition, nodal analysis, and just KVL, but I can't seem to figure out how to solve this circuit. Help would be greatly appreciated!

• The two diodes will be fried. You'll want some series resistors for them. If the simulation runs, the current through the diodes will be huge. – jippie Apr 30 '13 at 9:40
• @jippie Both diodes need not be fried :-) One will become a SED (Smoke Emitting Diode), then we can always analyze for the remaining one! – Anindo Ghosh Apr 30 '13 at 9:42
• @AnindoGhosh and the other one will not be a happy diode. – jippie Apr 30 '13 at 9:42
• If you drew the schematic yourself, I'd double check the reference to make sure you got the D1 polarity correct. In this situation, I would probably make a note of the problem with the circuit as drawn, possibly solve it assuming D1 died, and then additionally solve it assuming D1 was reversed (which was likely the intention). – darron Apr 30 '13 at 15:26
• When you encounter a situation like this, you should realize that some assumption you've made when simplifying isn't valid. In this case, it's the assumptions of the constant voltage drop model that are being violated. – DrFriedParts Apr 30 '13 at 15:50

The circuit as shown is not viable - or you could analyze it in two phases, if you must:

Phase 1:

• Each 1n4148 diode is rated for 200 mA continuous, 450 mA peak repetitive current.
• When wired as indicated, each diode will drop approximately 1 to 1.5 Volts (Fig.3 in datasheet) before the current exceeds absolute maximum rating
• As the supply voltage is 5 Volts, this far exceeds the maximum 3 Volts noted above, so one of the two diodes will burn out.

Phase 2.a: If D2 burns out and becomes an open circuit:

• There will be no voltage at Vout as D2 is now an open circuit
• Result: Vout = 0 Volts

Phase 2.b: If D1 burns out and becomes an open circuit:

• Vout = V1 - VD2 = ~ 4.4 Volts

Then there are the possibilities of D1 or D2 burning out to become short. That resultant analysis is left for you to do :-)

• What if the 'first' diode fails short? – jippie Apr 30 '13 at 9:44
• This is a worry... I have a number of diode circuits to analyze by hand and then test in the lab, I wonder why they would have put this one in if it is likely to burn the diodes out? Thankyou :) – Zimulator Apr 30 '13 at 9:49
• I would personally refuse to build the circuit as drawn in your question. – jippie Apr 30 '13 at 9:53
• @AnindoGhosh Do all your fried diodes work as a voltage source? "Vout = V1 - VD2 = ~ 5.4 Volts" – jippie Apr 30 '13 at 9:57

CircuitLab solves the circuit because it doesn't simulate effects like junction temperatures reaching beyond the limit, so that semiconductors melt.

A diode is not a fixed voltage drop. Current through a diode is related to voltage by an exponential equation. That exponential equation goes on forever: for any imaginable voltage, you can find a current. Actually, there is more than one equation because even the equations are idealizations of real behavior. A wortwhile read is the diode modeling Wikipedia article.

In the DC simulation, you forgot to add expressions for viewing the diode current, an important quantity that the designer must be concerned about. The DC solver reports that the current through the top diode is 2.755A, and through the bottom one, 2.750A (since the resistor takes 0.005 of it). Yes, the diodes are dropping 2.5V, but by means of drawing a very large current. Each diode dissipates 6.9W. Why don't you look up the datasheet for the 1N4148 to see what the actual limits are?

Perhaps the circuit is realizable. However, if so, it cannot be without some cryogenic cooling mechanism to keep the junction temperatures within limits! And even if it works, the results will not likely agree with CircuitLab's DC Solver: the voltage between the diodes will not be exactly halfway between 0 and 5.

One way you can solve the "impossible" circuit is by imagining that the diodes have a bulk resistance which is approximated by a tiny series resistor (and then continue to treat them as a fixed voltage drop):

This isn't physically correct, and still ignores the reality that the diodes will be destroyed, but it's one way to reproduce the results of the DC Solver. (The 650 $m\Omega$ figures are cooked up to make it work out to about the same values, while retaining the 700 mV assumption.)

simulate this circuit – Schematic created using CircuitLab

If you double-click on the diode symbol, you will see that the actual value CircuitLab uses for a series resistance in the diode (parameter R_S) is 0.568 $\Omega$. A little less than above, which means that CircuitLab calculated a higher voltage across the PN junction than 0.7. If we go with 0.568, it means that the voltage drop (V = IR) across this resistance is about 2.755A * 0.568, or about 1.56V. Two voltage drops of 1.565V leave 0.935V across each diode. I.e. CircuitLab applied some exponential formula to determine the forward voltage, which solved to 0.935V in consideration with R_S.

As far as your second circuit goes, it unsolvable because it is invalid. You cannot connect ideal voltage sources in parallel unless they have exactly the same voltage, in which case it is pointless because they are equivalent to a single voltage source with that voltage. If two unequal voltage sources are paralleled they short-circuit each other: their difference voltage faces a zero ohm impedance. Ideal voltage sources don't exist in the real world but devices that try to behave somewhat like ideal voltage sources will not like being connected together that way either.

Appendix: application of Shockley formula to CircuitLab figures.

$$I=I_S\left(e^{V_D/(nV_T)}-1\right)$$

We already know the final current $I$ is 2.755 A, which together with the series resistance R_S tells use that the voltage drop across the diode must be about 0.935. Let's see if that 0.935 works out back to the current. $V_D$ is just that voltage drop. The value $n$ (ideality factor) is given in CircuitLab's model for the diode. It is 1.752. Let's assume 26 mV for $V_T$, the thermal voltage. The $I_S$ value is also given: 2.92E-9.

Crunching the numbers, we get $I = 2.92\times10^{-9}\left(e^{0.935/\left(1.752\times0.026\right)}-1\right) = 2.397A$

This is in the ballpark of the 2.755 current value. Obviously, CircuitLab isn't using this formula, but a more advanced formula in which those other parameters of the diode come into play.