This is a problem with its solution:

What I don't understand is how he replaced the three diodes with the voltage sources in DC-analysis or with the resistances in AC-analysis. What assumptions did he make to do that?.

For DC-analysis: if it was one diode then it's ok for me but the problem is with the three connected in series. For AC-analysis I have the same problem as with DC plus two questions:

  1. Using small signal model, we replace the diode with a resistor rd and then we solve the problem and if the current id was negative then our assumption that the diode is on is wrong and it should be off. Is that right?
  2. If the diode is on in DC-analysis, does that mean that it should be on in AC-analysis?

1 Answer 1


The large signal model is used to find the bias point (8.95mA). It is assumed that small signal analysis will not result in large changes from that bias point.

You can check that assumption with the large signal model- the minimum current is a bit less than 8mA, and the maximum is 9.95mA so it's not too bad. The diodes are not only always on, but they're on with roughly 9mA +/- 10%.

So, given that, you may now replace the diodes with their small-signal model which is a resistor of resistance nVt/Id or about 2.8 ohms each, given the assumptions in this problem. The 20V source is replaced with a short and just the AC small signal remains. The idea is that we're looking at the effect of "small" changes from that bias point.

To get the final output voltage, he adds the bias point (2.1V) together with the changes from the AC small signal analysis, to get a small AC signal of 8.35mV peak added to a relatively large DC voltage (2.1V).

If you simulate this with a good simulator and accurate diode models you'll see that the assumption is only roughly correct and the sinusoidal waveform is a bit distorted with the higher voltages scrunched down as the diode dynamic resistance drops and the lower voltages expanded a bit. It's a linear approximation to behavior of a nonlinear system.

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