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Problems to find the correct load line

I have some problems about load line of this circuit...

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Usually, to find load line i calculate current in transitor when transistor is shorted and when is open circuit. In this case, when transtistor is shorted, vout=vss and current in transistor M1 shoud be Iq+(Vout/Rl): is this right? How can i obtain (Vdd-Vss)/Rl? Thanks in advance

Reply to
lionelgreenstreet
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Hints: Put in some perhaps arbitrary but possible numbers, like Vdd=3D

+5, Vss=3D-5, Rl =3D 1kohms. With the lower transistor shorted, the current in Rl is obviously 5mA. What happens as you vary Iq between, say, 0mA and 20mA? What can you tell me about Iq if both your equation and the equation in the picture are correct? Is the picture telling you that it is the only answer, or that it is a particular answer that may be interesting? You may also wish to consider the case where Vdd is not equal to -Vss, and how that changes the nice symmetrical picture.

(Extra credit: what does that capacitance do to the load line in the case of sinusoidal excitation?)

Cheers, Tom

Reply to
Tom Bruhns

What you actually need is two load poly-lines. Cutoff in one part will actually occupy much but a minority of the active load line in the other part.

Reply to
JosephKK

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