Electronics Circuit

[Juansta]

Hey all!
Just wondering if someone could give me a hand and stear me in the right direction...

How do I find the DC operating point in this situation?

Im really stuck in finding Vg mainly...
Any helpl would be greatly appreciated...

[moamps]

I(R2)=0
V(g-s)= - V(b-e)=-0.65V
V(g-ground)=V(g-s)+Id*Rs=Ie*Re

Regards,
Milan

[darkfenriz]

DC at gate is the same as at the emitter
with mosfet it results in both transistors are off

[Juansta]

Thanks for that people...

I wasnt saure whether I could disregard the resistor connected to the gate.... seems like I can for DC analysis, and then keep that into account when doing my samll signal?

How does Ib come into it? Just using Mesh analysis?

Im just really stuck on this, need a third perspective to help me see things that kinda get blurred when you look at the same problem over and over again...

[Juansta]

What about if the DC voltage source V1 was now reversed?

ie: someone drew it wrong...

[azira]

Originally posted by Juansta
Thanks for that people...
wasnt saure whether I could disregard the resistor connected to the gate.... seems like I can for DC analysis, and then keep that into account when doing my samll signal?

I think maybe you missed the point of what darkfenriz said.

M1 Vgs drop is ~4 Volts. Q1 Vbe drop is ~.7volts. That means the voltage at Q1:e is M1:g - 4.7 volts. Since no current flows into the gate of Q1, the voltage on both sides of R2 is the same.

So, since M1:g == M1:g - 4.7 is not possible, M1 must be off. If M1 is off, Q1:b is at gnd and therefore Q1 is off also.
--
Danny

[Juansta]

But this doesnt hold when I run a simulation in PSpice...

This is where Im getting very confused....