Norton Equivalent with Dependency on Open Circuit

Thread Starter

coolchriz90

Joined Jan 24, 2010
4
I have been struggling with this problem for the last couple hours and haven't been able to figure it out.

I know in general how to solve a Norton equivalent problem (short circuit current and Equivalent resistance) but having a the dependent source be dependent on the voltage drop across this open circuit has thrown me for a loop.

Thanks for any assistance.

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t_n_k

Joined Mar 6, 2009
5,455
What do you need to solve for? There is no problem statement - you've simply supplied the circuit.

Thread Starter

coolchriz90

Joined Jan 24, 2010
4
Sorry about that. I guess I've become slightly brain dead with how long I've been staring at this problem. I am trying to find the Norton Equivalent of this circuit.

t_n_k

Joined Mar 6, 2009
5,455
Sorry I didn't read your question carefully. You want the Norton Equivalent.

If you short terminals a and b you will have the Norton Current (=isc)

Vi=Vgs+isc*Rs
Vgs=Vi-isc*Rs -----(1)

Suggest converting the current source in parallel with ro to a Thevenin Equivalent - hence

Vth=gm*Vgs*ro
Rth=ro

isc=gm*Vgs*ro/(ro+Rs) -----(2)

Eliminate Vgs by solving (1) & (2) simultaneously to give isc in terms of Vi.

If you take Vab (open) as Vth you should be able to go from there to find the Norton resistance.

t_n_k

Joined Mar 6, 2009
5,455
In fact Rnorton is found by inspection rather than unnecessary calculation. What do you "see" looking into terminals a-b with the current source open?

Thread Starter

coolchriz90

Joined Jan 24, 2010
4
But the current source shouldn't be considered open when finding R_Norton because it is a dependent current source.

t_n_k

Joined Mar 6, 2009
5,455
But the current source shouldn't be considered open when finding R_Norton because it is a dependent current source.
Yep - agreed - my mistake. Well spotted.

You have to find Vab(open) and divide by isc.

I get Rnorton=ro+Rs+gm*ro*Rs