Hi all,

Studying for the FE Exam and I have found the "FE Electrical and Computer Practice Problems" study guide by Michael Lindeburg very helpful thus far. I am near the end, and this is the first problem I am a bit stuck on the solution and just want to make sure I am not misunderstanding, as these darn semiconductor circuits have been one of my weaker areas.

I have posted the full problem and solution for problem 9.

My hang up is on how they find the gate-to-drain voltage. I am fine up to VGD = VGS - VDS, but I also know VGS = VG - VS and VDS = VD - VS. VS is connected to the positive terminals of both sources. That to me seems to imply VGS would be 2V and VDS -1.9V, but the solution has that flipped and ends up with VGD = -3.9V. Could somebody help explain if the solution is correct, what I am overlooking.

Thanks!

 

i don't see why would anyone need Vgd. what matters is Vgs and Vds.

one can pick top side of the circuit as a reference and go from there. or, if negative values and p-type are confusing, flip everything and use what you are more comfortable with (flip circuit upside down, change fet for N-type, flip polarity of both sources).

does this make things more palatable?

 

Sort of, but now I'm also a bit more confused. If you go through the solution, VGD is needed to confirm what operating mode the circuit is in. I get VGD of -3.9V as does the solution when oriented like you've shown, but I'm still confused because for the sake of my original question, I'm trying to understand why VGS is -2V and VDS is 1.9V in the original circuit. IF VGS is VG - VS and in the original orientation VS is what's connected to the positive terminal, does that not imply VGS would be -(-2V) or +2V?

 

you are correct...

Convention is that Vab=Va-Vb so Va is the "positive"/"higher" potential.
But if you look at Vgs you see that Vgs = - E1, (negative terminal connects to gate, positive to source)
therefore Vgs=-(-2.0V)=+2.0V

when computing voltage in circuit one need to choose reference point.

assuming you choose Drain as "0V" (Vd=0V), we can choose how to get the Gate from your reference point:


i drew that path i chose (blue arrow) going through E2 and then through E1. Arrow goes in same direction as E2 (increases from - towards +) but then goes against E1 (from + towards -)

that means we get
Vgd= E2 - E1 =1.9V - (-2.0V) = 3.9V

Vdg is simply the opposite Vdg=-Vgd