where have i gone wrong in finding the values for Id and Vds in this transistor circuit, given the shockley curve?

 

Not sure what you expect in terms of answers ...

In any case, if ID≈IS=3.1 mA the drop across RS would be 6.2V, meaning the FET would be biased beyond pinch-off - which is inconsistent with there actually being current in the device.

I have ID=1.44 mA.

 

Were you taught about load lines?

 

Were you taught about load lines?

yes I was briefly explained to how load line analysis is set up, but still having trouble figuring out how Vgs and mA from my Shockley Curve is wrong?

 

yes I was briefly explained to how load line analysis is set up, but still having trouble figuring out how Vgs and mA from my Shockley Curve is wrong?

Your solution needs to satisfy VGS = -IS*2k You have written this out in your working, but seem to be ignoring it.

 

Your solution needs to satisfy VGS = -IS*2k You have written this out in your working, but seem to be ignoring it.

Okay, I see that if Id= Is=1.44mA, then my Vgs would be -2.88V. But im not seeing how there is if enough information to find these two values given Vgs= (-Id)(2k). Wouldn't I have to know either Vgs or -Id first?

 

Don't you also have another curve relating Id and Vgs? I would think that finding a common solution for the two would be significant.

What would happen if you plotted both curves on the same axes?

 

If both curves were plotted on the same axis you'd have a straight line. Im not sure I quite understand???

 

Only the equation VGS = -IS*2k gives a straight line. The other curve for Shockley's equation of ID versus VGS is, well, curved.


The curves will intersect if plotted on the same axes. Why not just try it - you already have the Shockley curve, plot the other one on top.

This is only a very rough approximation, by the way, and I've no idea if a graphical solution would cut any ice with your examiners.

 

If both curves were plotted on the same axis you'd have a straight line. Im not sure I quite understand???

I didn't say AXIS, (singular) but AXES, plural. Plot the curves on the same set of axes, that is, on the same chart and with identical scales.

 

I might have confused you in midst of my misunderstanding... Once I have the Shockley curve, I know that wherever "Vgs" crosses the curve you're able to find "Id"; and likewise, wherever Id crosses the Shockley curve you're able to find Vgs. Its just by an estimate. But what if you have the Shockley curve and neither "Vgs" or "Id" value to begin with. How is one or the other found?

 

Two curves cut each other. The Shockley curve and the source resistor bias curve.

Can you really not see this???? This gives a single possible working point.

 

SEE PAGE SIX.

http://ux.brookdalecc.edu/fac/engtech/andy/engi242/powerpoint/fet_dc_biasing.pdf

 

Plot the V-I line of your source resistor (Rs) on the graph with your Shockley curve.. The intersection of these two lines will give you Vgs and Id. This is because Id*Rs=Vgs.

 

SEE PAGE SIX.

http://ux.brookdalecc.edu/fac/engtech/andy/engi242/powerpoint/fet_dc_biasing.pdf

Visual aid always does the trick for me!!! Makes a whole lot more sense now that I actually see it. I understand the concept MUCH better now. THANKS A BUNCH!