Homework Statement
If the divisor P is a prime of the form 4q+1 then the number -1 or P-1 is certainly a residue.
Homework Equations
The Attempt at a Solution
First the book told me to prove that 1 and -1 are the only two remainders that are their own reciprocals modP...
wait wouldnt it just be a voltage divider and thus that would mean:
VL=Vout(Rout/Rout+RL)
Then simplifying, would get:
Rout=VLRL/(vout-vl)?
But if that is the case, do we just disregard the 1k resistor?
well I believe we would do the change in voltage over the change in current and that would give us the internal resistance... however, finding the current in this situation is a little more difficult :/
so then how could we measure the output impedance with the load resistor?
We could measure the voltage drop across the resistor, but I don't know how that could help us with the output impedance
so then my approximate output impedance would just be that divided by 200, which would give me 20.24 ohms?
Also, is that 1k resistor a load resistor? Because for the next part it says suppose we added in a load resistor. In order to do that, would I just put it in parallel with the 1k resistor?
I don't think that we have any equivalent circuit models...
The hint my teacher gave is that it is approximately equal to the output impedance of the circuit that supplies the base current divided by β.
But what is the output impedance of the circuit that supplies the base? is it that 4047.6...
i've never seen the pass transistor before.... but i've seen an emitter follower. I wasn't positive, but I thought that the input impedance would be the 10k and 6.8k resistors in parallel, which would give me about 4047.6. However, I'm not sure about the output impedance. Usually it is just...
Homework Statement
Calculate the output impedance of the emitter-follower circuit called a pass transistor. Assume that beta=200
**See attached diagram**
Homework Equations
The Attempt at a Solution
Not really sure how this works, I thought it would just be 1k cause that is the...
Thanks so much, that makes sense!!
However, am I right about the PNP being the other direction because then the current would be flowing in the opposite direction, so then the reverse current would be limited by a diode that is biased from B to E?
Well theoretically it should drop all of it :D, so the max should be zero with the diode
I realized that after I typed it....
So for the PNP would I just reverse the bias of the diode?
Well without the diode, wouldn't there have to be a bigger voltage drop in the base-emitter junction? With the diode, wouldn't it drop some of that voltage so the junction doesn't have to drop it all?