Output impedance of a Pass Transistor

  • #1
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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 resistor across Vout is measured.

Any help would be great!
 

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Answers and Replies

  • #2
gneill
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What circuit models have you been given for transistors? Or perhaps you have a set of characteristic curves for this particular transistor?
 
  • #3
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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 Rc or Re, but Vout is being measured in a different place this time.
 
  • #4
gneill
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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 Rc or Re, but Vout is being measured in a different place this time.

It's just a typical BJT (Bipolar Junction Transistor), nothing special. It's the particular circuit configuration that names it as a "pass transistor" or emitter follower.

Again, do you have any equivalent circuit models for the transistor?

One way to arrive at the output resistance is to divide the open circuit voltage by the short circuit current (find the output voltage of the circuit with no additional load, and then find the current if the 1K resistor is replaced by a short.

If you don't have an equivalent model for the transistor, or characteristic curves by which to determine the saturation current, you can only make a guesstimate for the current that will be flowing from the emitter.
 
  • #5
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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 ohms that I mentioned before?
 
  • #6
gneill
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Yes. You can replace the base bias circuit with a Thevenin equivalent which has a resistance comprised of the two bias resistors in parallel. So that's the output resistance of the bias network supplying the base.
 
  • #7
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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?
 
  • #8
gneill
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Yes, and yes.
 
  • #9
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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
 
  • #10
gneill
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What's the wording of the question for when the load resistor is added?

The 1K and load resistors will be in parallel with the transistor's output impedance.
 
  • #11
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This is for a lab so it says to find the output impedance by finding the change in voltage when the circuit is loaded by 150 ohms.
 
  • #12
gneill
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Okay, think back to when you were given problems where you had to find the internal resistance of a battery when you were given the unloaded voltage and loaded voltage (and/or current).
 
  • #13
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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 :/
 
  • #14
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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?
 
  • #15
gneill
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But if that is the case, do we just disregard the 1k resistor?

Good question. I think that in this case the 1k resistor is to be considered as part of the circuit for which you want to find the output resistance (imagine that it's included in the "black box" that you're examining for output resistance).
 

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