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Amplifiers - Voltage gain of a current controlled voltage source and current source.

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Homework Statement



I've been reviewing my electrical theory and I'm having trouble trying to calculate an expression for the voltage gain for these two specific circuits. (Images c and d of the question.png attachment). I already know the answers, however I am unable to derive the same solution. Answers are given with my solution attempt.



Homework Equations


Kirchoff's loop law and current junction law.

Voltage gain = Output voltage/Input Voltage

R_s is source resistance, R_L is load resistance.
r_o is amplifier output resistance and r_i is amplifier input resistance.


The Attempt at a Solution


I've attached my solution for b if it's unclear what I am asking. I'm pretty sure I've misunderstood something while working it out but haven't been able to figure out the correct steps. Any help is appreciated.

Edit: Attachments are now there.
 

Attachments

Last edited:

Answers and Replies

  • #2
berkeman
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Homework Statement



I've been reviewing my electrical theory and I'm having trouble trying to calculate an expression for the voltage gain for these two specific circuits. (Images c and d of the question.png attachment). I already know the answers, however I am unable to derive the same solution.



Homework Equations


Kirchoff's loop law and current junction law.

Voltage gain = Output voltage/Input Voltage

R_s is source resistance, R_L is load resistance.
r_o is amplifier output resistance and r_i is amplifier input resistance.


The Attempt at a Solution


I've attached my solution for b if it's unclear what I am asking. I'm pretty sure I've misunderstood something while working it out but haven't been able to figure out the correct steps. Any help is appreciated.
I'm not seeing any attachments. Try again?
 
  • #3
rude man
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Part (c): well, the input voltage is Vin and the output voltage is what if the controlled current source is μVin feeding an output impedance of ro?

What function, if any, does ri have?
 
  • #4
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r_i is a constant but I'll see where I get with your suggestions on C for now. It's just late here right now but I'm going back to it in the morning. Thank you very much.
 
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  • #5
rude man
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r_i is a constant but I'll see where I get with your suggestions on C for now. It's just late here right now but I'm going back to it in the morning. Thank you very much.
Yeah, ask yourself if ri has any significance at all ....
 
  • #6
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I thought r_i would have mattered as it is used in the voltage divider rule to figure out the input voltage of the amplifier and from there mu*v_in would have been the output voltage of the amplifier.

If I considered the first loop in C then we have:

Vs = VSource Resistance + VAmp input resistance.
Vs = I(Rs + ri)
I = Vs/(Rs + ri)

then multiply by r_i to get
VAmp input resistance = riVs/(Rs + ri)

The diagrams also aren't too clear on the symbols but I believe by vin and Iin is referring to the voltage/current across/through the amp. input resistor. I'll post the full picture for B that is given as a text example.
 

Attachments

  • #7
rude man
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Now that you've included the entire circuit - yes, it matters. It's a voltage divider as you said.

You're on the right track.
 
  • #8
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For C:

Then if I know the vin voltage, the amp. output voltage would just be μvin. But then by Kirchoff's loop law, since the load is in parallel with the amplifier output voltage, this would imply the load voltage, amplifier output resistor voltage and the amplifier voltage are equal.

vL = μvin = μriVs/(Rs + ri)

I just realized I forgot to put in Av for part C so:

And so gain is Av = vL/vs = ri/(Rs + ri)

However my text has the solution as Av = μrirsRL/[(ri + Rs)(ro + RL)] so I'm off by a factor of RL/(ro + RL).

The textbook seems correct only if the load ,amp. output resistance and amp. were in series since I'd be able to use voltage divider again to multiply in the extra factor I'm missing mentioned above or unless I've misunderstood something for the second part of the solution.
 
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  • #9
vk6kro
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The input voltage is given by the input current times the input resistance. Just like Ohms Law.

The output voltage is given by the output current (hfe* iin) times the parallel combined resistance of the load and the output z of the transistor.



If you keep the input current as your only variable, it will cancel when you calculate the voltage gain, which is just the ratio of the two calculations above.

Voltage gain = output voltage / input voltage
 
  • #10
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Gotcha, so referring to C:

Av = vL/vs

where:

By Ohm's Law:
vs = IRtotal = I(Rs + ri)

So far then I have:

Av = vL/I(Rs + ri)

Then vL = μvin = μIinReq

Where Req = RLro/(ro + RL)

Then vL = μIin*RLro/(ro + RL)

And now Av = μIin*RLro/vin[(ro + RL)(Rs + ri)]

But vin/Iin = ri by Ohm's Law thus:

Av = μri*RLro/[(ro + RL)(Rs + ri)]

Thanks so much for your help. Let me know if I had followed any of the above steps incorrectly. I'll try to work through D with what I now know.
 
  • #11
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Bah ignore this...I thought I had D right for a moment...


For D:

For Av = vL/vs

vs = iin(ri + Rs)

μiin = μvin/Req

where Req = RL + Ro

μiin = μvin/(RL + Ro)

Since the right loop of D is a series circuit, multiplying the output current on both sides by RL will yield vL on the left.

vL = RLμiin = μvinRL/(RL + Ro)

Av =
 

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