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  #1  
Old 11-15-2011, 12:33 AM
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jegues jegues is offline
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Question Series-Shunt Feedback Amplifier

I think I managed to get part a), but I'm stuck on part b).

I wrote all the equations I could think of in the circuit but I'm still a little stuck as to actually solving the DC emitter currents. Is there an assumption I can make about any of the other currents?

For example,

Can I assume,



Is this acceptable? Or should I be able to solve for everything?

I feel like I need to solve 1 value or make an assumption in order to get things rolling.

Thanks again!
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  #2  
Old 11-15-2011, 02:09 AM
t_n_k t_n_k is offline
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The closed loop gain should be ~100.
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Old 11-15-2011, 02:33 AM
t_n_k t_n_k is offline
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For the currents you can do an incremental approach.

First assume IB1≈0.

Then calculate the current in R1=0.7/1k=0.7mA. Hence current in R2=0.7mA and hence it follows that IE2=1.7mA.

Thus IB2=1.7mA/101=16.8uA.

This makes IC1=100uA-16.8uA=83.2uA. Thus IB1 would be better approximated as 832nA.

You could redo the process now using IB1=832nA rather than IB1=0A. But I doubt things would change that much.
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Old 11-15-2011, 02:58 AM
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Quote:
Originally Posted by t_n_k View Post
The closed loop gain should be ~100.
How did you draw this conclusion?

I've been reading in the book how to estimate the closed loop gain when we have a large loop gain but obviously I found the wrong relationship.

My reasoning was as follows,



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Old 11-15-2011, 04:24 AM
t_n_k t_n_k is offline
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Begging your pardon - it should be ~10 rather than ~100. I slipped an order of magnitude in my original "back of the envelope" estimate.

Basically, I'm looking at what happens at the output emitter (E2) as the input varies.

With the input 0V then VB1≈0.7V and hence VE2≈7.7V.

If the input increases from 0V to +10mV then VB1=0.71 V and hence

VE2=7.1+0.71=7.81V.

So the VE2 has changed by (7.81-7.7) 0.11V for a 10mV input change, which equates to a gain of Av=0.11/0.01=11.

If the input is -10mV then VB1=0.69V and VE2=6.9+0.69=7.59V

So for a change from 0V to -10mV the E2 voltage changes by 7.7-7.59=0.11V - again indicating a small signal gain of ~11.
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Old 11-15-2011, 04:29 AM
t_n_k t_n_k is offline
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I can run a simulation to check the result if you don't have access to simulation software.
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Old 11-15-2011, 04:45 AM
t_n_k t_n_k is offline
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Using the idea of feedback you can come up with the gain as follows.

If the open loop gain is Avo then



Re-arranging this gives



or



And if Avo is very large then



and finally



So Af=11 with the given values looks consistent.

Perhaps you have confused the transistor current gain beta values with the feedback beta factor.

i.e.


Last edited by t_n_k; 11-15-2011 at 04:52 AM.
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  #8  
Old 11-15-2011, 01:27 PM
Jony130 Jony130 is online now
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I was bored today so I decided to find open loop voltage gain.

I use this diagram



And I assume Ic1 = 100uA; Ic2 = 1.6mA

re1 = 26mV/100uA = 260Ω

re2 = 26mV/1.6mA = 16.25Ω

So the voltage gain of the first BJT is equal



And I'm not sure whether this (R1||R2)/(Hfe +1) should be take into account.

Voltage gain for Q2 is equal




So open loop gain

Ao = A1 * A2 = 4.12KV/V

The input resistance



So after we close the feedback loop the voltage gain will drop to:

Af = Ao/ ( 1 + Ao*β ) = 10.97 V/V

and

Rin_f = Rin* = 101KΩ

Rout_f = (R1+R2) /
(1+ Ao*β) = 29.3Ω But I also not sure about that.

So what do you think about my simplified analysis?
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  #9  
Old 11-15-2011, 08:07 PM
t_n_k t_n_k is offline
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Thanks for your useful analysis Jony130.

On the matter of which components to include in the open loop AC gain analysis one could perhaps assume that the feedback is 'killed' by shunting Q1 base to ground with a large capacitance. So the DC conditions would be 'preserved' but the AC gain would presumably more closely approach the open loop case.

Would that mean you then dispense with the (R1||R2)/(Hfe +1) term and replace the R1+R2 combination at Q2 emitter with just R2? Presumably yes.

In any event the open loop gain is highly dependent on the Q2 hfe value.

The addition of the load resistance RL to the circuit will also have an effect.
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  #10  
Old 11-15-2011, 11:41 PM
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Quote:
Originally Posted by t_n_k View Post
Perhaps you have confused the transistor current gain beta values with the feedback beta factor.

I think this the source of my confusion.

So the are the transistor current gain beta values NOT the feedback beta factor.

Thanks again!

Last edited by jegues; 11-16-2011 at 12:07 AM.
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