# Two stage BJT amplifier with feedback

Discussion in 'General Electronics Chat' started by The Electrician, Aug 15, 2009.

1. ### The Electrician Thread Starter AAC Fanatic!

Oct 9, 2007
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In another thread, a two stage BJT amplifier was analyzed. I'd like to carry it a bit further and add a single feedback resistor.

I want to show how just the addition of a single feedback resistor, RFB, between the two emitters substantially increases the difficulty of analysis.

The original circuit (first image) can be analyzed for voltage gain a piece at a time, but the addition of the feedback resistor (second image) pretty much requires the analysis to deal with the entire circuit at once. This means setting up a system of equations including all the variables.

In 1952, Jacob Shekel published a paper in the Proceedings of the IRE showing a systematic way of doing this with matrix arithmetic. His method allows calculation of the various gains and impedances from a single admittance matrix.

The schematic shows values for the intrinsic emitter resistances (re1 and re2) that are slightly different from what an exact analysis would give, but I chose to stick with the values that were in the original thread.

Using small signal nodal analysis, there are 5 nodes, numbered in red. I've labeled node 3 at both the collector of Q1 and the base of Q2 as a reminder that they are really the same node. Even though values are shown for the capacitors, assume that they are AC short circuits; zero ohms, in other words.

The transistors are assumed to have infinite intrinsic collector resistance (ro), and zero reverse voltage transfer ratio.

This circuit is a step above the elementary problems often seen in this forum, and I hope the more advanced members will find it interesting.

Let RFB have a value of 2344.06792623 ohms. Calculate the input impedance at node 1 for both circuits. Assume the source Vs (which is an ideal voltage source with zero ohms output impedance) applies a signal to node 1, and calculate the voltage gain from that input to the other 4 nodes.

As a hint, I'll tell you that the overall voltage gain for the circuit without the feedback resistor is 447.21755, and with RFB connected, the voltage gain to node 3 is -109.567.

Try out your analysis skills, using your favorite analysis method, and later I'll show how to do it using Shekel's method. If anyone wants a copy of his paper, post a disguised valid email address, and I'll send you a copy.

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2. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Hi Electrician,

I'm not sure if I'm an "advanced member" but I'm giving it a try. I have the same value of overall gain without feedback that you obtained but haven't started on the feedback case.

Did you also calculate any change in DC bias conditions due to the new DC path through Rfb or did you assume only the small signal AC values change? I guess one could add a series capacitor with Rfb to remove that condition but I'm not sure if the result would be the same.

Thanks for posing such an interesting problem - I'll probably request a copy of Shekel's analysis later on.

3. ### The Electrician Thread Starter AAC Fanatic!

Oct 9, 2007
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Sorry! I should have put a capacitor in series with RFB; assume AC coupling only.

4. ### Wendy Moderator

Mar 24, 2008
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I've been looking at the AAC book, it could use a section on amps and their classes. It has a few experiments in that direction already. You ever think of writing for something like this?

5. ### Jony130 AAC Fanatic!

Feb 17, 2009
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But this RFB isn't the positive feedback resistor?

Of course, there are amplifiers with both positive and negative feedback.

Oct 9, 2007
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Yes, it is.

7. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Hmm, maybe this simplified diagram give us the correct answer.

Maybe tomorrow (ups today 12:40PM ) I find the time to try solve this circuit.
But pspice will solve this circuit in a blink of an eye.

Last edited: Aug 17, 2009
8. ### Jony130 AAC Fanatic!

Feb 17, 2009
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We can write down the equations
(Node1) V1•(1/R2 + 1/R3 + 1/R8) - V3•(1/R8) = IC1 + Vin/R2
(Node2) V2•(1/R4 +1/R7) - V3•(1/R7) = -IC1
(Node3) -V1•(1/R8) - V2•(1/R7) + V3•(1/R5 + 1/R7 + 1/R8) = IC2
And
IC1 = β1•Ib1 = β1•(Vin - V1)/R2
IC2 = β2•Ib2 = β2•(V2 - V3)/R7
Solve this by a help of a Derive6 I get Ku=V(node3)/Vin=-90.50[V/V]
Rin=R1||R2||( (hfe+1)•re )/(1-V1)≈5.49KΩ

PS. THX Quarz

9. ### The Electrician Thread Starter AAC Fanatic!

Oct 9, 2007
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You haven't said whether your answer is in terms of the resistors in my original circuit, or in terms of the resistors in your model. However, since my original circuit didn't have R7 or R8, apparently you are giving answers in terms of your model.

I don't see R1 in your model, but your expression for Rin involves R1.

My first post asked the reader to "Calculate the input impedance at node 1 for both circuits. Assume the source Vs (which is an ideal voltage source with zero ohms output impedance) applies a signal to node 1, and calculate the voltage gain from that input to the other 4 nodes."

You only provided a value for Rin and the voltage gain to node 3 of the second circuit. Furthermore, when you say "I get Ku=V(node3)/Vin=-90.50[V/V]", it's not clear whether by saying "node3", you are referring to node3 in my posted circuit, or node3 in your model.

It would be good if your answer were in terms of the resistors in the circuit I posted if your answers are in symbolic form. If you use the resistors in your model, give their value in terms of the resistors in my posted circuit. In other words, what are the expressions for your R2, R4 and R7 in terms of the resistors in my posted circuit? And, where is R1?

Or, you could provide numeric values, in which case give results accurate to at least 5 digits.

Last edited: Aug 17, 2009
10. ### The Electrician Thread Starter AAC Fanatic!

Oct 9, 2007
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I think I see why I am confused. Where you say "(Node1) V1(1/R2 + 1/R3 + 1/R8) - V3(1/R8) = IC1 + Vin/R2", it appears that you are using R2 from your model.

But where you say "Rin=R1||R2||( (hfe+1)re )/(1-V1)≈5.49KΩ", you are using R2 from my circuit. It would be much better to use designators in your model that aren't the same as the designators for different components in my circuit.

11. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Well yes, sorry for confused you.
R2=(β+1)*re1
R3=re11
R4=Rc1||R3||R5
R5=re12
R6=Rc2||RL
R7=(β+1)*re2
R8=RFB

And accident, your circuit don't have more then one "correct" solution.

Last edited: Aug 17, 2009
12. ### The Electrician Thread Starter AAC Fanatic!

Oct 9, 2007
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I don't understand this comment. Should we expect that this circuit has more than one "correct" solution?
You haven't done all the parts of the problem. What about "Assume the source Vs (which is an ideal voltage source with zero ohms output impedance) applies a signal to node 1, and calculate the voltage gain from that input to the other 4 nodes."?

Also, I picked the particular value for RFB because it leads to an interesting condition. In fact, it was because I noticed this condition that I posed this problem. Do your formulas give the correct answer for all values of RFB?

Since you have already derived your formulas, change the value of RFB to 5000Ω and re-calculate and post your answers for that value.

13. ### Jony130 AAC Fanatic!

Feb 17, 2009
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I suspect that for the RFb=2344.06792623Ω the ac-voltage on node 2 (in your original diagram) is equal 0V.
And thus we can quite easily analysis this amplifier:

H11e=(β+1)*re1=7.7959kΩ

So

Rin=Vs/Is=R1||R2||H11e=5.50642943KΩ

Ku1=Node_3/Node_1=β1*( Rc1||R3||R4||Hie2) / H11e=-109.56742[V/V]

Hie2=(β2+1)*(re2+(re22||RFB)=13308.8874Ω

Rc1||R3||R4||Hie2=2847.2555Ω

Ku2=node_4/node_1=Ku1*(re22||RFB)/(re2+re22+RFB)=-90.5045529[V/V]

Ku3=node_5/node_1=Ku1*β2*(Rc2||RL)/Hie2=1646.53012[V/V]

And that's enough analysis for today.

Last edited: Aug 18, 2009
14. ### ELECTRONERD Senior Member

May 26, 2009
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Gosh Jony, where do you get all of these formulas? I'm not saying there incorrect, but I'd like to know in case I want to do more transistor design (which I will of course).

15. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Hi Electrician,

I've done some work on your challenging problem in the feedback case.

I can't get exactly the same values as you but approaching them.

It seems your design has adjusted Rfb to give the signal voltage at node 2 as 0V. Hence the "strange" value you give for Rfb to so many significant digits. As a first go I have Rfb=2344.089 to achieve this. I guess the "beauty" of forcing node 2 voltage to zero reduces the calculation burden. I think I'd have to put the equations in a spreadsheet, otherwise I end up always making a mistake somewhere - I basically did it longhand with a calculator. The overall gain looks like being somewhere around 1645 times, but I need to recheck.

The problem is a real mind cruncher!

16. ### The Electrician Thread Starter AAC Fanatic!

Oct 9, 2007
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Jony130 got the exact gains and posted them in post #13.

Jony130 also noticed the effect of the particular value of RFB, and used a shortcut method to solve the circuit.

I'm going to suggest a change in RFB that prevents the use of a shortcut.

17. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Yes
Non problem, for this schematics simplified diagram.

R2=(β1+1)*re1
R3=re11
R4=Rc1||R3||R4
R5=re12
R6=Rc2||RL
R7=(β2+1)*re2
R8=RFB

Node1 V1•(1/R2 + 1/R3 + 1/R8 +β1/R2) + V2•0 - V3•(1/R8) = Vin•(β1 + 1)/R2
Node2 -V1•(β1/R2) + V2•(1/R4 +1/R7) - V3•(1/R7) = - Vin•(β1/R2)
Node3 -V1•(1/R8) - V2•(1/R7 + β2/R7) + V3•(1/R5 + 1/R7 + 1/R8 +β2/R7) = 0
And if we solve this with help of a Cramer's rule and this applet we get the voltage node.
http://www.math.ubc.ca/~israel/applet/mcalc/matcalc.html

For RFB=5K
If I don't make any mistake in calculation here are the result:
V1=0.578180V---> node 2 in your diagram
V2=-46.318011V--->it is a Ku= node_3/node_1 in your diagram
and
Ku=node_5/node_1|=((V2-V3)/R7) *β2 *R6=689.000225[V/V]
Rin=R2/(1-V1)=18481.5798Ω ---> not take into account R1,R2 in your schematics.
And gosh I need a spreadsheet.

Well, these are basic equation which describe BJT in a CE amplifier

Last edited: Aug 18, 2009
18. ### The Electrician Thread Starter AAC Fanatic!

Oct 9, 2007
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Yes, that's why I selected that particular value for RFB.

You have in effect analyzed a circuit that has an AC ground connected to node 2 and and with RFB connected from node 4 to ground, rather than between nodes 2 and 4. Such a circuit has exactly the same gains from node 1 to all the other nodes.

Now, to extend the problem, imagine this. Build each circuit, my original one and the one you analyzed and enclose each one in a black box. Each circuit's reference node (ground) is connected to the metal of the box. The other 5 nodes are connected to terminals on the outside of the box. Since both circuits have exactly the same gains from node 1 to the other nodes, how can we tell them apart?

Also, I asked in post #12 for the gains when RFB is 5000Ω. The purpose of this is to eliminate the short cut analysis that can be done with the original value of RFB.

There is another value of RFB that gives an exceptional result. What is that value, and what is the particular result it gives?

Thanks to those participating in this exercise. I hope we'll hear from hgmjr and steveb.

19. ### The Electrician Thread Starter AAC Fanatic!

Oct 9, 2007
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If you use Jacob Shekel's method with a symbolic algebra software package, you can derive a formula for everything you might want to know about a given circuit.

20. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Thanks for pointing out my earlier oversight at not reading all of Jony130's posts. Apologies to him and well done to him for his persistence.

Looks like the gain notionally becomes infinite at Rfb=1.673425k or thereabouts (??). Does this mean it becomes an oscillator? I have to confess finding this by running a simulation of the circuit - several times. I'll see if I can use the equations to find the (exact?) value directly without 'cheating'.