# BJT Multistage DC analysis

Discussion in 'General Electronics Chat' started by Ronscott1, Dec 21, 2009.

1. ### Ronscott1 Thread Starter New Member

Nov 5, 2009
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Hi. I came across this guitar fuzz-distortion circuit. I am not sure how to start analyzing it. So far I have drawn the following four KVL loops:

Loop 1 (Green) = Vcc - Vce1 = R2*I2 - I3*R3
Loop 2 (Brown) = Vcc - Vbe = R1*I1 - R3*I3
Loop 2 (Pink) = Vcc - 2*Vbe = R1*I1 - R4*I4
Loop 3 (Blue) = Vcc - Vce2 = R1*I1

My unknowns are as follows:
Vce1, Vce2, I1, I2, I3, I4

Are there any other loops that I am missing? or what are two other equations I could use to set up a matrices? Also I really hate be a bother but could someone explain exactly how this circuit it working? I have been trying to get it but I do not understand, I really just want to figure out these kind of circuits are biased.
Best,
Ron

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2. ### Jony130 AAC Fanatic!

Feb 17, 2009
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To solve this circuit you need only two equations:
Vbe1 + (Ic1/β1)*R4 - [Ie2-(Ic1/β2)]*R3 = 0
Vcc - (Ic1+Ic2/β2)*R1 - Vbe2 - [Ie2-(Ic1/β2)]*R3=0

And if we assume β=∞ then
Ic1=(Vcc-2Vbe)/R1
Ic2=Vbe/R3

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3. ### Ronscott1 Thread Starter New Member

Nov 5, 2009
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I do know the beta, the beta is equal to 300.

4. ### hobbyist AAC Fanatic!

Aug 10, 2008
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Basic over view, without any math,
This is my personal analysis on it, don't know for sure,
but here goes,

Q2 is biased on through it's base resistor R1,
a voltage is developed across R3, this is the emitter voltage of Q2, also Q1 has it's base voltage established by Q2's emitter voltage, through R4. The collector voltage of Q1, is cross coupled to the base of Q2, so there is a feedback,

if Q2 begins to conduct less, then it's emitter voltage will begin to drop, this same voltage is applied to the base of Q1, so now Q1 base voltage drops, that makes it's collector voltage increase, so now the base voltage of Q2 increases, thereby Q2, conducts more which brings everything back into balance.

Now if Q2 conducts more, then emitter voltage of Q2, becomes higher so base of Q1 is higher, now Q1 conducts more and drops the base voltage of Q2, so Q2 conducts less then it comes back into balance..

It is a feedback regulating system.

Last edited: Dec 21, 2009
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5. ### Jony130 AAC Fanatic!

Feb 17, 2009
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So solving this equations
Vbe1 + (Ic1/β1)*R4 - [Ie2-(Ic1/β1)]*R3 = 0
Vcc - (Ic1+Ic2/β2)*R1 - Vbe2 - [Ie2-(Ic1/β1)]*R3=0

$\alpha=\frac{\beta}{\beta+1}$

For β=300; Vcc=9V; Vbe=0.65;
we get:
Ic1=730.114uA
Ic2=1.01411mA

And if I don't made any mistake in the matrix then the voltage gain is equal:

re1=26mV/Ic1
re2=26mV/Ic2

So for re1=35.6108Ω; re2=25.6382Ω and β=300
Voltage gain is equal:
Au=609.75 [V/V]

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Last edited: Dec 22, 2009
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6. ### The Electrician AAC Fanatic!

Oct 9, 2007
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Jony130, I see that you have started using Mathematica. I'm sure you are enjoying it!

In an earlier post some time ago, you said that this is only a hobby for you, but you seem to have a good understanding of circuit analysis. What do you actually do for a living? Maybe you should consider a change in vocation!

I get the same result you do for the voltage gain with the feedback resistor. I notice that without it, the voltage gain is nearly the same, 605.922.

This is because the voltage gain calculated by this method assumes the voltage source has zero output impedance, so the feedback current from Rf is shunted to ground, and Rf has little effect other than as a slight additional load in parallel with Re2 plus a feedforward path from the input to the emitter of Q2.

Once you have the Y matrix and its inverse, the Z matrix, you can determine anything you want to know about the AC performance of the circuit.

For example, suppose you inject a current at the input and you want to know the current gain from input to Rc2. It is given by Z[[4,1]]/Rc2 and the value is 142.42.

Suppose you want to know the impedances at each node (assuming the input is open circuited). They are simply the values on the diagonal of the Z matrix, namely:

537.214 Ω
520.04 Ω
2.984 Ω
2300 Ω

If you want to know these impedances with the input driven from a voltage source, you need only delete the first row and column of the Y matrix; this is equivalent to a short to ground at the input node. Then invert the remaining 3x3 Y matrix. The result is:

0 Ω (the input is shorted now)
9684.267 Ω (node 2)
55.5682 Ω (node 3)
2300 Ω (node 4)

We can see how much the feedback resistor reduces the impedance at nodes 2 and 3, if the input is driven from a high impedance source.

To find the gain if the circuit is driven by a non-zero source impedance, we need to add a row and column to the top left of the Y matrix and add a source resistance, Rs, in the appropriate positions of this augmented Y matrix. Then solve in the usual way.

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7. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Thx The Electrician I am pleased with Mathematica software.
And its thanks to You I start use this software.
And yes, I'm self teaching hobbyist (I start in '97). I current work in diesel engine factory in their maintenance unit (mechanical stuff and PLC/CNC).

By the way, can you explain in more detail the BJT matrix.
Because I don't understand the therm for collector β/(β+1)*re
And what about if I want more detail BJT model which consider H22 and H12 (reverse voltage ratio).
And what about JFET and MOSFET's

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Oct 9, 2007
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Jony130,

9. ### The Electrician AAC Fanatic!

Oct 9, 2007
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The 2x2 Y matrix for a transistor that has been specified in h parameters is:

$Y = \left[ \begin{array}{2}\frac{1}{h11} & \frac{-h12}{h11}\\ \frac{h21}{h11} & \frac{h11*h22-h12*h21}{h11}\end{array}\right]$

You can see that when h12 is zero, this is equivalent to:

$Y = \left[ \begin{array}{2}\frac{1}{(\beta+1)re} & 0\\ \frac{\beta}{(\beta+1)re} & \frac{1}{ro}\end{array}\right]$

This particular transformation is symmetric; the h parameters for a two-port for which the y parameters are known is:

$H = \left[ \begin{array}{2}\frac{1}{y11} & \frac{-y12}{y11}\\ \frac{y21}{y11} & \frac{y11*y22-y12*y21}{y11}\end{array}\right]$

10. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Ok I got this simple circuit

And I get this result

Voltage gain is ok but Zin and Zout is NG.

What I am doing wrong?

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11. ### The Electrician AAC Fanatic!

Oct 9, 2007
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Look at the first row of your active matrix; you have:

1/(β+1)*re -1/(β+1)*re

when it should be:

1/((β+1)*re) -1/((β+1)*re)

You got it right in post #5.

To avoid that kind of mistake, when you're typing fractions in Mathematica, if you type "control-/" instead of "/", you will get a mode that allows you to see what the result will be in standard mathematical format.

12. ### The Electrician AAC Fanatic!

Oct 9, 2007
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That should be "voila", not "walla". It's a French word that means something like "there you have it", or "there it is".

13. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Well, unfortunately there is still something wrong with this matrices.

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14. ### The Electrician AAC Fanatic!

Oct 9, 2007
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I forgot to deal with the output impedance problem.

Admittance (Y) matrices and impedance (Z) matrices have the following property:

The value of each element on the main diagonal of an admittance matrix is the sum of all the admittances connected to that node, with all other nodes short-circuited to ground.

The value of each element on the main diagonal of an impedance matrix is the impedance from that node to ground, with all other nodes open-circuited.

Given an admittance matrix, if you delete the Nth row and column, this will give you the admittance matrix of the network with the Nth node shorted to ground. (and, of course, this may change the numbering of the nodes; don't forget this important point)

So, for the circuit under consideration, if you invert the original 2x2 Y matrix, the 2,2 element is the output impedance of the circuit with the input open-circuited. If you delete the first row and column of the original Y matrix, and then invert it, the main diagonal of the derived Z matrix will contain the impedances of the nodes with the first node shorted to ground.

In this case, you will be left with only one element if you delete the first row and column of the original Y matrix. Then inverting that reduced matrix just gives the reciprocal of the 2,2 element of the original matrix, namely:

1/(1/R1+1/re+1/Re) which is just R1||re||Re

This is the output impedance with the input driven by a voltage source.

If the input is left open, or is driven by a current source, then the output impedance is Re.

If you want the output impedance with the input driven by a source having an impedance of Rs, then just add a term 1/Rs to the first element of the original Y matrix and invert it; the 2,2 element of the derived Z matrix will be the output impedance when the input is driven from a source impedance of Rs.