Electrical Engineering - Circuits - Transistors

[The Electrician]

Why shouldn't I include $R_{S}$ in my analysis? Don't I have to take into consideration the current flowing through it? I'm confused about just getting rid of them. I learned something called the node voltage method a while ago.

There won't be any current flowing in Rs because there won't be a voltage applied at Vs when you're calculating Vo/VI; the left end of Rs won't be connected to anything.

When the question asks for Vo/VI, it's saying in effect, "If some test voltage is applied to the base of the first transistor, what is the output voltage of the circuit?" If we've already calculated the ratio Av = Vo/VI, and now we have some new applied voltage Vin, we can find Vo due to this new applied voltage by just multiplying Vin by Av.

If the question asks for Vo/Vs, it's saying "If some test voltage Vs is applied to the left end of Rs, what is the output voltage?". But, when it asks for Vo/VI, it is assumed that no test voltage Vs is applied; only a test voltage VI is applied then.

The calculation of a voltage gain assumes that some node of the circuit is the place where the test signal is applied. That node is the denominator of the fraction defining Av; it would be Vs if we want Vo/Vs, or it would be VI if we want Vo/VI.

So, if what is wanted is Vo/VI, we want to know what the output voltage is when a test signal is applied to the base of the first transistor. Unless the left end of Rs is connected to something, it isn't involved at all. If you're applying a (mathematical) test signal at the base of the transistor, you don't want to also have an extraneous signal applied at Vs.

A choice must be made; is the gain wanted starting from the left end of Rs, or from the base of the transistor? You can't calculate the gain when two test signals are applied simultaneously, one at each point.

If the gain is wanted from the base of the transistor, then Vs and Rs shouldn't even be shown on the schematic; in that case, they're just confusion factors.

I also think it's a bit silly solving this question as it seems to be a question in algebra. I do have access to MATLAB at my school. My calculator is a TI-84 so it can perform those tasks, and during the test I didn't have access to a computer.

There's definitely a considerable amount of algebra involved, but I think you're having problems setting up the equations, without even considering the algebra to follow.

 

[rude man]

Alright well I'll attempt to solve this problem in this matter:

x Define an equation for $V^{~}_{I}$
x Write a KCL at node $V^{~}_{C}$, and solve for $V^{~}_{C_{1}}$
x Write a KCL at node $V^{~}_{O}$ and solve for $V^{~}_{O}$
x Combine these two equations by plugging in the equation for $V_{C_{1}}$ into the equation for $V_{O}$
x define the ratio $\frac{V^{~}_{O}}{V^{~}_{I}}$

I start off by defining $V^{~}_{I}$
$V^{~}_{I} = \frac{V^{~}_{S}R_{S}}{R_{Th}||βr_{e} + R_{S}}$

I know define $I^{~}_{B_{1}}$
$I^{~}_{B_{1}} = \frac{V^{~}_{I}}{βr_{e}} = \frac{V^{~}_{S}R_{S}}{βr_{e}(R_{Th}||βr_{e} + R_{S})}$

I know write KCL at $V^{~}_{C_{1}}$
$\frac{V^{~}_{S}R_{S}}{r_{e}(R_{Th}||βr_{e} + R_{S})} + \frac{V^{~}_{C_{1}}}{R_{C}||R_{Th}} + \frac{V^{~}_{C_{1}} - V^{~}_{O}}{βr_{e}} = 0$

I know factor out $V^{~}_{C_{1}}$ and move the term with $V^{~}_{O}$ to the other side of the equation.
$\frac{V^{~}_{S}R_{S}}{r_{e}(R_{Th}||βr_{e} + R_{S})} + V^{~}_{C_{1}}(\frac{1}{R_{C}||R_{Th}} + \frac{1}{βr_{e}}) = \frac{V^{~}_{O}}{βr_{e}}$

I know move the first term of the LHS to the RHS
$V^{~}_{C_{1}}(\frac{1}{R_{C}||R_{Th}} + \frac{1}{βr_{e}}) = \frac{V^{~}_{O}}{βr_{e}} - \frac{V^{~}_{S}R_{S}}{r_{e}(R_{Th}||βr_{e} + R_{S})}$

I now solve for $V^{~}_{C_{1}}$
$V^{~}_{C_{1}} = (\frac{V^{~}_{O}}{β} - \frac{V^{~}_{S}R_{S}}{R_{Th}||βr_{e} + R_{S}})\frac{1}{r_{e}(\frac{1}{R_{C}||R_{Th}} + \frac{1}{βr_{e}})}$

I now write a KCL at $V_{O}$
$\frac{V_{C_{1}} - v_{O}}{βr_{e}} + \frac{V_{C_{1}} - V_{O}}{r_{e}} = \frac{V_{O}}{R_{E}||R_{L}}$

I now move the RHS to the LHS
$\frac{V_{C_{1}} - V_{O}}{βr_{e}} + \frac{V_{C_{1}} - V_{O}}{r_{e}} - \frac{V_{O}}{R_{E}||R_{L}} = 0$

I now factor out $V_{C_{1}}$ and $V_{O}$
$\frac{V_{C_{1}}}{r_{e}}(\frac{1}{β} + 1) - V_{O}(\frac{1}{βr_{e}} + \frac{1}{r_{e}} + \frac{1}{R_{E}||R_{L}}) = 0$

I now move the second term to the RHS
$\frac{V_{C_{1}}}{r_{e}}(\frac{1}{β} + 1) = V_{O}(\frac{1}{βr_{e}} + \frac{1}{r_{e}} + \frac{1}{R_{E}||R_{L}})$

I now solve for $V_{O}$
$V_{O} = \frac{V_{C_{1}}(\frac{1}{β} + 1)}{r_{e}(\frac{1}{βr_{e}} + \frac{1}{r_{e}} + \frac{1}{R_{E}||R_{L}})}$

I now plug in my equation for $V_{C_{1}}$ into my equation for $V_{O}$ to get rid of $V_{C}$ as a unknown value.
$V_{O} = \frac{(\frac{V^{~}_{O}}{β} - \frac{V^{~}_{S}R_{S}}{R_{Th}||βr_{e} + R_{S}})\frac{1}{r_{e}(\frac{1}{R_{C}||R_{Th}} + \frac{1}{βr_{e}})}(\frac{1}{β} + 1)}{r_{e}(\frac{1}{βr_{e}} + \frac{1}{r_{e}} + \frac{1}{R_{E}||R_{L}})}$

At this point I will now clean up the numerator a bit by moving a term down to the denominator.
$V_{O} = \frac{(\frac{V^{~}_{O}}{β} - \frac{V^{~}_{S}R_{S}}{R_{Th}||βr_{e} + R_{S}})(\frac{1}{β} + 1)}{r_{e}(\frac{1}{βr_{e}} + \frac{1}{r_{e}} + \frac{1}{R_{E}||R_{L}})r_{e}(\frac{1}{R_{C}||R_{Th}} + \frac{1}{βr_{e}})}$

What is v₀ doing on both sides of the equation? You're after v_0//v_s so all terms in v₀ must be on the left-hand-side of the equation. You follow this mistake from here on, looks like.

Also: you're still using caps when you should be using lower-case.

Also: since you're after v_o/v_s, R_s is certainly a part of your answer. You are not after v₀/v_i.

 

[GreenPrint]

Removing $R_{S}$ and $v_{S}$ from the schematic does make this problem much easier to solve, however apparently I'm still doing something wrong.

x Perform DC analysis to solve for $r_{e}$
x write kCL at $v_{C_{1}}$ and solve for $v_{C_{1}}$
x write KCL at $v_{O}$ and solve for $v_{C_{1}}$
x combine the two equations and solve for the ratio $\frac{v_{O}}{v_{I}}$

DC Analysis
$R_{Th} = 24 KΩ||6.2 KΩ ≈ 4.927 KΩ$
$E_{Th} = \frac{R_{2}V_{CC}}{R_{1} + R_{2}} = \frac{(6.2 KΩ)15 V}{6.2 KΩ + 24 KΩ} ≈ 3.079 V$
$I_{B} = \frac{E_{Th} - V_{BE}}{R_{Th} + (β + 1)R_{E}} = \frac{3.079 V - 0.7 V}{4.927 KΩ + (150 +1)1.5 KΩ} ≈ 1.028x10^{-2} mA$
$I_{E} = (β + 1)I_{B} = (150 + 1)(1.028x10^{-2} mA) ≈ 1.552 mA$
$r_{e} = \frac{26 mV}{I_{E}} = \frac{26 mV}{1.552 mA} ≈ 16.753 Ω$

KCL at $v_{C_{1}}$
$\frac{v_{I}}{r_{e}} + \frac{v_{C_{1}}}{R_{C}||R_{Th}} + \frac{v_{C_{1}} - v_{O}}{βr_{e}} = 0$
factor out $v_{C_{1}}$ and move the first term on the LHS to the RHS
$v_{C_{1}}(\frac{1}{R_{C}||R_{Th}} + \frac{1}{βr_{e}}) = \frac{v_{O}}{βr_{e}} - \frac{v_{I}}{r_{e}}$
factor out $r_{e}$ on the RHS
$v_{C_{1}}(\frac{1}{R_{C}||R_{Th}} + \frac{1}{βr_{e}}) = \frac{1}{r_{e}}(\frac{v_{O}}{β} - v_{I})$
solve for $v_{C_{1}}$
$v_{C_{1}} = \frac{\frac{v_{O}}{β} - v_{I}}{r_{e}(\frac{1}{R_{C}||R_{Th}} + \frac{1}{βr_{e}})}$
multiply the denominator through by $r_{e}$
$v_{C_{1}} = \frac{\frac{v_{O}}{β} - v_{I}}{\frac{r_{e}}{R_{C}||R_{Th}} + \frac{1}{β}}$
multiply by $\frac{β}{β}$
$v_{C_{1}} = \frac{v_{O} - V_{I}β}{\frac{r_{e}β}{R_{C}||R_{Th}} + 1}$

KCL at $v_{O}$
$\frac{v_{C_{1}} - v_{O}}{βr_{e}} + \frac{v_{C_{1}} - v_{O}}{r_{e}} = \frac{v_{O}}{R_{E}||R_{L}}$
factor out all of the $v_{C_{1}}$ terms on the LHS and move all terms on the LHS to the RHS that contain $v_{O}$
$v_{C_{1}}(\frac{1}{βr_{e}} + \frac{1}{r_{e}}) = v_{O}(\frac{1}{R_{E}||R_{L}} + \frac{1}{βr_{e}} + \frac{1}{r_{e}})$
simplify a bit
$\frac{v_{C_{1}}(β + 1)}{βr_{e}} = v_{O}(\frac{1}{R_{E}||R{L}} + \frac{1 + β}{βr_{e}})$
solve for $v_{C_{1}}$
$v_{C_{1}} = v_{O}(\frac{1}{R_{E}||R{L}} + \frac{1 + β}{βr_{e}})\frac{βr_{e}}{β + 1}$
simplify
$v_{C_{1}} = v_{O}(\frac{βr_{e}}{(R_{E}||R_{L})(β + 1)} + 1)$

Combine the two KCL equations
$v_{O}(\frac{βr_{e}}{(R_{E}||R_{L})(β + 1)} + 1) = \frac{v_{O} - V_{I}β}{\frac{r_{e}β}{R_{C}||R_{Th}} + 1}$
split the RHS into two separate terms
$v_{O}(\frac{βr_{e}}{(R_{E}||R_{L})(β + 1)} + 1) = \frac{v_{O}}{\frac{r_{e}β}{R_{C}||R_{Th}} + 1} - \frac{V_{I}β}{\frac{r_{e}β}{R_{C}||R_{Th}} + 1}$
move the second term on the RHS to the LHS, move the term on the LHS to the RHS
$\frac{V_{I}β}{\frac{r_{e}β}{R_{C}||R_{Th}} + 1} = \frac{v_{O}}{\frac{r_{e}β}{R_{C}||R_{Th}} + 1} - v_{O}(\frac{βr_{e}}{(R_{E}||R_{L})(β + 1)} + 1)$
factor out $v_{O}$
$\frac{V_{I}β}{\frac{r_{e}β}{R_{C}||R_{Th}} + 1} = v_{O}(\frac{1}{\frac{r_{e}β}{R_{C}||R_{Th}} + 1} - \frac{βr_{e}}{(R_{E}||R_{L})(β + 1)} - 1)$
solve for the ratio $\frac{v_{O}}{v_{I}}$
$\frac{v_{O}}{v_{I}} = \frac{β}{\frac{r_{e}β}{R_{C}||R_{Th}} + 1}\frac{1}{\frac{1}{\frac{r_{e}β}{R_{C}||R_{Th}} + 1} - \frac{βr_{e}}{(R_{E}||R_{L})(β + 1)} - 1}$
simplify
$\frac{v_{O}}{v_{I}} = \frac{β}{\frac{r_{e}β}{r_{e}β + R_{C}||R_{Th}} - \frac{β^{2}r_{e}^{2}}{(R_{C}||R_{Th})(R_{E}||R_{L})(β + 1)} - \frac{r_{e}β}{R_{C}||R_{Th}} + \frac{1}{\frac{r_{e}β}{R_{C}||R_{Th}} + 1} - \frac{βr_{e}}{(R_{E}||R_{L})(β + 1)} - 1}$
simplify
$\frac{β}{-\frac{βr_{e}}{(R_{E}||R_{L})(β + 1)}(\frac{βr_{e}}{R_{C}||R_{Th}} +1) + r_{e}β(\frac{1}{βr_{e} + R_{C}|||R_{Th}} - \frac{1}{R_{C}||R_{Th}} - \frac{1}{(R_{E}||R_{L})(β+1)}) - 1}$
move the terms around in the denominator
$\frac{β}{r_{e}β(\frac{1}{βr_{e} + R_{C}|||R_{Th}} - \frac{1}{R_{C}||R_{Th}} - \frac{1}{(R_{E}||R_{L})(β+1)}) -\frac{βr_{e}}{(R_{E}||R_{L})(β + 1)}(\frac{βr_{e}}{R_{C}||R_{Th}} +1) - 1}$
factor $βr_{e}$
$\frac{β}{r_{e}β(\frac{1}{βr_{e} + R_{C}|||R_{Th}} - \frac{1}{R_{C}||R_{Th}} - \frac{1}{(R_{E}||R_{L})(β+1)} -(\frac{1}{(R_{E}||R_{L})(β + 1)})(\frac{βr_{e}}{R_{C}||R_{Th}} +1)) - 1}$
factor the denominator
$\frac{β}{r_{e}β(\frac{1}{βr_{e} + R_{C}|||R_{Th}} - \frac{1}{R_{C}||R_{Th}} - \frac{1}{(R_{E}||R_{L})(β+1)} (1 +\frac{βr_{e}}{R_{C}||R_{Th}} +1)) - 1}$
simplify
$\frac{β}{r_{e}β(\frac{1}{βr_{e} + R_{C}|||R_{Th}} - \frac{1}{R_{C}||R_{Th}} - \frac{1}{(R_{E}||R_{L})(β+1)} (2 +\frac{βr_{e}}{R_{C}||R_{Th}})) - 1}$
factor
$\frac{β}{r_{e}β(\frac{1}{βr_{e} + R_{C}|||R_{Th}} - (\frac{1}{R_{C}||R_{Th}} + \frac{1}{(R_{E}||R_{L})(β+1)} (2 +\frac{βr_{e}}{R_{C}||R_{Th}}))) - 1}$
plugging in some numbers
$\frac{150}{2512.95(\frac{1}{2512.95 + 2506.004} - (\frac{1}{25060.004} + \frac{1}{(937.5)(150 + 1)}(2 + \frac{2512.952}{25060.004}))) - 1}$
simplify
$\frac{150}{2512.95(1.992x10^{-4} - (3.990x10^{-5} + 7.064x10^{-6}(2.100))) - 1}$
simplify
$\frac{150}{2512.95(1.992x10^{-4} - (5.473x10^{-5})) - 1}$
simplify
$\frac{150}{2512.95(1.445x10^{-4}) - 1}$
simplify
$-\frac{150}{0.637}$
simplify
$-235.479$
Is this what you got?

 

[GreenPrint]

I thought I was after $\frac{v_{O}}{v_{I}}$ and not $\frac{v_{O}}{v_{S}}$. Sorry about using caps. I was taught to do a caps and a ~ above it in circuits 1, which is probably wrong but it doesn't show up when I use the equation format, I changed it to lowercase in the post above.

 

[The Electrician]

You are not after v₀/v_i.

He says he is after v₀/v_i. I pointed out to him that this seems a little strange given that Vs appears to be the input in the schematic shown in post #1, but he says in post #35 "My textbook has defined $A_{V} = \frac{V_{O}}{V_{I}}$.".

 

[The Electrician]

GreenPrint, you are going about this in a cumbersome ad hoc manner. You will get much easier and more error free results if you approach the problem in a systematic manner.

You have 4 unknown node voltages; vb1, vc1, vc2 and ve2. I'm assuming you are still going with the notion that the Av you want is vo/vi, not vo/vs. You don't really need vc2 to calculate Av and Ai, so eliminate vc2 as an unknown; that leaves you with 3 unknowns. (If you want vo/vs, you will have 4 unknown node voltages.)

The way to proceed is to write KCL equations for your 3 unknowns, vb1, vc1 and ve2. Don't try to solve the equations one at a time as you write them; wait until you have all 3 and then solve them as a system of simultaneous equations.

Rearrange each equation so that it is in this form (the unknowns are shown in red):

Eq1) vb1*y11 + vc1*y12 + ve2*y13 = Itest

Eq2) vb1*y21 + vc1*y22 + ve2*y23 = 0

Eq3) vb1*y31 + vc1*y32 + ve2*y33 = 0

Itest is a test current (a signal; an AC current) injected into the b1 node; we'll make it unity--1 amp. You couldn't actually inject 1 amp into the base of a real amplifier, but this is mathematics.

Your first KCL equation should be at the b1 node. The equation is:

vb1*(1/(β re) +1/Rth) + vc1*(0) + ve2*(0) = 1

You should be able to see that y11 = (1/(β re) + 1/Rth), y12=0 and y13=0.

The parameters y11, y12, y13, etc., are the coefficients of the unknown voltages vb1, vc1, ve2 in the first equation.

Your second KCL equation is:

$\frac{v_{I}}{r_{e}} + \frac{v_{C_{1}}}{R_{C}||R_{Th}} + \frac{v_{C_{1}} - v_{O}}{βr_{e}} = 0$

Rearranging terms, we get:

vb1*(1/re) + vc1*(1/(β re) + 1/Rth + 1/Rc) + ve2*(-1/(β re)) = 0

You should be able to see that y21 =1/re, y22=(1/(β re) + 1/Rth + 1/Rc) and y23=-1/(β re).

The parameters y21, y22, y23, etc., are the coefficients of the unknown voltages vb1, vc1, ve2 in the second equation.

Now see if you can get the third KCL equation into the form I've shown.

When you do, you will have a system of 3 simultaneous equations. You can then solve them for the 3 node voltages with Matlab or some similar method; maybe even with your calculator. Because you injected 1 amp (mathematically that is), you will have rather high voltages for the 3 nodes, but that's ok. If you divide ve2 by vb1, you will have Av.

I'll discuss calculating Ai after you get Av.

 

[rude man]

If you do it the way I suggested, by solving for i_b1(v_s) first, then there are only 2 independent nodes left: v_c1 and v_e2.

Since I think you already have i_b1(v_s) you should write your equations around the remaining two independent nodes.The 1st current source βi_b1 is the input to the rest of the circuit.

The problem pretty clearly states to get A_v = v_{o[/SUB/vs.The 1K source resistor Rs makes about a 4:1 difference in gain.}

 

[The Electrician]

The 1K source resistor R_s makes about a 4:1 difference in gain.

This doesn't seem quite right. The input resistance at b1 is approximately (β re)||Rth = (150*16.753)||4927 = 1664 ohms. Adding 1k in series with the input forms a voltage divider of 1664/(1664+1000) ≈ .62 which doesn't even cut the gain by a factor of 2.

 

[GreenPrint]

I think you may actually be correct. I have asked my professor and I am indeed looking for $\frac{v_{O}}{v_{S}}$

I have tried resolving the problem using the following method
x define $i_{B_{1}}(v_{S})$
x perform KCL at $v_{C_{1}}$ and solve the equation for $v_{C_{1}}$
x perform KCL at $v_{E_{2}}$ and solve the equation for $v_{C_{1}}$
x set the two KCL equal to each other and solve for the ratio $\frac{v_{O}}{v_{S}}$

I define the current $i_{B_{1}}$ as a function of $v_{S}$
$i_{B_{1}}(v_{S}) = \frac{v_{S}(R_{Th}||βr_{e})}{(R_{S} + R_{Th}||βr_{e})βr_{e}}$

solving the KCL equation at $v_{C_{1}}$ and solving for $v_{C_{1}}$ provides me with
$v_{C_{1}} = (\frac{v_{E_{2}}}{βr_{e}} - \frac{v_{S}(R_{Th}||βr_{e})}{(R_{S} + R_{Th}||βr_{e})βr_{e}})\frac{1}{\frac{1}{R_{C}||R_{Th}} + \frac{1}{βr_{e}}}$

solving the KCL equation at $v_{E_{2}}$ and solving for $v_{C_{1}}$ provides me with
$v_{C_{1}} = \frac{v_{E_{2}}(\frac{1}{βr_{e}} + \frac{1}{r_{e}} + \frac{1}{R_{E}||R_{L}})}{\frac{1}{βr_{e}} + \frac{1}{r_{e}}}$

setting these two equations together and solving for $\frac{v_{O}}{v_{E_{2}}}$ provides me with
$\frac{v_{O}}{v_{E_{2}}} = \frac{(R_{Th}||βr_{e})(R_{C}||R_{Th})β(\frac{βr_{e}}{R_{C}||R_{Th}} + 1)(\frac{1}{β} + 1)}{(\frac{1}{β} + 1 - (\frac{βr_{e}}{R_{C}||R_{Th}} + 1)(\frac{1}{β} + 1 + \frac{r_{e}}{R_{E}||R_{L}}))(R_{S} + R_{Th}||βr_{e})(r_{e}β + R_{C}||R_{Th})}$

plugging in the following values
$R_{Th} ≈ 4.927 KΩ$
$r_{e} ≈ 16.753 Ω$
$βr_{e} = 2512.95 Ω$
$R_{Th}||βr_{e} ≈ 1664.165 Ω$
$R_{C}||R_{Th} ≈ 2506.004 Ω$
$\frac{βr_{e}}{R_{C}||R_{Th}} ≈ 1.003$
$\frac{1}{β} + 1 ≈ 1.007$
$R_{E}||R_{L} = 937.5 Ω$
$\frac{r_{e}}{R_{E}||R_{L}} ≈ 1.787x10^{-2} Ω$

I get
$\frac{v_{O}}{V_{E_{2}}} ≈ -90.214$
saving all values to a variable on my calculator and getting a final answer provides me with
$\frac{v_{O}}{V_{E_{2}}} ≈ -90.267$
which is pretty close to what you got. $\frac{v_{O}}{v_{E_{2}}}$

Does this look better?

 

[rude man]

This doesn't seem quite right. The input resistance at b1 is approximately (β re)||Rth = (150*16.753)||4927 = 1664 ohms. Adding 1k in series with the input forms a voltage divider of 1664/(1664+1000) ≈ .62 which doesn't even cut the gain by a factor of 2.

Whatever. It makes a big difference. There is no justification in ignoring it and settling on v_o/v_i when what is required is v_o/v_s.

My main point has been to solve for i_b1(v_s) first since that computation can be made independent of the rest of the circuit. You just showed how easy that is. That simplifies the rest to just 2 independent nodes and an easy second computation. There is no need to solve the four-node problem simultaneously as you have indicated.

 

[rude man]

I think you may actually be correct. I have asked my professor and I am indeed looking for $\frac{v_{O}}{v_{S}}$

I have tried resolving the problem using the following method
x define $i_{B_{1}}(v_{S})$
x perform KCL at $v_{C_{1}}$ and solve the equation for $v_{C_{1}}$
x perform KCL at $v_{E_{2}}$ and solve the equation for $v_{C_{1}}$
x set the two KCL equal to each other and solve for the ratio $\frac{v_{O}}{v_{S}}$

I define the current $i_{B_{1}}$ as a function of $v_{S}$
$i_{B_{1}}(v_{S}) = \frac{v_{S}(R_{Th}||βr_{e})}{(R_{S} + R_{Th}||βr_{e})βr_{e}}$

solving the KCL equation at $v_{C_{1}}$ and solving for $v_{C_{1}}$ provides me with
$v_{C_{1}} = (\frac{v_{E_{2}}}{βr_{e}} - \frac{v_{S}(R_{Th}||βr_{e})}{(R_{S} + R_{Th}||βr_{e})βr_{e}})\frac{1}{\frac{1}{R_{C}||R_{Th}} + \frac{1}{βr_{e}}}$

solving the KCL equation at $v_{E_{2}}$ and solving for $v_{C_{1}}$ provides me with
$v_{C_{1}} = \frac{v_{E_{2}}(\frac{1}{βr_{e}} + \frac{1}{r_{e}} + \frac{1}{R_{E}||R_{L}})}{\frac{1}{βr_{e}} + \frac{1}{r_{e}}}$

setting these two equations together and solving for $\frac{v_{O}}{v_{E_{2}}}$ provides me with
$\frac{v_{O}}{v_{E_{2}}} = \frac{(R_{Th}||βr_{e})(R_{C}||R_{Th})β(\frac{βr_{e}}{R_{C}||R_{Th}} + 1)(\frac{1}{β} + 1)}{(\frac{1}{β} + 1 - (\frac{βr_{e}}{R_{C}||R_{Th}} + 1)(\frac{1}{β} + 1 + \frac{r_{e}}{R_{E}||R_{L}}))(R_{S} + R_{Th}||βr_{e})(r_{e}β + R_{C}||R_{Th})}$

plugging in the following values
$R_{Th} ≈ 4.927 KΩ$
$r_{e} ≈ 16.753 Ω$
$βr_{e} = 2512.95 Ω$
$R_{Th}||βr_{e} ≈ 1664.165 Ω$
$R_{C}||R_{Th} ≈ 2506.004 Ω$
$\frac{βr_{e}}{R_{C}||R_{Th}} ≈ 1.003$
$\frac{1}{β} + 1 ≈ 1.007$
$R_{E}||R_{L} = 937.5 Ω$
$\frac{r_{e}}{R_{E}||R_{L}} ≈ 1.787x10^{-2} Ω$

I get
$\frac{v_{O}}{V_{E_{2}}} ≈ -90.214$
saving all values to a variable on my calculator and getting a final answer provides me with
$\frac{v_{O}}{V_{E_{2}}} ≈ -90.267$
which is pretty close to what you got. $\frac{v_{O}}{v_{E_{2}}}$

Does this look better?

Certainly, your answer is in the right ball park and close to my earlier approximate computation of -100.

And, good for you to have solved for i_b1(v_s) first. You saw how easier that is than bulling your way thru a 4-node problem.

 

[The Electrician]

You are supposed to be solving for vo/vs, yet in numerous places you have the ratio vo/ve2; isn't ve2 the same as vo?

You seem to have done something different from what you show. For example, you have two KCL equations for vc1:

$v_{C_{1}} = (\frac{v_{E_{2}}}{βr_{e}} - \frac{v_{S}(R_{Th}||βr_{e})}{(R_{S} + R_{Th}||βr_{e})βr_{e}})\frac{1}{\frac{1}{R_{C}||R_{Th}} + \frac{1}{βr_{e}}}$

and

$v_{C_{1}} = \frac{v_{E_{2}}(\frac{1}{βr_{e}} + \frac{1}{r_{e}} + \frac{1}{R_{E}||R_{L}})}{\frac{1}{βr_{e}} + \frac{1}{r_{e}}}$

and then you say that you set these two together and solved for vo/ve2. Of course, you meant that you solved for vo/vs.

The problem I see is that if you do in fact use these two equations, you don't get what you say you got. Setting these two equations, as you have shown them, together and solving for vo/vs gets this result:

$\frac{v_{O}}{v_{E_{2}}} = \frac{(R_{Th}||βr_{e})(R_{C}||R_{Th})(\frac{βr_{e}}{R_{C}||R_{Th}} + 1)(\frac{1}{β} + 1)}{(\frac{1}{β} + 1 - (\frac{βr_{e}}{R_{C}||R_{Th}} + 1)(\frac{1}{β} + 1 + \frac{r_{e}}{R_{E}||R_{L}}))(R_{S} + R_{Th}||βr_{e})(r_{e}β + R_{C}||R_{Th})}$

You'll notice that there is a missing β in the numerator. One wonders, how did you get the result you posted? I can't have come from the two KCL equations you posted for vc1.

The answer is that the first KCL equation you show has an extra β. It should be:

$v_{C_{1}} = (\frac{v_{E_{2}}}{βr_{e}} - \frac{v_{S}(R_{Th}||βr_{e})}{(R_{S} + R_{Th}||βr_{e})r_{e}})\frac{1}{\frac{1}{R_{C}||R_{Th}} + \frac{1}{r_{e}}}$

Somehow in bulling through all of this, you got a good result, even though your first KCL equation, as you have shown it here, has an error.

If you used the first KCL equation as you show it in post #44, your result for vo/vs would be off by a factor of 150.

The advantage of setting up a nodal solution as I showed is that you let the computer (or calculator) do all the mistake-prone algebra. The elements of the 4 KCL equations are individually very simple. You don't solve them individually; you just set them up, which is almost trivially easy.

They can be solved in various ways, but formatting them as an array gives a compact representation. Solving them as a linear system of 4 simultaneous equations gives all 4 node voltages at once. There is no "bulling through" required. The computer or calculator does all the algebra for you, without any mistakes, like this:

The answer is an array of the 4 node voltages, (vs, vb1, vc1, vo). Since the input vs was set to 1, the output voltage at vo, -90.22, is the gain Av.

You can instantly see that the gain from vs to vb1 is .62, and from vs to vc1 is -91.83

This problem was of moderate complexity. As soon as you have a few more nodes, probably with feedback, there will be no possibility of getting a mistake free solution doing all the algebra by hand. This method is what simulators like pSpice use.

 

[rude man]

The advantage of setting up a nodal solution as I showed is that you let the computer (or calculator) do all the mistake-prone algebra. The elements of the 4 KCL equations are individually very simple. You don't solve them individually; you just set them up, which is almost trivially easy.

They can be solved in various ways, but formatting them as an array gives a compact representation. Solving them as a linear system of 4 simultaneous equations gives all 4 node voltages at once. There is no "bulling through" required. The computer or calculator does all the algebra for you, without any mistakes, like this:

The answer is an array of the 4 node voltages, (vs, vb1, vc1, vo). Since the input vs was set to 1, the output voltage at vo, -90.22, is the gain Av.

You can instantly see that the gain from vs to vb1 is .62, and from vs to vc1 is -91.83

This problem was of moderate complexity. As soon as you have a few more nodes, probably with feedback, there will be no possibility of getting a mistake free solution doing all the algebra by hand. This method is what simulators like pSpice use.

The point of this exercise for the OP is not to plug numbers into a math program, nor even to get the right answer, but to understand circuitry, equivalent circuits, ac vs. dc analysis, etc. etc. You learn nothing from plugging raw numbers into math software. Even less by running pspice. I've used pspice for over 30 years but first I made sure I understood circuits. That's what the circuits courses are for.

 

[The Electrician]

To say more about all this. Your method of solution reduced the 4 unknowns to 2 by taking advantage of the fact that you didn't want vc1 anyway, and that vs is a known since it is the node where we want to inject our input.

Setting your 2 KCL equations equal to each other is how you solved 2 simultaneous equations., with a considerable amount of algebra.

When you use a computer to solve systems of simultaneous equations, it's no harder to solve 20 simultaneous equations than it is to solve 2.

As an example, imagine that we want to improve the linearity of your circuit. We add a 50k ohm feedback resistor from vo to vb1; let there be a large capacitor in series with it to avoid upsetting the DC bias. You still have 4 nodes, but you will find that the problem becomes very much more complicated using the method you did.

But, if you have already got the system of 4 KCL equations set up, it's a trivial matter to add the feedback resistor, and let the computer do the additional work.

Here's the system with the feedback resistor added (in red); you can see that the additions to the equations are trivial. Solving the system with a linear solver by computer easily gives the result shown. Av is reduced by about a factor of 3. The only extra work to do is to add 4 elements to the system:

I know you're a busy student, but if you give a try to solving the circuit with feedback by hand, you'll appreciate just how much the addition of one more resistor complicates things.

 

[GreenPrint]

Ya I think I made a mistake in my post, I met to say $\frac{v_{O}}{v_{S}}$.

I don't see how there is an error in my KCL equation for node $v_{C_{1}}$. I'll assume that all of the currents are flowing out of node, giving me
$\frac{v_{s}(R_{Th}||βr_{e})}{R_{S} + R_{Th}||βr_{e}} + \frac{v_{C_{1}}}{R_{C}||R_{Th}} + \frac{v_{C_{1}} - v_{E_{2}}}{βr_{e}} = 0$
I factor the terms with $v_{C_{1}}$
$\frac{v_{s}(R_{Th}||βr_{e})}{R_{S} + R_{Th}||βr_{e}} + v_{C_{1}}(\frac{1}{R_{C}||R_{Th}} + \frac{1}{βr_{e}}) - \frac{v_{E_{2}}}{βr_{e}} = 0$
move the terms without $v_{C_{1}}$ to the RHS
$v_{C_{1}}(\frac{1}{R_{C}||R_{Th}} + \frac{1}{βr_{e}}) = \frac{v_{E_{2}}}{βr_{e}} - \frac{v_{s}(R_{Th}||βr_{e})}{R_{S} + R_{Th}||βr_{e}}$
solve for $v_{C_{1}}$
$v_{C_{1}} = (\frac{v_{E_{2}}}{βr_{e}} - \frac{v_{s}(R_{Th}||βr_{e})}{R_{S} + R_{Th}||βr_{e}})(\frac{1}{(\frac{1}{R_{C}||R_{Th}} + \frac{1}{βr_{e}})})$

I don't really see what I'm doing wrong. the current through the second resistor of resistance $βr_{e}$, is the voltage across the resistor divided by the resistance.
$i_{B_{2}} = \frac{v_{C_{1}} - v_{E_{2}}}{βr_{e}}$

Are you saying that the current through this resistor is $βi_{B_{2}}$? and not just $i_{B_{2}}$. I'm not exactly sure why this would be the case.

I agree that the goal of the exercise is to understand the circuitry, but as a student I should try and get the right answer. This problem has made me realize that the way I was solving these problems was just down right wrong and have opened my eyes to a better way to solve these problems.

Believe it or not, this question appeared on of my examinations which I got wrong. I know that during a test I would rather much not solve this problem using algebra and that solving it using matrix would be much easier and allow for less errors. I could plug the values into my TI-84 and get the answers. I think perhaps it's a good idea to solve circuits like this by hand, but when it gets more complex it's just down right a waste of time, and when during an examination it doesn't make much sense to plug through the algebra.

I don't know just my thoughts.

 

[GreenPrint]

To say more about all this. Your method of solution reduced the 4 unknowns to 2 by taking advantage of the fact that you didn't want vc1 anyway, and that vs is a known since it is the node where we want to inject our input.

Setting your 2 KCL equations equal to each other is how you solved 2 simultaneous equations., with a considerable amount of algebra.

When you use a computer to solve systems of simultaneous equations, it's no harder to solve 20 simultaneous equations than it is to solve 2.

As an example, imagine that we want to improve the linearity of your circuit. We add a 50k ohm feedback resistor from vo to vb1; let there be a large capacitor in series with it to avoid upsetting the DC bias. You still have 4 nodes, but you will find that the problem becomes very much more complicated using the method you did.

But, if you have already got the system of 4 KCL equations set up, it's a trivial matter to add the feedback resistor, and let the computer do the additional work.

Here's the system with the feedback resistor added (in red); you can see that the additions to the equations are trivial. Solving the system with a linear solver by computer easily gives the result shown. Av is reduced by about a factor of 3. The only extra work to do is to add 4 elements to the system:

I know you're a busy student, but if you give a try to solving the circuit with feedback by hand, you'll appreciate just how much the addition of one more resistor complicates things.

Also where did you get $R_{f}$ from? In my book it uses something like $r_{O}$ every now and then as a output resistance of the transistor that is very large that it sometimes includes in problems and sometimes doesn't. Is this the same resistor?

 

[NascentOxygen]

Also where did you get $R_{f}$ from?

The Electrician decided he would add a little spice to this rather bland problem, so he added an extra resistor to give feedback to (according to theory) improve the performance of your basic amplifier.

He changed the circuit just for fun. R_f is his feedback resistor ("We add a 50k ohm feedback resistor from vo to vb1").

 

[The Electrician]

Ya I think I made a mistake in my post, I met to say $\frac{v_{O}}{v_{S}}$.

I don't see how there is an error in my KCL equation for node $v_{C_{1}}$. I'll assume that all of the currents are flowing out of node, giving me
$\frac{v_{s}(R_{Th}||βr_{e})}{R_{S} + R_{Th}||βr_{e}} + \frac{v_{C_{1}}}{R_{C}||R_{Th}} + \frac{v_{C_{1}} - v_{E_{2}}}{βr_{e}} = 0$
I factor the terms with $v_{C_{1}}$
$\frac{v_{s}(R_{Th}||βr_{e})}{R_{S} + R_{Th}||βr_{e}} + v_{C_{1}}(\frac{1}{R_{C}||R_{Th}} + \frac{1}{βr_{e}}) - \frac{v_{E_{2}}}{βr_{e}} = 0$
move the terms without $v_{C_{1}}$ to the RHS
$v_{C_{1}}(\frac{1}{R_{C}||R_{Th}} + \frac{1}{βr_{e}}) = \frac{v_{E_{2}}}{βr_{e}} - \frac{v_{s}(R_{Th}||βr_{e})}{R_{S} + R_{Th}||βr_{e}}$
solve for $v_{C_{1}}$
$v_{C_{1}} = (\frac{v_{E_{2}}}{βr_{e}} - \frac{v_{s}(R_{Th}||βr_{e})}{R_{S} + R_{Th}||βr_{e}})(\frac{1}{(\frac{1}{R_{C}||R_{Th}} + \frac{1}{βr_{e}})})$

I don't really see what I'm doing wrong. the current through the second resistor of resistance $βr_{e}$, is the voltage across the resistor divided by the resistance.
$i_{B_{2}} = \frac{v_{C_{1}} - v_{E_{2}}}{βr_{e}}$

Here's what you got in post #44:

$v_{C_{1}} = (\frac{v_{E_{2}}}{βr_{e}} - \frac{v_{S}(R_{Th}||βr_{e})}{(R_{S} + R_{Th}||βr_{e})βr_{e}})\frac{1}{\frac{1}{R_{C}||R_{Th}} + \frac{1}{βr_{e}}}$

And here's what you get in this post:

$v_{C_{1}} = (\frac{v_{E_{2}}}{βr_{e}} - \frac{v_{s}(R_{Th}||βr_{e})}{R_{S} + R_{Th}||βr_{e}})(\frac{1}{(\frac{1}{R_{C}||R_{Th}} + \frac{1}{βr_{e}})})$

Why the difference?

One problem I see is that the first subexpression at the beginning of your derivation:

$\frac{v_{s}(R_{Th}||βr_{e})}{R_{S} + R_{Th}||βr_{e}}$

Doesn't have units of current. You might want to fix that and solve for vc1 again.

Believe it or not, this question appeared on of my examinations which I got wrong. I know that during a test I would rather much not solve this problem using algebra and that solving it using matrix would be much easier and allow for less errors. I could plug the values into my TI-84 and get the answers. I think perhaps it's a good idea to solve circuits like this by hand, but when it gets more complex it's just down right a waste of time, and when during an examination it doesn't make much sense to plug through the algebra.

I don't know just my thoughts.

You would be surprised how fast the solutions for network problems like this grow with the addition of just one or two more nodes, or additional parameters. The current problem is just about the limit of what you can reasonably do without the algebra becoming so cumbersome that it can't be done without mistakes.

Single transistors are often described at low frequencies with 4 parameters, hie, hfe, hre and hoe. In student problems it's almost universal that hre is taken to be zero (ignored, in other words), and typically, hoe (also known as ro) is taken to be zero. If, in your problem, hre had not been zero, your algebra would have been totally out of hand.

 

[GreenPrint]

Start off by defining $v_{B_{1}}$
$v_{B_{1}} = \frac{v_{S}(R_{Th}||βr_{e})}{R_{S} + R_{Th}||βr_{e}}$
define the current $i_{B_{1}}$
$i_{B_{1}} = \frac{v_{S}(R_{Th}||βr_{e})}{(R_{S} + R_{Th}||βr_{e})βr_{e}}$
write KCL at node $v_{C_{1}}$
$\frac{v_{S}(R_{Th}||βr_{e})}{(R_{S} + R_{Th}||βr_{e})r_{e}} + \frac{v_{C_{1}}}{R_{C}||R_{Th}} + \frac{v_{C_{1}} - v_{E_{2}}}{βr_{e}} = 0$
on the LHS move terms without $v_{C_{1}}$ to the RHS and factor terms on the LHS that contain $v_{C_{1}}$
$v_{C_{1}}(\frac{1}{R_{C}||R_{Th} + \frac{1}{βr_{e}}}) = \frac{v_{E_{2}}}{βr_{e}} - \frac{v_{S}(R_{Th}||βr_{e})}{(R_{S} + R_{Th}||βr_{e})r_{e}}$
solve for $v_{C_{1}}$
$v_{C_{1}} = (\frac{1}{\frac{1}{R_{C}||R_{Th}} + \frac{1}{βr_{e}}})(\frac{v_{E_{2}}}{βr_{e}} - \frac{v_{S}(R_{Th}||βr_{e})}{(R_{S} + R_{Th}||βr_{e})r_{e}})$
factor out $r_{e}$
$v_{C_{1}} = \frac{1}{r_{e}}(\frac{1}{\frac{1}{R_{C}||R_{Th}} + \frac{1}{βr_{e}}})(\frac{v_{E_{2}}}{β} - \frac{v_{S}(R_{Th}||βr_{e})}{(R_{S} + R_{Th}||βr_{e})})$
simplify
$v_{C_{1}} = (\frac{1}{\frac{r_{e}}{R_{C}||R_{Th}} + \frac{1}{β}})(\frac{v_{E_{2}}}{β} - \frac{v_{S}(R_{Th}||βr_{e})}{(R_{S} + R_{Th}||βr_{e})})$

I don't see how the β cancels out but believe that the answer above is correct and don't see how it's not. Thanks I didn't realize that I forgot about the $r_{e}$ previously in the current generator.

 

[The Electrician]

Start off by defining $v_{B_{1}}$
$v_{B_{1}} = \frac{v_{S}(R_{Th}||βr_{e})}{R_{S} + R_{Th}||βr_{e}}$
define the current $i_{B_{1}}$
$i_{B_{1}} = \frac{v_{S}(R_{Th}||βr_{e})}{(R_{S} + R_{Th}||βr_{e})βr_{e}}$
write KCL at node $v_{C_{1}}$
$\frac{v_{S}(R_{Th}||βr_{e})}{(R_{S} + R_{Th}||βr_{e})r_{e}} + \frac{v_{C_{1}}}{R_{C}||R_{Th}} + \frac{v_{C_{1}} - v_{E_{2}}}{βr_{e}} = 0$
on the LHS move terms without $v_{C_{1}}$ to the RHS and factor terms on the LHS that contain $v_{C_{1}}$
$v_{C_{1}}(\frac{1}{R_{C}||R_{Th} + \frac{1}{βr_{e}}}) = \frac{v_{E_{2}}}{βr_{e}} - \frac{v_{S}(R_{Th}||βr_{e})}{(R_{S} + R_{Th}||βr_{e})r_{e}}$
solve for $v_{C_{1}}$
$v_{C_{1}} = (\frac{1}{\frac{1}{R_{C}||R_{Th}} + \frac{1}{βr_{e}}})(\frac{v_{E_{2}}}{βr_{e}} - \frac{v_{S}(R_{Th}||βr_{e})}{(R_{S} + R_{Th}||βr_{e})r_{e}})$
factor out $r_{e}$
$v_{C_{1}} = \frac{1}{r_{e}}(\frac{1}{\frac{1}{R_{C}||R_{Th}} + \frac{1}{βr_{e}}})(\frac{v_{E_{2}}}{β} - \frac{v_{S}(R_{Th}||βr_{e})}{(R_{S} + R_{Th}||βr_{e})})$
simplify
$v_{C_{1}} = (\frac{1}{\frac{r_{e}}{R_{C}||R_{Th}} + \frac{1}{β}})(\frac{v_{E_{2}}}{β} - \frac{v_{S}(R_{Th}||βr_{e})}{(R_{S} + R_{Th}||βr_{e})})$

I don't see how the β cancels out but believe that the answer above is correct and don't see how it's not. Thanks I didn't realize that I forgot about the $r_{e}$ previously in the current generator.

In post #47, I pointed out that you had a mistake in your first KCL equation for vc1 in post #44.. What you had was this:

The β in red shouldn't be there.

The derivation you show in post #54 obtains this result:

$v_{C_{1}} = (\frac{1}{\frac{r_{e}}{R_{C}||R_{Th}} + \frac{1}{β}})(\frac{v_{E_{2}}}{β} - \frac{v_{S}(R_{Th}||βr_{e})}{(R_{S} + R_{Th}||βr_{e})})$

This result is correct; it's not the same as the one I complained about. See how easy it is to make mistakes when doing so much algebra?

My point in post #47 was, if you had in fact used the incorrect expression for vc1, you wouldn't have obtained the correct final result for vo/vs. Yet, somehow you did. Obviously you used a correct expression for vc1, not the incorrect one you showed.

Now all you have to do is get the correct current gain, Ai.

 

[GreenPrint]

Oh ok thanks.

I can just use
$A_{i_{T}} = -A_{v_{T}}\frac{Z_{i_{1}}}{R_{L}}$

The input impedance of the first stage is just $R_{Th}||βr_{e}$

So

$A_{i} = -A_{v}\frac{Z_{i}}{R_{L}} = 90.214 \frac{4.927 KΩ||2512.95 Ω}{2.5 KΩ} ≈ 60.052$

Does this look correct?

 

[The Electrician]

Oh ok thanks.

I can just use
$A_{i_{T}} = -A_{v_{T}}\frac{Z_{i_{1}}}{R_{L}}$

The input impedance of the first stage is just $R_{Th}||βr_{e}$

So

$A_{i} = -A_{v}\frac{Z_{i}}{R_{L}} = 90.214 \frac{4.927 KΩ||2512.95 Ω}{2.5 KΩ} ≈ 60.052$

Does this look correct?

I thought your instructor had said that Av was to be calculated from the Vs point. To be consistent, you should also calculate Ai from that point, so the input impedance should be $R_S+R_{Th}||βr_{e}$

 

[GreenPrint]

So then when I take this into consideration I get about 1.066 so the current again is essential 1. Does this sound correct?

 

[The Electrician]

So then when I take this into consideration I get about 1.066 so the current again is essential 1. Does this sound correct?

I get:

$A_{i} = -A_{v}\frac{Z_{i}}{R_{L}} = 90.214 \frac{1.0 kΩ+4.927 KΩ||2512.95 Ω}{2.5 KΩ} ≈ 96.138$

 

[GreenPrint]

$(\frac{1}{4927} + \frac{1}{2512.95})^{-1} ≈ 1664.165$
$1664.165 + 1000 ≈ 2664.165$
$\frac{2664.165}{2500} ≈ 1.066$?