Engineering Electrical Engineering - Circuits - Transistors

[GreenPrint]

Homework Statement

Given the configuration below

http://img844.imageshack.us/img844/3364/c5e8.png

Find A_{V} and A_{I}

Homework Equations

The Attempt at a Solution

I start off by drawing the small signal equivalent r_{e} model.

http://imageshack.us/a/img268/7201/9e.png

I start off by calculating R_{th}
R_{Th} = 24 KΩ||6.2 KΩ ≈ 4.927 KΩ

I then find the voltage at the base V_{B} of the first transistor
V_{B} = \frac{R_{2}V_{CC}}{R_{1} + R_{2}} = \frac{6.2 KΩ(15 V)}{6.2 KΩ + 24 KΩ} ≈ 3.079 V

I then find the current through the base I_{B} of the first transistor assuming that V_{BE} ≈ 0.7 V
I_{B} = \frac{V_{B} - 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 then find the current through the emitter I_{E} of the first transistor
I_{E} = (β + 1)I_{B} = (150 + 1)1.028x10^{-2} mA ≈ 1.552 mA

I then find the current through the collector I_{C} of the first transistor
β = \frac{I_{C}}{I_{B}}, I_{C} = βI_{B} = 150(1.028x10^(-2) mA) ≈ 1.542 mA

I then find r_{e}
r_{e} = \frac{26 mV}{I_{E}} = \frac{26 mV}{1.552 mA} ≈ 16.753 Ω

I then find βr_{e}
βr_{e} = (150)16.753 Ω = 2512.95 Ω

I then find the voltage across the resistor with resistance of βr_{e}
V_{βr_{e}} = I_{B}βr_{e} = 1.028x10^{-2} mA(2512.95 Ω) ≈ 2.583x10^{-2} V
This is also the input voltage into the amplifier V_{I}

I then find the current through the Thevenin resistance
V_{I} = R_{Th}I_{Th}, I_{Th} = \frac{V_{I}}{R_{Th}} = \frac{2.583x10^{-2} V}{4.927 KΩ} ≈ 5.242x10^{-6} A

I then find the input current into the amplifier I_{I} using Kirchhoff's Current Law
I_{I} = I_{Th} + I_{B} = 5.242x10^{-6} A + 1.028x10^{-2} mA ≈ 1.552x10^{-5} A

I then find the output voltage of the first stage V_{O_{1}}
V_{O_{1}} = -I_{C}R_{C} = -1.542 mA(5.1 KΩ) ≈ -7.864 V
This is also the input voltage into the second stage V_{I_{2}}

Alright. This is where I start getting slightly confused on how to proceed from here. I can't find I_{B}, r_{e}, I_{C}, I_{E} from what I have solved for so far. I believe I have to make some approximations which I'm not entirely insure are completely valid.

I see that the input impedance Z_{I} of the second transistor is
Z_{I} = R_{Th}||(βr_{e} + (β + 1)R_{E}||R_{L})
Here I make the assumption βr_{e} << (β + 1)R_{E}||R_{L}, which I'm not sure is entirely valid, but I don't see any other way to solve this problem. Using this assumption,
Z_{I} = R_{Th}||(β + 1)R_{E}||R_{L}
I'm not really sure that this quantity has any use in this problem.

From here I'm a bit confused on how to proceed. I know that the voltage at the base of the second transistor V_{B} is equal to the output voltage of the first transistor V_{O} which we found earlier. Using the path from the base to the resistors of the emitter, the total impedance is
Z = (β + 1)R_{E}||R_{L}
Here I made the same assumption in the paragraph right above this one which I'm not sure is valid.

From here I'm not really sure what to do. Thanks for any help that anyone can provide me in solving this problem. I thought about applying the voltage divider law to find the voltage at the emitter of the second transistor V_{E}, the only problem is that I don't know the value of βr_{e} and am not exactly sure I can solve for it.

 

[rude man]

It looks like you're mixing ac and dc analysis which you should not do.

This is an ac problem. Ignore things like bias points unless those dc voltages and/or currents directly affect ac gains. You assume the circuit is properly biased. If you're not sure. do a dc analysis first.

Why do you assign a dc current to each of your two controlled current sources? They are controlled by your input voltage. They can be either beta*i_b or g_m*V_be. In the latter case, g_m is a function of dc value V_be.

 

[GreenPrint]

This is interesting. I thought you were supposed to solve this type of problem in this matter. My book solves these types of problems in this way. I'm not exactly sure how to solve this problem using AC. I solved for beta*I_{B}, I know beta and I know I_{B}, at least for the first transistor anyways.

I thought to solve these type of problems your supposed to draw the short circuit equivalent, replace all the capacitors with shorts, dv voltage sources you just ignore? I don't know there's much I don't understand about this. I'm not exactly sure why the values of the capacitors are not given, in these types of problems we just ignore them.

I have solved previous problems in this matter, drawing a small signal equivalent circuit and finding the dc values.

 

[GreenPrint]

Could I just use the formula to find I_{B} for a common emitter configuration, and instead use R_{E}||R_{L}?

I_{B} = \frac{V_{B} - V_{BE}}{R_{Th} + (β + 1)(R_{E}||R_{L})}

If I could do this I could see a clear path on how to solve this problem. Is this a valid formula for the second stage?

I can't find this configuration in my book or the internet any where.

 

[rude man]

The capacitor values aren't given because you're supposed to assume that the frequencies are high enough so that the capacitors look like ac shorts.

This is a small-signal, ac problem. What do you mean, you've never solved an ac problem? You do see that the input has to be ac, otherwise the capacitors won't let any signal thru, right?

You don't ignore Vdc power supply voltages (in your case, +15Vdc), you let them = 0. All capacitors are zero ohms (to you, for the time being).

Your book does not assign constant currents to controlled current sources, that I can guarantee! You need to fully appreciate the difference between ac and dc analysis.

 

[rude man]

This is interesting. I thought you were supposed to solve this type of problem in this matter. My book solves these types of problems in this way. I'm not exactly sure how to solve this problem using AC. I solved for beta*I_{B}, I know beta and I know I_{B}, at least for the first transistor anyways.

You know the dc I_B but you don't know the ac i_B. You're only interested in the latter, since the ac i_B is controlled by the input ac voltage.

Do not compute any dc values! (Unless you have reason to suspect that the circuit is not properly biased. I can tell you right now that it's OK).

Another thing: your controlled current sources are ac currents, not dc. They're called "controlled" because their value depends on the voltage or current somewhere else in the circuit. In your case the current sources are either g_mv_be or βi_b. Take your pick, either one works. I myself prefer the latter.

 

[GreenPrint]

Could you give an example of solving a transistor configuration using ac values? I think I may be confused because this is how I have been solving these types of problems.

 

[rude man]

Change the first controlled current source from 1.542 mA to βi_b1.
Change the second current source to βi_b2.

Using your equivalent circuit ONLY, compute i_b1 based on v_s and the three resistors (1K, 4.927K, βr_e. Label them first. All your resistors should be labeled, and your current sources labeled separately as I have done. You put in the numbers at the very end only.

Now use KVL or KCL or whatever you like to compute v_c2. USE ONLY THE EQUIVALENT CIRCUIT. DO NOT COMPUTE ANY DC VOLTAGES OR CURRENTS.

Get into the habit of using lower case letters for ac quantities and upper case for dc.

You did a good job of putting the equivalent circuit together. You just need to distinguish between ac and dc quantities.

 

[GreenPrint]

Alright. Well I start off by first finding the voltage at the base of the transistor, also equal to the input voltage of the first stage
V^{~}_{b_{1}} = \frac{(R_{Th}||βr_{e_{1}})V_{S}}{R_{S}+R_{Th}||βr_{e_{1}}}

I can then find the current through the base of the first transistor
I_{b_{1}} = \frac{V_{b_{1}}}{βr_{e_{1}}} = \frac{(R_{Th}||βr_{e_{1}})V_{S}}{(R_{S} + R_{Th}||βr_{e_{1}})βr_{e_{1}}}

I can then find the output voltage of the first stage
V_{O_{1}} = -βI_{b_{1}}R_{C_{1}} = \frac{-(R_{Th}||βr_{e_{1}})V_{s}R_{C_{1}}}{(R_{S} + R_{Th}||βr_{e_{1}})r_{e_{1}}}

I can then find the input impedance of the second stage
Z = R_{Th}||(βr_{e_{2}} + (β + 1)R_{E_{2}}||R_{L})

I can then find the current through the base of the second transistor using that the output voltage of the first stage is the input voltage of the second output voltage and the voltage at the base of the second transistor.
I_{b_{2}} = \frac{V_{O_{1}}}{Z} = \frac{(R_{Th}||βr_{e_{2}})V_{S}R_{C_{1}}}{(R_{S} + R_{Th}||βr_{e_{1}})r_{e_{1}}(R_{Th}||βr_{e_{2}} + (β + 1)R_{E_{2}}||R_{L}}

I can then find the voltage at the collector of the second transistor
V_{C_{2}} = -βI_{b_{2}}R_{C_{2}} = \frac{β(R_{Th}||βr_{e_{1}})V_{S}R_{C_{1}}R_{C_{2}}}{(R_{S} + R_{Th}||βr_{e_{1}})r_{e_{1}}(R_{Th}||βr_{e_{2}}||βr_{e_{2}} + (β + 1)R_{E_{2}}||R_{L}}

I believe that this is correct, however I'm not exactly sure how this helps. I don't know r_{e_{1}} or r_{e_{2}} and can't find these values withoutI_{e}. I need to find V_{e} the voltage at the emitter of the second transistor in order to find the output voltage.

These equations seem a bit messy and I assume that I need to make some approximations? This has opened my eyes to what I was doing all along was down right wrong. I questioned just assuming that r_{e_{2}} and r_{e_{1}} are equal to each other, which they are obviously not. I would like to fully understand this problem, I'm just not sure where to go from here.

Could I find the current through the base of the second transistor and add it to the current through the collector of the second transistor to find the current through the emitter of the second transistor and find the output voltage in this matter?

Thanks for all of the help!

 

[rude man]

1. Show how you obtained r_e1 = 2513 ohms. I got around 1600 ohms.

2. you need to redraw your equiv. ckt. Give a unique symbol to each component including sources and transistors. Do not use numbers until the very end.

3. Obtain a value for both current sources in either β and i_b or v_be and g_m. I prefer the former since then the two current sources have the same expression.

4. Write kcl, kvl or whatever equations to the rest of the circuit. The rest cannot be divided up like the first part. I would take the two independent nodes v_c1 and v<sub>e2 and sum currents to zero at each node.

Hint: the collector resistor for Q2 does not affect your computations.</sub>

 

[GreenPrint]

Well this is how I got that value

http://imageshack.com/a/img443/6140/5kw5.png

 

[GreenPrint]

Alright. Simplifying the equation at the collector of the first stage provides me with

V_{c_{1}}(\frac{1}{R_{C}} + \frac{1}{R_{Th}} + \frac{1}{βr_{e_{2}}}) + \frac{V_{b_{1}}}{r_{e_{1}}} = \frac{V_{e_{2}}}{βr_{e_{2}}}

Simplifying the equation at the emitter of the second stages provides me with

V_{c_{1}} = \frac{V_{e_{2}}r_{e_{2}}}{(R_{E}||R_{L})(\frac{1}{β} + 1)}

I believe these equations are correct. How would you suggest I proceed from here?

Can I find V_{b_{1}} using

V_{b_{1}} = \frac{V_{CC}R_{2}}{R_{1} + R{2}}

Even though this is a dc voltage?

 

[The Electrician]

I questioned just assuming that r_{e_{2}} and r_{e_{1}} are equal to each other, which they are obviously not.

re1 and re2 are the same. The capacitor between the collector of the first transistor and the base of the second transistor prevents the DC voltage at the base of the second transistor from being affected by the DC voltage at the collector of the first transistor.

The DC bias conditions of both transistors are identical.

The biggest problem this circuit has is that the second transistor has a 5100 ohm collector resistor. The AC gain to this second collector is very high, and the second transistor will be a very bad emitter follower, because its collector voltage is not at all constant. The collector of the second transistor should be connected directly to 15 volts.

 

[rude man]

Here's how you get r_e: r_e = dV_be/dI_b
but dI_c/dV_be = β dI_b/dV_be = I_c/V_T

with β = 150, V_T = 0.026V, I_c ~ I_e = 1.59 mA

So r_e = β*V_T/I_e = 150*0.026/1.59e-3 = 2453 ohms. Hmm, looks like I got about the same as you after all ... . Sorry, I computed a lower number 1st time around.

I hope you understand that the quantities r_e and g_m are ac quantities but they are found from dc analysis. β = g_mr_e. For you, β is an ac and a dc parameter, and about the same number.

What about your two controlled current sources?

OK, so now I await your fixing your equiv. ckt. up so it can be used effectively.

 

[rude man]

The biggest problem this circuit has is that the second transistor has a 5100 ohm collector resistor. The AC gain to this second collector is very high, and the second transistor will be a very bad emitter follower, because its collector voltage is not at all constant. The collector of the second transistor should be connected directly to 15 volts.

I would not call that a big problem. This is a second-order effect unless of course the 5.1K forces saturation. Not much difference between 1V and 10V V_ce, operation-wise. It's true that it would be better to short the 5.1K.

 

[The Electrician]

It drastically limits the swing that can be obtained at the emitter of the second transistor.

 

[GreenPrint]

These posts have made me question a couple of more things.

So then my DC analysis for I_{B_{1}} and r_{e_{1}} is correct. Based off your posts, transistors have two betas β and β^{~} an AC value and a DC value. Interesting. Thus far I have assumed they are the same.

In my AC equivalent circuit, I need to find βi_{B_1}^{~} I can't use I_{B_{1}} in this calculation because I need the AC value i_{B_1}^{~} which I'm supposed to get from my small circuit equivalent drawing?

I don't see how r_{e_{2}} and r_{e_{1}} are the same. I know understand that r_{e} = \frac{26 mV}{I_{e}}, where I_{e} is the DC current and not the AC current. For the first transistor I used the formula I_{B} = \frac{V_{E} - V_{BE}}{R_{Th} + (β + 1)R_{E}} and then found I_{E} using I_{E} = (β + 1)I_{B}, thereby allowing me to calculate r_{e}. I don't see how the DC current through the base could be the same for the first transistor and the second transistor because the second transistor has R_{L}, which the first transistor does not. Because I believe the transistors have different currents through the base based off this they would have different currents through the emitter and as a result different r_{e}. I guess these conclusions are wrong but I'm not exactly sure why.

Here's my new drawing with the stages separated from each other.
http://img706.imageshack.us/img706/8138/jv89.png
Thanks for the help.

 

[rude man]

It drastically limits the swing that can be obtained at the emitter of the second transistor.

Not drastically. It can handle nearly 1V rms at the output which is the standard max. voltage for an A/V signal, for instance.

No, what is bad about this circuit is its linear dependence on beta of the first transistor. In addition to very uncertain gain over temperature and from transistor to transistor, even within one batch. Also, g_m varies significantly with input voltage, causing large disortion of the input sine wave at the output. That limits the input to very small voltage fluctuations so the output limitation imposed by the second collector resistor is almost negligible.

Any design, the gain of which depends significantly on beta, is bad. A good design assumes beta → ∞, exactly analogous to the gain of an op amp being assumed to have infinite gain.

 

[rude man]

These posts have made me question a couple of more things.

So then my DC analysis for I_{B_{1}} and r_{e_{1}} is correct. Based off your posts, transistors have two betas β and β^{~} an AC value and a DC value. Interesting. Thus far I have assumed they are the same.

Yes, you had the right r_e1 all along.

There are two betas but the second beta is unimportant. It's OK to assume they are the same, and that the ac and dc betas are also the same.

Your equiv. ckt. is a huge improvement! But use v for voltage, not V. V is a dc voltage, v is an ac voltage.

I would not split it as you have though:

1. find i_b1 as a function of v_s. That computation is independent of the rest of the circuit.

2. you now have 2 independent nodes, v_c1 and v_e2. Sum currents to zero at each node or use KCL or whatever to solve for v_e2 which is your output voltage.

In my AC equivalent circuit, I need to find βi_{B_1}^{~} I can't use I_{B_{1}} in this calculation because I need the AC value i_{B_1}^{~} which I'm supposed to get from my small circuit equivalent drawing?

Sure! Very simple. v_s generates what current i_b1?

I don't see how r_{e_{2}} and r_{e_{1}} are the same. I know understand that r_{e} = \frac{26 mV}{I_{e}}, where I_{e} is the DC current and not the AC current.

Not the correct expression for r_e. r_e = βV_T/I_c.
Since β is assumed the same and both I_c are the same, r_e1 = r_e2.

For the first transistor I used the formula I_{B} = \frac{V_{E} - V_{BE}}{R_{Th} + (β + 1)R_{E}} and then found I_{E} using I_{E} = (β + 1)I_{B}, thereby allowing me to calculate r_{e}. I don't see how the DC current through the base could be the same for the first transistor and the second transistor because the second transistor has R_{L}, which the first transistor does not. Because I believe the transistors have different currents through the base based off this they would have different currents through the emitter and as a result different r_{e}. I guess these conclusions are wrong but I'm not exactly sure why.

Look again at the dc situation for the two transistors. They're identical! And even if R_L1 ≠ R_L2, that would make no difference unless V_be drops below about 1V. You problem is again that you're not separating ac and dc parameters. The equiv. ckt. is for ac only; don't try to use it for dc. And r_e is a dc computation as you know.

 

[The Electrician]

Not drastically. It can handle nearly 1V rms at the output which is the standard max. voltage for an A/V signal, for instance.

No, what is bad about this circuit is its linear dependence on beta of the first transistor. In addition to very uncertain gain over temperature and from transistor to transistor, even within one batch. Also, g_m varies significantly with input voltage, causing large disortion of the input sine wave at the output. That limits the input to very small voltage fluctuations so the output limitation imposed by the second collector resistor is almost negligible.

Any design, the gain of which depends significantly on beta, is bad. A good design assumes beta → ∞, exactly analogous to the gain of an op amp being assumed to have infinite gain.

Yes, drastically. The max output swing with the 5100 ohm resistor in place is theoretically .823 volts RMS, compared to 5.3 volts RMS with collector connected to 15 volts. That's the limitation which is drastic, and which I referred to, not a comparison with an A/V signal.

It's not a second order effect if substantially more than .8 volts RMS output capability is needed, which the output stage could provide without that 5100 ohm collector resistor in there.

I didn't say that this is the only problem, but often academic circuits use a simple, non-feedback common emitter stage to illustrate circuit analysis, with its attendant distortion, but there's no reason for the 5100 ohm resistor

Also, g_m varies significantly with input voltage, causing large disortion of the input sine wave at the output. That limits the input to very small voltage fluctuations so the output limitation imposed by the second collector resistor is almost negligible.

Any given circuit may be subject to several design criteria. If it's required that the output be large in amplitude with moderate distortion, then the output limitation due to the 5100 ohm resistor is not negligible. If lower distortion and less β dependency is needed than a better input stage would be required but the output stage would still have that drastic limitation.

 

[The Electrician]

Not the correct expression for r_e. r_e = βV_T/I_c.

Every reference I have says that re = V_T/I_e

 

[NascentOxygen]

Every reference I have says that re = V_T/I_e

Correct.

 

[GreenPrint]

Well my equation

i_{b_{1}} = \frac{v_{s}(βr_{e_{1}}||R_{Th})}{βr_{e_{1}}(R_{S} + R_{Th}||βr_{e_{1}})}

KCL at v_{c_{1}}

\frac{v_{s}(βr_{e_{1}}||R_{Th})}{r_{e_{1}}(R_{S} + R_{Th}||βr_{e_{1}})} + v_{c_{1}}(\frac{1}{R_{C}} + \frac{1}{R_{Th}} + \frac{1}{βr_{e_{2}}}) - \frac{v_{e_{2}}}{βr_{e_{2}}} = 0

KCL at v_{e_{2}}

βr_{e_{2}} + βi_{B_{2}} = \frac{v_{e_{2}}}{R_{E}||R_{L}}

My text says that r_{e} = \frac{26 mV}{I_{E}}. I don't see how in this example r_{e_{1}} = r_{e_{2}} because I still don't see how I_{E_{1}} = I_{E_{2}}. You say that I_{E_{2}} doesn't depend on R_{L}, but I don't see how this can be so. Also thus far in DC analysis I have assumed V_{BE} ≈ 0.7 V, which would be less than 1.0 V.

I'm not sure how to solve for i_{B_{2}} which I assume I need to plug into my two KCL equations at the two nodes.

Thanks for any help.

 

[NascentOxygen]

A capacitor isolates the DC conditions on each side of it because DC does not pass through a capacitor (except to cause as a brief transient). As you can see, each of the transistor stages has identical base biasing, and identical collector and emitter resistors, so the DC conditions are the same for each transistor. This means I_E will be the same for each, therefore r_e is going to be the same for each.

Yes, that's 0.026/I_E at room temperature.

 

[GreenPrint]

Ah that makes much more sense. I now understand why r_{e_{1}} = r_{e_{2}}. So for my equation for v_{e_{2}} I plug in i_{b_{2}} = \frac{v_{c_{1}} - v_{e_{2}}}{βr_{e}} and get KCL at v_{e_{2}} or v_{O} and placing v_{S} with v_{I}

v_{O} = \frac{r_{e}v_{C_{1}}(\frac{1}{β} + 1)}{\frac{1}{r_{e}}(β + 1) + \frac{1}{R_{E}||R_{L}}}

I also have this equation
\frac{v_{I}(βr_{e}||R_{Th})}{r_{e}(R_{S} + R_{Th}||βr_{e})} + v_{C_{1}}(\frac{1}{R_{C}} + \frac{1}{R_{Th}} + \frac{1}{βr_{e}}) - \frac{v_{O}}{βr_{e}} = 0

This looks like a mess. Am I really supposed to take the equation right about this line and solve for v_{C_{1}} and plug this into the equation for v_{O} and try to solve for \frac{v_{O}}{v_{I}}?

 

[GreenPrint]

I don't see really what was the point but I get

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

Is this what you get?

 

[rude man]

Correct.

I was thinking of r_π which is the ac input resistance to the hybrid-pi model, which = βr_e. You're right, r_e = V_T/I_c. Sorry about the confusion.

 

[rude man]

Ah that makes much more sense. I now understand why r_{e_{1}} = r_{e_{2}}. So for my equation for v_{e_{2}} I plug in i_{b_{2}} = \frac{v_{c_{1}} - v_{e_{2}}}{βr_{e}} and get KCL at v_{e_{2}} or v_{O} and placing v_{S} with v_{I}

v_{O} = \frac{r_{e}v_{C_{1}}(\frac{1}{β} + 1)}{\frac{1}{r_{e}}(β + 1) + \frac{1}{R_{E}||R_{L}}}

I also have this equation
\frac{v_{I}(βr_{e}||R_{Th})}{r_{e}(R_{S} + R_{Th}||βr_{e})} + v_{C_{1}}(\frac{1}{R_{C}} + \frac{1}{R_{Th}} + \frac{1}{βr_{e}}) - \frac{v_{O}}{βr_{e}} = 0

This looks like a mess. Am I really supposed to take the equation right about this line and solve for v_{C_{1}} and plug this into the equation for v_{O} and try to solve for \frac{v_{O}}{v_{I}}?

Well, I won't check your math in detail. But if you got i_b1(v_s) and v_e2/v_c1 right then you're OK. You did a nice job cleaning up your equiv. ckt. labeling.

If it's any consolation:
1. this circuit would never be encountered in a professional environment, as I explained.
2. the small-signal equiv. ckts. are there just for learning and are not typically used. Things get very simplified if you assume β → ∞, on which any good circuit is based.

 

[GreenPrint]

I would like to check though to make sure that my answer is correct. I have checked online and checked my textbook and I can't find this configuration any where. I don't want to think my answer is correct if it's not.

 

[rude man]

I would like to check though to make sure that my answer is correct. I have checked online and checked my textbook and I can't find this configuration any where. I don't want to think my answer is correct if it's not.

You won't get many points off if you made a math error. Anyway, I doubt that anyone has the perseverance to check all your math. Maybe.

BTW a very rough number for the gain is around 100, based on the original circuit, not the equiv. ckt. But it should be approximately right. Maybe you could compare that number with your detailed calculations. You can get this by ignoring everything to the right of R_TH2 and assuming v_c2 = v_b1 is your output (basically, an emitter follower has close to unity voltage gain without presenting a significant load at the base).
).

 

[The Electrician]

I don't see really what was the point but I get

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

Is this what you get?

If I evaluate what I got using the indicated values for the components, the result is:

If I evaluate your expression, the result is:

I can't tell you where your mistake is, but there is apparently at least one.

 

[The Electrician]

I would like to check though to make sure that my answer is correct. I have checked online and checked my textbook and I can't find this configuration any where. I don't want to think my answer is correct if it's not.

One mistake I see early on is in post #8 where you have:

I can then find the output voltage of the first stage
V_{O_{1}} = -βI_{b_{1}}R_{C_{1}} = \frac{-(R_{Th}||βr_{e_{1}})V_{s}R_{C_{1}}}{(R_{S} + R_{Th}||βr_{e_{1}})r_{e_{1}}}

The load resistance seen by the collector of the first transistor is not just RC1. You have to include the input resistance of the second transistor.

 

[GreenPrint]

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}})}

I now simplify the denominator by combining the two r_{e}
V_{O} = \frac{(\frac{V^{~}_{O}}{β} - \frac{V^{~}_{S}R_{S}}{R_{Th}||βr_{e} + R_{S}})(\frac{1}{β} + 1)}{r^{2}_{e}(\frac{1}{βr_{e}} + \frac{1}{r_{e}} + \frac{1}{R_{E}||R_{L}})(\frac{1}{R_{C}||R_{Th}} + \frac{1}{βr_{e}})}

I can now define the voltage gain
A_{V} = \frac{V_{O}}{V_{I}} = \frac{\frac{(\frac{V^{~}_{O}}{β} - \frac{V^{~}_{S}R_{S}}{R_{Th}||βr_{e} + R_{S}})(\frac{1}{β} + 1)}{r^{2}_{e}(\frac{1}{βr_{e}} + \frac{1}{r_{e}} + \frac{1}{R_{E}||R_{L}})(\frac{1}{R_{C}||R_{Th}} + \frac{1}{βr_{e}})}}{\frac{V^{~}_{S}R_{S}}{R_{Th}||βr_{e} + R_{S}}}

I now simplify this fraction a bit
A_{V} = \frac{(\frac{V^{~}_{O}}{β} - \frac{V^{~}_{S}R_{S}}{R_{Th}||βr_{e} + R_{S}})(\frac{1}{β} + 1)}{r^{2}_{e}(\frac{1}{βr_{e}} + \frac{1}{r_{e}} + \frac{1}{R_{E}||R_{L}})(\frac{1}{R_{C}||R_{Th}} + \frac{1}{βr_{e}})}\frac{R_{Th}||βr_{e} + R_{S}}{V^{~}_{S}R_{S}}

At this point it doesn't seem as if V_{S} will cancel out, which is a problem. Is this how you went about solve for the voltage gain? I think solving in this way make sense. I also don't see how you got R_{S} in your final answer because it looks like it would have to cancel out some how with V_{S} when it cancels out.

Thanks for any help.

 

[The Electrician]

At this point it doesn't seem as if V_{S} will cancel out, which is a problem. Is this how you went about solve for the voltage gain? I think solving in this way make sense. I also don't see how you got R_{S} in your final answer because it looks like it would have to cancel out some how with V_{S} when it cancels out.

Thanks for any help.

I got Rs in my final answer because the circuit you showed in post #1 asked for Av without specifying whether Av was defined to be Vo/Vs, or Vo/VI, where you later took VI to be the (small signal; AC) voltage at the base of the first transistor.

If Av is Vo/Vs, then Rs will necessarily appear in the final result.

If what you want is Vo/VI, then you shouldn't start your analysis with Rs included; get rid of Rs and start your analysis at the base of the first transistor.

An assumption that is made when analyzing amplifier circuits like this, is that the source driving the input is a voltage source, with an output impedance of zero ohms. If the source is to have a non-zero output impedance, that must be added explicitly. That's what Rs may be for in your circuit; Rs would stand for R-source. The source Vs is assumed to be ideal, and Rs adds a Thevenin output impedance to give a non-ideal source.

So, what is the problem asking for? Vo/Vs, or Vo/VI? If it's Vo/Vi then eliminate Vs and Rs and start your analysis without them. Doing it the way you did just adds unnecessary complication; you don't need Rs and Vs if you just want Vo/VI.

You obviously have learned KCL; have you learned how to use KCL to formulate equations using the nodal method of circuit analysis?

Do you have access to Matlab, Mathcad or some similar math program at your school? Or, do you have a calculator that can solve systems of simultaneous equations?

 

[GreenPrint]

This question actually appeared on of my past tests and I got the question wrong. The questions asked me to

1.) What is the amplifier configuration
2.) perform dc analysis to determine dc current flows and small signal parameters
3.) Draw the small signal equivalent circuit and determine the overall voltage gain
4.) Determine overall current gain

My textbook has defined A_{V} = \frac{V_{O}}{V_{I}}. I didn't realize that it could be defined as A_{V} = \frac{V_{O}}{V_{S}}, but think the question is askng about A_{V} = \frac{V_{O}}{V_{I}}. 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.

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.

 

[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.