# Homework Help: Electrical Engineering - Circuits - Transistors

1. Oct 29, 2013

### GreenPrint

1. The problem statement, all variables and given/known data

Given the configuration below

http://img844.imageshack.us/img844/3364/c5e8.png [Broken]

Find $A_{V}$ and $A_{I}$

2. Relevant equations

3. 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 [Broken]

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.

Last edited by a moderator: May 6, 2017
2. Oct 29, 2013

### 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.

3. Oct 29, 2013

### 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.

4. Oct 29, 2013

### 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.

5. Oct 29, 2013

### 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.

6. Oct 29, 2013

### rude man

You know the dc IB but you don't know the ac iB. You're only interested in the latter, since the ac iB 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 gmvbe or βib. Take your pick, either one works. I myself prefer the latter.

7. Oct 29, 2013

### 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.

8. Oct 29, 2013

### rude man

Change the first controlled current source from 1.542 mA to βib1.
Change the second current source to βib2.

Using your equivalent circuit ONLY, compute ib1 based on vs and the three resistors (1K, 4.927K, βre. 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 vc2. 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.

9. Oct 30, 2013

### 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 without$I_{e}$. 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!

10. Oct 30, 2013

### rude man

1. Show how you obtained re1 = 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 ib or vbe and gm. 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 vc1 and ve2 and sum currents to zero at each node.

Hint: the collector resistor for Q2 does not affect your computations.

11. Oct 30, 2013

### GreenPrint

Well this is how I got that value

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

Last edited by a moderator: May 6, 2017
12. Oct 30, 2013

### 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?

Last edited: Oct 30, 2013
13. Oct 31, 2013

### The Electrician

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.

14. Oct 31, 2013

### rude man

Here's how you get re: re = dVbe/dIb
but dIc/dVbe = β dIb/dVbe = Ic/VT

with β = 150, VT = 0.026V, Ic ~ Ie = 1.59 mA

So re = β*VT/Ie = 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 re and gm are ac quantities but they are found from dc analysis. β = gmre. For you, β is an ac and a dc parameter, and about the same number.

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

15. Oct 31, 2013

### rude man

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 Vce, operation-wise. It's true that it would be better to short the 5.1K.

16. Oct 31, 2013

### The Electrician

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

17. Oct 31, 2013

### 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 [Broken]
Thanks for the help.

Last edited by a moderator: May 6, 2017
18. Oct 31, 2013

### rude man

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, gm 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.

19. Oct 31, 2013

### rude man

Yes, you had the right re1 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 ib1 as a function of vs. That computation is independent of the rest of the circuit.

2. you now have 2 independent nodes, vc1 and ve2. Sum currents to zero at each node or use KCL or whatever to solve for ve2 which is your output voltage.
Sure! Very simple. vs generates what current ib1?

Not the correct expression for re. re = βVT/Ic.
Since β is assumed the same and both Ic are the same, re1 = re2.
Look again at the dc situation for the two transistors. They're identical! And even if RL1 ≠ RL2, that would make no difference unless Vbe 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 re is a dc computation as you know.

20. Oct 31, 2013

### The Electrician

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

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.

21. Oct 31, 2013

### The Electrician

Every reference I have says that re = VT/Ie

22. Nov 1, 2013

### Staff: Mentor

Correct.

23. Nov 1, 2013

### 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.

24. Nov 1, 2013

### Staff: Mentor

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 IE will be the same for each, therefore re is going to be the same for each.

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

25. Nov 1, 2013

### 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 $\frac{v_{O}}{v_{I}}$?