# BJT Cascade amplifier w/ temperature consideration.Homework for school.

Discussion in 'Homework Help' started by electrogirl, May 10, 2010.

1. ### electrogirl Thread Starter Member

May 2, 2010
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0

Hi I need help with my homework for school.
I've solved 2 stage BJT cascade amplifier before but this one is different because you have to consider the temperature. Unfortunately this type of problem was not discussed to us yet. So I posted it here to get some help.

Disregarding the temperature this is how I would solve it.

DC analysis
VE = VB - VBE = 0.82 - 0.7 = 0.12V
IE = VE/RE = 0.12V/1k = 0.12mA
re = 26mv/IE = 216Ω

AC analysis
Avtotal = Vo/Vi
Vo = β2Ib2(1K)
Vi = Ib1β1re + β1Ib1(100Ω)
Expressing Ib2 in terms of Ib1
Using Current Divider theorem
Ib1 = -β1Ib1(5k)/5K + (β2re2+β2(1k)
thus,
Avtotal = [β(1K)[-βIb1(5K)/5K +β(re2 + 1K)]/[Ibβ1(re1 + 100)
Avtotal = -12.75
Vo = Avtotal X Vi
Vo = -12.75 X 0.1 sin1000t
Vo = -1.275 sin1000t, but the real answer should be Vo = -4.08 sin100t V

Last edited: May 11, 2010
2. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,635
1,299
But 0.82V is the voltage on base of Q2 not Q1
So Icq2 = 1.2mA then re2 = 26mV/1.2mA = 21.7Ω
So gain of a first stage is equal:

Au1 = ( 5KΩ||(201*1KΩ) ) / (21.7Ω+100Ω) = 4.87kΩ/121.7Ω = 40.08 [V/V]

Icq1 ≈ 5.3mA
So re1 = 26mV/5.3mA ≈ 5Ω
And the voltage gain of a second stage is equal
Au2 = 1KΩ / (1KΩ+ 5Ω) = 0.995[V/V]

So the overall gain is equal
Aus = 40.08 * 0.995 = 39.87 [V/V]

Vo ≈ - 4sin100t

Last edited: May 10, 2010
3. ### electrogirl Thread Starter Member

May 2, 2010
47
0
Jony thank you for responding.Btw about Au1 why 201*1K and not βre+1K?
How did you get Icq1 = 5.3mA?

4. ### PRS Well-Known Member

Aug 24, 2008
989
36
I may be wrong, but it appears the emmiter resistor at Q1 is 1 kilohm, not 100 ohms. Otherwise it's not biased properly. Could you recheck this?

5. ### electrogirl Thread Starter Member

May 2, 2010
47
0
Hi...
the emitter resistor Q1 is 1K and emitter resistor Q2 is 100Ω

6. ### electrogirl Thread Starter Member

May 2, 2010
47
0
How is the DC base voltage of Q1 computed? If I've known how to compute for the voltage at the base of Q1, I should be able to compute for the overall voltage gain and thus compute the the value of Vo(which is the required parameter to this problem).

overall voltage gain is simply Avt = Vo/Vi
Vo ≈ β2Ib2(1K)
expressing Ib2 in terms of Ib1
Ib2 = - βIb1(5k)/[5K + (β2re2 + β(1k)]
Vo = β2(1K)[- β1Ib1(5K)/5K + β2(re2 + 1K)

Vi = Ib1(β1re1) + β1Ib1(100)
Vi = Ib1β1(re1 + 100)

thus,
Avt = {[β2(1K)-β1Ib1(5K)/5K + β2(re2 + 1K)]/[Ib1β1(re1 + 100]}

I've tried substituting to this formula the values given by Jony130 and the result is 39.887 the same result as computed by jony130.

But the problem is, I don't know how Jony get the value of Icq1 = 5.3mA?
Is this because of the temperature?Can explain anyone to me this please?

Last edited: May 11, 2010
7. ### PRS Well-Known Member

Aug 24, 2008
989
36
To get to Q1 the signal has to get through the first stage, Q2. Q2 is not biased in the linear amplifier mode. The official answer suggests Q2 is not being operated as a switch. It's supposedly a linerar amplifer. But with the resistor values you have given, it just doesn't work. I'll show you how the Q2 stage is usually solved when I get back. I've got to mow some lawns. But, briefly, Thevinize the input and write a current loop to obtain Ie. Later....

8. ### PRS Well-Known Member

Aug 24, 2008
989
36
I included an attachment showing the way this circuit should be solved. The first amp, which I renamed Q1 is a CE amp with dual-ended supply. You need to Thevinize the input and a do a loop as shown. This gives you Ie1 and the rest of the analysis goes from there. See attachment.

The problem is that, given the resistors biasing the first stage, the stage simply isn't biased right for a linear amplifier. You need to check that you wrote down the circuit right. I just doesn't work as shown.

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9. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,643
468
This problem has what seem to me to be a number of ambiguities.

Notice the very first thing it says "Solve for Vo. a) From (+)terminal to (-) terminal b) From (+) terminal to Ground". We see that they distinguish the (-) terminal from ground.

Then they say "Vb = 0.82 V @ 10 degrees Celcius". They don't say Vb with respect the the (-) terminal; they just say 0.82 volts which implies that means with respect to ground. But if we take it with respect to ground, then the circuit is very badly biased. The only thing that makes sense is that they mean Vb is at -5.18 V with respect to ground, or 0.82 V with respect to the (-) terminal (as they call it).

Furthermore, they say Vb, not Vbe. I'll take them at their word; they mean Vb.

One must wonder why they would tell us what Vb is at 10 degrees, and then say that Room temp = 25 degrees if they don't want us to make use of that information.

They have failed to tell us if they want the analysis to be performed at 10 degrees, or at 25 degrees (room temp). I'll assume they want the analysis at room temp.

If we do an accurate analysis of the various currents and voltages, we find that if we assume Vbe to be .696 V, then Vb is at .8199939 V with respect to the (-) terminal. And this is said to be at 10 degrees.

Using the rule of thumb that the voltage drop across a forward biased silicon diode decreases at 2 mV/degree, we would have a Vbe of .696 - 15*.002 = .666 V at 25 degrees (room temp).

Setting Vbe to .666 V, we have Vb = .816336 V, Vc = 4.52 V, Ie = 1.5034 mA, re = .025/.0015034 = 17.29Ω and the voltage gain for the first stage is Av2 = -41.386.

With Vc = 4.52 V and Vbe of the emitter follower the same .666 V, we have Ve of the emitter follower as 3.85 V and Ie1 = 3.85 mA which gives re of the emitter follower equal to 6.745Ω. Then the gain of the emitter follower would be Av1 = 1000/(1006.7) = .993 for an overall gain Vtot = -41.111.

The quiescent output voltage of the emitter follower of 3.85 V will not allow for an unclipped output voltage of 8.16 V peak-to-peak. So, it looks like assuming Vbe = .666 V at room temp is not working out.

However, if we ignore all the temperature numbers and just solve the circuit with Vbe = .7, we will have Vc = 6.006 V. If we take Vbe to be .696 V, which gives Vb = .82 V (with respect to the (-) terminal), we will have Vc = 5.83 V. Either of these will be able to give 8.16 V peak-to-peak output voltage.

I don't get a voltage gain of exactly 40.08 with either of these assumptions for Vbe, but assuming Vbe = .7 V gets the closest.

10. ### PRS Well-Known Member

Aug 24, 2008
989
36
Hi Electrician! I'm glad you came in on this. I had the same questions you brought up and the way I see it, this circuit doesn't really work at all. But now I'm curious enough to breadboard it and test it. I've got to mow some lawns first, but I'll let you know what happens.

11. ### PRS Well-Known Member

Aug 24, 2008
989
36
I breadboarded the first amplifier in the circuit. The collector is at -6 volts and this is due to excessive current via the bias arrangement. A sinewave at the input comes out at the collector with the bottom clipped at -6 volts. The top looks okay until the gain is raised such that the output reaches almost +6 volts, then it clips, too.

Either the OP diagram is wrong due to the professor or to electrogirl's mistake.

12. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,643
468
The problem with this circuit is that the emitter resistor is only 100Ω. This makes the biasing behave like the simple resistor-from-the-supply-to-the-base type of bias. Totally dependent on β, and very sensitive to Vbe.

But, if you pick things just right, it will work, especially in simulation/analysis where there isn't any temperature drift to worry about!

13. ### electrogirl Thread Starter Member

May 2, 2010
47
0
The diagram that i've posted was really the circuit diagram given to us (me and my classmate have the same diagram).