# Amplifier General Questions

Discussion in 'Homework Help' started by howartthou, Jul 9, 2009.

1. ### howartthou Thread Starter Active Member

Apr 18, 2009
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0
Hi All

Some homework questions that I need help with are:

Q1.
The attached pic shows a "constant current source". The question is: Whats the emitter current in the circuit?

Note: The transistor is silicon.

My attempt:

I = 5.6v / 5000 = 1 mA

Questions:
Is this correct If so, why? I ignored the 1K resistor, Vin and RL. So why are they ignored? If not then, help!

Also, how does knowing the transistor is silicon help determine the emitter current

Q2.
The output stages of operational amplifiers are often:
1. Differential amplifiers
2. a-c amplifiers
3. Darlington amplifiers
4. Complimentary amplifiers

My attempt:
I know it isnt 1 because differential amplifiers are usually input to op amps.

My guess is 4 but I have no idea really

Q3.
In the second pic, the question is: What is the purpose of C3?

Options given:
1. It passes a-c and blocks d-c, this biasing the emitter Q2.
2. It passes a-c and blocks d-c, the keeping the emitter Q2 at a-c ground.

My attempt.
I guess 2. But I am not sure why?

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2. ### millwood Guest

simple:

1) you just need to figure out the voltage drop on that resistor. it is the zener voltage drop - the Vbe of the transistor;

2) typically the last stage is a complimentary stage.

3) none of the answers is correct. capacitors have lower impedance at higher speed. so all C3 does is to lower the impedance on the emitter and allows the amplifier to have a higher ac gain without sacrificing its dc stability.

3. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
Q1.

The zener diode sets the base voltage as 5.6V. The 1K simply sets the zener bias current for the zener diode - it would be about 19.4mA. If the transistor is silicon type, the base-emitter forward bias voltage would be about 0.6 to 0.7 volts - let's say it's 0.65 volts for the given conditions. So the actual voltage across the 5K emitter resistor Re would be VRe = 5.6-0.65 = 4.95 Volts. The current in the emitter resistor would be 4.95/5 mA = 0.99mA. RL is presumably adjusted so that the transistor doesn't go into saturation. It shouldn't particularly effect the sourced current value - I'm assuming the 5K is the load which sees the "controlled" current.

If the 5K was set to another value the current would be different - so in that sense the circuit isn't strictly a constant current source - rather a constant volatge source for the load or emitter resistor.

4. ### millwood Guest

it is a CCS in the sense that the collector current is (largely) independent of supply voltage.

5. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,993
1,116
Can you tell me why answer 2 is wrong.
Assume that Ie=1mA; R8=1K and C3=470uF
For me the answer 2 is OK. Its all true
Of course this answer don't show the "full picture" of C3 destination.

6. ### millwood Guest

the word "ground" means "shorted" or very very load resistance / impedance.

in the case of C3, you can find a low frequency a/c where C3's impedance is still substantial. so in that sense, it does not "ground" an ac signal.

obviously, answer #2 is truer as the frequency goes higher.

7. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,993
1,116
From my point of view you are going to far into details in this simple question.
This is a audio pre-amplifier so C3 will by act as a "short" for as-signals from audio band.

8. ### howartthou Thread Starter Active Member

Apr 18, 2009
96
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Thanks everyone, much appreciated.

Tnk very insightful answer! My God, I think I actually understand what you said with the 1k setting the bias for the zener and what followed. Could you explain why the +25v and RL are also "ignored". Is that because their affect on the base-emitter current is already determined by the voltage shown on the Zener? Arrggghh!

millwood, thanks for your input. I am a real noob though, so I need more details, nothing is obvious to me yet, I am really struggling with things that are "obvious" to you guys, but thanks.

Jony, masterful, thanks, you are really good. But, as much as I have learned here, I am still really dumb. For Q3, please explain if C3 wasnt there wouldn't the emitter still go to ground???? I mean I just don't get the purpose of C3 and how it takes it to ground. If you removed C3 ground is still there!

9. ### millwood Guest

+25v/RL: they do have an impact on the current going through the load and on the circuit overall. the 1K resistor and +25v together determines the current going through the zener. the voltage drop over the zener IS impacted by the currernt going through it, though the impact is quite small in general, until the point where the zener is damaged.

a very high RL will push down Vce on the transistor and force it to stop working.

so what have been articulated here are true to the extent those parameters are "reasonable".

C3: think of C3 as a variable resistor whose "resistance", Zc3, goes down as frequency goes up. the ac gain of the amplifier is determined by R9/(R8//Zc3), assuming no load, and R10 is sufficiently large.

so as Zc3 goes down, the amp's gain goes up with frequency.

10. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,301
339
The gain of the "amplifier", whose input is the left terminal of C1, and whose output is applied to R11, is determined by more than R9/(R8//Zc3), even if R10 is infinite.

That expression is an approximate expression only for the gain of the second stage of the "amplifier".

11. ### millwood Guest

you are correct on both fronts.

the gain formula will get you 90%+ there, through, and usually is the upper limit.

given that the 1st stage is a follower (gain =~1x), the formula above will give you the open loop gain of the amp, under the same assumptions.

the close-loop gain of the amp is about (R10+Zc5)/R4, assuming of course the open loop gain is sufficiently large.

12. ### millwood Guest

on 2nd inspection, the feedback (by r8/c5) is positive feedback, rather than negative feedback.

and I was wrong about the 1st being a follower earlier. its gain is R3/R4.

13. ### Audioguru New Member

Dec 20, 2007
9,411
896
No.
It is negative feedback.
If the feedback went to the base of the first transistor then it would be positive feedback.
But a positive swing to the base is the same as a negative swing to the emitter, so the feedback is the opposite.

14. ### millwood Guest

yeah, it is negative feedback.

15. ### howartthou Thread Starter Active Member

Apr 18, 2009
96
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hi millwood
Even if I think of c3 as a variable resistor (great way to teach, thanks), if it wasn't there wouldnt Q2 still go to ground (see Q3) ??????????

16. ### millwood Guest

Q2's emitter is grounded through a resistor (R8).

17. ### howartthou Thread Starter Active Member

Apr 18, 2009
96
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Ummm, yes, that was my point but thats doesn't explain the "purpose of C3" if this is answer:

"It passes a-c and blocks d-c, thus keeping the emitter Q2 at a-c ground"

How can C3 be keeping the emitter at a-c ground is the a-c is grounded through R8 anyway

18. ### millwood Guest

it goes back to what the question meant by "ground". to me, to ground something is to tie that thing to the ground -> with negligible resistance / impedance.

so I would not call the emitter being rounded in this design, dc or ac wise: for DC, the emitter is tied to ground via R8; for ac, via C3.

but there are people who do indeed "ground" a pin via a resistor / capacitor and I don't think the term "ground" is used consistently.

the question here appears to be in the 2nd camp.

19. ### howartthou Thread Starter Active Member

Apr 18, 2009
96
0
Sorry, I just cannot see how a-c is via c3 when the a-c also runs through R8

20. ### millwood Guest

if C3 is sufficiently large, its ac impedance will be much lower than the dc resistance of R8 - a typical design in this case.