# Question: DC analysis of a transistor amplifier?

Discussion in 'Homework Help' started by samy555, Aug 16, 2015.

1. ### samy555 Thread Starter Active Member

May 24, 2010
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I could not complete the analysis of this circuit.

This is what I managed to do:

IB1 = (VCC - VBE1)/R3 = (3 - 0.7)/100K = 23uA

If β = 100, then IC1 = 2.3mA

VCE1 = VCC – IC1 * R1 = 3 – (2.3m* 2.2K) = - 2.06 V (Error)

The Voltage between points A & B (VAB) = 0.7 volt

So, IC1 = VAB/ R1 = 0.7/2.2K = 0.318 mA

Then, VCE1 = 3-0.7 =2.3 V

I stopped here. I wish I couldcalculate IC2 and VCE2 to complete the picture.

Thank you.

Aug 21, 2008
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3. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Notice that Ic1 = IR1 + IB2 and also you know that Ic1 = 2.3mA and IR1 = 0.318mA So IB2 is equal to ??

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4. ### DickCappels Moderator

Aug 21, 2008
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Picking up from your calculation of the current through R1, which you found of be 0.318 ma, the current through the base of Q2 is the difference between the current from the collector of Q1 and the current through R1, which is 2.3 ma - 0.138 ma = 1.98 ma.

If Q2 also has a beta of 100 then the collector could source up to 198 ma. You can only get 47 ma through a 64 ohm resistance with a 3 volt supply, so Q2 must be saturated. Check the collector-emitter saturation voltage curve on the datasheet in the 47 ma region to get your approximate VCE figure (as close as you can come because of variation from device-to-device) and you have everything you need to know in order to find IC2.

Last edited: Aug 16, 2015
5. ### samy555 Thread Starter Active Member

May 24, 2010
116
3
No.... IR1 = IC1 + IB2

Everyone uses LTspice, then come here and tell me the IC2= 47 mA
I want to see equations by which I can do the calculations
Thank you alot

6. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Are you kidding me ?? How can IR1 be the sum of a Ic1 and Ib2 ??
As for Ic2, because Q2 will be in saturation the Ic2 current is equal to :
Ic2 = (Vcc - Vce(sat))/R5 ≈ Vcc/R5 = 46.8mA

7. ### WBahn Moderator

Mar 31, 2012
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I still can't see the circuit that the TS is working with. Is the circuit posted by Jony130 in Post #3 the circuit?

8. ### Jony130 AAC Fanatic!

Feb 17, 2009
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It is very strange but if you quote TS first post you will see the circuit. And Yes I reposted the circuit.

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9. ### WBahn Moderator

Mar 31, 2012
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So say you!

I didn't see the circuit, but I did see that they were links to content on another forum. When I copied the link into a new browser window I got the registration page for that forum. So I'm guessing that I would have to be a member in order to see that content. Are you by chance a member?

Thanks!

Apr 5, 2008
15,806
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Hello,

The circuits of the TS are not shown as they are images from edaboard.
You will need to be logged in on edaboard to see them.
I do not have an account over there, so I can not see them.

Bertus

11. ### WBahn Moderator

Mar 31, 2012
18,093
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Where does LTSpice come into play?

You are making some assumptions about your transistors, namely that Vbe = 0.7 V and β = 100. You might as well make the usual additional assumption that Vcesat = 0.2 V.

You should also run your analysis with Vbe = 0.6 V and β = 300 and Vcesat = 0.25 V (or some combination of those values) to see if the circuit is overly sensitive to any of those parameters (most likely β).

12. ### Jony130 AAC Fanatic!

Feb 17, 2009
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So the mystery is solved, and by accident I'm also a member on edaboard. And TS post the same question on edaboard.

13. ### WBahn Moderator

Mar 31, 2012
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Okay so far, though you need to learn to track your units properly.

This assumes that all of Ic1 flows through R1. But there is an alternate path for it to come from, namely Ib2.

If Ib2 wasn't connected (so that Ic1 did have to flow through R1) what this would tell you is that the assumption that Q1 was in the linear region was incorrect and that, in fact, it would be in saturation.

Assuming Q2 is in the linear region.

Again, you are assuming that Ic1 is equal to the current in R1, which is simply not the case. The base current in Q2 HAS to go through Q1 as well.

Again, assuming that Q2 is in it's linear region.

Because of the clamping effect of Q2, it is reasonable to assume that Q1 is, indeed, in it's linear region. This isn't guaranteed, but it is likely the case.

But this means that only 0.3 mA of the 2.3 mA collector current is going through R1 and that the rest is Ib2. Note that if β=200, then Ib2 goes from 2 mA to 4.3 mA, meaning that the circuit is very sensitive to transistor beta.

If Ib2 = 2 mA, then if the circuit is in the active region with β = 100, then Ic2 would be 200 mA. But that would result in a voltage drop across R5 of 12.8 V. So you know that Q2 is not in the active region but, instead, in saturation. The saturation voltage is going to be in the ballpark of 0.2 V, so the collector current is going to be around 2.8V/64Ω which is 44 mA. Looking at Fig 14 of

This is a pretty reasonable estimate of Vcesat.

14. ### samy555 Thread Starter Active Member

May 24, 2010
116
3
Yes,you're rightI'm sorry

What did you mean by:" the clamping effect of Q2 "?

Thank you very much, explanation was very cool and I benefited a lot.

15. ### WBahn Moderator

Mar 31, 2012
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Simply that it will hold (i.e., clamp) the base-emitter junction voltage at about one diode drop. In heavy saturation that may well be significantly more than 0.7 V, but (provided it isn't so much as to destroy Q2) it will be much lower than the ~2.8 V needed to saturate Q1.

16. ### samy555 Thread Starter Active Member

May 24, 2010
116
3
Thank you very much

I'll re-design the circuit so that (Ic1) is very small (say 0.4 mA) and I'll replace Q1 fixed bias by voltage divider bias. After finishing the design I will present it here to receive the necessary corrections.

Thank you again

17. ### WBahn Moderator

Mar 31, 2012
18,093
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What is the goal of the design?

18. ### samy555 Thread Starter Active Member

May 24, 2010
116
3
The goal is I feelt hat connecting a PNP transistor as a direct coupling to a preceding stage have many advantages although some considered that the circuit was wrong or have problems. The important feature of the circuit is that it can handle light loads like speakers or antennas.

19. ### samy555 Thread Starter Active Member

May 24, 2010
116
3
Hi

First I redesigned the same configuration but I tried to adjust R1 (RB of Q1) untill the value shown (750K),,, In practice you may get another value. The secret lies in that this R1 to adjust to get low IC (0.4 mA or so) and half VCC for VCE when Q2 is not connected

Look:

Connecting Q2

VCE1 increased (as expected) from 1.56V to 2.2V.

If we apply 10mV peak signal to the input:

The o/p signal is amplified more than 90 times and was not distorted

Tomorrow I will offer a stronger and much better design.

Thank you

Last edited: Aug 17, 2015
20. ### JoeJester AAC Fanatic!

Apr 26, 2005
3,403
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Make the load closer to a speaker ... 8 ohms or 4 ohms.

I don't think I've ever heard "speakers" described as "light loads" before.