Question: DC analysis of a transistor amplifier?

Thread Starter

samy555

Joined May 24, 2010
116
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.
 

DickCappels

Joined Aug 21, 2008
6,185
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.
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.
 
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Thread Starter

samy555

Joined May 24, 2010
116
Notice that Ic1 = IR1 + IB2 and also you know that Ic1 = 2.3mA and IR1 = 0.318mA So IB2 is equal to ??
No.... IR1 = IC1 + IB2

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.
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
 

WBahn

Joined Mar 31, 2012
25,223
I still can't see the circuit that the TS is working with. Is the circuit posted by Jony130 in Post #3 the circuit?
 

WBahn

Joined Mar 31, 2012
25,223
It is very strange but if you quote TS first post you will see the circuit.
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?

And Yes I reposted the circuit.
Thanks!
 

bertus

Joined Apr 5, 2008
20,317
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.

@TS, please upload the pictures to THIS board.

Bertus
 

WBahn

Joined Mar 31, 2012
25,223
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
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).
 

Jony130

Joined Feb 17, 2009
5,145
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?
So the mystery is solved, and by accident I'm also a member on edaboard. And TS post the same question on edaboard.
http://www.edaboard.com/thread342703.html
 

WBahn

Joined Mar 31, 2012
25,223
This is what I managed to do:

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

If β = 100, then IC1 = 2.3mA
Okay so far, though you need to learn to track your units properly.

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

The Voltage between points A & B (VAB) = 0.7 volt
Assuming Q2 is in the linear region.

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

Then, VCE1 = 3-0.7 =2.3 V
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

http://www.onsemi.com/pub_link/Collateral/2N3906-D.PDF

This is a pretty reasonable estimate of Vcesat.
 

Thread Starter

samy555

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

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

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

http://www.onsemi.com/pub_link/Collateral/2N3906-D.PDF

This is a pretty reasonable estimate of Vcesat.
Thank you very much, explanation was very cool and I benefited a lot.
 

WBahn

Joined Mar 31, 2012
25,223
What did you mean by:" the clamping effect of Q2 "?
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.
 

Thread Starter

samy555

Joined May 24, 2010
116
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.
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
 

Thread Starter

samy555

Joined May 24, 2010
116
What is the goal of the design?
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.
 

Thread Starter

samy555

Joined May 24, 2010
116
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
 
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JoeJester

Joined Apr 26, 2005
4,184
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.
 
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