Two Stage Common-Emitter BJT amplifier

Thread Starter

Petrucciowns

Joined Jun 14, 2009
62
I am having some problems finding certain values on the following circuit:

http://img268.imageshack.us/img268/5766/questionyyw.jpg
I started the circuit by finding the base voltage of both stages by using the equation: VB 1 or VB1= VCC x R2 / (R1 + R2)

I then found Ve for both stages by subtracting .7 from VB
I found IE of both stages by= (VB -.7)/ (RE + other resistor between emitter and ground)

I found the AC emitter resistance of both stages by using the following equation: 25 mv/IE


After this point I had some difficulties. I realized the the equations I was able to use with a single stage common emitter BJT were not working such as:


ZIN BASE = hfe x r’e --- which I used for single stage was no longer working, but I don't think the value of hfe is given. What would Q1/hfe mean?

ZIN = R1 II R2 II ZIN (Base) -- I used this for single stage as well and no longer seems to work for two stage.

There are several voltage gain values needed, can anyone explain the difference between them, what they are for, and how to calulate them. For single stage I would use the equation: AV= rC/ r’e

Also how is the voltage across Rload found?



So as a summary I am unable to find the following values:
Zin (base)
rc1
Zin
Av (all of them)
VRload


I would really appreciate any help. You guys have been been great so far.:)

 

t_n_k

Joined Mar 6, 2009
5,455
I think the two notations Q1/hfe will refer to the beta values of each transistor.

It should probably have read Q1/hfe=300 (as shown) and Q2/hfe=200 for the right hand transistor - looks like a typo error. The use of lowercase (small signal values) may be inconsistent with the use of the values for DC conditions, but in the absence of other data you would have to use these to calculate the DC bias conditions.
 

t_n_k

Joined Mar 6, 2009
5,455
Here's how I'd find Zin (base1)

1. Find the Q1 emitter current (I calculated 960uA)
2. Find the total Q1 effective emitter resistance Re = 68Ω + 26/Ie1(mA)
3. Find the base input Zin = (hfe*Re) // 25 k // 75k

where // means "in parallel with"

I got a slightly higher value for Zin [~11.3kΩ]

The 26/Ie(mA) term is a "rule of thumb" I use to find the the equivalent emitter resistance - which I think is your r'e1 value - and this gives me 27Ω.

But it's all within the reasonable range of variation. The value of Vbe you choose is also a factor 0.7V is fine. I'd normally use 0.65V - but it's all academic really.
 

t_n_k

Joined Mar 6, 2009
5,455
Answer 12) is given as rc1=1k.

This should be approx 2.87k, which fits with the stage gain being approx 30.5 as as shown in the diagram....

I found Av1=2870/(27+68)=30.2
 

Thread Starter

Petrucciowns

Joined Jun 14, 2009
62
Answer 12) is given as rc1=1k.

This should be approx 2.87k, which fits with the stage gain being approx 30.5 as as shown in the diagram....

I found Av1=2870/(27+68)=30.2

I had a feeling my rc1 was wrong. I have the equation Rc // Zin2 and I get the same answer. So for rc2 would it be Rc2 // Zin1?


Also what would the equation for Av1 be. Where are you getting the values from? Do you know the difference between the different Av's?
 
This is where I am having the problem how do you find hfe? What does does Q1 and Q2 refer to?
Q1 refers to the first transistor and Q2 refers to the second.

Just to the right of the first transistor you see "Q1/hfe=300"; that's how we know that the hfe of the first transistor is 300.

Just to the right of the second transistor you see "Q1/hfe=200"; obviously they meant to say "Q2/hfe=200".
 
I had a feeling my rc1 was wrong. I have the equation Rc // Zin2 and I get the same answer. So for rc2 would it be Rc2 // Zin1?
rc2 would be Rc2||Rload = 1k


Also what would the equation for Av1 be. Where are you getting the values from? Do you know the difference between the different Av's?
Av1 = rc1/(re1+re11) = 30.5

Av2 = rc2/(re2+re12) = 14.8

And, of course, Av(system) = Av1*Av2 = 451.4
 

Thread Starter

Petrucciowns

Joined Jun 14, 2009
62
rc2 would be Rc2||Rload = 1k




Av1 = rc1/(re1+re11) = 30.5

Av2 = rc2/(re2+re12) = 14.8

And, of course, Av(system) = Av1*Av2 = 451.4

Oh I see Av(system) is a combination of the gain from both stages.

Using the equation for the individual gain you have above I don't seem to be getting the correct answer.


I have 2.86k/(1.8k+68) =1.53

Or did you mean the collector resistor? Either way I don't get a correct answer.
 

t_n_k

Joined Mar 6, 2009
5,455
Electrician's equations are correct - you've just plugged in the wrong values.

rc1=2.86 kohm
re1=25.9 ohm
re11=68 ohm

rc2=1 kohm
re2=11.52 ohm
re12=56 ohm

Voila!
 

Thread Starter

Petrucciowns

Joined Jun 14, 2009
62
Q1 refers to the first transistor and Q2 refers to the second.

Just to the right of the first transistor you see "Q1/hfe=300"; that's how we know that the hfe of the first transistor is 300.

Just to the right of the second transistor you see "Q1/hfe=200"; obviously they meant to say "Q2/hfe=200".

Ok so if that is hfe shouldn't these equations work for Zin and Zin(base)

ZIN BASE = hfe x r’e -

ZIN = R1 II R2 II ZIN (Base)


How come they dont? They work with single stage
 

t_n_k

Joined Mar 6, 2009
5,455
Ok so if that is hfe shouldn't these equations work for Zin and Zin(base)

ZIN BASE = hfe x r’e -

ZIN = R1 II R2 II ZIN (Base)


How come they dont? They work with single stage
They do - you just need to know which resistances are included in the analysis.

Zin base = hfe x (re total)

re total = "intrinsic" emitter resistance [25/Ie(mA)] + external unbypassed (mid-band AC) emitter resistance

Reisitors Re1 in both stages are not included as they are considered fully bypassed by their parallel caps at mid-band frequency. Their sole function is for operating point biasing. Of course they play a part in gain at very low frequencies as the effectiveness of the capacitor bypass decreases.
 
Ok so if that is hfe shouldn't these equations work for Zin and Zin(base)

ZIN BASE = hfe x r’e -

ZIN = R1 II R2 II ZIN (Base)


How come they dont? They work with single stage
Zin(base) = (1+hfe) * (r'e1+re11)

(Use a factor of (1+hfe) instead of hfe for ultimate accuracy, but for cases like this where hfe is large, it won't make much difference. If hfe were really low, it could be important.)

then you will have:

ZIN = R1 II R2 II ZIN (Base)
 

Thread Starter

Petrucciowns

Joined Jun 14, 2009
62
Electrician's equations are correct - you've just plugged in the wrong values.

rc1=2.86 kohm
re1=25.9 ohm
re11=68 ohm

rc2=1 kohm
re2=11.52 ohm
re12=56 ohm

Voila!
Ah I see. I keep mixing up Re and the emitter resistance, because they look so similar.
 

Thread Starter

Petrucciowns

Joined Jun 14, 2009
62
Zin(base) = (1+hfe) * (r'e1+re11)

(Use a factor of (1+hfe) instead of hfe for ultimate accuracy, but for cases like this where hfe is large, it won't make much difference. If hfe were really low, it could be important.)

then you will have:

ZIN = R1 II R2 II ZIN (Base)
Cool, but when I use Zin(base) = (1+hfe) * (r'e2+re12) for the second stage it's a bit off by about .5 Is the equation different for the second stage?

When I use the ZIN equation: ZIN = R1 II R2 II ZIN (Base) the answer is off as well.

How is VRload calculated. What is the current used?

 

t_n_k

Joined Mar 6, 2009
5,455
Cool, but when I use Zin(base) = (1+hfe) * (r'e2+re12) for the second stage it's a bit off by about .5 Is the equation different for the second stage?

When I use the ZIN equation: ZIN = R1 II R2 II ZIN (Base) the answer is off as well.

How is VRload calculated. What is the current used?

The same applies to both stages.

Small variations in calculated values are acceptable - depends on what assumptions are made.

As Electrician pointed out some assumptions are OK depending on the circumstances- e.g. I used hfe instead of (1+hfe) as the multiplier, where the latter is strictly correct.

Other values in the analysis you posted will differ from one person's attempt to another.

For instance, Vb for the the first stage is given as 2.5V when it's closer to 2.44V .... and so on. In the final washup it probably isn't that significant. If you built & tested the circuit you would probably get just as much variation from one case to another.
 
Top