# Homework Help: BJT amplifier - calculate resistances

1. May 30, 2014

### etf

Hi!
Calculate RC1, RC2, RE2 for maximum undistorted symmetrical signal at output if DC collector currents are IC1=IC2=1mA. Calculate voltage gain Av.
Vcc=9V, RE1=1.8kΩ , VBE=0.7V, VCES=0.2V, VT=26mV, beta=100, VA→∞.
Solution: RE2=700Ω, RC1=5.724kΩ, RC2=3.141kΩ, Av=gm1*gm2*rpi2*RC1*RC2/(rpi2+RC1).

After assuming that BJT's operate in forward active mode and solving circuit bellow I got RE2=700Ω, RC1=5.782kΩ. Is it ok? How to find RC2? I have no problems calculating Av.

2. May 31, 2014

### Jony130

Re1 looks good
But you are slightly off the true Rc1 value.
What is the current that is flow Rc1 ? What is the voltage across Rc1?

As for the Rc2 maybe this will give you some hint
But for your circuit Vc is not equal to 4.5V. You must include DC emitter voltage and Vce(sat).
9V - (2.518 + 0.2) = 6.282V.
So to get maximum undistorted signal Vc2 must be equal to ??

P.S

What it is ??
https://www.physicsforums.com/attachment.php?attachmentid=70181&d=1401493487

Last edited: May 31, 2014
3. May 31, 2014

### etf

Thanks, I will take a look on post... My second scheme is scheme for DC analysis, I just used models for BJT's in forward active mode (http://people.senecac.on.ca/john.kawenka/EDV255/images/bjt05.png [Broken]) instead of usual BJT symbol.

Last edited by a moderator: May 6, 2017
4. May 31, 2014

### Jony130

Your DC model is incorrect. Try to draw it again.

5. May 31, 2014

### etf

Whats wrong with model?

6. May 31, 2014

### Jony130

The model is ok. The overall diagram is wrong. Why T1 base is short to ground?
Why Vcc is connect is not connected to ground ?

7. May 31, 2014

### etf

E1 should be connected to ground, instead of B1. I don't know why I connected B1 to ground :)

8. May 31, 2014

### Jony130

Do you managed to find Vc2 voltage ?

9. May 31, 2014

### etf

I can't figure it out :(

10. May 31, 2014

### Jony130

What is the voltage at Q2 emitter ??
And notice that voltage at collector can swing from Ve2 + Vce(sat) up to Vcc.
So to get maximum undistorted symmetrical signal at collector we must chose Vc at the middle of this swing.

11. May 31, 2014

### etf

VE2=VC1E1-VB2E2=2.518V so voltage at collector (VC2) can swing from 2.518V+0.2V=2.718V up to Vcc=9V (6.282V swing). Now I should find Vc and after that RC2?

12. May 31, 2014

### Jony130

Exactly, and you should pus Vc at the middle of the swing.

13. May 31, 2014

### etf

2*(vc2e2-0.2)=6.282 → vc2e2=3.341v.
Vc2e2+rc*ic2-rc1*ic1-vb2e2=0 → rc2=(rc1*ic1+vb2e2-vc2e2)/ic2
rc2=3141Ω
Thanks a lot!

14. May 31, 2014

### Jony130

The answer is good, but you over complicate this.
All you need to find Rc2 is to use a II Kirchhoff law for the output loop.

Vcc = VRc2 + Vce2 + Ve2

Ve2 = Ie2*Re1 + Ic2*Re2 = 1.01mA*1.8kΩ + 1mA*700Ω = 2.518V

The swing

Vcc - (Ve2 + Vce(sat)) = 9V - (2.518V + 0.2V) = (9V - 2.718V) = 6.282V

From this VRc2 = Vce2 = 6.282V/2 = 3.141V

And

Rc2 = VRc2/Ic2 = 3.141/1mA = 3.141kΩ

And Rc1 = (Vcc - (Ve2 + Vbe2))/( Ic1 + Ib2 ) = 5.782V/(1mA + 10μA) = 5.724kΩ

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15. Jun 1, 2014

### etf

I have another example.
Calculate resistances R1, R2, RC, RE for maximal symmetrical undistorted voltage on output, if total maximal collector current is Icmax=4mA. Vcc=10V, load resistance RL=10kΩ, β=100, VBE=0.7V, VCES=0.2V, VT=26mV, VA=100.

In DC analysis, I calculated VCE=VCC-(beta*RC/(RE*(beta+1))+1)*((VCC/R1-VBE/R1-VBE/R2)/(1/R1+1/R2+(1-beta/(beta+1))/RE)). For maximal swing Vce shoud be on middle of load line. I'm not sure I understand what total maximal collector current means
Here is solution:
RE=2*beta*(Vcc-VCES)/(Icmax*(1+beta))-(beta/(1+beta))*(RC*RL/(RC+RL)+RC)
R2=R1*(2*beta*VBE+(1+beta)*RE*Icmax)/(2*beta*(Vcc-VBE)-[R1+(1+beta)*RE]*Icmax)

Last edited: Jun 1, 2014
16. Jun 2, 2014

### Jony130

OMG, how do you manage to derive the formula for RE and Vce ??

17. Jun 3, 2014

### Jony130

I'm also not sure what total maximal collector current is.
I see two options.
First:
Ic_max = Icq - quiescent DC current or maybe Ic_tot = Icq + Ic_ac ?

And to find components values we can use ohms law.
For example assume Veq = 1V and Icq = 1mA and from there we have
Re = 1V/1.01mA = 990Ω = 1K.
Rc = 4.5V/1mA = 4.4KΩ ≈ 4.3KΩ and voltage at base Vb = Veq + Vbe.
So you design voltage divider to get Vb at output. And the current that is flow through divider should be much larger than the Ib

http://eelinux.ee.usm.maine.edu/cou....BJT Amps. for Undistorted VoltageSwing-X.pdf