# Common Base Amplifier calculation.

Discussion in 'Electronics Homework Help' started by majdi, Mar 5, 2014.

1. ### majdi

33
0
Jul 10, 2010

Given alpha = 0.99

1.How to calculate Re When Ieq = 1.1mA
Should it be KVL
Re = Vee/Ieq?

2.Find Rc when Vceq = -9
3.Find re when Ieq = 1.1mA and Vt = 26mV

i cannot found PNP example around my note and google. Hope someone can give guide on this question.

2. ### Harald KappModeratorModerator

11,196
2,559
Nov 17, 2011
Only if Ve (Emtiter voltage of the transistor =0V.

For the calculation of the DC operating point consider all capacitors as open circuit. You then have Vee between Re and the base of the transistor. Assuming a typical Vbe for a bipolar transistor, you find the voltage across Re and from the voltage across Re and current through Re you arrive at the value of Re.

Let us assume Vceq=-9V (not millivolts or something else).
From the voltage across Re, calculated in the previous step, and Vce you find V(rc). From V/rc) and Icq you can calculate Rc. The problem is: we do not know Icq. YOu may be allowed to make a reaonable assumption, namely Ibq<<Ieq. In that case Icq ~Ieq and you can calculate Rc.
Note that in reality Icq=Ieq-Ibq!

Which transistor model do you know? Insert the parameters from above calculations and solve for rbe.

Aggreed, many, if not most, examples use NPN transistors. But the calculations are exactly the same, only the polarities of voltages and currents are reversed.

3. ### majdi

33
0
Jul 10, 2010
1. Vre = Vee - Vbe
Re = Vre/Ieq

Vbe for Si BJT = 0.7v

Vre = 5v - 0.7v
= 4.3v

So Re = 4.3/1.1
= 3.91R ----> corrected 3.91K

right?

Im trying another 2 and 3..

Last edited: Mar 6, 2014
4. ### Harald KappModeratorModerator

11,196
2,559
Nov 17, 2011
Do you realize why I emphasized 9V?

Your calculation scheme is correct, but by omitting the relevant units you made a mistake on the order of 10^3. What is 4.3V/1.1mA?

5. ### majdi

33
0
Jul 10, 2010
I realize that, but my mistake miss look at it.

Q2.

Vcc = Vrc + Vcb
Vcb = Vceq - Vbe

Vcb = -9V - (-0.7V)
= -8.3V

Vrc = Vcc - Vcb
= -15V - (-8.3V)
= -6.7V

Icq = Alpha x Ieq
= 0.99 x 1.1mA
= 1.089

Rc = 6.7/1.089
= 6.152K

Im always confuse about the units

Last edited: Mar 6, 2014
6. ### Harald KappModeratorModerator

11,196
2,559
Nov 17, 2011
Looks good, only
- you don't tell where you get Alpha from and why you set it to 0.99
- you still write your equations without units.

Mathematically seen
is wrong. 6.7/1.089=6.152, not 6.152k

You've added the "k" because I've made you aware of your first mistake. But how do you ensure you will never forget to correct the numerical value for the decade and correct unit?
I strongly suggest you always keep the correct units throughout your full calculation. This will help you to get the correct numerical value (including the decade) and get you the correct unit (e.g. V/A=Ohm). This is also a good check for plausibility. If your result gives a wrong unit (e.g. V^2/A instead of V/A), you know that something is wrong.

7. ### majdi

33
0
Jul 10, 2010
**Alpha was given = 0.99 in question.

I will follow your advise to improve my calculation skills. Be honest i always have problem to decide either answer should be in K, Ohm or mA.

3. Find re when Ieq = 1.1mA , Vt = 26mV
Where re and Vt stand for, in the circuit it show Re with big Re, and Vt non in the circuit. Should i draw the equivalent simplified, low frequency hybrid-pi small signal model first?

Here my draw:

Last edited: Mar 7, 2014
8. ### Harald KappModeratorModerator

11,196
2,559
Nov 17, 2011
Sorry, an oversight on my part.

It will come out automatically if you use the units throughout your calculations.

3) If you don't have Vt in the equivalent circuit, how are you going to use it in the calculation of re?
My line of tjinking is more like: Decide which transistor model to use. You should have at least one in your training documents, that is probably the right one. From tasks 1 and 2 you have defined the DC operating point of the transistor. Insert the parameters (those that were given and those you calculated) into the model equation and solve for re.

9. ### majdi

33
0
Jul 10, 2010
3.) re = Vt / Ieq

re = 26mV / 1.1mA
= 23.63 Ohm

And the small signal current gain is :
Av=Vout/Vin or Av = Ic x RL / Ie x Re

10. ### majdi

33
0
Jul 10, 2010
Now i understand, re stand for internal resistance emitter, Vt for thermal voltage.

small signal current gain:

Ai = Icq / ieq
= 1.098mA / 1.1mA
= 0.998mA

Last edited: Mar 11, 2014
11. ### Harald KappModeratorModerator

11,196
2,559
Nov 17, 2011
Congratulations.
Even it may have been more difficult than you expected, I'm sure you've learned more by going this way yourself with only a little help from me

12. ### majdi

33
0
Jul 10, 2010
Thank you Harald Kapp, for me this is big help and i really appreciate that. And the unit thing im also have little improvement on how to decide the unit. ::THumbUp::