# BJT common collector amplifier design.

Discussion in 'Homework Help' started by Max Kreeger, Oct 1, 2013.

1. ### Max Kreeger Thread Starter Member

Oct 1, 2013
94
0
Hey guys,

So I'm required to design a BJT common collector amplifier. We were told to use the circuit from the textbook (picture provided below). It has to have a current gain of 20.

Beta = 100
ic= 10mA
Vce= 1.0V

What I've done so far is used all the formulas on the other page (pic below) and pretty much worked backwards playing around with the numbers until I got a current gain of 20 (Ai).. I ended up with these values:

R1=R2= 80k
Rb=40k
R'L= 0.39
RL = 1k
Re=0.4
rπ= 0.65
Av= 0.999
Zit= 40k
Zi= 20k
Ai= 19.98
G= 19.96

But when I try actually apply it all...nothing checks out. I'm probably approaching it all wrong.

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Last edited: Oct 1, 2013
2. ### Efron Active Member

Oct 10, 2010
81
15

If Vce (operating point) is equal to 1.0V then VE = 19V and VB is almost 20V!!

In this situation the input signal will have distortion (out of linear range).

If Vcc is 20V, Vce should be 10V. With that, you can go on and calculate VE and RE because you know IC.

In your exercise, can you choose RL and Rs or have they to be those indicated in the picture?

3. ### Max Kreeger Thread Starter Member

Oct 1, 2013
94
0
We can choose any value for everything, although I just left them as they were to see what would happen. As for Vcc I just stuck with the 20V that was shown but again. That can be anything. I did manage to get a common emitter working, I used 9V for that one.

4. ### Efron Active Member

Oct 10, 2010
81
15
Let's stick with Vcc = 20V and RL = 1K.

If Ai = 20, then Zi = 20K because RL = 1K.

Now rπ = beta*25/Icq and you know Icq = 10mA. >> rπ = 250 ohm.

In this kind of amplifiers, R1 = R2, then Vbq = 10V >> Veq = 9.3V >> now you have RE.

With all these known values, you should be able to calculate Rb from the definition of
Zi = Rb // ( rπ + (1 + beta)*RL' ).

However, at the end, Vce should be about half Vcc ~10V for the amplifier to work in its linear region.

Last edited: Oct 3, 2013
5. ### Max Kreeger Thread Starter Member

Oct 1, 2013
94
0
Thank you, very much! I'll give that a go and report back.

EDIT:

Ok so for Re I used the formula: Re= VT/Ie = rπ/(1+Beta) = 2.475
R'L= 1/((1/1000)+(1/2.475)) = 2.469

So then Zi = Rb // ( rπ + (1 + beta)*RL' )
>>> 20K= 1/((1/RB)+(1/866.268))
>>> RB ~ 20K

Is that right?

Last edited: Oct 3, 2013
6. ### Efron Active Member

Oct 10, 2010
81
15
Nop!

RE is the resistance at the emmitter = Vemitter / Ie = 9.3V / 10mA ~1K (rough calculation).

Note: Ie = Ic = 10mA in your spec.

So RL' = RE // RL = 0.5K.

So, what should be the value for Rb?

7. ### Max Kreeger Thread Starter Member

Oct 1, 2013
94
0
I get the same result :\

( rπ + (1 + beta)*RL' ) >> 250+(1+100)*500 = 50750

Zi = Rb // ( rπ + (1 + beta)*RL' )
>>> 20K= 1/((1/RB)+(1/50750))
>>> RB ~ 20K

EDIT:

SORRY..I see what I did wrong.

Rb ~ 33K ?

Last edited: Oct 3, 2013
8. ### Efron Active Member

Oct 10, 2010
81
15
Bingo; same as mine.

9. ### Max Kreeger Thread Starter Member

Oct 1, 2013
94
0
Now how would I go about working out the capacitor values?
Would R1 and R2 = 66K ?

Last edited: Oct 4, 2013
10. ### Efron Active Member

Oct 10, 2010
81
15
R1 = R2 = 66K (theoretical value because you won't find a 66K in the shop - 68K will be the more practical choice).

For the capacitor, it is easy to understand;

The capacitor is seen as an impedance by an AC signal. This impedance depends on the C value and on the frequency (f) of the AC signal and it's given by 1/(2*pi*f*C).

For a practical amplifier, the value of C must be such that the resistance seen by the AC signal is almost 0, otherwise it will introduce equivalent resistor to the formulas you just used before.

So you will choose the highest value of C that you can upon the needs, but not too much either because it will decrease the global gain at lower/higher frequencies.

We don't know the frequency range required by your circuit, but if you remain at KHz, usual value for these capacitors is 100uF.

11. ### Max Kreeger Thread Starter Member

Oct 1, 2013
94
0
I had a bit of a play around with numbers. This is what I got:
When I used 100uF I got: 1.59

You said "almost zero", so when I used 1.7mF I got
1/ (2*π*1000*(1.7*10^-3) = 0.09 - would that decrease the gain?

Thanks again for your time and help man. Very much appreciated.

12. ### Efron Active Member

Oct 10, 2010
81
15
You will not easily get 1.7mF component in real life (too big).

100uF is good enough, 470uF is even better.

You have to consider the two extremes, at very low frequency (almost DC) and at very high frequency.

For example, 100uF is suitable for 1KHz or above.

If you're using a 1HZ AC signal, the capacitance of C will be 159 ohms and your gain will decrease because it will increase Zi.

On the other side, if you use a 1MHz AC frequency, the parasite capacitors between base and emitter of the BJT will almost become shortcut when compared to rπ (equivalent resistance between base and emitter at Ieq) and the gain will be decreased to 0 --> only the current going through rπ will be amplified.

For the external capacitors in your circuit, the lowest the value of impedance for C, the less intrusive the capacitor and therefore the less impact on gain.

13. ### Max Kreeger Thread Starter Member

Oct 1, 2013
94
0
Right, I understand now. I'll hopefully get lab access at university sometime next week and be able to test these values.

Once again thank you very much

Max

14. ### Max Kreeger Thread Starter Member

Oct 1, 2013
94
0
Ok so I soldered the circuit..I used 470uF electrolytic capacitors. Since the largest green/red cap I could find was a 2uF...C1 was placed "+ -" and C2 was placed "- +" when I tested it though..input and output signals were coming out the same with some slight distortion. We were told to measure current gain by dividing the output voltage by output resistance..which confuses the hell out of me. Could the problem be the polarized caps? and would a 2uF cap work as well? Considering I'd have to change the freq to 1Ghz which the equipment can't produce..

Last edited: Oct 16, 2013
15. ### Efron Active Member

Oct 10, 2010
81
15

With this type of capacitors, it is very important to respect the polarity of the leads, otherwise their behavior becomes unpredictable.

At every instant, the TOTAL voltage on positive lead MUST BE GREATER than the voltage in the negative lead. TOTAL means DC + AC.

Take a look at your output pin (the emitter). Whatever the voltage value at the load, the emitter will always have a voltage value greater than the negative lead of the capacitor, so the correct placement is "+ -".

At the base side, whatever the voltage of the input source (assumed small with respect to VB in DC), the TOTAL voltage of the base will be greater than the input source. The correct placement is "- +".

One simple way to look at it is to make the analysis in DC (assumed Vin is small enough).

* In DC (Vin=0V), the leads of the input capacitor will have, from left to right, 0V - VB >> correct placement is "- +".

* In DC (Vin=0V), the leads of the output capacitor will have, from left to right, VE - 0V >> correct placement is "+ -".

------------------------------------------------
This is normal between Vout and Vin because the voltage gain is almost 1 (typical behavior for common collector amplifiers).

So the question is what is the relationship between Iout and Iin?

16. ### Max Kreeger Thread Starter Member

Oct 1, 2013
94
0
Ok, I'll switch them around and give it another test tomorrow but I'm still confused as to how we test it. All our lecturer has told us is what I just told you previously "dividing the output voltage by output resistance". Hopefully when I change the caps around I'll see a difference.

Just out of curiosity, if you don't mind me asking. Are you an Elec engineer? Your knowledge is impressive.

Thanks again.

17. ### LvW Active Member

Jun 13, 2013
674
100
Hi Max Kreeger, I think, at least regarding this problem I can help you.
I suppose you are mixing output and input resistance with each other.
The current gain of the amplifier stage (not for the BJT alone) is

Vi=Iout/Iin.

we get

18. ### Max Kreeger Thread Starter Member

Oct 1, 2013
94
0
The only bit that confuses me is the Rin, so input resistance. On the equipment it says there will be a resistance of 50 Ohms. Is that what you're referring to?

19. ### LvW Active Member

Jun 13, 2013
674
100
No - the resistance of app. 50 ohms is the input resistance measured into the emitter node. In an amplifier circuit, this resistance appears as the input resistance of a common base stage and/or as the output resistance of a common collector stage.
However, this value is not to be used for current gain calculation (as above) because the output signal is produced by the product emitter current times load resistor Rload only.
But this low output resistance allows application of the common collctor stage as a buffer (unity gain is nearly unaffected by changing external loads.)

Oct 1, 2013
94
0