# simple Common emitter design

Discussion in 'Homework Help' started by simpsonss, Oct 8, 2010.

1. ### simpsonss Thread Starter Active Member

Jul 8, 2008
173
0
hi,

Below is my Common emitter amplifier. I used PSPICE to generate the output signal but it seems like it is wrong in phase shift. Because what i read from book is common emitter will have a 180 out of phase output. But my result is a 90 degree out of phase. Can anyone point out what's wrong with my circuit?

thank you.

2. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,294
482
Take out capacitor C1 and resistor R5.

3. ### Wendy Moderator

Mar 24, 2008
20,735
2,498
Check the phase between base and collector. That is where it will be 180°. You are also not counting the RC phase shifts, you are measuring after the capacitor C1. You are also measuring before C2, another RC phase shift.

He needs C1 to maintain his bias on the transistor.

4. ### Ghar Active Member

Mar 8, 2010
655
72
The decoupling caps add a high-pass characteristic.

You have them set such that your 1 kHz isn't well into the pass band yet, giving extra phase shift.

Try plotting the frequency response (AC sweep).

The poles are approximately;
Output cap -
$f_1 = \frac{1}{2\pi (R_3 + R_5) C_1} \simeq 4 Hz$

Input cap -
$f_2 = \frac{1}{2\pi (R_1|| R_2) C_2} \simeq 290 Hz$

Emitter bypass cap-
$f_5 = \frac{1}{2\pi (r_e + \frac{R_1||R_2}{1 + \beta}) C_5} \simeq 1.5 kHz$

Your signal needs to be well above those frequencies to not be phase shifted or attenuated...

5. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,294
482
So just move probe that looks at output to point Q1?

6. ### simpsonss Thread Starter Active Member

Jul 8, 2008
173
0
hi thanks for the replies,

i'm new in PSPICE. Can someone advice me on how to make the plot graph in a more readable view? i mean the setting for axis.

7. ### JoeJester AAC Fanatic!

Apr 26, 2005
3,311
1,108
Is there a way to "seperate" the outputs to two graphs in the same window?

I'm not familiar with PSpice, but it is an option in the software I use.

8. ### Ghar Active Member

Mar 8, 2010
655
72
Yeah, you can "add Y-axis" or you can "add plot to window"

9. ### simpsonss Thread Starter Active Member

Jul 8, 2008
173
0
ok.thanks for the guidance to add another plot. and i get this. can someone teach me how to read the Y-axis? what is the Vpp for both sine wave?

10. ### Georacer Moderator

Nov 25, 2009
5,142
1,266
See the little 2, on the left of the top of the rightmost axis? That propably denotes that this axis refers to signal number 2, which as we read on the bottom is the collector voltage. That leaves the first axis to read the input values.

11. ### hgmjr Moderator

Jan 28, 2005
9,030
214
I believe you need to revisit the selection of the two base bias resistors. They seem to be at least an order of magnitude smaller than needed.

hgmjr

12. ### hobbyist Distinguished Member

Aug 10, 2008
764
56
Your questions have been answered for the phase shift,

But as an add on, for future reference
if you want to learn proper transistor stage biasing, then
For a CE amp. it is common practice to put the voltage at the collector terminal. at around 1/2 the supply voltage. (somewhere close to it)

That is proper biasing, assuming no leakage currents into the base divider, you have
6.54V. at the base and subtracting the Vbe, your emitter voltage sits at 5.84V.
Your emitter current is around 584uA. which puts a voltage drop across R3 of 584mV.
which sets your collector voltage at around 11.4V.

This will give you very little amplification without distortion, also it limits the input voltage before introducing distortion.

I hope this helps in other ways in your learning this, even though it is irrelevent to your original question.
These are things youll learn as you go further in this work.

13. ### simpsonss Thread Starter Active Member

Jul 8, 2008
173
0
As i refer to the reference book i should measure the input before the capacitor and output after the output's capacitor. But Bill mention i need to look at the base and collector point to have a 180 phase shift. so which way should i go for?

thanks.

14. ### simpsonss Thread Starter Active Member

Jul 8, 2008
173
0
hi,

ok now i come out with all the resistors value in this way.The input source is 60mVp-p 1Khz sine wave. And below is the two plotting graph that i get for base VS collector and source VS Vout. For base VS collector i can see the signal is 180 out of phase but no for source VS Vout. so is it i'm suppose to get the from the base VS collector? Now it makes me confuse of which point is the correct point to look at so that it is the real output signal. From the reference book they always label the signal at signal source and Vout. Which way suppose it be?

15. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,294
482
I think Base vs. Collector is the right one.

I think that when you put the capacitors between the points where you take measurements, you introduced additional phase shift.

In the Base vs. Collector you are a measuring voltage from base to ground, and from collector to ground. You do not have any additional elements to influence your measurements.

Last edited: Oct 11, 2010
16. ### Ghar Active Member

Mar 8, 2010
655
72
You're interested in signal source vs Vout because that is the actual amplifier input and out. And again, your problem is your frequency response.
You made the situation worse by changing R4 and R5, you need to increase the size of your capacitors or resistors if you want to have 1 kHz in the passband of the amplifier.

I showed you some approximated equations for the pole frequencies in my earlier post. You need all of those to be well below 1 kHz.

17. ### simpsonss Thread Starter Active Member

Jul 8, 2008
173
0
ok. Now i get what u mean. After the calculations my circuit is as below.

And then i get the input signal Vs Output signal with input= 1Khz 40mVp-p

And then input = 2Khz 40mVp-p

It seems like there is still delay on the output signal. And another problem is if i increase the input signal Vp-p slightly higher which is around 50mVp-p i'll get a attenuated signal at the output. Any idea about this?

thanks.

18. ### simpsonss Thread Starter Active Member

Jul 8, 2008
173
0
And then after trying an error on all the resistors and input signal and capcitor value i get this. And it look more like a 180 out of phase common emitter amplifier. Am i right?

But it still have problem in attenuation when i increase the Vp-p to 100mV. i want the input signal to be higher is because that my function gen is not able to tune to as low as 40 or 60mVp-p. Any idea?

thanks.

19. ### Georacer Moderator

Nov 25, 2009
5,142
1,266
OK, here is a brief analysis of your circuit. It shows the relation of the Voltage gain with the circuit components.

Operation Point Calculation:
$I_B=\frac{E_{th}-V_{BE}}{R_{th}+(\beta +1) \cdot R_E} \approx 5.73uA\\
\Leftrightarrow I_C\approx0.001mA\ taking\ \beta\ \text{equal to 200, as typicaly mentioned in the datasheet}\\
\text{Using the hybrid pi model we have }g_m=\frac{I_C}{V_T}\approx 0.0385\\
and\ A_v=-(R_L||R_C) \cdot g_m \approx 45.05$

which is a good description of what is going on now in the circuit.

In order to increase the voltage gain you need to increase $R_C$ and decrease $R_E$ considerably. Try it out (since I didn't ).

If someone else thinks I 'm wrong please say so. This is uni knowledge speaking, not exprerience.

20. ### Ghar Active Member

Mar 8, 2010
655
72
Are you sure the gain is actually decreasing or are you just distorting your waveform?

Check out the positive vs negative peaks... you have your collector biased at around 10 V.
Your output signal can never go higher than 12 - 10 = 2V.
With 2V and a gain of 45 (using Georacer's number) your input signal is limited to < 44mV... it's actually much less because the signal starts distorting well before that. All your plots show significant distortion on the positive peaks (flattened)