# First Post: BJT CE Amp Design resistor in series with bypass capacitor?

Discussion in 'Homework Help' started by mwarre25, Apr 24, 2011.

1. ### mwarre25 Thread Starter New Member

Mar 22, 2011
12
0
Hey guys,

This is my first post. I am wrapping up a BJT CE amplifier project. Here are the parameters:

DC Power Consumption <= 30mW
Overall Gain (open circuit gain= no load resistor, right?) = 80 V/V
Less than 10% change in Icq if Beta varies between 100 to 200
Overall Swing of 8Vp-p with rails (GND to 15V)

*I'm using a 2N3904

I pretty much have all of the parameters met(let me know if I don't plz) but what I'm confused about is why I had to but a small resistor in series with my bypass capacitor.

I came to the conclusion that I needed this from students previously taking the class. I was getting serious clipping without it. But I'm frustrated that I can't calculate it.

What I'm asking for is just a basic description of why this works and if there is a way I can calculate this.

Thanks for this forum,

Mike

File size:
83 KB
Views:
59
2. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,481
1,264
This resistor set the voltage gain to be equal 80V/V
Av ≈ Rc/ ( re + RE||R3)
re = 26mV/Ic ≈ 17Ω ( I assume that your collector current is equal 1.5mA)

RE||R3 = 5KΩ/80 - 17Ω = 45Ω

R3 = (45*100)/( 100 - 45) = 81Ω

Last edited: Apr 24, 2011
3. ### mwarre25 Thread Starter New Member

Mar 22, 2011
12
0

I think I was confused about what I have as RE(100 ohms)

So I know that for AC that resistor is bypassed so its only really used for bias right?

What I tried to do was use Av=-RC/RE with my known value of -80 for Av and backtrack.

In my initial calculations I set IE to 2ma and I set a VE( to about .15V).

That resulted in an RE of 75ohms. I then got the resulting values for RC(5k) and did all the calculations necessary for VB, VC, VCE, IB, IC etc.

But then when I tested I was clipping so I changed RE and R3 to get my gain right and make sure I wasn't clipping.

So to get a better calculation should I just change my circuit to a CE with an emitter resistor with out the bypass cap?

4. ### mwarre25 Thread Starter New Member

Mar 22, 2011
12
0
Thanks Jony130 for the help.

I want to contribute to the forum by contributing my mistake haha.

I achieved a better circuit than my first circuit above using the "common-emitter with emitter resistor" design. The first mistake I made was a foolish one. When I tested the circuit for proper gain, waveform, etc, I had a bypass capacitor in parallel with the emitter resistor.

This shorts the emitter resistor pretty much and the gain is closer to AV=-RC/re than AV=-RC/re+RE. Thats why I had to add the resistor in series with the bypass capacitor because it added to re to about 62 ohms (*see Jony130s calcs) to get my -80 gain.

So for all that trouble I realized that the bypass cap was not even needed. Still using Av=-RC/re+RE I just biased the bjt by setting:

VCC=15V
VC=7.5V
IC=2mA

That gave me:

RC = 3.75K
IE=2.013mA
IB=13.333uA

I plugged RC into the gain equation and got:
(RE+re)=46.875Ω

Knowing Vt=.025V(I used .03739057V, long story) and IE:

re=Vt/IE = 18.57Ω

then RE=(RE+re)-re=28.3035233Ω

thus satisfying my gain.

then the only things left were VE, VB

One of the "rules of thumb" I read was to make VCE as close to (1/3)VCC as possible so I messed around with different values of VB(using a spreadsheet) and came up with:

VB=0.7693V
VE=VB+VBE=.0569V (I used .71231557V for VBE, measured)

So with this, with the coupling caps, I got a pretty close gain (-79.138)
(100mvp-p input)

I then chose some close standard resistor values because I need to make this. These in turn changed my calculations but I havent calculated that stuff yet.

Hope this helps somebody

File size:
59.5 KB
Views:
39
5. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,481
1,264
Well I make very simply mistake in the gain equation. But I corrected the error already.

6. ### atferrari AAC Fanatic!

Jan 6, 2004
3,018
1,122
Good!

Please, could you move the instrument that is hiding the emitter?

7. ### Audioguru Expert

Dec 20, 2007
10,512
1,168
Your new circuit with a very low value emitter resistor has poor thermal stability:
If the transistor is warm then its base-emitter voltage becomes less so it will saturate.
If the transistor becomes cool then the base-emitter voltage will increase and it will be cutoff.

The circuit will not work with many transistors:
If the current gain is high then the transistor will be saturated.
If the current gain is low then it will be cutoff.
Also, different transistors have different base-emitter voltages.

Use a DC emitter resistor that is 1/10th to 1/20th the collector resistor value so it swamps the effects listed above. Then use another resistor in series with a capacitor to ground at the emitter for more AC voltage gain.

8. ### mwarre25 Thread Starter New Member

Mar 22, 2011
12
0
Hey,

Sorry atferrari. The only thing behind the instrument is ground. I have attached my final report and if you look at figure 8 you can see the entire circuit. This is my first IEEE report so its a little rough. Don't bash me too hard guys please haha

Thanks for all of the help though everyone I appreciate it.

Thanks audioguru for the info. So the configuration I started with is common then? The thing that swayed me at first was that i was having a dumb moment trying to get the gain equation right with my values but now I know what I was doing wrong.

File size:
2.8 MB
Views:
36
9. ### hobbyist AAC Fanatic!

Aug 10, 2008
884
85
I didn't read through your whole report, just skimmed over it, so I don't know if the circuits are correct or not, but that is not why I'm posting this.

The reason I am posting to this thread is to say,

EXCELLENT workmanship on your paper, you did a VERY NICE report on this assignment.

Keep up the GREAT work...
Excellent paper work.....

10. ### mwarre25 Thread Starter New Member

Mar 22, 2011
12
0
Thanks alot sir I really appreciate it.