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
    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,

  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
    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
    the IC is about 2mA(1.98675mA).

    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
    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:


    That gave me:

    RC = 3.75K

    I plugged RC into the gain equation and got:

    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:

    VE=VB+VBE=.0569V (I used .71231557V for VBE, measured)
    This made my VCE=7.443V

    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
  5. Jony130

    AAC Fanatic!

    Feb 17, 2009
    Well I make very simply mistake in the gain equation. But I corrected the error already.
    And I check your answer tomorrow.
  6. atferrari

    AAC Fanatic!

    Jan 6, 2004

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


    Dec 20, 2007
    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

    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.
  9. hobbyist

    AAC Fanatic!

    Aug 10, 2008
    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
    Thanks alot sir I really appreciate it.