Common Emitter Amplifier

Discussion in 'General Electronics Chat' started by Mashly, Apr 8, 2014.

  1. Mashly

    Thread Starter Member

    Apr 4, 2014
    33
    1
    Hi Everyone, sorry if this question has been asked lots of times but I am at a loss of what to do next.

    I have designed a common emitter amplifier using the turtorial on this website found here,

    http://www.learnabout-electronics.org/Amplifiers/amplifiers20.php

    and The Art of Electronics book and cannot see why it is not working.

    I am using a NPN BC108 transistor, but do not seem to be able to get any amplification.

    I am supplying the circuit with 9V from a bench power supply (I have also tried a battery as this is what I want to use eventually).

    I chose Iq to be 0.03A, calculated Rl to be 300 Ohms, Re to be 36 Ohms, R1 and R2 to be 4.8 kOhms and 1.2 kOhms respectively.

    On the input I am putting 3V at 1KHz in and all I get out is either a saw tooth wave with a bounce where it should go negative. Is this because I need to supply it with positive and negative 9V to get swing both ways?

    If I shift the input wave up I just get the same out I put in.

    If you need anymore info just let me know and I will post it if I know it.

    Thanks
     
  2. MrChips

    Moderator

    Oct 2, 2009
    12,421
    3,357
    I tested your circuit with your values on a breadboard and it works fine.
    Reduce your input signal to 100mV peak-to-peak.

    What is your value of C1?
     
  3. Rolf Zetterberg

    Member

    Sep 20, 2008
    14
    2
    As far as I can see,it cannot work.
    0.03 amps through the 300 ohms collector resistor drops 9V.Nothing left for the transistor.
    Mashly,why not use the old-fashioned method and measure the voltages?
     
  4. PRS

    Well-Known Member

    Aug 24, 2008
    989
    35
    Zetterberg is right. Your biasing is bad. Nor can you put a 3 volt signal into the input of a common emitter such as this without serious harmonic distortion. What is it you want to do? Maybe we can help you design something that works.
     
  5. Mashly

    Thread Starter Member

    Apr 4, 2014
    33
    1
    Hi there everybody, thanks for your help with this.

    I have reduced the signal to 100mv and still nothing. I see what your saying Rolf I get a very small voltage on the collector when I measure it (in the area of mV). How would I design the biasing for this transistor as I am mis-understanding something obviously.

    PRS I am trying to build a guitar pedal with diode clipping eventually, but I thought I would walk before I can run, so I started with this amplifier. Is the reason I cannot put 3 volts in because of the base emitter voltage saturation?

    Thanks again for your help

    Mashly
     
  6. Veracohr

    Well-Known Member

    Jan 3, 2011
    550
    75
    Say we go with the 30mA you've set up. You ideally want the collector voltage to be centered on the midpoint of the supply voltage, so that you can get maximum signal swing. So assuming you want the collector at 4.5V, you need to drop 4.5V across the collector resistor, so calculate it: Rc = 4.5/.03 = 150Ω.
     
  7. MrChips

    Moderator

    Oct 2, 2009
    12,421
    3,357
    You must be doing something wrong. I tested your circuit on a simulator and in real life and both work.

    The bias voltage is a bit high.

    R1 = 7.5kΩ works better.
     
  8. PRS

    Well-Known Member

    Aug 24, 2008
    989
    35
    I once designed an audio amplifier and noticed that when I turned the supply voltage down the output was clipped. I played my guitar through it on a hunch and sure enough it sounded like a fuzz box with the intensity of the fuzz directly proportional to the amount of clipping. So I think I know what you're talking about.

    What you are building is a preamp. Right? And you're planning to include special affects like fuzz. The first thing to do then is to build a small signal amplifier, just as you were doing. You only need 10mV at the output if your going to feed it into a guitar amplifier. And your guitar pickups are going to put out something in the same range of voltage. You don't need much gain. Measure the voltage amplitude of your guitar using an oscilloscope if you've got one.

    Make the input resistance high (about 10 kohms) and the output resistance low (100 ohms or less). Your guitar (input) will like that, and your amplifier (output) will like it. Go for a gain of from 10 to 100. Tomorrow I'll draw up a diagram and post it.
     
  9. PRS

    Well-Known Member

    Aug 24, 2008
    989
    35
    MrChips, the original design is gone from the board, but in that both me and Zetterberg saw a problem in the biasing I don't think we were wrong. Did you actually use the original design? Care to list the resistor values you used?
     
  10. MrChips

    Moderator

    Oct 2, 2009
    12,421
    3,357
    I used the same values as stated in post #1.

    R1 = 4.8kΩ
    R2 = 1.2kΩ
    R3 = 300Ω (RL)
    R4 = 36Ω (RE)
     
  11. Veracohr

    Well-Known Member

    Jan 3, 2011
    550
    75
    First, decide what gain you want.

    Second, get rid of C4. I don't know why it's there. It just creates a lowpass filter.

    Third, increase R1 and R2 by a factor of about 10. These determine the input impedance, and you want that to be high.

    Fourth, use a lower collector current. Just a few mA should do. I used 1mA in simulation.

    Fifth, set the collector DC voltage as I mentioned above, according to the collector current you choose.

    Sixth, you can set your gain by putting a resistor in series with C3. If the amplifier is properly designed, your gain will be approximately RL/R (R = the new resistor). You can then vary the gain if you make R a potentiometer.

    Seventh, be sure to use capacitors that are large enough to have low impedance at the frequencies you're using. If you're shooting for guitar-range frequencies, something above 10μF should be good. Using large capacitors with low impedances means you don't have to account for them (as much) in gain and frequency response calculations.
     
  12. Mashly

    Thread Starter Member

    Apr 4, 2014
    33
    1
    Hi guys, I am still struggling with this.

    Veracohr thanks for your info, its simple really not sure why I didn't see it. But I took this on board and what PRS said about having about 100Ω output impedance and upped the Iq to 0.05A to make the Rc 90Ω.

    From that, how would I design it to have 10mV at the input as I thought the transistor would drop 0.6V across the base-emitter junction?

    Thanks

    Matt
     
  13. BobTPH

    Active Member

    Jun 5, 2013
    782
    114
    You are not going to get much voltage amplication with a Rc of 90 Ohms. You are moving in the wrong direction. Try 4.7K.


    Bob
     
  14. Veracohr

    Well-Known Member

    Jan 3, 2011
    550
    75
    It does drop (about) 0.6-0.7V, but that doesn't mean it's attenuating your input signal. R1 and R2 cause the base voltage to be at some positive value. If the base-emitter drop is 0.7V, and you choose R1 and R2 so that the base voltage is centered on 1.7V, that means the emitter will have approximately the 10mV input signal alternating around 1V (1.7V - 0.7V).

    Choosing a low Rc isn't the best way of getting low output impedance. A common-emitter amplifier typically has high output impedance, it's just a consequence of the configuration. To get low output impedance, put an emitter follower after the amplifier to buffer it. The output impedance of an emitter follower is about (RE || 25mV/Ie).
     
  15. Mashly

    Thread Starter Member

    Apr 4, 2014
    33
    1
    By the way does anyone know a good circuit simulator that works on Linux (Mint 16)? Would be good to test the circuits with out having to build them every time.

    Thanks

    Matt
     
  16. Mashly

    Thread Starter Member

    Apr 4, 2014
    33
    1
    So I seem to have this kind of working now. I designed it for 10mV but if I put that in the voltage at the collector just hits the rail at +9V. If I increase the voltage in to 1V and DC shift it up 1.6V I get 4.5V out. So I am amplifying the signal just not how I meant to.

    Am I have to DC shift due to the fact I need to put in + and - voltage for it to be able to swing the other way or is that only on op-amps?

    These are the revised values.
    Rl=4.5k
    Ic=0.001A
    Ve=1V
    Re=1k
    R1=45K
    R2=10K

    Thanks everyone,

    Mash
     
  17. Veracohr

    Well-Known Member

    Jan 3, 2011
    550
    75
    What test frequency and what capacitor values are you using? Simulating with those resistor values and 100μF capacitors (excluding C4 from your original link), I see a maximum gain of 311 and peaking at 6V, with a lower cutoff frequency of 57Hz.
     
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