common emitter amp output distorted, can't figure out why, need help please

Discussion in 'General Electronics Chat' started by count_volta, May 15, 2010.

  1. count_volta

    Thread Starter Active Member

    Feb 4, 2009
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    Hi guys, I am working on a common emitter amplifier for a lab experiment. It seems to be working but I'm getting a distorted output waveform and don't understand why.

    First of all, here is my schematic.

    [​IMG]

    Ce = 338 uF
    Cc1 = 20 uF
    Cc2 = 8 uF

    The output voltage is taken across the 1k load resistor near Cc2. My design specs said to create an amplifier that has a 5 volt peak to peak ac swing on the output. So I used the characteristics from the curve tracer and selected an operating point Q that meets these specs. Below is a screenshot of the operating point.

    [​IMG]

    From this I figured out all the resistors necessary to get me to that operating point and did so successfully. Made all my measurements and the dc bias is pretty much perfect.

    Now I applied the small signal ac sinusoid which is to be amplified. Below is a screenshot from the oscilloscope of what I got.

    [​IMG]

    As you see the output is 180 degrees out of phase with the input like its supposed to be. Its also more or less as large as its supposed to be with a gain of around 50. But its very distorted. I don't understand why. I spent half the day trying to figure this out and ended up nowhere.

    Can anyone please give me some clues as to what might be wrong here?
     
    Last edited: May 16, 2010
  2. Bychon

    Member

    Mar 12, 2010
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    try raising the ohms in the load resistor.
     
  3. count_volta

    Thread Starter Active Member

    Feb 4, 2009
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    Ah, I can't, the specs said to use a 1kΩ load resistor.

    Also guys, please explain the reason for your suggestions. I heard enough stuff like, "stick in a 2uF capacitor". And when I ask why, the reply is "Hmm dont know, it works".

    My professor more or less said something like that earlier and it made me very mad in all honesty. I'm trying to specialize in electronics and I want to learn how to do this stuff and WHY certain things work over others.
     
  4. Bychon

    Member

    Mar 12, 2010
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    When your collector resistor is 1k and your load resistor is 1k, the load resistor has quite a lot of effect on the current through the transistor. The basic equations don't work. You have to calculate in the load resistor to get the math right, and I don't remember that formula right now.
     
  5. Bychon

    Member

    Mar 12, 2010
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    It would help me if you named the sizes of the capacitors.
     
  6. count_volta

    Thread Starter Active Member

    Feb 4, 2009
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    Allright sure. We have some equations given to us to figure out the capacitors. I just used those equations.


    Ce = 338 uF
    Cc1 = 20 uF
    Cc2 = 8 uF

    I can't really change the load resistor. Its set by the design specs. In fact let me paste the design specs.

    Design a single-stage common emitter BJT amplifier to drive an AC coupled load of 1kΩ with an undistorted peak-to-peak voltage swing of 5V. The output impedance of the stage is to be 1kΩ.

    Also, I tried varying the capacitors, they don't really have much effect on the distorted waveform. Why could this be happening? According to my design I should be in the active mode the whole time.



     
  7. Bychon

    Member

    Mar 12, 2010
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    You need to set up the transistor to use way more current than the load resistor will use. You are supposed to make the load current insignificant compared to the transistor's current. "Rule of thumb" would be 10 to 1.

    Peak voltage = 2.5, Rl = 1k 2.5/1000 = 2.5 ma. If the collector current was .25 amps, the load would become less significant.

    I'm glad this is a simulation. A quarter amp of class a amplifier is a wasteful animal.
     
  8. count_volta

    Thread Starter Active Member

    Feb 4, 2009
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    I thought the whole point of these things is to have a high input impedance and low output impedance.

    Anyway, how would my load resistor cause the output waveform to be distorted? I tried it with no load resistor at all, just connected my scope to the capacitor Cc2 and ground. The waveform looks the same.

    The only thing I know that could cause it to be distorted is if the transistor goes to either saturation mode or cutoff mode. You can see in the oscilloscope display that its amplifying and more or less correctly, it just messes up the input signal while doing so.
     
  9. Bychon

    Member

    Mar 12, 2010
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    I guess I was barking up the wrong tree.
     
  10. rjenkins

    AAC Fanatic!

    Nov 6, 2005
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    Think of your circuit without the transistor, just the 1K resistors and the coupling cap.

    What happens when you turn the power on?
    You initially get equal voltage, half supply, across the two resistors, which reduces as the capacitor charges.
    While the coupling capacitor is working (ie. the voltage across it is near enough constant), the collector voltage can never rise above half supply.

    When you get to low enough frequencies that the collector can follow the waveforrm, it's only because the coupling cap is too small to pass that frequenct to the load resistor without loss.

    Common emitter amps are not specifically low output impedance.

    Either reduce both the collector and emitter resistors by a factor of 10 to give better matching to the load, or add an emitter follower stage to give a low impedance output.
     
  11. Audioguru

    New Member

    Dec 20, 2007
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    It is normal for a single transistor without any negative feedback to have severe distortion as you showed.
    Try removing the emitter caspacitor which will add a lot of negative feedback. Of course then the gain wilkl be reduced a lot.
    Here is my simulation:
     
  12. R!f@@

    AAC Fanatic!

    Apr 2, 2009
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    or may be use better caps at input and output, like low ESR or a better quality, plus use a transistor with proper bandwidth. U have made common mistakes
    Guru. It's an AC amp and it's suppose to have a emitter cap.

    Better caps improves frequency response and driver bandwidth also helps in better waveform reproduction in the desired range.
    Wonder why some are so incomplete?
     
    Last edited: May 16, 2010
  13. Audioguru

    New Member

    Dec 20, 2007
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    The emitter capacitor increases the AC gain by eliminating the negative feedback. Then the distortion is very high as is shown in my simulations.

    The distortion is 40% and the gain is 150 or more (it is so distorted that it is hard to measure).
    When the gain is reduced to 9.2 with the unbypassed emitter resistor providing negative feedback, the distortion is "only" 3%.
     
  14. count_volta

    Thread Starter Active Member

    Feb 4, 2009
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    I can't really remove anything. I pretty much have to use the design I showed. They practically gave it to us, I just had to calculate all the resistors and capacitors.

    Nobody explained yet exactly WHAT causes the distortion. Its in the active mode. Why would there be distortion?

    We have not learned about negative feedback for BJT's yet. It looks like I will have to torture my professor and TA until they explain it to me. I paid good money for this course. Also I want to go into the electronics field. I want to know how this works.
     
  15. Audioguru

    New Member

    Dec 20, 2007
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    A transistor has a current input to its base. A current input produces a low distortion output current.
    But your circuit feeds the transistor a varying voltage. The base-emitter is a diode that has an exponential or logarithmic relationship to its output current which causes distortion.
     
  16. Wendy

    Moderator

    Mar 24, 2008
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    Basically the gain of the circuit is too high. You are clipping. The emitter capacitor has a heck of a lot of effect on the gain.

    Going through the math. I've helped people with this specific type of problem before, so I have a drawing in my library I'll use for parts designations.

    [​IMG]

    Your beta shows as 176 (of course, this is only for DC). The means the resistance through the transistor from base to ground is around 89.8KΩ. This value is in parallel with R2, which means R2║Q1 is 10.6KΩ.

    This means the base voltage is 3.05V, the emitter is 2.45V, setting up a current of 4.80ma. R3 is dropping 4.8V, subtracting the power supply voltage the collector of Q1 is 9.2V.

    RL also has a dramatic effect on the signal though, as far as the signal can swing and the total gain of this circuit. Gain is Rc/(Re║Ce), give or take. If Ce is effectively 0Ω then the gain is whatever the transistors beta at that frequency. A common trick to drop AC gain is to put a resistor in series with Ce.

    OK, we have the DC values set, now for the AC analysis. If C2 is treated as 0Ω, and for gain calculations Ce is 0Ω. Re still has to be considered in terms of voltage swing on the output. With Q1 off the most voltage you will get out of this circuit (the positive AC peak) is 7V. Re║RL is 338Ω. With Q1 fully on the most negative AC peak is 14V*338Ω/(338Ω+1KΩ), or 3.54V. The max voltage swing with 1KΩ as the load is 7V-3.54V, or 3.46VP-P.

    Hope this helps.
     
  17. Audioguru

    New Member

    Dec 20, 2007
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    No.
    The transistor is not clipping. It is severely distorting because the positive swing at the collector is compressed by the exponential or logarithmic relationship of the varying input voltage signal to its collector current.

    No.
    The beta of a transistor is its current gain, not its voltage gain.
     
  18. Wendy

    Moderator

    Mar 24, 2008
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    Hmmm, and here I thought when a circuit hit the edge of its range it was clipping, which this circuit is doing. It is flattening out because it can go no further. Reduce the input, or reduce the gain, either will put the transistor back into linear mode.

    Yep, and if you bothered to read any of my math I was talking current gain. The math also matches what is happening very closely if you observe. If you want to criticize, stay on track.

    To Count Volta:

    Having Ce so high also means the input impedance of this circuit is extremely low, less than 10Ω. You might want to add Rinput as shown in my schematic as an alternate solution.
     
    Last edited: May 16, 2010
  19. R!f@@

    AAC Fanatic!

    Apr 2, 2009
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    Clipping is distortion if you really know what it means.
    Clipping from top or bottom or rather flattening results is too much gain or lower supply.
    In class A crossover is not there since one tr is used
    And transistors are current devices, so V or I or whatever, it is gain. When said it means V and I for amps.
    Guru, I think you should revise wht you say
     
  20. Wendy

    Moderator

    Mar 24, 2008
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    Gain of this circuit Rc/(R3║Ce). You can drop Ce dramatically (it will affect frequency response, but cure the problem at that frequency) or you can drop R3 (I'm assuming RL is required to be at that value). The latter option will require you to rebias your circuit, but is probably the best solution.

    If you want to see what I'm talking about remove Ce altogether, the waveform will mostly clean up since the gain will approximately be unity.

    With R3 and Re at the values they are you are stuck with a max swing of 4.4V, so bumping R3 down is probably a good idea.

    If you want to go through how to calculate a voltage swing for an AC output I'll go through the math with you.
     
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