Common Emitter Design

Discussion in 'Homework Help' started by Sioux, Jul 2, 2009.

  1. Sioux

    Thread Starter Member

    Apr 17, 2009
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    Hey Folks,

    First post and already a question. Sorry for the entrance.
    I'm taking an Electronics Course and we are assigned to design a Common Emitter Amplifier. Now I searched the site but no question has the same property as mine. We aren't given a operating point, so no Q current.
    I've attached the Basic Schematic that we are given and are supposed to work with.

    The Question Asks,
    We are to drive an 8 Ω speaker with a max of 5 Watts RMS. Choose the Quiescent state to optimize dynamic range. The Transistors beta is 100. Resistors are to be chosen to give an output impedance of 0.8Ω within 10%. The Caps are to give a rolloff at 10Hz.


    Now while I started thinking and working on this, I figured that I should have a Quiescent Voltage of 0 V at the emitter. Thus I get 0.6 Volts at the base. I can figure out the ratio of R1 and R2 to be \frac{0.6-30}{-30-0.6}. So if i choose R1 or R2 i can get a value for the other. Now what about Re though? Which way should I approach thinking about this?

    Thanks for any help.

    EDIT: The 2N222 should be a generic transistor and not the 2N222 specifically. I used this in a SPICE simulation
     
    Last edited: Jul 2, 2009
  2. R!f@@

    AAC Fanatic!

    Apr 2, 2009
    8,754
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    In typical amplifier the quiescent state is normally known as the amps idle state, in which no signal is given with no out put but a steady quiescent current or idle current flows through the transistor when it is properly biased. Which means there will be a CE voltage drop and a collector current plus an emitter together with the base current.

    By the way who designed that circuit

    Rifaa
     
  3. hobbyist

    Distinguished Member

    Aug 10, 2008
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    56
    Is the instructions for a common emitter amp?

    I don't think that is a common emitter amp.
    I'm pretty sure that is a common collector amp.

    A common emitter has the load on the collector terminal.
    This is on the emitter side.
     
    Last edited: Jul 2, 2009
  4. Sioux

    Thread Starter Member

    Apr 17, 2009
    15
    0
    Unfortunately I don't know who designed the circuit. Why do you ask?

    I see that the voltage drop has to be 30 V for Vce. I guess I'm choosing the quiescent voltage at the emitter to be 0V, but i don't see how this will help me in figuring out Re. Is this just an arbitrary choice?
     
  5. Sioux

    Thread Starter Member

    Apr 17, 2009
    15
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    hobbyist: Yes, sorry about the mislabel.
     
  6. steveb

    Senior Member

    Jul 3, 2008
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    I was getting ready to offer some advice on how to choose Re, but then I saw an issue with the specification. I don't see how you can achieve the 0.8 ohm output impedance spec and the 5 W power into an 8 ohm load at the same time.

    Output impedance is approximately Hib in parallel with Re since the input source looks like a short to ground (taking the bias resistors out of consideration). At room temperature, Hib is about 0.03 Volts divided by the quiescent current Iq. Hib must be greater than 0.8 ohms which means that Iq must be less than 38 mA. However, you can't drive an 8 ohm load at 5 W with only 38 mA. The quiescent current must exceed the maximum peak current needed to drive the load, otherwise the transistor will go into cutoff.

    Maybe someone can see a mistake in my logic, but I don't know how to resolve this discrepency.
     
    Last edited: Jul 2, 2009
  7. Sioux

    Thread Starter Member

    Apr 17, 2009
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    Ok this may be another mix up on my part. The Input should not be shorted to ground at all. This is the schematic i used for my SPICE simulation and required the ground. The Actual Schematic provided doesnt show the source, but only C1. The lead to C1 is labeled as Vin in my schematic.

    Sorry about that, hope that makes more sense now.
    I've uploaded a new image.
     
    Last edited: Jul 2, 2009
  8. steveb

    Senior Member

    Jul 3, 2008
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    Actually, it still doesn't make sense to me because they would need to specify an input source impedance of around 0.8 ohms in order to resolve my concern. The way you wrote it seems right to me unless they mentioned an input source resistance.
     
  9. Sioux

    Thread Starter Member

    Apr 17, 2009
    15
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    Wow, I seem to riding the Fail train...
    Anyways, yes, the input has an impedance of 5kΩ.

    I didnt think that this was required knowledge at first, but here is the rest, Vin has 5kOhms impedance and runs at 1 Volt RMS. This is the whole deal now
     
  10. steveb

    Senior Member

    Jul 3, 2008
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    Ah, ok. It makes sense now. Make sure you include that source impedance in the SPICE simulation.

    OK, so the output impedeance becomes (note that || means parallel combination)

    Rout=( (R1 || R2 || Rsource)/beta + Hib ) || Re

    I think you can choose Re based on the power requirements. Remember that power calculations will use RMS values, but the quiescent current will need to consider the peak values. Your Iq must be greater than the peak current the load will need for 5 W RMS operation.
     
    Last edited: Jul 2, 2009
  11. steveb

    Senior Member

    Jul 3, 2008
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    OK, now I see another issue. If the input voltage is 1 V RMS, then the output voltage will be a little less than that because the common collector does not amplify voltage. A 1 V RMS voltage into a 8 ohm load is only 1/8 of a Watt, not 5 W. I'm confused again. ;)

    EDIT: Maybe, the 1 V spec is not important for the 5 W power spec.
     
    Last edited: Jul 2, 2009
  12. Sioux

    Thread Starter Member

    Apr 17, 2009
    15
    0
    I think the question may mainly concern itself with impedance matching.
    So I guess we dropped the impedance using amplifier to 0.8Ω rather then the previous 5kΩ. A previous question had us calculating the Power dissipated in the Speaker if it were directly attached to the input, which of course is super small.

    So if i took your hint correctly, i can find that the The max output voltage across C2 should be E=\sqrt{P*R}E=\sqrt{40} of course this is RMS so peak can be found. Likewise, i could find the current required to make this happen. I'm still not sure what to do about Re.
    Should I forget about the power rating and just make the output impedance work out to 0.8Ω +- 10%. And if so, any suggestions on guidelines here?

    One more note, I think the question wants me to devise an amplifier, which is capable of 5W RMS on the Load not knowing the source. The question later goes on to ask what the Wattage across the Speaker is once the source is connected
     
    Last edited: Jul 2, 2009
  13. steveb

    Senior Member

    Jul 3, 2008
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    Well, my point was that once you find the peak AC current needed by the load, you need to allow the Q-point DC current to be at least that much, or the transistor will hit cutoff. Re is what controls the Q point current. The bias resistors control Vb, then Ve is 0.6 V less than that, then you know the voltage drop on Re, and the Q point Ie is determined by Re.
     
  14. Sioux

    Thread Starter Member

    Apr 17, 2009
    15
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    What if I change the Bias voltage at the base to give me 1 Volt below \sqrt{80}=V_{p}, where V_{p} is the peak voltage across Rload to give me 5 Watts? Do I really need to bias at 0 Volts?
    In other words, could i not superimpose the signal on a larger DC voltage?
     
    Last edited: Jul 2, 2009
  15. steveb

    Senior Member

    Jul 3, 2008
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    I'm not sure if you need 0 volt bias. You don't need it to get 5 W power, but there is a mention of optimizing dynamic range. This may mean to have maximum signal swing which is near 0 volt bias.
     
  16. steveb

    Senior Member

    Jul 3, 2008
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    By the way, did you mean 1 volt ABOVE this voltage
     
  17. Sioux

    Thread Starter Member

    Apr 17, 2009
    15
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    That is what I suspected. At least that's what I thought dynamic range implies myself. The maximum possible Voltage swing would be at 0V right? So i can swing to -30 V (cutoff) or 30 V(saturated).
    Hrm, I'm currently playing with some values, I determined Re to be 17Ω which lead to R1 and R2 being 165 Ωand 172 Ω respectively. This would give me 0.79Ω output impedance at the Quiescent point. This is also neglecting Hie, since the current is so large. (Hie i calculated to be 0.017 Ω).

    I still really am not sure if this is correct, since I simply chose some values to make it work out to 0.8Ω output impedance. No way I'm getting 5 Watts this way. By the way, the quiescent current Ie is 1.77 Amps.

    Also, thanks for all your help so far, particularly steve :)
     
  18. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    As was discussed in another thread a while ago, shouldn't this be:

    Rout=( (R1 || R2 || Rsource)/(beta+1) + Hib ) || Re

    A very small difference as we both know, but in the interests of accuracy....

    Also, it seems a little strange to me to have a common base h parameter in this expression; I would probably have put it:

    Rout=( (R1 || R2 || Rsource)/(beta+1) + re ) || Re
     
  19. Sioux

    Thread Starter Member

    Apr 17, 2009
    15
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    I meant 1 Volt below after the diode drop. So it can swing up by 1 volt as the input goes up.
     
  20. steveb

    Senior Member

    Jul 3, 2008
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    Yes, most definitely. :)
     
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