Common Emitter Amp Design

Discussion in 'Homework Help' started by embi, Oct 7, 2012.

  1. embi

    Thread Starter New Member

    Oct 7, 2012
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    Hi,

    I have to design a Common Emitter Amplifier using a 2N2222 NPN transistor and an 18V DC supply.

    The specifications are as follows:

    • The input signal to the amplifier is a 0.2 Vp sinusoidal signal, at a frequency of 50 kHz.
    • The amplifier should have an undistorted, symmetrical output with an 8 Vpp swing.
    • The output should be maintained over a bandwidth of at least 1 kHz – 10 MHz.
    • The quiescent current is to be 6 mA, at a quiescent collector-emitter potential of 8 V
    • Assume the load impedance is much larger than the output impedance of your amplifier.
    • Ensure the power dissipation of the transistor is kept below maximum operating limits.
    I've done a few hours of research but I'm really blacked out at the moment. Can anyone tell me steps of how to design the circuit. Where to begin and what formulas to use.





    I'm drawing the circuit in Micro-Cap and have attached a screenshot of the initial design (without values for the components). Does it look right so far?


    Thank you so much.
     
  2. Audioguru

    New Member

    Dec 20, 2007
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    Your schematic has the emitter resistor completely bypassed with a capacitor so the voltage gain will be more than 180. Add an unbypassed resistor in series with the emitter to reduce the gain and reduce the distortion.

    I want to show you and your teacher that a single transistor with a voltage gain of 20 and having a fairly high output level produces obvious distortion and its output waveform is not symmetrical. The distortion is about 2% and the top of the waveform is compressed a little.

    But I will not do your homework for you so I didn't post the details of my simulated circuit.
     
  3. bountyhunter

    Well-Known Member

    Sep 7, 2009
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    • IMHO those two specs are incompatible with a single transistor amplifier using a 2N2222. I hope he meant to say a bandwidth of 1kHz to 10kHz. If he thinks it will put out an undistorted sine wave at 10 MHz, he shouldn't be teaching electronics.
     
  4. embi

    Thread Starter New Member

    Oct 7, 2012
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    But in your schematic you also have the capacitor parallel to the emittor resistor. How does that work?

    Also I'm not asking for the values for the resistors, just how do I go about finding them. And where do I start?

    Thank you
     
  5. Audioguru

    New Member

    Dec 20, 2007
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    I said to add an unbypassed resistor in series with the emitter. It is R5 in my schematic. Then the voltage gain is about 23 without a load and is 20 with a load.

    You use simple electronics theory and simple arithmatic to find the values for the resistors.
     
  6. embi

    Thread Starter New Member

    Oct 7, 2012
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    Sorry, I missed that one.

    This is where I'm lost.

    Given the spicifications:
    How do I find the values for firstly the resistors, and secondly the capacitors. I know it involves Quiescent point, β value and such.. but no idea how to apply these.


    Could you give an example maybe using another transistor?

    my knowledge on this is quite low and thus I know I'm asking a lot, but could you explain the purpose of each resistor so eventually I can do this all by myself.

    Thank you
     
  7. Audioguru

    New Member

    Dec 20, 2007
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    Ask your teacher to explain it again or show which pages in your text book.

    I wrongly used a completely different transistor, a 2N3904. I tried the simulation with your 2N2222 and the output was exactly the same.

    I think your teacher or your text book should teach simple transistors theory to you, not me.

    If you do not have a teacher nor a text book then look in Google.
    I learned about it in a short course (one or two half-days) about transistors at a college.
     
  8. embi

    Thread Starter New Member

    Oct 7, 2012
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    I've looked at this website but I'm having trouble relating it to my amplifier.

    Their circuit looks different to mine.

    I still don't know where to start. So far Google hasn't been any help, and I can't contact my teacher either.
     
  9. MrChips

    Moderator

    Oct 2, 2009
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    Start with the information given to you.

    You have 6mA collector current. You know the supply voltage and the Vce voltage.
    Hence you can calculate the collector and emitter resistances.
     
  10. Audioguru

    New Member

    Dec 20, 2007
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    It is exactly the same circuit as yours except it is missing an output coupling capacitor and load resistor. Mine is also the same except I assumed the beta of the transistor is 200, not 100. Therefore my base voltage divider resistor values are twice as high. I added R5 to it to reduce the gain to the needed 23 times and to reduce the distortion.
     
  11. embi

    Thread Starter New Member

    Oct 7, 2012
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    Ic=(Vcc-Vce)/Rl
    6mA=(18-8)/Rl
    Rl=1.67k

    That seems quite far off what Audioguru has..

    and how to go from there
     
  12. MrChips

    Moderator

    Oct 2, 2009
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    Well that's a start but remember the 6mA is only at the Q-point.
     
  13. embi

    Thread Starter New Member

    Oct 7, 2012
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    Im still completely lost.

    the quiescent current is 6mA, Vce is 8V. I'm pretty sure I used the correct formula. So do I now have the correct value for Rl (which is collector resistance)

    So in relation to the attached circuit, I have now found R3.
    What is the next step?

    It would be a great help if you could guide me through it please.

    Thanks so much
     
  14. Audioguru

    New Member

    Dec 20, 2007
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    On purpose I planned the currents to be too low so I do not give away the homework. Just now I changed the values so the collector current is 6mA and the output is the same as before.

    I didn't try it at 10MHz but I think a peaking capacitor must be added parallel to the unbypassed emitter resistor to get the response flat up to 10MHz.
     
  15. Audioguru

    New Member

    Dec 20, 2007
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    1) The emitter current is almost the same as the collector currrent so plan about 0.8V across both emitter resistances using Ohm's Law.
    2) You should look at the datasheet of the 2N2222 to see the typical base-emitter voltage at a collector current of 6mA then add it to the emitter voltage.
    3) Calculate a voltage divider to have 10 times the base current which is 1/10th the collector current. The voltage divider sets the base voltage.
    4) Calculate the value of the unbypassed emitter resistor to set the unloaded gain to about 23. Use subtraction to calculate the bypassed emitter resistor value.
    5) Calculate the load resistor value to reduce the gain to 20.
    6) Calculate the input capacitor, output csapacitor and emitter resistor bypass capacitor values for a response down to 1kHz.
     
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  16. MrChips

    Moderator

    Oct 2, 2009
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    There is no single "correct" answer. We are giving you some range of values to work with.

    You say you are completely lost. What have you attempted so far? I am reluctant to show you more without some work on your part. The rules of Homework Help is to guide you, not give you the answers.

    The Q-point of a class-A amplifier is the mid-point DC bias point.
    Look back at that Electronics Tutorials example and draw your load line.
    What happens when Vce = 0 ?
     
  17. embi

    Thread Starter New Member

    Oct 7, 2012
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    the due date for the question is already passed but it's OK I just want to make sure in the future I can do these easily.

    I'm just trying to crack it. I still don't really know the way to start. I might have overlooked it or something but it's not clear to me.
    I am given:
    *Vin=0.2V 50kHz
    *Vcc=18V
    *Vce=8V

    I'm too confused to make a proper start, i did try to calculate Rload however in the next part.

    When Vce=0 there is saturation. Ic=Vcc/Rl
    according to the datasheet:
    collector-emitter saturation voltage is 400 mV, Ic=150 mA, Ib=15mA
    so, 150mA=18V/Rl -> Rl=18V/150mA=120Ω

    but then again I had this formula for Rl too..
    Ic=(Vcc-Vce)/Rl
    6mA=(18-8)/Rl
    Rl=1.67kΩ

    which one (or neither) is right?

    edit: according to http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/loadline.html (also gives an edited version of the formula)
    Ic=9mA
    so Ic=(Vbb-Vce)/(Rc+Re)
    Rc+Re=18V/9mA=2kΩ
    is this resistance shared between the Rc and Re, how do I divide them up?


    Maybe u can make an example for a similar question but different values?
    the initial problem was:
     
    Last edited: Oct 8, 2012
  18. MrChips

    Moderator

    Oct 2, 2009
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    As a guideline, put the Q-point at the mid-point of the DC load line.
    Make Ic = 12mA when Vce = 0V
    Calculate Rc + Re under these conditions. What do you get?
     
  19. embi

    Thread Starter New Member

    Oct 7, 2012
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    that gives Rc+Re=1.5kΩ, but why 12mA? Did u see my edit in the previous post?

    edit: according to http://hyperphysics.phy-astr.gsu.edu.../loadline.html (also gives an edited version of the formula)
    Ic=9mA
    so Ic=(Vbb-Vce)/(Rc+Re)
    Rc+Re=18V/9mA=2kΩ
    is this resistance shared between the Rc and Re, how do I divide them up?
     
  20. MrChips

    Moderator

    Oct 2, 2009
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    Why did you choose 9mA?
    12mA is twice 6mA. This puts the Q-point at the mid-point of the DC load line.

    18V/12mA = 1.5kΩ

    Audioguru has given you some very helpful hints. Did you give that a try?

    Your Vcc = 18V and you only need 8V. Hence you have a lot of room to work with.
    You need to apply negative feedback at Ve using Re.
    AG suggests 0.8V as a starting point. There is a lot of room here.
    Ve can be in a range from 0.5V to 2V.
    Set Ve = 1V and calculate Re. Now determine Vb.
    Select β = 200 and calculate Ib.
    Set the current through R1 and R2 to 10 times Ib.
    Determine R1 + R2 and R1 and R2 separately.

    Plug these values into a circuit simulator and see what happens.
    Put C3 across Re and see what happens.
    Split Re for gain of 40.
     
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