BJT Common Base Amplifier

Discussion in 'Homework Help' started by wendys, Sep 12, 2013.

  1. wendys

    Thread Starter New Member

    Sep 2, 2012
    6
    0
    Hey everyone, I am having trouble with BJT amplifier design and needed help designing a common base with certain specifications. Can anyone help me by guiding me through where exactly to start. Any help is appreciated. Thanks.

    Design a common base amplifier to work between a 50 ohm driver and a 50 K ohm load:
    1. Output impedance 10K or greater
    2. Gain at least 10
    3. Frequency Response 20 - 100 KHz +3db
    4. Capable of driving the load to +10V
     
  2. WBahn

    Moderator

    Mar 31, 2012
    17,715
    4,788
    Where exactly to start is to show YOUR best attempt at solving YOUR homework problem. That will give us some insight into where you are, what you are thinking, and where you might be going astray. Then we are in a better position to try to lead you toward a solution.
     
  3. wendys

    Thread Starter New Member

    Sep 2, 2012
    6
    0
    My approach was as follows.

    First I pick a value 50k ohm for the load resistor as specified.
    Since the output impedance should be at least 10k, I know that Rc||RL will need to be at least 10k. I selected a value of 13k ohm for Rc.

    Since I am using a 2n2222 I know that Vt is = 25mV.

    Also, I want my input impedance to be 50 ohms as specified (driver).
    to get this 50 ohm input impedance i use the equation Re = Vt/IE.

    I get a value for IE = 520uA.

    VE = RE*IE , So i picked an arbitrary value of RE = 10k ohm.

    VE=(10k)(520uA) = 5.2V.
    VB = VE + 0.7V = 5.9V

    To get the 5.9V at the base I set up a voltage divider using Vcc of 15V. I used R1 = 5K and R2 = 3.3k to give me a voltage division of 5.9 volts from 15v Vcc.

    I tried to simulate this in PSPICE but I do not understand the results very well.
    I attached images of the circuit and the simulation output.
    http://imgur.com/a/3F0oI#3ldDEPh

    I am not sure how I can get the circuit to drive the load to + or - 10 volts...

    Thanks
     
    Last edited: Sep 19, 2013
  4. Efron

    Member

    Oct 10, 2010
    81
    15
    Some indications.

    You got a very large gain in your circuit and your transistor goes to saturation and cut off.

    Try to guess your voltage gain and it will help you.

    Note 1: for me, a driver of 50 ohm doesn't imply that you have to have an input resistance of 50, but a 50 ohm resistance put in series with Vin. For better performances, you should look to increase you input resistance so that the Vin source is not loaded by your input circuitry.

    Note 2: to have +/-10V at the load, you need at least 20V range at collector point. If you choose a 15VDC power supply, it will not work.
     
  5. wendys

    Thread Starter New Member

    Sep 2, 2012
    6
    0

    Thank you, I will work on what you mentioned and report back what I am able to find.

    I really appreciate your input.
     
  6. wendys

    Thread Starter New Member

    Sep 2, 2012
    6
    0
    So, I reviewed my design as per Efron's help and this is what I found.

    I inserted a 50 ohm input resistor after vin,

    Since I want a gain of 10, and I am using a collecter resistance Rc of 13k I found that 13k/10 = re input impedance is 1300 ohms.

    I used this 1.3k re to find Ie emitter current by using re = Vt / Ie which is 20 uA.

    I chose an Re of 10k to find Ve.
    20uA = Ve/Re . = .2 volts which gives me a base voltage of .2 + .7 = .9 volts.

    I used voltage division R1 = 22k, R2 = 1k to get a Vb of .9 from 20 volts Vcc.

    My voltage swing is from 0 to -20 volts. with a 1v input. I am not sure how I can get this to be +- 10 volts.


    Link to sim and circuit
    http://imgur.com/a/psSSZ
     
  7. WBahn

    Moderator

    Mar 31, 2012
    17,715
    4,788
    Having a DC offset at your output makes no sense, but that doesn't mean that you can't have a transient response that has to die away.

    You have a 50kΩ resistor in series with a 1000μF capacitor. What is the resulting time constant? How does that compare with the duration of your simulation?

    Oh, and you need to give yourself some voltage headroom. You can't reasonably expect a single-transistor linear amplifier to support a 20Vpp swing from a 20V supply. See what happens if you use a 30V supply.
     
  8. wendys

    Thread Starter New Member

    Sep 2, 2012
    6
    0
    The problem is we do not have a power supply capable of 30v. we have to use one that is a max of 20v. The professor said that as long as it is close to the +- 10v than it is fine for this design.

    Thank you for mentioning the time constant which i completely ignored.

    I was able to find a capacitor value that works.

    How does this look?
    http://imgur.com/a/0hlzZ
     
  9. WBahn

    Moderator

    Mar 31, 2012
    17,715
    4,788
    Okay, I understand your constraint and as long as the instructor is willing to accept the distortion, you should be okay.

    But do run a simulation with 30V and see if it significantly improves the quality of the output signal. You might also look at trying to characterize the distortion, equal just qualitatively, as you raise the supply voltage above 20V. My guess is that there will be a point at which the waveform improves significantly with just a fairly small increase and then it will continue improving but at a much slower rate.
     
  10. Efron

    Member

    Oct 10, 2010
    81
    15
    Remember that the equivalent load resistance in small AC signal seen at the collector side is Rc (13K) in parallel with Rload (50k), not only Rc. That is about 10K and not 13K.

    I have the feeling that the emitter current is quite low! but your maths seem correct.

    For this kind of Frequency Response 20 - 100 KHz, usual values for coupling capacitors should not overpass 10u. Have you checked the frequency response of your circuit? Is it OK with respect to the spec?
     
  11. Efron

    Member

    Oct 10, 2010
    81
    15
    Other thing,

    You're introducing a 1V AC amplitude input signal. Ignoring the 50 ohm resistor in series with Vin, you should allow your circuit for 1V amplitude at the emitter side. However, you're considering 0.2V as nominal voltage (you chose a 10K RE).

    With that, your input signal will be distorted at the emitter side.

    I suggest you to also display the input signal at emitter.
     
  12. wendys

    Thread Starter New Member

    Sep 2, 2012
    6
    0

    So does this mean that if in the lab I apply the vin of 1v. it should not work properly because i calculated for Ve being. .2V . so should i recalculate with Ve = 1 v?
     
  13. Efron

    Member

    Oct 10, 2010
    81
    15
    In general, you must first analyse the extreme voltage limits of your bjt-based amplifier as well as the operating point (DC analysis).

    In your circuit, this shall be done for the collector (output) and also for the emitter (input). If the operating point is not in the middle of maximum limits, the result will mostly be distortion of the signal at that point (assuming gain is big enough).

    If you need a +/-10V range at the output (total of 20V), you should have a collector operating voltage of about 10Vdc, assuming Vcc = 20V (DC analysis).

    If you input 1V amplitude at the emitter - total of 2V range, you should look for an emitter operating voltage of 1V or more - so that in any case, the absolute voltage at the emitter remains positive (it cannot go below 0V in your circuit).
     
    Last edited: Sep 27, 2013
Loading...