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Help designing class A amplifier.

Discussion in 'Electronics Homework Help' started by CiaranM, Oct 18, 2012.

  1. CiaranM

    CiaranM

    74
    1
    May 19, 2012
    Hello! One of my college assignments requires me to design a class A amplifier. I'm having trouble with it.

    NOTE: forget about the resistor values. They have nothing to do with the question, they're only there as Multisim required it.

    There is supposed to be 1mA through R1, and the voltage across R2 is supposed to be 1/10Vs, which is 0.9V in this instance. The base voltage is supposed to be 0.7V higher than this, so there should be 1.6V at the base. The Hfe can be assumed to be 200.
    Can anyone explain how I go about getting all the resistor values?
     

    Attached Files:

  2. Harald Kapp

    Harald Kapp Moderator Moderator

    9,129
    1,842
    Nov 17, 2011
    The emitter current (through R2) is the sum of the base current of the transistor and the collector current (through R1, this is a given value from the assignment).

    The base current is the collector current divided by Hfe (assuming Hfe is the same as the DC current gain - which is not necessarily so, but lacking more information is a valid assumption).

    From the emitter current and the voltage across R2 you can calculate R2. R3 and R4 make a voltage divider to set the base voltage (you already know that value). You have one degree of freedom here since an infinity of possible resistor combinations can set the base voltage. You have to make a reasonable assumption. For example you can state that the current through R3 should be on the order of 10*base current. Thus the base current has only a small influence on the divider.
    Having thus defined a current for R3 you can get the current through R4 by taking the base current into consideration. From the current through R4 and the base voltage you get R4. And finally it is easy to find the missing value for R3.

    Give it a try.
     
  3. CiaranM

    CiaranM

    74
    1
    May 19, 2012
    hello, thanks for your help!
    How is that I through R3 could be higher than Ib? I thought that the resistance would match Ib, since it is in series with another resistor.

    I got some values:
    Ie = 1m + 1m/200 Ie = 1005u
    R2 = 0.9/1005u R2 = 895.52

    R1 = (9 - 0.9)/1m R1 = 8k1

    R3 + R4 = 9/5u = 1M8
    (1M8/9) x 1.6 = 320k for R4
    which leaves 1480000 for R3
    [Vs = Vin x R2/(R1+R2) 9 x 320k/(320k + 1480000) = 1.6]

    is this right?
     
  4. Harald Kapp

    Harald Kapp Moderator Moderator

    9,129
    1,842
    Nov 17, 2011
    And where would the current through R4 come from iF I(R3)=Ib?

    As for your values:
    R1 = (9 - 0.9)/1m R1 = 8k1
    R1 cannot be grater than that value (8k1). But fi you make it exactly 8k1 there is no headroom left for the amplifier to actually amplify. Assum the base voltage rises due to a positive input signal. Then the base curent rises, too. From this follows an increased collector current. But since the transistor is already saturated (Vce~0V) you cannot drive this current. The input signal is clipped.
    A change in input voltage (Delta(V2)) changes (in a first approximation) the voltage across R2 in the same way.
    Therefore Delta(V2) ~ Delta(V(R2)).
    This change in voltage across R2 changes the current in the same way: Delta(Ie) = Delta(V(R2))/R2=Delta(V2)/R2.
    Neglecting IB (Ib<<Ie) this same change in current accors at R1: Delta(I(R1))=Delta(Ie).
    This change in current produces a change in voltage across R1: Delta(V(R1))=Delta(I(R1))*R1=Delta(Ie)*R1
    Since Delta(Ie) = Delta(V2)/R2 it follows that Delta(V(R1))=Delta(V2)/R2 *R1
    Solve this for Delta(V(R1)/Delta(V2)= R1/R2
    The expression Delta(V(R1)/Delta(V2) is the gain of the amplifier, therefore gain=R1/R2 (ever assuming the transistor is not saturated).

    The gain is the missing variable in defining R1. You need to know the gain that shall be achieved by the circuit and from that you can calculate R1.

    No. I told you that "that the current through R3 should be on the order of 10*base current." With respect to the first part of my answer: If you set the current through the divider to the same value as the expected base current, the real base current will be much lower due to the current through R4.
    Re-calculate that part following my hints.

    And sorry, the last part of your calculation I don't understand. Maybe it will become clear once the obove tips have been followed?

    Regards

    Harald
     
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