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Homework Help: Common-Base Amplifier Current Gain

  1. May 27, 2013 #1
    1. The problem statement, all variables and given/known data
    Problem statement and solns. in attachment.

    2. Relevant equations
    Can anyone explain the equation for the current gain Ai? I do not understand where the divider temr comes from.

    3. The attempt at a solution

    Attached Files:

    Last edited: May 27, 2013
  2. jcsd
  3. May 27, 2013 #2

    The Electrician

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    I don't see an attachment.
  4. May 27, 2013 #3
    My apologies, it is there now.
  5. May 27, 2013 #4
    I have figured it out, it's just a current division.
  6. May 28, 2013 #5

    rude man

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    I disagree with the given expression.

    We are talking variational currents here. That means ii ~ 1 uA or less.

    With ic = 25 uA, the emitter input resistance re ~ 25mV/25uA ~ 1K
    so summing currents at the emitter we get
    ii = ie + reie/R4. Remember, ie ~ 1 uA or less;

    finally giving
    Ai = ic/ii = α/(1 + re/R4) ~ α.

    This is the same expression as the given one if you replace Rl with re, or 50 ohms with 1000 ohms. Doesn't make a whole lot of difference numerically but the given expression is still wrong.
  7. May 28, 2013 #6
  8. May 28, 2013 #7


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    ii = ie + αreie/R4

    Negligible difference, but the algebra looks neater.
  9. May 28, 2013 #8

    The Electrician

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    The book you referenced uses the symbol Rs for the source resistance; your problem uses RI for that resistance. The book uses Ri for what your problem calls Rin.

    I think you may be confusing symbols.

    The resistor labeled RI in your problem can't have any effect on Ai because the current entering the left end of RI is the same as the current leaving the right end of RI. This means that Ai should be the same for RI of 2200 ohms as it is for RI of 50 ohms.

    Furthermore, the problem in the book is using an h-parameter model, and gives the parameters in those terms. Calculations based on h-parameters may, in some cases, not be the same as the same as for other models.

    The referenced book problem has a non-zero value for hrb, the reverse voltage transfer ratio. Your problem has apparently taken this parameter to be zero, which is typical for problems not using the h-parameter model. This means that you can't directly use a formula from the book that includes hrb.

    You have not used ro except for your Rout calculation. Strictly speaking, you should use it for all your calculations, although it may have only a small effect on numerical results. The problem schematic doesn't show it, but I would assume ro is to be taken in parallel with RL.
  10. May 28, 2013 #9
    Thank you for your input. The use/lack of use of ro was a source of confusion for me, and I asked about that somewhere else. I was told that not only would it make a very messy equation, but in fact it had little effect on the results. (I wish the author had explicitly stated this in the text, but that is another issue)

    As far as confusing symbols, yes I am aware that the symbols are not the same, so let me be more specific. In the attachment that shows my actual problem statement, my small signal drawing, and the given solutions, how does one arrive at the divider term R4/(R4 + Ri)? Ri in that problem is the resistance from the source, not input resistance, which is labeled Rin.

    Once again thank you for taking the time to reply.
  11. May 28, 2013 #10

    rude man

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    It may make the algebra look neater, but it's incorrect.

    The basic summing of currents at the emitter, keeping in mind we are dealing with small variations about I_c = 25 uA, is

    i_i = i_e + v_e/R4 = i_e{1 + (25mV/25uA)}/R4 = i_e(1 + r_e/R4) = (i_c/α)(1 + r_e/R4)

    A_i = i_c/i_i = α/(1 + r_e/R4).

    This is also what the OP posted if you replace his/her R_l with r_e.
  12. May 28, 2013 #11

    rude man

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    The reason you don't understand it is that it's wrong. Read my post #5 again.

    I looked at your link. It does not contradict what I wrote.

    Remember that the circuit you gave us is not the complete circuit. It is merely a small-signal representation of the complete circuit of the link.
    Last edited: May 28, 2013
  13. May 28, 2013 #12
    Yes I see that it is wrong to compare the two, I realize that the two "Ri's" are not the same. My apologies for not clearer, when I said "I don't understand it" I meant this entire problem, not specifically referring to h-parameter model.

    I was not posting that to attempt to contradict what you wrote originally, I hadn't analyzed your post properly, and at that point I don't think I had realized that the two "Ri's" were different. I was confused by the fact that the form of the equations was the same.

    In my post #9 I was attempting to refer only to my actual problem statement because I had realized this:

    I see now that my error was attempting to explicitly solve for ie and ii and calculate the ratio, rather than summing currents at the emitter. I often stare so hard at a problem that I get lost in it and get stuck looking at it one way, not seeing the trees for the forest as it were. You have been an immense help, and I thank you for having the patience and taking the time to reply. Also thank you to everyone else who has contributed.

  14. May 28, 2013 #13

    The Electrician

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    If you want the element in blue to be the source resistance, then it should be RI with a capital I subscript, not Ri. The term Ri doesn't appear anywhere in your attachment.

    In the attachment to your post #1, it looks to me like the source resistance is RI, not Ri.

    You have a current divider with RI as the series resistance, and the parallel combination R4||Rin as the shunt element.

    You can always use a voltage divider technique to solve a current divider.

    You have an input current of Ii; this current flows through RI and then into the parallel combination R4||Rin, causing a voltage drop of Ii*(R4||Rin) across the parallel combination. Then the current into Rin (this is the current into the base) will be (Ii*(R4||Rin))/Rin. Simplifying this gives R4/(R4+Rin).

    The source resistance RI is the wrong thing to use. It's not involved in the current division between R4 and Rin.

    Edit: I see that while I was composing this, other posts occurred that apparently cleared up the confusion. But, I'll leave this for what it's worth.
  15. May 28, 2013 #14

    rude man

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    Very good, JC. You are doing fine.
  16. May 28, 2013 #15

    The Electrician

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    The current gain with ro included isn't too bad, actually:

    [tex]Ai = \frac {ro}{R_L+ro}\frac{\alpha \text{ R4}}{R4+re}[/tex]

    A person could almost have guessed that. The same expression as for the case without considering ro, multiplied by another current divider ratio made up of ro and RL.
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