# Amplifier theoretical analysis problem

I have this problem in a book and I've did good through the entire problem until I had to calculate current gain. I've been stuck for the past 30 minutes. The formula I get is different from the one in the book and I'm wondering why. This is probably some simple mathematical thing and I'll probably end up embarrassing myself but I just don't know why I get different current gain. I can't continue if I don't understand this.

What I get is that the second term in current gain (circled red on the picture) is reverse, that is numerator and denominator are reverse. I just need an explanation for that term in the book.

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The text as supplied is consistent. The line you say is inverted is essentially the same as the line above it. If you have trouble as shown, why do you not say that th eline before that is also "reverse"? And thje top line in the line above the one you ring = iL which derives correctly from further up the page. SO - starting at the very first line, verify you are happy with each line until you find one that seems wrong. Which one is it? IThe line you circle CANNOT be the only one that seems incorrect. –  Russell McMahon May 27 '12 at 4:59

What you see in the the textbook is perfectly correct. I think the confusion comes from the fact that the ratio $i_b/i_{in}$ is not spelled out in a very obvious form. $i_b$ is clearly in the numerator, but $i_{in}$ is in the denominator and must stay there. I made a kind of 'slow motion calculation' to show, how the ratios are rerranged:

$\dfrac{\dfrac{\beta R_C}{R_C+R_L}i_b}{i_{in}}=\dfrac{\dfrac{\beta R_C}{R_C+R_L}}{i_{in}}i_b=\dfrac{\beta R_C}{R_C+R_L}i_b\dfrac{1}{i_{in}}=\dfrac{\beta R_C}{R_C+R_L}\dfrac{i_b}{i_{in}}$

Afterwards, $i_b/i_{in}$ is simply available by applying the formula for the current divider:

$I_{BRANCH 1} = I_{TOTAL} \times \dfrac{Z_{BRANCH 2}}{Z_{BRANCH 1} + Z_{BRANCH 2}}$

which clearly yields:

$\dfrac{R_B}{R_B+r_{be}}$

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I had to edit the edit. :) The numerator is clearly $Z_2$, that is, the branch through which the current to be calculated does NOT flow. –  Count Zero May 27 '12 at 10:25
Yes it makes sense but I did that as you shown. The problem is my $$i_{in} = i_b* \frac{R_B}{R_B+r_{be}}$$ when coming into $$A_i = \frac{i_L}{i_{in}}$$ formula cancels out the $$i_b$$ and we are left with $$A_i = - \frac{\beta * R_C}{R_C + R_L} * \frac{R_B + r_{be}}{R_B}$$. The term $$\frac{R_B + r_{be}}{R_B}$$ is reversed in the book. What am I doing wrong? –  aarnes May 27 '12 at 11:12
It's $i_b=i_{in}\dfrac{R_B}{R_B+r_{be}}$! The current in the branch is the total current current multiplied by the ratio $\dfrac{R_B}{R_B+r_{be}}$. :) (See the left hand side of the circuit, where it branches off before point b.) –  Count Zero May 27 '12 at 11:24
I see. That was my mistake. Thank you! –  aarnes May 27 '12 at 11:34