therocks
6th December 2007 - 07:59 PM
hello,
would like some help with the following question.
an amplifier with a gain of 8 has 10% of its output fed back to the input. Determine the gain of the stage with a) negative feedback and
positive feedback.
ive read my books and cant seem to shed any light on where to start with this. can someone please point me in the right direction.
thanks keith
mr_homm
7th December 2007 - 03:35 AM
You just have to think about what the amplifier does: it takes whatever is fed into it and multiplies that by 8 to generate its output. Let's say the input is x. Then with no feedback, the output is y = 8x.
With negative feedback, the output is subtracted from x and then fed in as the input. This means that if the output is y, then the input is x-y. But then that gets multiplied by 8, so the output is 8(x-y). But that must be the output you started with, which is y. This gives the equation 8(x-y) = y. Solving for y gives 8x = 9y, i.e. y = 8/9 * x. This means that the gain (defined as the final output y over the original input x)is now 8/9 instead of 8.
With positive feedback, it's the same, but you get 8(x+y) = y, so 8x = -7y, so y = -8/7 * x, for a gain of -8/7. The - sign means that if x is a sine wave, then y is down when x is up.
Hope that helps!
--Stuart Anderson
meBigGuy
8th December 2007 - 10:24 AM
I get different numbers. Am I missing something?
Input = I
Output = O
Gain = 8
Negative feedback of 10% (subtract 10% of output from input)
O = 8(I - .1O) which gives O = 8I - .8O which gives 1.8O = 8I
Effective gain = O/I = 8/1.8 = 4.44444...
Checking, with 1V in, and 4.44... out
1 - .444... = .55... * 8 = 4.444...
With positive feedback
O= 8(I+.1O) = 8I +.8O which gives 0.2O = 8I
Effective gain = O/I = 8/.2 = 40
mr_homm
9th December 2007 - 05:56 AM
No! I was missing something. I was too busy explaining the idea and forgot to put the 10% factor in. Your numbers are correct.
The general formula, with a feedback factor f and gain G is:
y = G(x + f*y) = Gx + Gfy, so y - Gfy = Gx, so (1-Gf)y = Gx. This gives the effective gain g = y/x = G/(1-fG). Positive feedback occurs when f>0, negative when f<0. As you can see, negative f reduces the gain, small positive f increases it, but when f = 1/G, the gain goes to infinity. Larger positive f than that actually causes the sign of g to reverse and the magnitude to decrease again.
Hope that helps, and sorry for the error!
--Stuart Anderson