Common-emitter oscillator

1. Nov 23, 2014

suv79

1. The problem statement, all variables and given/known data

estimate the value of turn-ratio n

R1=4.7 kΩ, R2=24 kΩ, Rl=2.7 kΩ, hfe=250, hoe=10^-5, hie=4 kΩ

2. Relevant equations

loop gain=1/n[hfe/hie*R'l]

R'l=Rl||hoe||[n^2(R1||R2)]

3. The attempt at a solution

loop gain=1

2. Nov 24, 2014

rude man

1. Open the feedback loop and calculate the complex voltage gain (amplitude and phase. In doing so, the transformer secondary must be loaded wit the same complex impedance that the closed loop sees.) Assume k=1 for the transformer.
2. Invoke the Barkhausen criterion or whatever it's called these days: O/L gain = 1.0,
phase = 0.

3. Nov 24, 2014

suv79

yes loop gain is 1 to make it oscillate, and the feedback is positive, so there is no phase shift. :)

iam having a problem with my maths, to work out 'n'
in my attempt i have got stuck

4. Nov 24, 2014

rude man

5. Nov 24, 2014

suv79

6. Nov 24, 2014

rude man

Stand by, am looking at this some more.

7. Nov 24, 2014

suv79

ok :) so i can ignore the siemens Hoe, reciprocal which is very large,

i was thinking it was quadrature, :) but was not sure

8. Nov 24, 2014

rude man

Right, hoe can be ignored.
What do you mean by quadrature? What would shift the phase of hoe by 90 degrees?
I'm still looking at this. It does look (right now, to me), as though we need to solve an equation of the form
1/nRL + 1/n3R + 1/n3hie = hfe/hie. Maybe Wolfram Alpha can help.
You should not have introduced numbers until the very end.
I'll double-check this some more soon.

9. Nov 24, 2014

suv79

10. Nov 24, 2014

suv79

cancel out top and bottom ?

11. Nov 24, 2014

rude man

You could cross-multiply in the denominator, then cance l out n2(R1 + R2). Tha's assuming your equation is right to begin with.

Looking at your 1st eq'n in post 9, this is a lot like what I have:
change your R'L into a conductance 1/R'L:

so 1/R'L = 1/RL + 1/ n2R
where I define R = R1||R2.
Now, what I notice is you didn't include hie as a load on the transformer secondary. So the above changes to
1/R'L = 1/RL + 1/ n2R + 1/n2hie
And your 1st equation in post 9, being

(ignore 2nd & 3rd equations)
makes your equation 1 = (1/n)(hfe/hie)/(1/R'L) with the new 1/R'L
which wouild then agree with what I came up with.
So, bottom line, include conductance 1/n2hie and write the equation in conductances rather than admittances, and you have a cubic expression for n that goes as an3 + bn2 + c = 0.

EDIT: I looked up the solution for this equation with b = -1 and it is horrendous. You would definitely need some kind of math software to solve for n. I would leave the equation as is and hand it in with the answer being "the real solution of an3 - n2 + c = 0." :) (Defining a and c of course in given parameters).
(The other two solutions are imaginary).

Last edited: Nov 25, 2014
12. Nov 25, 2014

suv79

i dont think i need 1/n2hie
because equations given in the question dont not have it in for R'L

13. Nov 25, 2014

rude man

hie is in parallel with R1||R2. You included R1||R2; why not also hie? And hie and R1||R2 are almost equal so you can't say hie >> R1||R2.

14. Nov 25, 2014

suv79

R'L is effective resistive load, R1||R2 feeds the transformer
hie is the input impedance

15. Nov 26, 2014

rude man

You can make a tentative assumption that n >> 1 and solve for n. This makes the math trivially simple. Then, if it turns out that n >> 1 you were justified in the assumption.

16. Nov 27, 2014

The Electrician

It looks to me like neglecting hoe makes a much larger difference than neglecting the loading effect of hie

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17. Nov 27, 2014

rude man

Your values for n correspond very well with mine ( n = 164 to 169 range; I came up with n = 169. As I said, you can make life much simpler if you assume n >> 1, then all you're left with is R'L = RL.