help needed with wien bridge oscillator.

Discussion in 'Electronic Basics' started by LynneyDee, Jan 6, 2006.

1. LynneyDeeGuest

hi everyone,

I hope there's somebody here who can help me. here's the schematic of the 440Hz reference oscillator of a minimoog synth:

http://users.telenet.be/Lynneymoog/reference.bmp

I've simulated this circuit in Electronics Workbench and (believe it or not) I got the same result : 440Hz. now I'd like to calculate the theoretical central frequency. I used this formula: fo = 1/(2.Pi.R.C)

I've been working on this for a couple days now, but I just don't come to 440Hz...

Lynn

2. Charles SchulerGuest

Units? R in ohms, C in farads.

3. Pooh BearGuest

It doesn't look like a real Wein bridge to me. Unmatched arms. That value for R1 looks *very* wrong too btw.

Are you using 4k75 for R ? Try again with ~ 13k5 ( 4k75 + 4k75 + 1k/2 + 16k9 ) /2

Graham

4. Andrew HolmeGuest

Look at the Wien Bridge circuit here
http://www.ecircuitcenter.com/Circuits/opwien/opwien.htm

R1, C1 are in parallel
R2, C2 are in series

Normally, R1=R2=R and C1=C2=C and the frequency is 1.0 / (2*pi*R*C)

The capacitors are equal in your circuit, but the resistances are not - so
we don't quite get the right answer using the standard formula. We need to
analyse the network. Relating the above symbols to component values in your
circuit:

C = 0.03uF || 0.6nF = 30.6e-9

R1 = R50 || R55 = 92k || 16.9K = 14.3e3
R2 = R17 + R68 + R71 = 10e3

Now we need to plot the transfer function of the network, and find the
frequency where the phase shift passes through 180 degrees. I use a maths
from http://www.scilab.org/

If you run the following code in SCILAB, you get a graph which shows phase
shift passing through 180 degrees at 440Hz

s = poly(0, 's');
R1 = 14.3e3;
R2 = 10e3;
C = 30.6e-9;
xp = 1 / (1/R1 + C*s);
f = xp / (R2 + 1/C/s + xp);
xbasc(0);
xselect();
bode(syslin('c', f), 10, 10e3, .01);

SCILAB has functions for calculating the exact frequency, so you don't have
to read it off the graph.

Or, you can get a very rough approximation of the frequency using the
standard formula with R=10k and C=30nF.

5. Andrew HolmeGuest

Or increase the value of R68 to 4k8, to "balance" the bridge:

R1 = R50 || R55 = 92k || 16.9K = 14.3e3
R2 = R17 + R68 + R71 = 4750+4800+4750 = 14.3e3

i.e R = R1 = R2

Then you can use the standard formula.