Looking to modify this circuit for 30Hz operation

Jul 18, 2008 7 Replies

I was given the parts to build this sine wave generator. As far as I can tell it's a Wein Bridge with the two LEDs serving the function originally reserved for the incandescent lamp.



formatting link



Unfortunately it's fixed frequency, the 5K variable only trims one "side" of the wave, not selecting the fundamental as I'd hoped. It looks liket the 68K resistors and 0.047uF caps determine the period of oscillation (one pair for each half).



Does anyone know the mathmatical relationship between the values of R5/ R6 and C1/C2 and the frequency of oscillation? I'd like to get this working in the 30Hz range.



Thanks, Steve


To change the frequency, you have to (I think) change the values of R5 and 6, simultaneously. Double both of these and the frequency should fall to about half.

If you want to make a variable oscillator, you need a ganged pair of pots that track each other well.

Regards, John Popelish

Thanks John,

Any idea of the approx values to get this near to 30Hz? I could just make it up with sockets and keep plugging away, but I wondered if the math was relatively simple.

** Yep.

  • It acts as a fine frequency trim only.
** The cap is actually 0.0047 uF.

Makes the frequency just on 500Hz.

F = 1 / 2.pi.R.C

** Change C1 & C2 to 0.082 uF ( 82nF)

Gives you about 29 Hz.

... Phil

If I recall, the basic formula is f=1/(2*pi*R*C) for each of the RC pairs in the positive feedback loop, so with C being

4.7nF and f being 30 Hz, R would be 1.13 meg. It might be better to use a larger pair of capacitors so that the resistors could be proportionately smaller.

Any RC pairs that have an R*C product of 1/(2*pi*30)=0.00531 seconds would produce 30 Hz. For instance

1.13meg*4.7nF=0.00531 but so does 53.1k*100nF. 100nF=0.1uF.
Regards, John Popelish

That's great thanks - I ordered some caps close to your value, and a ganged 200K dual pot - I should be able to get pretty close now.

The frequency of oscillation of a Wien Bridge is: f = 1/(2*pi*(R5*(R6 + R9)*C1*C2)^1/2).

The R's and C's are the same sub numbers as your schematic. There is no requirement that R5 = R6 or C1 = C2. That's just a convenience. When those components are equal, the above formula simplifies as can be seen and was posted. When the components are equal, the gain must equal exactly 3. The purpose of the feedback network with the LED is to allow the gain to automatically adjust to the right value for oscillation.

If the R and C values are not equal, a different gain than 3 will balance the bridge for stable oscillation. The gain is critical, too much and the output slams into the amplifier limits and too little and the oscillation quickly dies out. The LED's help stabilize the gain by lowering the gain when the drive gets too high at which point the LED's conduct more.

Because the LED's conduct at the top (and bottom) of the sine cycle, this method of stabilization adds some distortion to the sine wave which may be ok depending what you intend to do with the signal.

** You will have a large range of frequencies available now.

From about 10 Hz up to 200 Hz or more.

..... Phil

Join the Discussion

Have something to add? Share your thoughts — no account required.

Didn't find your answer?

Ask the community — no account required