# Sine Wave Generator Circuit - Bipolar

I have this sine wave generating circuit working. It is based on an application note from TI (http://www.ti.com/lit/wp/snoa839/snoa839.pdf).

we are driving an inductive sensor with a 2KHz sine wave, and right now I am putting a filter and amplifier to make the the generated signal have zero DC offset, and have the amplitude we need. I was wondering what changes to the circuit would be needed to change this circuit to a bipolar (+-5V), or is there an alternative layout that would be better for bipolar? • (1) connect -ve rail (pin 4) to -5V. (2) connect R2 to 0V instead of +5V. (Alternatively, delete it and reduce R1 to 500R. (3) verify that your chosen opamp works on +/-5V before switching on. – Brian Drummond Jan 10 '18 at 13:48
• The DC midpoint is set by R2 and R4 at V+ /2. Connect R2 to ground placing R2 and R4 in parallel. Then the opamps need a negative supply voltage as well so connect the opamp's Vss pin to a -5 V supply rail. – Bimpelrekkie Jan 10 '18 at 13:51

## 1 Answer

I was wondering what changes to the circuit would be needed to change this circuit to a bipolar (+-5V)

I'd start by powering both op-amps with bipolar supplies, something like +/- 9 volts. This gives enough supply overhead to produce a sensible output level of a 5 volt peak sine wave. The split bipolar supply also naturally gives you a centre point of 0 volts (desireable). R4 would also need to connect to the negative rail.

Next I would ensure that the output from the first stage (the relaxation oscillator) fed the 2nd stage (a 2nd order low pass filter) via a decoupling capacitor and resistor to ground. Those two components are there to remove any DC offset from the oscillator because that sort of oscillator is going to produce an offset that is generally undesireable due to asymmetries in the output limiting.

I might also consider that to give a reasonably constant output amplitude, the output from the oscillator is limited by precision voltage shunts (possibly LT1389 or SPX4040) to restrict the output to +/- 2.5 volts. This would then mean that the 2nd stage filter will need a gain of 2 to obtain +/- 5 volts peak to peak. That gain can be tweaked to set the output precisely at +/- 5 volts.

If you choose rail to rail op-amps and have a stable + and - 5 volt rails then you can probably rely on the square wave output's amplitude from the first stage.