Can I use a voltage divider and/or voltage follower to produce 1V from 9V for an op-amp oscillator?

Question

For a circuit I'm building, I'm using an op-amp oscillator to generate a sine wave. The sine wave will have a maximum voltage of 1V and a minimum voltage of -1V. The circuit will be powered by a 9V battery, and I already have a selection of chips to pick from for converting 1V to -1V. How should I regulate 9V down to 1V? I'd rather not buy a chip to do this, as I'm pretty certain that I have everything I need.

If I use a voltage divider (with resistors) to generate 1V, can I then feed that into a voltage follower to eliminate the loading effects? The page mentions that doing so will produce stability issues, so I'm not sure.

Thanks,

Edits

Some details about my oscillator:

• It will be a sine wave
• I need at least 8 kHz. In the range of 8 to 10 kHz would be fine.
• I haven't drawn up the circuit yet, but I'm thinking of using a Wein bridge oscillator. I'll take other suggestions though, because I have never designed an oscillator before
• Powered by a 9V battery

Answer 1

What you are describing will give you a 1 V output, but is probably not the best way to do it. Using a regulator designed as such will probably be more efficient and be able to handle higher loads. But if the parts you have on hand let you do it this way, and you don't need a lot of current at 1 V, there's no reason not to go ahead.

It's not difficult to build a working voltage follower, but you should make sure the op-amp you choose to implement it is a "unity-gain stable" one. If the op-amp has this feature, it will say so in the datasheet, usually in the first few paragraphs or in the bullet-list of features.

For a 1 V output, if you are powering the op-amp from 9 V and Ground (not +/- 9 V) you'll also need to check that the output can swing within 1 V of the minimum supply. Most recent parts should be able to do this, but if you aren't careful you could get caught by this.

Also, realize your op-amp will be burning 8 V times the output current in power. If you are drawing more than a few 10's of mA, you will want to check that the package can handle the power load, just as if this were a regulator circuit.

Answer 2

If you want a +/- 1 V wave from a circuit that runs on a 9 V or so supply, then I would recommend that it generate a voltage swinging from 3.5 V to 5.5 V (i.e. 4.5 +/- 1 V).

The trick is what you do with that voltage next. Already I see a red flag because you mention only a positive voltage source but you expect your output voltage to go below 0 V. There are two ways to do this:

1. DC Blocking

In this case, you take the voltage and pass it through an RC high-pass filter (i.e. series C, followed by R to 0 V), then run that to your load. The resistor should be high to keep it from loading down the amplifier, and the capacitor should be big enough to keep the filter corner well below your oscillator's band. R = 150k and C = 1000 pF (NPO/C0G) (f_c = 1061 Hz) should be fine.

2. Redefine Ground

"Ground" is a simplification. There's nothing magic about the node you choose as your reference voltage (0 V). In a lot of applications you want to minimize its impedance, but again, it's still a node that carries current and has a voltage drop if the impedance is high enough.

So in this case, you just redefine your 4.5 V offset voltage as "ground". To lower its impedance, you can even drive it with a unity-gain op-amp. The rest of your circuit treats the 9 V battery as a 4.5 V rail and a -4.5 V rail. This option lets you drive the output directly from the amplifiers.

I'd recommend option 2, but it needs more parts. It's hard to make a better recommendation with so little information.