Generating microvolt signals
With a very low output you will probably not need buffering as the bottom resistor in the divider would be only a few ohms which is then the effective output impedance.

I normally generate the signal at a few volts p-p so you can easily display it on a scope then divide it down with resistors that are in the kilohm range.  You may need to divide it in more than one stage to avoid the wire resistance being an issue. The following diagram shows a 1000:1 attenuator.  To get 1 million to 1 you could replace the upper resistor with a 1megohm resistor or put two attenuators in cascade.  For low frequencies a single stage would be ok but for frequencies above a few 10's of kilohertz two stages would avoid capacitative parasitics being a problem.  I wouldn't advise going below a few ohms for the lower resistor or wire resistance could cause errors.







simulate this circuit &ndash; Schematic created using CircuitLab

Rather than making the attenuator you may be able to use off the shelf Audio or RF attenuators to do the job, they will often go down to DC.  A 1,000 to 1 ratio is 60dB, 1,000,000 to 1 ratio is equivalent to 120dB.

For example you could generate the signal at 5v p-p then attenuate it by 120dB to get 5uV p-p.  Be careful about shielding and grounding or you may compromise the accuracy of the output.