Best means for a dual vref?

[william_teter]

I have a circuit that needs two voltage references, lets say 5v and 2.5v

What would be the best way to ensure the lower value was exactly 1/2 the value of the higher voltage (e.g. accounting for % tolerances on the 5v voltage reference)? Temperature variances won't be too much of a concern, but if a solution accounted for it, that would be nice.

I've thought of:
1. Using an individual  5v and 2.5v vref IC
2. Using a resistor divider using high precision resistors

I think the latter idea is a better one since two resistors are far cheaper AND will account for the 5v reference tolerances. The downside would be no current which I shouldn't need much at all of (it is the vref for a dac with microamps current specs on the pin)

Just a theory question!

[Hero999]

It's impossible to say without any additional information such as a schematic or detailed description of the circuit.

[william_teter]

This circuit takes the output of a dac (min of 0v and max of ~4.096v.. [depending on % off]) that comes in through the Y port and creates a signal used to drive a galvo amplifier. The galvo amplifier requires a balanced bipolar differential signal between -10v and 10v.

In this circuit the ILDA Y + line outputs between -5v to 5v and the ILDA Y - line outputs between +5v to -5v.

I think this circuit suffers from the following:
-Dead zone caused by op-amp's dc offset(?)
-Independent voltage references (both appear to be sourcing about 1.2mA on the vRef lines) can be off by 0.5% (and that's ignoring temperature differences!). That means the center could be shifted a full 1% off!
-Off-shifted center will mean a dac step above the center value will not be the same size as a dac step below the center value of the dac on the ILDA Y+ and ILDA Y- outputs.


Edit: It's kinda hard to read, but that top left voltage says 2.048V

[Hero999]

To answer the origional question. It's definitely better to use a potential divider and a 5V reference. If more current is required, a non-inverting buffer could be used to reduce he output impedance.

[william_teter]

Fair enough. Thank you.

Any ideas on how to get rid of the offset issue? I have tried multiple solutions, but this has worked the best sofar

[Guest MonkeyGate]

Not really possible.  You can however, build a stable 7.5v circuit and then do what you want.  There is a special circuit that splits a signal exactly in half, but I can't remember what its called...maybe a splitter!!