Voltage signal multiplying\dividing

[Alex30]

So I have a circuit where a 10 volt reference signal comes in and then I have an inline 10K resistor and a 10K pot going to ground in order to vary the signal between these two resistors from 0->5 V to control a comparator.

This 0->5V signal actually refers to a 0->3A current limit I am setting up.

So I want to take that 0->5V signal and multiply it by a factor of (3/5) so I can throw that number on a voltmeter to display what I have set the current limit to. Obviously this means I cannot degrade the signal at all otherwise it will screw up the comparator. What is the best way of accomplishing this?

Thanks

[Rerouter]

Use an op amp, one as a buffer so you do not load your signals, driving a voltage divider, and your voltmeters impedance should be high enough that you could throw a resistor dividor on the output for the voltmeter, or use a second op amp to properly divide the buffered signal down if you want the lowest possible power consumption,

[Alex30]

Excellent thanks that worked

[Alex30]

Say there isn't a way to subtract 20mV from that signal without using more opamps is there?

[T3sl4co1l]

Just another voltage divider; compensate for its loading effect in the previous divider, or buffer the signal.

Subtraction can be done by raising ground, if you can do this with your voltmeter.  Otherwise, op-amp I guess (you don't need another, just tweak the inputs with a little bias current).

Tim

[Alex30]

Doh, why didn't I think of that? I'm already using a voltage reference in my circuit so I should be able to just divide that down to 20mV and run the voltmeter across these two signals. I think the mark of experience is just seeing these things in the simplest way. Thanks for that

[macboy]

Say there isn't a way to subtract 20mV from that signal without using more opamps is there?

You might also consider taking a good look at your grounds... where are you connecting the low side lead of the measuring meter? Do you have any resistive losses (V = I*R) contributing an extra 20 mV on that line? It should have a dedicated connection to the low side of the 0-5V voltage divider.

You should also consider using a "precision" opamp. These are designed to introduce less DC offset for applications like this.