Help with voltage divider

jeuhyis

Hi first time poster. I'm a little stuck, through trial and error on Mentor graphics I was able to create the attached circuit. Using +&- 9V supply rails I managed to output -10mV, which I required. Now I have come round to try and work it out by hand i'm struggling to remember how! I know that the first part is basically a potential divider but how does this work with a positive and negative voltage?

Any help greatly appreciated

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willem2

Treat it as a divider between 18V and 0V and then subtract 9V from the output voltage.

Phrak

The two supplies on the left and the two resistors can be replaced with an equivalent single supply and single resistor.

R = 10*8.87/(10+8.87) KOhms

V = (9+9)8.87/(10+8.87) - 9 Volts

jeuhyis

So therefore using this
R = 10*8.87/(10+8.87) KOhms = 4.7KOhms

V = (9+9)8.87/(10+8.87) - 9 Volts = -0.53895V

So then is there some calculation required using the 500K and 1K utilising the -0.53895V to reach the 10mV?

Thanks for your help so far

Phrak

That's an odd combination of components. What's it for?

Vout = -0.539[10/(500 + 4.7 + 10)]

not 10mV, but -10mV

jeuhyis

Its for an input offset voltage on the non-inverting input of an op-amp to give a zero voltage output. I spent ages designing an RMS circuit for a sound level meter and then I hit this problem. Managed to figure out what offset was required simulated that circuit but then I could remember how to hand calculate!

berkeman

Input offset voltages vary with temperature, and from part-to-part. What opamp are you using? There may be a more reliable way to deal with the input offset voltage. Can you post your schematic?

jeuhyis

Here's a screen shot of what I've been designing. Im happy with the overall performance of the circuit, its being used within a group to create a sound level meter, and i have this working as a Type 0 with a 60dB dynamic range.

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