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# Op-Amp with ref voltage

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#### drkidd22

##### Member
So I have this op-amp circuit I can't get my head around.
It has a reference of 10V on the non-inverting input and an input voltage of 9.98V.
Wouldn't my output then be (Vin-Vref) * R2/R1. Which would be an output of 6.06V? But I get 16.06V.

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16.06 is correct. Just think about - you have a lower voltage at negative output - so therefore the output voltage HAS TO BE higher than Vref.

Perform the next two simulations;
1 make both supplies exactly 10 volts

2 make both supplies exactly 9.5 volts.

In both instances measure Vout

16.06 is correct. Just think about - you have a lower voltage at negative output - so therefore the output voltage HAS TO BE higher than Vref.

So my equation turns into [(Vin-Vref) * R2/R1] + Vref?

So my equation turns into [(Vin-Vref) * R2/R1] + Vref?
Because Vin is an inverting input it's actually [-(Vin-Vref) * R2/R1] + Vref

So I've built 2 circuit similar to the original. The only difference is that one is built with an OP279 and the second circuit is built with a OPA2340.
Now with the same imput voltage the output is significantly different. Could this be due to the input offset voltage?
I get a wider swing with the Op279 than the OPA2340.

"Significantly different" meaning exactly what, 1 volt? 100 millivolt? 10 millivolt?

More than half volt.
See simulation, I see similar results on bench test.

Have you compared, on their individual datasheets, what is the Voh for each device with a load of well over 10 mA?

Have you compared, on their individual datasheets, what is the Voh for each device with a load of well over 10 mA?
The OP279 has Voh of +4.8V and Vol of 75mV. I can't find the Voh or Vol on the datasheet of the OPA2340.
The OP279 does have a supply range of 4.5V to 12V while the OPA2340 is 2.5V to 5.5V.

Ok I fixed an error on the simulation. But still you can see a difference of about 250mV.

Its there, but shown in not explicit way, on the chart: Figure 15. Output Voltage Swing vs Output Current

Do something: significantly reduce the load current, at least by a factor of 10X, and measure/simulate again.

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