Op amp as max voltage function

1. Dec 16, 2005

matejhowell

Ok, I just want to see if this would work (but never tried it).

Given a standard (though not ideal, let it be real world) op-amp, it has 5 main inputs -- I+, I-, O, V+, V-. If no loops were formed, and V+ was connected to I+ and V- was connected to I-, would the output O always be the maximum voltage on those two pins? What are some real world constraints?

Normally, this is a comparator, letting V+ pass to O if I+ > I-, and V- pass to O if I+ < I-. But just thought of this twist, didn't know if anyone had some thoughts...

MLH

2. Dec 16, 2005

mezarashi

Yes, in theory and generally in practice, because most op-amps have an open-loop gain of atleast >100. This means that any differential voltage seen across its input will be amplified 100 over times (over the saturation voltage). Since the output cannot go beyond saturation, it will always be in saturation. Usually, some bias currents (due to the non-ideal nature) is enough to make the output go to saturation. This also means that, it is never recommended to turn on an op-amp without negative feedback (unless you're using it as an oscillator which will require postive + negative feedback). Op-amps in saturation == very hot.

3. Dec 16, 2005

Staff: Mentor

Nah. The only thing that will make an opamp hot is if its output current spec is exceeded, which you're not going to do when you pin the output open circuit one way or the other.

As for the OP's question, yes, the output will stick high given the input biasing that you describe, but it will be "high" minus some voltage that depends on the output transistor structure. Take a look at the equivalent circuit for a typical opamp (usually included on the datasheet), and you'll see basically what the output stage looks like. Many CMOS opamps will drive rail-to-rail at the output.

Keep in mind also that especially for many comparators, the input voltage range may not include the rails. So if you pull one or both of the inputs too close to the rails, you may get a nonsensical output voltage.