- #1

JJBladester

Gold Member

- 286

- 2

## Homework Statement

What exactly is op-amp saturation?

## Homework Equations

[tex]V_{out}=A_{v}V_{in}[/tex]

## The Attempt at a Solution

I am doing a lab where we use an LM339A quad voltage comparator and a voltage divider network to create a voltmeter. A single red LED is connected to each of the four output terminals. When the input voltage exceeds V

_{ref}of each comparator, the LED is forward biased, conducts current and lights up.

The lab asks us to find the saturation voltage of the comparator by "measuring any comparator output terminal when an LED is lit". I did this and found

[tex]V_{sat}=261mV[/tex]

The datasheet has a saturation voltage [itex]V_{OL}=250mV[/itex] (typical) and a voltage gain [itex]A_{V}=200V/mV[/itex] (typical).

So, my measured saturation voltage is about 4% off from the datasheet. Not bad... but I'm having a hard time describing exactly what is meant by saturation voltage. I read this PF post but I'm still slightly confused.

I believe that saturation occurs when the open-loop input differential voltage exceeds some small voltage and that since the op-amp has a high open-loop gain, the output is driven to ±V

_{out(max)}.

Here's what I wrote in my lab... Does it make sense?

The saturation voltage was measured to be 250 mV which is a 4.4% error from the datasheet value of 261 mV. The saturation voltage is the input voltage at which the output voltage [itex]\left ( A_{OL}\cdot V_{sat} \right )[/itex] will be at its maximum positive or negative value. So, if we have a voltage difference of 250 mV between the inverting and non-inverting inputs, the op-amp's high open-loop gain will force the output to ±V_{out(max)}.