# Question about the output voltage of an op-amp. (ua741)

When i hook up my opamp to a 9 volt battery (like in the image below), i get 8 volts from the output pin. It gives the same reading when i attach both the inverting and non-inverting pins to the 9 volt rail. Shouldn't it be closer to 0 volts or few milli volts?

Why is the opamp behaving like this? Is it broken?

• step 1: take ua741; step 2: place on rock or piece of scrap wood; step 3: put on protective eyegear; step 4: find hammer; step 5: hit ua741 with hammer until pins are connected to separate pieces; step 6: go buy a better op-amp – Jason S Jan 15 '12 at 15:32
• lol lm348 it is then. – drdr Jan 16 '12 at 3:16

As Oli said, in the first case you are operating it out of spec since the input common mode range does not extend to the positive supply. In the second case the inputs are just floating so you don't know what voltage they end up at.

However, even if you were to short the two inputs together and hold them somewhere within the valid range, like 4.5V, you would still likely not get 0V out. This is because the opamp is not perfect and has some input offset voltage. This is essentially a small error voltage added to one of the inputs. Let's say for example this opamp this day at this temperature at this input voltage has 2 mV input offset. When you tie the two inputs together, the opamp will see this 2 mV input instead of the real 0V input. That apparent input will be multiplied by the opamp's open loop gain, which is probably at least 100k or so. 2mV * 100k = 200V. Obviously this amp can't produce 200V, so it will drive it's output as far as it can towards the positive supply rail.

Note that the input offset could just as well have been -2mV, in which case the output would drive as far as it can towards the negative supply rail.

Input offset voltage is a reality of opamps, and is one of the issues that need to be considered in designing a circuit with them.

In both cases the output will be undefined since you are operating it out of spec.

For it to operate correctly as an amplifier you need to provide negative feedback. As you have it it will operate as a comparator (a valid configuration) but you still need to keep within the common mode input range. As far as I know the 741 is not rail to rail input so setting both inputs at one rail (e.g. 9V) will produce undefined results.
Leaving the inputs floating as in the second diagram will also produce undefined results since there will be slight offset between them which will cause the output to saturate (in either direction, which one is unpredictable)
So in either case since there is no negative feedback to control the huge gain of the opamp, it will saturate. In your case it is saturating at the positive rail (or just under since it's not rail to rail output)

To check the specs for yourself here is the 741 datasheet. You will need to learn about what each parameter means to use opamps successfully. It may seem a little daunting at first but you quickly pick up which ones are important for different circuit requirements.

I suggest you pick up a good book on opamps and study the common configurations and how they work. All about circuits has a good section on opamps for starters.

Oli and Olin have explained well the fact that your output is influenced by the offset voltage; the reason because your Opamp saturates at 8V instead of 9, is that it's an old opamp, not designed for low voltage, is not a rail-to-rail opamp.

More in detail, it means that the inner circuit has an output stage (transistor Q14, Q15, Q20 in the figure) with resistors that offer an output protection from short circuiting. Basically the current on R6 determines the voltage between base and emitter of Q15, that drains current from the base of Q14 so basically limiting the output current to avoid burning components.

But that also means that the maximum output voltage achievable is

+V - Vcesat - Vbe that is about +V - 1V

The same applies for low voltage, and that means that it's not rail-to-rail.