# Op-amp gain too big, can't figure out why

We are using a 741 opamp using an inverting input. My students built a circuit, and the gain was way too high. They had taken it apart. So I had them build it in front of me. The gain is still too high, and we're not sure what's going on.

As a sanity check, I had my students build a voltage divider. An AA battery, with a voltage divider 1 MΩ and 220 Ω. It should be approximately 0.2 mV across the 220.

We take that and connect it to the opamp circuit of the 741, +Vin to ground. On the Op amp circuit, we have 1 MΩ feedback from output to where the signal comes in, and a 2.2 kΩ going to -Vin.

Since the input impedance of the Op Amp circuit with our two resistors is on the order of the voltage divider, we expected it to mess with the voltage. But we measured 0.3 mV or so.

Yet the output is -10.37 V, nothing like 1M/2.2k.

Any clue what's going wrong? Would a schematic help? A picture of the breadboard with the wires as is?

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Please add a schematic - it eliminates the ambiguity in your question. The more you post, the better. – W5VO May 21 '12 at 20:09
Your hand waving description of the circuit is confusing, and also relies on your interpretation of things. Show the schematic of how you actually hooked things up. You do have negative feedback around the opamp, right? Saying things like "using the inverting input" puts everything else you say in question. – Olin Lathrop May 21 '12 at 20:45
16 hours and question still not fixed, time to close this mess. – Olin Lathrop May 22 '12 at 12:32
Olin: 16 hours is not long enough. Many people do not visit this site multiple times a day. If you said 3 days and question still not fixed, and there was evidence that the OP had logged in a few times but not made any effort to clarify, I'd understand. We need to give less frequent visitors a break, though. – Jason S May 23 '12 at 11:21
@Olin: We definitely disagree, then. "He refused, so now it should definitely be closed. He's just jerking us around at this point." You ascribe far too much maliciousness to the OP. – Jason S May 24 '12 at 1:44

Would a schematic help?

Yes

First thing, the divider by itself gives about 330 µV, because the AA battery should be 1.5 V, not 1V. And, including the input resistance of the amplifier, it shouldn't be that affected since it's 2.2k in series with the input resistance of the op-amp and the feedback resistor.

So, when connected to the amplifier, the divider gives 300µV.

Now, the amplifier's gain should be -1000/2.2 = -455, which multiplied by 300 µV gives -136 mV. If you run the simulation, you will see -56 mV: this is due to the input bias current, which for the 741 is high (80 nA) and unbalances the feedback divider. If you change the amplifier to the TL082 (in the simulation at least) you'll see the output going to the expected 136 mV.

Since your amplifier gives -10V, it's likely that the op-amp is saturated, and that may be caused by a missing connection somewhere: are you sure that the non inverting input is grounded? To what? The divider, being connected to the battery, is also referenced to ground, isn't it?

UPDATE: as Markrages pointed, the 741 has a maximum rated input offset voltage of 5 mV, which is one order of magnitude bigger than the signal you are amplifying. You can solve this with another op-amp (741 is one of the worst) or try to work on it with the offset nulling circuit, as suggested on the datasheet.

## EDIT:

I was thinking again about the choice of the 741: there's only one reason to use it for didactic purposes (the only use that makes sense). And it's to show the non-idealities of op-amps, as it has every flaw that you can expect from one. I've also been taught about its flaws, like slew rate, saturation range, bias current, offset and low gain.

If you want just to demonstrate how op-amps work, use a better one. And make simpler circuits (small gain and good input signal, maybe a waveform generator), to remain on the sweet spot where everything works.

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lm741 has 5mV max offset voltage. Using it to amplify a 300µV signal is not likely to work well. – markrages May 23 '12 at 7:08
@markrages true. But looking at the datasheet, there is the possibility to use an offset nulling pot – clabacchio May 23 '12 at 7:17
@stevenvh rearranging your math, given a 10V output it's sufficient 10/200'000 = 50+ µV offset to saturate it – clabacchio May 23 '12 at 7:57
Joking aside, it will give you an output offset voltage, but in your schematic even that 5mV wouldn't saturate it. I have no explanation for it either. The 1000V would be an explanation if the feedback connection is broken for some reason. – stevenvh May 23 '12 at 8:12
@stevenvh I think the offset explanation can work, together with bias current: consider that the feedback resistor is 1M, 10µA is enough to saturate it. A pot in the feedback could help in finding the issue – clabacchio May 23 '12 at 8:26

If you look at the diagram in one of the previous answers:

You need to introduce a buffering amplifier to isolate the two circuits. The bufffer would separate the voltage divider from the OpAmp. The output resistance of the voltage divider is affecting the input resistance to the opamp.

Recalculate Vin/Vout for the COMPLETE circuit (Shown above( and you'll see that the voltage gain is not what you anticipate it to be.

Edit:

$$\frac{V_{out}}{R_4} = \frac{V_x}{R_3}$$

$$\frac{R_3V_{out}}{R_4} = V_x$$

Eq2: $$\frac{V_{in}-V_x}{R_1} =\frac{V_x}{R_2} + \frac{V_x}{R_3}$$

$$\frac{V_{in}}{R_1} = V_x( \frac{1}{R_1}+\frac{1}{R_2} +\frac{1}{R_3})$$

$$V_x = \frac{Vin}{ 1+\frac{R_1}{R_2} +\frac{R_1}{R_3}}$$

$$\frac{R_3V_{out}}{R_4}= \frac{Vin}{ 1+\frac{R_1}{R_2} +\frac{R_1}{R_3}}$$

$$V_{out}= \frac{R_4 Vin}{R_3 ( 1+\frac{R_1}{R_2} +\frac{R_1}{R_3})}$$

Not a simply voltage divider any more now is it?

Plugging in all the values you end up with 136mV on the output.