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

Question

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?

Answer 1

Would a schematic help?

Yes

This is your circuit as I read it:

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.

Answer 2

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.

you said you measured 300mV. Chuck your op amp. Its bad.