I built this with a simple breadboard and using the following op-amp:
http://en.wikipedia.org/wiki/File:Op-Amp_Non-Inverting_Amplifier.svg

OPA244:
http://www.farnell.com/datasheets/9250.pdf

The supply is 12V regulated and V- is connected to ground.

The resistors are R2=10k and R1=100.

Vin is generated with a function generator, frequency is a few Hz.

With Vin = 50mV I get about 2V for Vout (measured with oscilloscope)
Gain 40

With Vin = 200mV I get about 6V for Vout
Gain 30

Naturally I should be getting a gain of 101. I've tried two different op-amps and get the same value.

I've also tried with a 1k resistor instead of the 10k. Here the gain varied between 3 and 2 instead of 11.

Any idea what could be going wrong? Is the op-amp not suited for my application?

Edit:
Seems like I jumped the gone on this. For a single supply op-amp r1 should not be connected to ground but to V+/2?
http://www.eng.yale.edu/ee-labs/morse/compo/sloa058.pdf

So I just use a voltage divider with like 2 10k resistors from my source to get this and everything will be fine? I'll try tomorrow at least.

 

Yes, the idea is to bias the amp to produce a quiescent output voltage of Vcc/2. This way the output signal will swing above and below this point.

 

This is my circuit.
The source is 12.11V

I input 10mV and get back 300mV still when the gain should be 100.

Could it be the op-amp acting up or is there something wrong with my circuit?

 

Your gain should be 101. With a noninverting opamp configuration, you calculate 1+(R2/R1); so 1+(10k/100) = 1+100 = 101.

You're amplifying the input offset along with the input voltage. Although the "typical" input offset is +/-0.7mV with a +/-7.5v supply, you're using a single supply.
Even then, 0.7mV * 101 = 70.7mV out even with the input grounded and a dual supply. The input offset can go as high as +/-2mV over temperature.

Another problem is how the Vcc/2 reference has been established. If you use two equal resistors from Vcc to GND, their junction will be at a potential of Vcc/2. However, it is high impedance. Unless you establish that junction as the common ground point for both your input signal and the reference for your output, the feedback path will cause the junction voltage to go all over the place, which will throw off your readings.

One way to keep the Vcc/2 reference pretty stable is to use an opamp as a voltage follower/buffer. The junction of the divider is connected to the + (noninverting) input, the - (inverting) input is connected to the output. The output of the opamp tracks the voltage at the junction of the divider, plus the input offset of the opamp.

 

Your gain should be 101. With a noninverting opamp configuration, you calculate 1+(R2/R1); so 1+(10k/100) = 1+100 = 101.

You're amplifying the input offset along with the input voltage. Although the "typical" input offset is +/-0.7mV with a +/-7.5v supply, you're using a single supply.
Even then, 0.7mV * 101 = 70.7mV out even with the input grounded and a dual supply. The input offset can go as high as +/-2mV over temperature.

Another problem is how the Vcc/2 reference has been established. If you use two equal resistors from Vcc to GND, their junction will be at a potential of Vcc/2. However, it is high impedance. Unless you establish that junction as the common ground point for both your input signal and the reference for your output, the feedback path will cause the junction voltage to go all over the place, which will throw off your readings.

One way to keep the Vcc/2 reference pretty stable is to use an opamp as a voltage follower/buffer. The junction of the divider is connected to the + (noninverting) input, the - (inverting) input is connected to the output. The output of the opamp tracks the voltage at the junction of the divider, plus the input offset of the opamp.

Many thanks for your reply.

I do use that virtual ground as the ground for both the input and the output so the secondary op-amp should not be necessary right? I appreciate that the op-amp will introduce some small error that does get amplified. Still 300+-70mV or even 200mV for +-2mV is nowhere near the expected 1V, so there must be another large error. What I do note is that as I lower the input voltage the gain increases severely. I just tried again. With 35mv I get 900mv. With 3.5mV I get 500mV. So yeah it really is all over the place.
Maybe I do need a stable Vcc/2 after all...

 

Well, 2mV in should give you 202mV out plus the input offset x 101, not 1v out.
35mV in should give you 3.535 plus the input offset x 101 out.

Why don't you start by just determining what your input offset x gain is?
Connect the noninverting input to your Vcc/2 reference point, and measure Vout using Vcc/2 as the 0v reference point.

Are you certain that your resistor values are accurate? Did you by chance use much lower values than you think you did?

Are you certain that your signal generator's output is isolated from the ground of the power supply? If they are not, that would explain some of these problems.

 

Can you use 2 batteries which will create a true split supply?

Another way is to use a "rail splitter" IC like this:

http://focus.ti.com/lit/ds/symlink/tle2426.pdf

 

This is my circuit.
The source is 12.11V

I input 10mV and get back 300mV still when the gain should be 100.

Could it be the op-amp acting up or is there something wrong with my circuit?

You have biased the wrong input pin.

The non-inverting input pin (+) should be at half rail, not the inverting input pin (-).

 

You have biased the wrong input pin.

The non-inverting input pin (+) should be at half rail, not the inverting input pin (-).

Your statement is not correct.

Have a look at the attached circuit and simulation, and the resulting output plots.

The top plot of V(out) is referenced to ground; the other two plots are referenced to Vcc2.

 

What is the application of your opamp, DC Amp, AC Amp?

 

Your statement is not correct.

Have a look at the attached circuit and simulation, and the resulting output plots.

The top plot of V(out) is referenced to ground; the other two plots are referenced to Vcc2.

Ahh...sorry. I should have read the previous posts properly. I thought that the input was with reference to real ground (0v) rather than virtual ground.

I am more used to using op amps in ac circuits like this one :-

http://www.falstad.com/circuit/#%24+1+5.0E-6+16.817414165184545+61+5.0+43%0Aw+400+176+464+176+0%0AR+512+80+560+80+0+0+40.0+12.0+0.0+0.0+0.5%0AR+528+384+576+384+0+0+40.0+0.0+0.0+0.0+0.5%0Aa+256+176+400+176+1+12.0+0.0+1000000.0%0Aw+512+80+208+80+0%0Ar+208+80+208+160+0+10000.0%0Ar+208+160+208+384+0+10000.0%0Aw+400+384+208+384+0%0Aw+256+160+208+160+0%0Ar+400+176+400+240+0+10000.0%0Ar+400+240+400+304+0+100.0%0Ac+400+304+400+384+0+0.0010+5.996904471275544%0Aw+256+192+256+240+0%0Aw+256+240+400+240+0%0Aw+208+160+176+160+0%0Av+128+160+80+160+0+1+10.0+0.02+0.0+0.0+0.5%0Aw+96+384+176+384+0%0Aw+80+160+80+384+0%0Aw+80+384+96+384+0%0Aw+176+384+208+384+0%0Ac+128+160+176+160+0+9.999999999999999E-5+-6.00071503604615%0Aw+400+384+512+384+0%0Aw+512+384+528+384+0%0Ao+15+64+0+290+0.0390625+9.765625E-5+0+-1%0Ao+0+64+0+290+10.0+9.765625E-5+1+-1%0A