# INA126: Amplifying a very small A/C voltage

Part Number: INA126

Hi, I'm trying to build an EMF detector which can detect very low levels of EMF.

I have a coil which outputs approx 0.2V AC when subjected to very high levels of EMF from a demagnetiser (50Hz?). But I want to amplify the coil's output so that I can easily detect very low levels of EMF with it.

I am using an instrumentation amplification IC (INA126PA) to amplify the 0.2V AC signal and an AC voltmeter to measure the output. I want this to be about 100V AC, but at the moment I am getting OV, with or without a 100 ohm external resistor (RG)

I have included a circuit diagram below. The circuit is currently on a breadboard. My 3V supply is from 2 x 1.5V batteries.

I'm sure I have done something really silly. Can anyone please tell me (simply) what it is?

Regards,

Alan.

• In reply to Peter Iliya:

Alan,

Another thing to consider is to AC couple your signal since you have to be careful about amplifying the input offset voltage of your INA. You can remove this DC offset by putting an integrator in the feedback into the reference pin. Although you will want the integrator's cutoff frequency (1/2piC1R1) to be less than the frequency of the coil signal you are trying to amplify. You also want this integrator amplifier to have low input bias current.

• In reply to Peter Iliya:

Hi Peter,

I honestly have no idea what frequency will be introduced into the coil. I guess it could be anything from 50Hz upwards, including radio frequencies and mobile phones. In fact I hope to use a frequency counter in series with the output (via a very small capacitor) to find out the frequencies of signals obtained.

If you think that coupling it with an op-amp will work, I will certainly try it. The circuit diagram you have provided in your second post looks very interesting. Am I right in thinking that Vref and Vout2 are the amplified outputs, while Vout1 and earth are the current ones? Does V29 and V19 mean 29V and 19V respectively? Only I was hoping to keep my supply voltage down to 18V (2 x 9V batteries) as it is meant to be a portable device.

Regards,

Alan Morris

• In reply to Alan Morris:

In the previous circuit, Vref is the DC offset of the output from the INA. It is created by integrating the output of the INA at a frequency of 1.59 Hz and is fed into the reference pin. Effectively the gained up input voltage offset of the INA is then subtracted from the output. Vout1 is the signal after its first amplification from the INA. This is fed into the OPA388 with respect to ground and amplfied again to result in the final amplified signal Vout2. "V2 9" and "V1 9" are both 9V batteries labeled as nodes V2 and V1.

The noise of the amps you choose will be critical, but it seems you will need some bandwidth if you plan on amplifying radio and mobile phone frequencies.

Here is another circuit that's using the INA163, a very low noise INA that will reduce your noise by half compared to the INA126. You can run the simulation in TINA-TI and see how it can amplfy a 1uVpp/50Hz signal to a 1Vpp at Vout2.

ina163e2e.TSC

• In reply to Peter Iliya:

Very many thanks Peter,

I will start ordering the parts right away.

Regards,

Alan.

• In reply to Peter Iliya:

Hi Peter, 17.5.17

Re. the circuit diagram you sent me on 5.5.17.

I bought the components and assembled them on a breadboard to check performance with my AC voltmeter connected to Vout and Vref. No meter response was obtained, but the OPA388 chip got extremely hot instantly, so I had to disconnect.

I attach the diagram you sent me with added terminal numbers for OPA192IDR and OPA388IDR, which I deduced from their technical data sheets. I had to assume that the vertical stripe in front of the lettering on the chips was the equivalent to the large dot on the INA163UA/2K5, so presumed the tag just underneath it was no.1. The other diagram I attach is of my actual wiring and so may explain the position of this stripe better.

NB : As the amps were all Surface Mounts, I had to solder them to IC adaptors, which meant that on the breadboard, the chips are actually at 90o to the adaptors. But to avoid confusion, I have shown them as if they were the 'though hole' type.

Could you please tell me if I have misinterpreted your diagram or incorrectly labelled the tags on the chips?

Or could the OPA388 chip be faulty?

Regards,

Alan Morris

• In reply to Alan Morris:

Hello Alan,

The reason the OPA388 is overheating is because the supply +/-9V is above the absolute maximum rating of +/-3V or 6V single supply. I would recommend reducing the supply voltage to +/-2.5V.

Best,

Errol Leon
Texas Instruments
Precision Op Amp Applications
• In reply to Errol Leon6:

Hi Errol & Peter,

Thanks for this information. By using the INA163 circuit Peter supplied (see previois post) and putting a 100 ohm resistor between the +9v and the OPA388, and another between the -9v and the OPA 388, I was able to reduce the voltage supply fto the OPA388 from +/-9v to +/- 1.8v.

I know this is short of the optimum +/- 2.5v, but I still expected to see a big (100x) improvement over the amplification I got with the INA126 on its own (see below),. But instead I got much the same as before. Is this because we can't use +/- 9v source on the OPA388 as planned? In which case could it be replaced by a suitable OP amp that can take this voltage?  Or can you suggest another solution?

I haven't yet tried the earlier design of Peter's where INA126 was used in combination with OPA192 and OPA388, but since OPA388 seems to be the source of the problem, I doubt if that would be any better.

Regards,

Alan Morris

• In reply to Alan Morris:

Alan,

I aplogize for the confusion about supplying the OPA388 with a +-9V supply in earlier schematics, but simply connecting a 100-Ohm resistor in-series with the 9V battery and the supply rails of the party will not bring down the voltage supplied to the OPA388. It will only increase the source resistance which will affect output distortion and present possible stability issues. Maybe I am misinterpreting this. The power supplied to the OPA388 needs to be +2.5V on V+ and -2.5V on V-. You could accomplish with a resistor divider network and buffer this output into the OPA388.

If you don't want to do this, you can use another part that can accept +-9V supply like the OPA192.
• In reply to Alan Morris:

Hello Alan,

To add to Peter's response you can use OPA2192 which is the dual channel of the OPA192.

Best,

Errol

• In reply to Peter Iliya:

Thanks to both of you,

I had measured the potential difference between tab 4 and tab 7 and got 3.6v with the two resistors in place, then assumed this meant +/-1.8v supply to the OPA388, but I admit to not being very sure of myself in my understanding of this. It seems I have 3 options available :-

1) build a resistor divider - but this will surely be a drain on my batteries
2) replace the OPA388 with another OPA192
3) replace both the OPA192 and OPA388 with an OPA2192

Of these, no.3 sounds the simplest. Is there any chance of one of you updating my circuit diagram for this option? I can get the OPA2192 from Farnell, but I would also want to get any other resistors or capacitors at the same time.

Sorry to take up so much of your time, but I am very keen for this to work.

Best Regards,

Alan Morris