Simple Instrumenatation Amplifier Circuit

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The INA118 instrumentation amplifier works perfectly from a 5V supply. But the inputs (the patient) must be biased at half the supply voltage which is +2.5V.
The opamps and +2.5V reference voltage in the schematic biases the patient at +2.5V.

I think I have this setup properly now, thanks for the clarification.

The function of pin 5 on the INA118 and INA326 are completely different so you need to modify the original circuit for it to work with the INA118.

Yeah I've looked at the two datasheets now and I can see the problem - unfortunately I was under the impression that the two components were very similar.

My understanding from the datasheet of the INA118 is that pin 5 is the reference that sets the output offset, while pin 5 on the INA326 sets the value of gain for the second internal stage of the Instrumentation amplifier.

However, a couple of things in the 326's datasheet seem to suggest that the two aren't all that different:-

INA326 Datasheet said:
Offset voltage and drift for other circuit values can be estimated from the following equations:
VOS = 10μV + (50nA)(R2)/G (3)
dVOS/dT = 0.12μV/°C + (0.16nA/°C)(R2)/G

and

**broken link removed**

Based on this it doesn't seem like I'd need to make a whole load of adjustments to get my circuit working with the INA118 - can you offer any insight into what I might need to do?

Thanks
 
hi,
This is in line with what I would expect, the inst amp 'cancels' the common mode input voltage of ~ +2.5V and so the amp output should be near zero.
This is what you have measured.

Ok, good to hear this - sounds like I've got something approaching a correct setup now thankfully...

Taking Ron's point, which inst amp are you working with.???
Please advise and if necessary post the correct datasheet.

Apologies for the confusion caused again - the schematic I'm working off is based on the INA326, however the Instrumentation Amplifier I have to build the circuit with is the INA118...


Understood, this all sounds ok compared to the measurements I've taken...


I actually have electrodes and leads that I could hook myself up to the circuit with if that would be a more suitable way of testing? Otherwise I've try out your suggestion with a potentiometer...

This will give ONLY a dc level change in the output of the amp..

Could you just clarify this point for me please? Varying the resistance of the potentiometer on the input will only alter the DC level of the output, correct? This is because the gain is set by the impedance between pins 1 & 8 I assume...

The remainder of the circuit expects a 'pulse' [ac] signal from the patients heart...
To fully test it you will need to inject low level pulses into the input bridge.

Again, would just attaching the electrodes and hooking myself up be the easiest way to go?

DONT use +/-12V on this circuit as some has suggested.

Don't worry, the last person I'm going to be taking advice from is Mister "Instrumentation Amps are impossible"!:

Do you follow.?

Besides the multiple areas in which I've asked for clarification I follow perfectly! Thanks for all your help so far...

Why have you taken on such a complex project.?

I'm doing a project based on Digital Signal Processing of ECG signals. The signal processing is the most important part and that's all in hand, but being able to take a real signal from a patient would be very impressive compared to simply using code-generated datapoints or the like...
 

As you can understand the ECG expects 'electrical potentials' between the body electrodes, that are generated when the heart muscles are working.

With that simple test with a pot, its just to check that the instamp is working at a 'dc' level, using a 'dc' signal due to an imbalance in the 'bridge'.

I would NOT connect any electrodes to your body UNTIL the unit has been approved by a competent engineer..... do you have a cat.?
 

Cool, I'll give that test a ty tomorrow so, and I won't hook myself up until i can confirm that small animals aren't fried by the circuit .

In the meantime I've been having a look at the ECG Amplifier circuit from the INA118 datasheet, shown here:-

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I'm going to try and build that as well as it's quite a bit more basic and it's based on the right IC. I'm confused a little about the cylinder encircling the two arm inputs though. This looks like insulation or something, but what does that mean with regard to the output of the right-most opamp?

Also, I'm assuming this circuit is dual supply - am I right in thinking this?
 
Eric, the only reason changing one of the 40k resistors will create a differential voltage on the input is due to input bias current. That doesn't sound like a great idea.
 
I had already deleted that post when you replied, because I saw your INA118 ECG schematic.
 
Pins 2 & 3 for some reason show a voltage of around 0.48V that I don't know the reason for.

This is your problem. The common mode input range on the INA118 only extends down to 0.98V for single supply operation. With a common mode of .48V it won't work in single supply, but it will for dual supply. Second paragraph, page 10 of the data sheet for reference.
 

Thanks for the tip. If I were to set up a bridge circuit like this one would it help?

**broken link removed**

 
The cylinder is the shield over the cable (you must use shielded twisted pair cable to minimize noise pickup). The shield is driven by the buffered average of the two signals to reduce the attenuation of the signal by the shield capacitance.

Yes, that circuit will only work with a dual supply.
 
If you are using an A/D that can't handle negative input voltages, there is no need for dual supplies. Since you are essentially measuring an AC voltage, absolute DC accuracy is not required, so a single-supply system, with the input common mode voltage set at approximately the middle of the input common mode range, and the amplifier output set at approximately the A/D's midscale voltage, is sufficient. If both those voltages are the same (probably ≈Vcc/2), then you only need one reference.
 
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The cylinder is the shield over the cable (you must use shielded twisted pair cable to minimize noise pickup). The shield is driven by the buffered average of the two signals to reduce the attenuation of the signal by the shield capacitance.

Cool that makes sense, thanks. Is there an alternative to using a shielded twisted pair cable for this application? I don't have access to one at the minute...
 
Cool that makes sense, thanks. Is there an alternative to using a shielded twisted pair cable for this application? I don't have access to one at the minute...
Not really...

Also, see my post just above yours, which you might have missed.
 
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Thanks, I had missed this post originally - I'll keep what you say in mind when prograaming the ADC.

As far as the twisted pair shielded cable goes, would you guys be able to suggest a suitable one for this application? I'll probably order one tomorrow...

Thanks for all the help so far by the way - I appreciate it...
 
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Eric, the only reason changing one of the 40k resistors will create a differential voltage on the input is due to input bias current. That doesn't sound like a great idea.

Its only as amp check, the post explains it will ONLY create a diff voltage, after the check its will be removed.
 
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hi spkrguy,
Ive been telling him this requirement for a couple of days, as you say its in the datasheet
 
Its only as amp check, the post explains it will ONLY create a diff voltage, after the check its will be removed.
I understand that. But the differential voltage created is low and totally unpredictable, as the input offset current is ±10nA max. You could get +400uV, -400uV, or any voltage in between, including zero.
 
I understand that. But the differential voltage created is low and totally unpredictable, as the input offset current is ±10nA max. You could get +400uV, -400uV, or any voltage in between, including zero.

I am trying suggest a quick test in some way, other than the OP strapping himself to the pads.

I did ask if he has a cat.

This is one of the many problems that 'hobbyists' run into when building gear like this, they have not got any suitable test gear.
 
Ok, so I have the entire circuit working now (thank god). I'm getting an ECG signal out of roughly 1V amplitude which is what I was looking for. All I need now is a quick method of isolating the patient from the power supply for safety.

Is there a quick, easy way to do this? Like with diodes/capacitors maybe?
 
I have seen this same reply on two other websites.
You need to use a medically-approved isolation part. A cheap unapproved transformer might short and electrocute the patient. A opto device might also short.
 
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