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I am using the LT1168 instrumentation amplifier for my project and I am trying to figure out how to make it work with a Li-Poly 3.7 Volts battery. I am not able to understand the single supply operation theory from the datasheet. I have given +3.7 volts to the V+ pin and the V- pin is grounded but I am unsure what has to be given to the Ref pin of the opamp.

Here's the link to the datasheet: http://cds.linear.com/docs/en/datasheet/1168fa.pdf

It would be really helpful if you guys can kindly help me with the connections to make it work with a single supply. Thanks.

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Your main problem: -

Wide Supply Range: ±2.3V to ±18V

This means that the minimum single supply voltage is 4.6 volts i.e. twice 2.3 volts. Trying to run it from a 3.7 volt battery is always going to cause anguish. Even if you suppied it 4.6 volts you have to be aware of what the output can deliver under load. Take a look at this in the data sheet: -

enter image description here

If you tried to drive a load that took peaks of (say) 10 mA, the output voltage range is 1.3 volts to 3.3 volts.

I think you have chosen the LT1168 without too much thought.

Of course you could always boost the 3.6 volts up to (maybe) 6 volts or more using a boost converter.

Pin 5 (reference input) would normally be provided with a mid-rail voltage. This can be achieved with a resistor potential divider. Use low value resistors (i.e. 1 kohm) because the ref input has an impedance of 60 kohm.

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  • \$\begingroup\$ Yes I didn't think that it could not operate on single supply with low voltages. Thanks for your detailed explanation. I will have to buy the voltage inverter ic to provide negative supply now. \$\endgroup\$ – Nkvk Jan 30 '18 at 3:42

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