Function generator cannot generate differential voltage?

Question

I am using the AD620 as instrumentation amplifier to amplify the body's ECG signal. An ECG is a differential signal (measured with two electrodes) and I want a differential gain of \$G_d = 1000\$.

The datasheet has this formula: \$ G_d = 1 + \frac{49.4 \: \text{k}\Omega}{R_G}\$, so using two resistors of \$27 \: \Omega\$ gives me a gain of \$916 \$ which is good enough.

I want to test the amplifier, by generating a 5 mV peak-peak sinusoid, send it into the in-amp and expect around 5 V peak-peak on the output. So I thought of using this setup: -

^{simulate this circuit – Schematic created using CircuitLab}

My instructor suggests generating a 15 mV peak-peak sinusoid and using this setup instead: -

^{simulate this circuit}

Question: Why is the second setup needed to generate a differential signal? What does the second setup do that the first one doesn't?

Answer 1

The first circuit should work ... if you have the "good" power supply.

A generator can't produce "differential" output if not able to do so.

Here is a schematic which can be used to simulate it.

Or this of your instructor.

Note the Vref used ...