# Function generator cannot generate differential voltage?

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?

• Probably nearly all of the signal generators I've ever seen don't have a differential output. Mar 8 at 15:22
• @Andyaka Okay, but does that matter? Won’t the first setup do the trick if all I want to check is the differential gain?
– Carl
Mar 8 at 15:44
• It will but, it won't prove that a common mode level of non-zero has an unwanted effect maybe. Mar 8 at 15:46
• Check your signal generator. The usual BNC connector is grounded on some but not on others. If it is grounded then the network suggested by your instructor could be used. Not grounded does not imply fully differential though you can still investigate common mode effects. Mar 8 at 16:41