Hooking unbalanced test equipment to balanced circuit

[twelvetone]

I've got a question about unbalanced test equipment being hooked to balanced gear.

For example:

If I send a balanced 0dBu 1kHz sine wave to my gear on an XLR connector, that'll be 2.19V p-p. This means it would be 1.095V on each leg of the balanced connector, right? (pin 1-pin2, and pin1-pin3 on XLR).

That way, when an unbalancing circuit sums them (and inverts the pin1-pin3 signal), they'll sum to 2.19V p-p, or .775V rms.

Right?

So, if I have an unbalanced signal generator, and I hook it up at 2.19V p-p between pins 1 and 2, that will give essentially the same result, because there will be no input on pin 1-pin3, so there will be no addition to the signal; we just won't get the benefit of CMRR.

Does that sound right? Anybody know?

Thanks!
Mike

[tpryan]

If I send a balanced 0dBu 1kHz sine wave to my gear on an XLR connector, that'll be 2.19V p-p. This means it would be 1.095V on each leg of the balanced connector, right? (pin 1-pin2, and pin1-pin3 on XLR).

Maybe, maybe not, and it doesn't matter to the differential input anyway. All that matters is the difference in voltage between pins 2 and 3.

So, if I have an unbalanced signal generator, and I hook it up at 2.19V p-p between pins 1 and 2, that will give essentially the same result, because there will be no input on pin 1-pin3, so there will be no addition to the signal; we just won't get the benefit of CMRR.

Connect your generator to pins 2 and 3, hot to pin 2 and ground to pin 3. It's not necessary for basic testing, but for better common mode performance you can insert a resistor between the pin 3 connection and ground at the generator end of the cable. The R value should match the output impedance of the generator.