Hi,
I want to change the circuit of my load resistor to this:
In my opinion it would be possible to measure normal and bridged amps in this way, correct?
That circuit makes the outputs ride at 1/2 the actual amp output voltage, so can only increases the headroom by 6dB.
Perhaps use high power load resistor, but two indepedent identical low power resistive dividers to ground to bring the output voltages down by a factor of 10 or so. However that means there are ground currents from the dividers that have to return to the amplifier ground somehow. So 8 or 4 ohm dummy load, and say 0.1% 8k2 : 910R dividers to drop by 20.0dB (those would dissipate around 1W each with 100Vrms in, with ground currents around 10mA with negligible Johnson noise)
I just built an attenuator like this, but I get too much hum with a resistor of 1000 ohms.
Thats with shortened input cable of the QA403:
And I get this with a resistance of 1000 ohms:
Are all your leads shielded? I get nothing like that with 6k8 resistor and short cables, either single-ended or differential.
What about using a 10x oscilloscope probe to the QA40x input?
As far as I know an oscilloscope probe needs an input impedance of 1M. The QA403 has an input impedance of appr 100k, I don‘t believe that it works.