0
\$\begingroup\$

I have a d-type amplifier board which has "floating" outputs, i.e. the -ve output terminals are not connected to ground.

I want to measure the power output and clipping point of this board with an oscilloscope.

With a common ground amplifier I would connect my probe to the +ve terminal of my load resistor, and the ground clip to the -ve terminal of the load resistor, adjust volume until I hit the clipping point, and then use VRMS to calculate my power.

I was going to do the same thing with this floating output amplifier, but I remembered from oscilloscope 101 that my ground clip should ALWAYS be connected to ground.

To measure power, should I connect my ground clip to the board's GND terminal, or to the -ve terminal of the load resistor? Are both ways "safe"? What is the scope actually measuring in these scenarios?

I have a USB powered scope which I believe is grounded via the USB plug (if that makes a difference).

\$\endgroup\$
1
  • \$\begingroup\$ Note that the power you measure using a resistive load may not be the power you will see going into an inductive load like a speaker. \$\endgroup\$
    – ocrdu
    Commented Sep 10, 2022 at 10:36

1 Answer 1

2
\$\begingroup\$

Your caution is wise. I would recommend that you measure the peak-to-peak voltage of one output relative to ground. The load will see double that voltage as the two outputs will be in anti-phase.

So if I measure a Vpp of 20 V, my Vrms will be approx 7 V, so my power would be 72/8 Ω = 6.125 W and I would multiply that by 2 to get my true power output per channel of 12.25 W.

No, it's better than that. The beauty of the bridge design is that while one side goes up, the other goes down so the voltage seen by the load is 40 Vp-p = 14 VRMS. Power will then be V2 / R = 142 / 8 = 24.5 W.

The error you made was to double the power. You should have doubled the voltage which will quadruple the power.

\$\endgroup\$
2
  • \$\begingroup\$ Ok, so if I measure a Vpp of 20V, my Vrms will be approx 7V, so my power would be 7^2/8ohm = 6.125W.. and I would multiply that by 2 to get my true power output per channel of 12.25W ? \$\endgroup\$
    – pigfrown
    Commented Sep 10, 2022 at 11:01
  • \$\begingroup\$ Good news. See the update! \$\endgroup\$
    – Transistor
    Commented Sep 10, 2022 at 12:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.