Good morning everyone,
I’m getting a Quantasylum QA403 Audio Analyzer next week. Here’s my question for the experts on the forum:
I want to measure the THD of an audio amplifier with 50 watts per channel.
I wanted to terminate the speaker outputs with 8 ohm load resistors (instead of speakers).
How can I connect the QA403 now?
Tap the voltage at the resistors in parallel?
I’m worried that the voltages are too high for the QA403 input.
Greetings from Germany
Hi @Diskus. I believe that at this link you will find all the information you need to use the QA403 correctly:
50w into 8ohms is 20V which is well under the max voltage input to the QA403, so you do not need to do any tapping.
Nevertheless a 70V or 80V RMS capability would would be great, no further problems with attentuator-management.
There are a lot guys out there with amps over 500W and more.
It’s 50W, not 500W he’s asking about. And sure, 80V RMS would be wonderful, but you won’t get the noise figures and you certainly won’t get it all in a USD $599 box.
The QA403 is in a very sweet, sweet spot for performance and price. You want to feed more voltage in- build a simple outboard attenuator. If you don’t want to do that, shell out 20-30x more and buy an AP.
You’re right, the QA403 is a very great piece of gear at a very good price and I’m very satisfied.
It’s more of a dream in terms of possible overloads when measuring high-performance amps.
Hi @cfortner, I think the outboard attenuator is the right path. There’s a section below about building an atten for the type of amp you are measuring.
Hi,
I stumbled over the same limitation measuring power amplifiers with higher wattage and higher output voltages that the QA403 can safely handle. I designed and successfully built this attenuator. It either passes the signal as is, or attenuates iby precisely 6dB or 10dB (neat figures that simplifies the following math). With the higher attenuation selected, you can measure voltages up to 177 Vpeak with the QA403. That equals already continuous 2kW into 8 ohms. The input resistancesof the QA403 are taken into account.
This is a neat alternative if you want to build it yourself, and it uses no exotic parts:
Enjoy,
Christian
Christian,
Thanks for posting this schematic. I don’t understand the 3 modes. (1) you always add the load of the 17K8 resistor. (2) you add a series resistance of (a) 17K8 or (b) 17K8+15K. I only see 2 modes this way, since I don’t see the surpass switch.
Regards, Gerrit
Hi Gerrit,
If you read the text in the schematic, it mentions that the switch is a 3-position version, officially of the type ON-OFF-ON. So there is a center position (-10dB) where R1/R3 (R2/R4) are connected in series. That should answer your question 
Perhaps the problem is that I measure single ended instead of balanced? R5 / R6 are always in place and will affect (load) the input signal the way I see it. These resistors are not bypassed by a switch, so it’s not straight through. I still don’t get it from a single ended point of view.
A 17k8 load can be totally ignored when you are measuring an amplifier with sub-ohm output impedance that is capable of driving 2,4 or 8 ohm loads. Your 8 ohm load will reduce to 7.996 ohm due to the R5/R6 resistor(s)… Can be completely neglected.
IF you measure single-ended then you can completely ignore the lower half of the attenuator, either terminate the analyzer minus input to GND (or via a 50 ohm terminator), or alternatively connect that input to the amplifier output (speaker) GND. This will improve your measurements by removing any common mode signals in your setup. I usually always measure differentially to get the most precise results.
