A note on measuring very high power amps (~2 kW) in both single-ended and bridged mode. This includes some discussion and simulations on attenuators as well as amp topologies.
Many thanks for the wiki post. I learned a lot and appreciate the effort you put into this. I do have a question.
What determined the resistors values I have circled in the schematic below? I understand that the ratio between them dictates the attenuation levels, but there are many values to choose from that give the same attenuation. e.g. 10.2K/1.36K/10.2K etc. Is 5.1K/680/5.1K some how better?
Thanks
Hi @Buckskin, it’s a good question. The top-level aim was to make it quick and easy to measure a commercial 1000W amp into 4 ohms. This came from the limits of the desired loads (LPR100 series allows 500W for 5 seconds). Two 8 ohm parts in parallel would give 4 ohms with a 1000W 5 second rating.
Next, at those power levels, the assumption is the amp will be bridged/balanced.
Then we can do the numbers on everything else. 1000W into 4 ohms means 63.2Vrms = 36 dBV. We want to get to the 18 dBV full scale input range (for noise reasons–once the atten is engaged the analyzer noise goes up) and have a bit of margin. So, 24 dB atten would give us 12 dBV input @ 1000W, which gives us some room when using the 18 dBV full scale input.
the QA403 input is an OPA1612. Roughly, you want to keep your source Z around 1kohm or less because the OPA1612 has phenomenal voltage noise, it’s current noise is a bit high. 2kohm source probably isn’t an issue for amps, but 10k definitely is. Additionally, as your input Z climbs, your susceptibility to picking up power line hum increases.
Additionally, if we target 24 dB of max input attenuation, there will be other levels (maybe 12, maybe 18 dB of atten) that will need a higher value resistor.
So, if we pick the small R in our 24 dB attenuator to be 680 ohm, that is well below 1k with room to grow for the other atten levels.
Now, to get 24 dB atten with a small resistor of 680 ohms, that means the big resistors will be 5.11k each. Remember, the value of those doesn’t really matter for noise. It’s all set by the 680 ohm.
Given the Vrms and the big/small resistors, we can calc the current through the series resistors at 6 mA, and the dissipation is 172mW for the 5.11k, and 22mW for the 680 ohm.
OK, so the above there set our peak.
Next, we know the pair of loads is good for 200W continuous per the manufacturer (if properly heat sinked). We can run through similar numbers. 200W continuous means 28.3Vrms = 29 dBV, and we’d want 17 dB to get to 12 dBV full scale. Let’s thus aim for 18 dB atten.
The 5.1k are already determined, so we calc the divider at 1.47k for 18 dB atten. Still pretty good for the 1k noise target.
And then, we run the power numbers again for the unbalanced case. Since we only have a single 5.1k in unbalanced mode, that effectively doubles the current and quadruples the power. The 5.1kohm unbalanced will see 610 mW at 1kw. That sets the 1W requirement on the schematic.
That said, you could probably double the R as you note without an issue. You would measure a bit more noise and see 50/60 Hz just a bit higher. And the reason for doing that would purely be power dissipation in the big divider R. But since it’s easy to find 1W resistors in this package, it makes sense to target the values I did.
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Wow. Great reply. A clear line of thinking that clarifies a lot previous posts here on the forum. Let me unpack this and apply it to my specific case. I suspect things will work out better than my previous attempts.
Thank You.