Once again, thank you.
Life is getting in the way so I won’t be able to run the suggested tests for a few days.
My configuration has remained constant over the course of this thread. No scopes. Just the QA402, the attenuation network and my BNC cables with mini grabber adaptors. A DVM was added when you requested some voltage measurements, but immediately removed after that.
OK, thanks. If the system is totally at rest (not running acquisition, the DACs are off) and you connect things up, there’s no overload clicking right? And then if you do a control+space to run a single trace, the overload clicking starts even though the value into the analyzer is 18 dB - 40 = -22 dBV and the input range is 6 dBV?
I haven’t had time to run your previously suggested tests but I will. The one suggested in your last post was easy. So here are the results.
Correct. No clicking.
Correct. Clicking only starts at 6dBV and below. Single acquisition time domain trace shown below.
And just for grins, single acquisition trace at 12dBV input range for comparison. No clicking.
While it sounds like Buckskin may have some protection response issue within his QA402, I need to return to the basics of the attenuator shown and it’s appropriateness for power amplifier measurements. The circuit shown is is a floating attenuator, with no ground connection.With these values it provides 40 dB of attenuation to the differential mode signal, but 0 dB attenuation to the common mode signal. This means an automatic 40 dB reduction of CMRR. Not good. When driven with a single-ended source, where one of the input terminals is grounded, the full common mode signal, which is 1/2 of the total voltage, will appear on each output terminal with a tiny (-40dB) difference between them. This really doesn’t protect the analyzer input from the large amplifier output voltage swings. If the attenuator is fed from a symmetrical opposing voltage source, such as a bridged mono amplifier, then there is little to no common mode signal to begin with and the analyzer is well protected.
It’s much better to split the 200 Ohm resistor into two 100 Ohm resistors with a ground connection between them, connected to the analyzer ground. This provides equal attenuation to common mode and differential mode signals. While the CMRR is more dependent on precisely matched resistors, the net effect if using reasonably precise resistors, is still better than not attenuating common mode voltage at all. Everyone likes to brag on their CMRR but no one wants to talk about their common mode voltage range, except for the transformer people.
With modern power amplifiers it’s best to determine the output stage configuration with an oscilloscope before proceeding further. If it’s Class D there will be some clocking frequency residue which you should take into consideration in any measurements. The speaker output marked - may be grounded or driven, where a single channel is actually a bridged pair internally. You might even find the + side grounded and only the - driven, typically on even numbered channels while the odd numbered channels are the opposite.(check the amplifier manual on how to wire a bridged output) You may find a significant level of common mode DC voltage riding on both outputs. None of this affects the loudspeaker in any way, since it’s a fully floating differential device. It’s the wild west out there, it’s not safe to assume anything.
Dale
@matt
Sorry about the delay on these tests.
Got them all except the last one. I’m a bit confused on what is meant my balanced loop back with only a 10K and 200 ohm resistor. However, I think your subsequent request for a single cycle covered it adequately. Single cycle with a smoking gun spike posted above. Correct me if I’m wrong.
DAC outputs are identical and inverted.
Done. All good.
Done. No clicking in either case.
I’m confused. See the first few lines in this post.
Yes, @Buckskin, I agree this is indeed the smoking gun. Can you contact support at quantasylum and Jan will get you a USPS label to return the product (assuming you are in the US).
thank You. Its on its way.
Hi,
I’m measuring power amps (valve amps, transistor amps, digi amps, grounded and bridged ones up to 800-1000W).
I use a power resistor switchable 8 Ohm 500W, 4 Ohm and 2 Ohm 1000W.
Until now, I use a differential probe connected to the load with an attentution 1:50/1:500 (Micsig DP10013), max input voltage 130/1300V, followed by an UAD Volt 2 interface connected via USB to a laptop with Windows 10.
The whole measurement side is without any connection to ground (diff probe, Volt 2, laptop).
It works, but with low signals at the inputs I get poor results due to the fix attentuation 1:50 (sn ratio).
If I want to measure amps up to 1000W, I need an input voltage capability of 100-120V RMS.
Now I’m thinking about a QA403 instead of the diff probe.
My schematic:
Diff input L+ L- to the load, no ground connection to DUT and load
Between load and diff input of the QA403 a symmetrical attentuation network 1:3 like that one shown in this thread (3 resistors)
The DUT will be feeded symmetrical by the L+ and L- output of the QA403, no ground connection to load or DUT but cable shielding connected to the QA403 ground
Will this work properly without any problems?
I painted a simplified schematic of my planned setup:
Of cause it is a bit different with a bridged amp.
No comments? I want to know if it works in this way before I buy the QA403.
Thanks
Chr
Hi @cfortner, overall this looks like a solid setup. But there is a bit of a wrench lurking inside the setup that isn’t surmountable, but needs to be understood since you are working towards a universal setup for all topologies.
Let’s start with SPICE to understand the external divider. In the picture below, I’m using a voltage controlled current source to generate a balanced signal. This is TI’s preferred way of nudging SPICE to do what is needed. I wish I could find the TI engineer’s name that did the blog post on this, but I cannot. But in short, it’s straightforward. We take a voltage source with a peak voltage of 1.41V. And then we use two voltage controlled voltage sources to generate two sines of half amplitude AND opposite polarity (VCVS2 has a gain of MINUS 500m)
For the simulation, we’ll drive at 100Vrms (141Vp) and the attenuator should give us a factor of 100 attenuation. When we run the transient simulation, we can clearly see the stimulus at 141Vp:
Now, the divided signals are too hard to see, so let’s just plot those. Here we can see the differential signal is precisely as expected (divided by 100) with a peak of 1.41V. And we can see each leg of the divider is half of that and out of phase by 180 degrees. This makes perfect sense, and feeding this signal into the QA403 balanced inputs would agree.
OK, now let’s short one side of the divider as you have in your schematic.
And now run the simulation again:
And we can see the gold trace (VM1) is what we’d expect at 0.5Vrms. And that’s well within the range of the QA403 balanced input. HOWEVER, look at each leg of the balanced input. One leg is swinging about +/-35.6Vp, and the other leg is swinging +/- 34.77Vp. And so, the gold difference signal (VM1) is correct. But this requires the QA403 inputs to handle massive swings. 36Vp=25.5Vrms=28 dBV. And it can, but you need to be at a much higher full scale input level, which hurts your noise. It’s much better to let the attenuator do the heavy lifting outside of the QA403.
So, the summary here is this: If you want to measure a balanced amplifier that is actually driving push pull, then your setup (minus the DUT ground on the output) is correct. In that case, you’d want to establish the ground on the input of the DUT.
If you want to measure a single-ended amp (one where the + is driven and the - is ground), then you’d want to short/bypass the lower R3 resistor. In that case, the - output should read as a short to the input ground of the amp.
And finally, be aware that the BNC shells on your scope are usually at ground. That is, you can touch your DVM in continuity mode to the shell of the scope and to the ground prong in the wall and read continuity. This means if you clip your scope ground lead to the - of a push-pull amp, some bad things can happen. The class D amps will usually cope with the faults. But other topologies may not.
Can be tricky stuff, so sorry if too verbose!
Hi Matt,
thanks a lot for your explaination! But naturally I have some more questions.
This is the schematic measuring a bridged amp:
Grounding only affects the wire from left balanced minus output (BNC pin) to the input of the DUT.
With a not bridged, normal amp (speaker minus on DUT ground) it looks like that, right?
What I don’t understand: For the signal from QA403 to amp input I’m using the symmetrical output signal from L + and L -.
The wires in beetween the red line on the next picture are on DUT ground level:
The QA403 ground isn’t on DUT ground level except L minus (BNC pin). In my understanding the input of the QA403 input is seeing a symmetrical signal independent from ground, because the ground of the whole measurement site isn’t on DUT-ground. Am I wrong? Where is my fault?
The DVM and the oscilloscope in parallel of the load are fix installed and connected and batterie powered without ground connection. They will not been used for other things.
I now changed the attentuation from 3:1 to 10:1. The input signals at lower levels will be in the range of some mVrms. Isn’t it to low for a reliable groundfloor? If it is enough level, does it make sense to attentuate 100:1 and use well shielded cables from the attentuator to the QA403?
Thanks for your time
Christian
Another thought:
What would happen, if I use an external ground free signal generator instead of the internal of the QA403?
This is the schematic measuring a bridged amp:
Yes, this makes sense to me
With a not bridged, normal amp (speaker minus on DUT ground) it looks like that, right?
In this config, R3 should be 0 ohms to ensure you don’t have the imbalance outlined above. And yes, the wires going into the QA403 inputs should be shielded. But that shield connection comes from the BNC shells. And all BNC shells (input and output) are tied together on the QA403 (as you have drawn).
The QA403 ground isn’t on DUT ground level except L minus (BNC pin). In my understanding the input of the QA403 input is seeing a symmetrical signal independent from ground, because the ground of the whole measurement site isn’t on DUT-ground. Am I wrong? Where is my fault?
You don’t want the L- output driving into a short to ground (as you have drawn). This will result in excessive heating of the output opamp. Instead, if you have an unbalanced input to your amp, tie the minus to ground (as you have done), and use the QA403 L+ output to drive the + input on the DUT. And then establish a ground connection between the QA403 USB shells and the DUT via this same cable. So, that usually means and RCA from QA403 L+ out to the DUT. The center pin of the RCA carries the signal, and the shield of the RCA establishes the ground reference between the DUT and QA403.
Isn’t it to low for a reliable groundfloor?
That depends on your amp. For example, a LM3886 has an SNR of 92 dB. 60W into 4 ohms is 16Vrms = 24 dBV, so the noise floor will be -68 dBV at full power (check my math!). The LM3886 has a noise floor of 2uV when sitting idle, which is -114 dBV. That you’d want to measure on the 0 dBV input range for sure (no atten) if you are building an LM3886 with a gain of 1.
A TPA3255 (class D) has a noise floor of 85 uV, which is -81 dBV. Easy to measure in a single shot, even with a 20 dB attenuator.
And so, your measurement strategy will probably need to be multi-tiered and based on the amp quality. That is, be ready to make your next measurement based on the results of your current measurement. And at each step, know how far away you are from the limits of the test setup.
What would happen, if I use an external ground free signal generator instead of the internal of the QA403?
You will probably want to establish a ground for sure. The QA403 and your test setup will be immersed in power line energy. If you don’t establish a ground, then you are really relying on the hope that both the + and - inputs are seeing that energy equally AND that the CMRR of the QA403 and setup will be sufficient to knock that down. Often times, if you have a floating setup and are looking at strong power-line components, you can touch an earth ground to the BNC shells and see the power line component drop by 30 or 40 dB! In these cases, your measurement is probably capturing a lot of power line energy (suggesting a higher-z measurement) AND you’ve introduced an imbalance someplace (unequal source impedances) which will hammer the CMRR of the analyzer.
I like to have these where I’m working. You plug it into a wall outlet. The hot and neutral pins aren’t connected (they are plastic). The earth lug is broken out to a banana connector. Then, you have an easy way to touch the banana wire around your circuit to see where the grounding point wants to be. The ground clip on a scope probe connected a scope (even if the scope is off) works too.
Usually, it makes the most sense to establish a ground on the input side (and the RCA input to an amp usually has its shield tied to ground, so you pick it up there whether you want it or not).
Wow, it’s getting clearer.
This is a drawing with a switch grounded amp / bridged amp (I removed the DMV and Osci for better survey):
I connected the QA403 out L+ to DUT input + and the BNC shell of L+ to DUT- /ground. Is that correct? Should I terminate BNC L- with a 50 Ohm-plug?
I left the QA403 input from the load symmetrical to Input L+ and L-.
Because I want to measure normal and bridged amps, I put a switch to be able to set the negative part of the attentuator to ground, closed for asym amps and opened for bridged amps.
Is that correct?
Edit: + means hot, - means cold at amp and load
Hi @cfortner. The unused outputs of the QA403 must never be terminated, but must be left open. Only the unused inputs of the QA403 must be terminated at 50/75 Ohm-plug.
Thanks for the advice, thats logical…
Here a schematic of my measurement construction with a switch asymm/symm and a second switch attentuation 10:1/100:1
Will this work like expected?
Thanks a lot
Christian
Hi @cfortner, yes, I think so. Your output impedance of your divider can be quite high–about 10K it looks like. That will have a pretty high noise floor, especially with the bipolar input amps on the QA403.
I’ve built a lot of load boards over the years. My current favorite config is something like this:
This has two Arcol 8 ohm loads in parallel, which gives 4 ohms. I’ve written a lot about these loads before. Per the load spec, this gives 200W continuous, and up to 1000W for 5 seconds. The mounting plate is 5mm. This board has a single divider.
Another version of the of the board (which hasn’t been built yet) adds jumpers and a fuse.
Jumpers off 0, 18 and 24 dB of attenuation. Note the resistors are 1W. But the output Z is around 1K or so, which is about where you want to be for the OPA1612 input amps on the QA403. There’s also a fuse that is good for up to 16A.
1000W into 4 ohms is 63Vrms = 36 dBV, and the 24 dB attenuator will knock that down to 12 dBV. That would let you sit on the 12 dBV input for THDN measurements, and the 18 dBV input for THD measurements.
If you put everything into excel, and figure you want a max of 500W into 8 ohms, and 1000W into 4 ohms, and 2000W into 2 ohms, then you have the same voltage for all 3 configs. More generality is better, of course. But it’s fairly hard to find a 1000W into 8 ohm amp. The pro stuff seems to fall into line with the configs I outlined.
In any case, everyone’s needs are different! I think your circuit will serve you well for years to come. You can play with the noise penalties you are thinking about by sticking resistors on the input of the Qa403 and looking at noise floor. I think you’ll find for several K input Z, it gets really hard to get rid of the 50/60 Hz.
Thanks a lot, Matt!
I’ll change the divider to lower values, even if I don’t want to measure ultra high quality amps.
Today I ordered the QA403 here in Germany, it‘s offered by the Electronic Magazine Elektor.
