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Hypex NCx500 Class D Amplifier Review

Rate this amplifier:

  • 1. Poor (headless panther)

    Votes: 3 0.7%
  • 2. Not terrible (postman panther)

    Votes: 6 1.3%
  • 3. Fine (happy panther)

    Votes: 55 12.0%
  • 4. Great (golfing panther)

    Votes: 394 86.0%

  • Total voters
    458

boXem

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I think that thermal management will also play a major role, especially with NADs.
Thermal management affects how much time the max power can be sustained, but not it's value. There is no fancy stuff with it, module temperature is monitored, once it's too high, module shuts down until a reasonable level is reached back.
 

pogo

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And hasn't NAD built in headroom right here to get over 800W at 2ohm?: Link
 
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boXem

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And hasn't NAD built in headroom right here to get over 800W at 2ohm?: Link
Different power supply, different measurement method (burst here). And as written above, accuracy of the short circuit detection limiting the power in 2 Ohm.

Plus one thing that I suspect, even if I have no proof of it: reviewers buy a 2 ohm load but do not measure it. These loads are accurate at +/- 10%. If the measurement is done with a 2.2 Ohm load but the calculation done as if the load was 2 Ohm, a 730 W amp is measured at 800 W.
 

Megaken

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Can someone please explain the whole buffer thing? Gain without buffer is 26db, gain with buffer is something else... What does that actually mean?
 

DrCWO

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Can someone please explain the whole buffer thing? Gain without buffer is 26db, gain with buffer is something else... What does that actually mean?
By default the NCX500 has 10.5dB Gain.

This is not enough for many preamplifiers to get maximum output wattage of the NCx500 modules.
Using a RME ADI-2/4 as DAC preamp for example that delivers high output voltage the 10.5dB are sufficient.
Using a more "normal" preamp you need extra amplification to get up to about 26dB.

This is normally handled by a buffer that resides on the input board of the amplifier manufacturer.
The downside is that this may decrease the SINAD as it adds noise.

Hope that helps
Best DrCWO
 

Julf

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So the buffer stage provides additional gain and a higher input impedance.
 

DrCWO

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I like refer to my post earlier in this thread.

Quick summary:
I listened to my new NCx500 Amp from Rouge-Audio and it was extremely good with a "normal" three way speaker and very poor with my horns and their full-range drivers.

As I am exhibiting at the High End in Munich (Hall 3/M03) I didn't find the time for measurements until now. What I did by curiosity (following the advice of @Rob Fens) was inserting a 5.6 Ohm resistor in front of the drivers. After adjusting gain this improved the result a lot! I did not find time for serious listening but this looks very promising!

I already posted this link to the article from Nelson Pass with measurements on high efficient full-range drivers. Sure, a bit of marketing for FirstWatt too. I googled a bit more and found this scientific paper with very interesting measurements regarding Current and Voltage Drive amplifiers. In his conclusion the author wrote:

"It is shown by measurements with single- and two-tone signals that nonlinear distortions above the fundamental resonance region in moving-coil speakers can be momentously reduced by CC. In an ordinary cone driver, the greatest and most audible modulation products in the midrange and higher typically diminish to a fraction of what they are on VC. Over 30-dB reductions in these products are achieved with drivers having nonconductive coil formers."
CC and VC means Current Control and Voltage Control.

Great findings I guess especially for my setup with a sigle driver (with nonconductive coil formers) connected to the amp without a crossover!

Now my suggestion: Let's modify the NCx500 to be a current drive amp. My idea how this might be done:
The NCx500 modules have a bridged output stage. On each part of this bridge there is a feedback line on the connector. Hypex recommends to connect it to the outputs to eliminate the effects caused by the resistance of the connector pins. Maybe these feedback lines may be used to convert the NCx500 from voltage to current drive.

Any ideas how this can be accomplished?

Best DrCWO
 

Megaken

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By default the NCX500 has 10.5dB Gain.

This is not enough for many preamplifiers to get maximum output wattage of the NCx500 modules.
Using a RME ADI-2/4 as DAC preamp for example that delivers high output voltage the 10.5dB are sufficient.
Using a more "normal" preamp you need extra amplification to get up to about 26dB.

This is normally handled by a buffer that resides on the input board of the amplifier manufacturer.
The downside is that this may decrease the SINAD as it adds noise.

Hope that helps
Best DrCWO
Thank you for the explanation.

I was going to ask how I would know if my preamp is enough, but then noticed that you said "to get maximum output"; as I will never need more than 50watt, can I assume that any preamp will give me that and I don't need to worry about it?
 

Megaken

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So the buffer stage provides additional gain and a higher input impedance.
Then what is this stuff? Sounds like it does more than just db level?

Screenshot_20230515-111331.png
 

Julf

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Then what is this stuff? Sounds like it does more than just db level?

View attachment 285615
If you believe marketing and sales people, even network cables for streamers make a huge difference, not to mention shakti stones on top of your speakers. Notice the usual audiophile marketing thing - it is not stating facts or making claims, just "customers say" or "customers notice" or "customers hear". If you call them on the stuff, they can say "well, our customer told us so".
 

CDMC

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Thank you for the explanation.

I was going to ask how I would know if my preamp is enough, but then noticed that you said "to get maximum output"; as I will never need more than 50watt, can I assume that any preamp will give me that and I don't need to worry about it?
To perform the calculation, you need to not be looking at watts, but the voltage you want the amplifier to hit. This is because amplifiers do not output wattage, they output voltage into an impedance, which results in wattage (Amplifier wattage = voltage squared divided by resistance). For example, to get 50 watts output at 8 ohms requires your amplifier output 20v, but at 4 ohms, it is 14.14v and 2 ohms it is 10v, but at 16 ohms you need 28.28v. Put another way, to maintain the same 28v that gets you 50watts at 16 ohms into 4 ohms, requires 196 watts.

So lets say you need to hit 28v output on your amplifier with 10.5 db of gain. The question then is what input voltage do you need. The answer is 8.4 volts. This of course assumes that your signal being fed to your preamplifier is at -0dbfs, with your preamplifier at maximum gain.
 

Megaken

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To perform the calculation, you need to not be looking at watts, but the voltage you want the amplifier to hit. This is because amplifiers do not output wattage, they output voltage into an impedance, which results in wattage (Amplifier wattage = voltage squared divided by resistance). For example, to get 50 watts output at 8 ohms requires your amplifier output 20v, but at 4 ohms, it is 14.14v and 2 ohms it is 10v, but at 16 ohms you need 28.28v. Put another way, to maintain the same 28v that gets you 50watts at 16 ohms into 4 ohms, requires 196 watts.

So lets say you need to hit 28v output on your amplifier with 10.5 db of gain. The question then is what input voltage do you need. The answer is 8.4 volts. This of course assumes that your signal being fed to your preamplifier is at -0dbfs, with your preamplifier at maximum gain.
Wait, how do I know what voltage I need the amp to hit?
 

Julf

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Wait, how do I know what voltage I need the amp to hit?
Power (P) = voltage (U) * current (I). Current = Voltage / Impedance (R). Thus P = U^2 / R, so U^2 = P * R and thus U = sqrt(P * R). So voltage is the square root of the power you want multiplied by the impedance of your speaker. @CDMC actually stated that, in saying "Amplifier wattage = voltage squared divided by resistance".
 

Megaken

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Power (P) = voltage (U) * current (I). Current = Voltage / Impedance (R). Thus P = U^2 / R, so U^2 = P * R and thus U = sqrt(P * R). So voltage is the square root of the power you want multiplied by the impedance of your speaker. @CDMC actually stated that, in saying "Amplifier wattage = voltage squared divided by resistance".
"voltage is the square root of the power you want multiplied by the impedance of your speaker"

Ok let's say I want 100watt.
Sqrt(100*8ohm) = 28.28volt

Do I have that?
And what does the db have to do with it?

Sorry I know a lot of basic questions but I smoked weed in high school and did liberal arts afterwards.
 

AdamG

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"voltage is the square root of the power you want multiplied by the impedance of your speaker"

Ok let's say I want 100watt.
Sqrt(100*8ohm) = 28.28volt

Do I have that?
And what does the db have to do with it?

Sorry I know a lot of basic questions but I smoked weed in high school and did liberal arts afterwards.
Go create a separate Thread for your basic amp questions. This is a review thread and we try to keep these on topic. There are also plenty of reference threads and libraries here where you can read and learn about this stuff. But not here in this thread.


Thank you for your cooperation and understanding.
 

Megaken

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Go create a separate Thread for your basic amp questions. This is a review thread and we try to keep these on topic. There are also plenty of reference threads and libraries here where you can read and learn about this stuff. But not here in this thread.


Thank you for your cooperation and understanding.
You are absolutely correct, I apologize, feel free to delete, and thank you
 

Julf

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"voltage is the square root of the power you want multiplied by the impedance of your speaker"

Ok let's say I want 100watt.
Sqrt(100*8ohm) = 28.28volt

Do I have that?
And what does the db have to do with it?

Sorry I know a lot of basic questions but I smoked weed in high school and did liberal arts afterwards.
Decibel calculator

And @CDMC already stated "So lets say you need to hit 28v output on your amplifier with 10.5 db of gain. The question then is what input voltage do you need. The answer is 8.4 volts." as well.
 

Rick Sykora

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You are absolutely correct, I apologize, feel free to delete, and thank you

btw, a quick ASR search does not reveal an obvious good article on amplifier gain. For something fairly simple, see here:

 

Megaken

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By default the NCX500 has 10.5dB Gain.

Wait, buckeye page on it says default gain is 26.8db.

Or am I confusing something?

Screenshot_20230515-141728.png
 
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