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Power Amplifier [150Euros]
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Old 21st August 2019, 04:26 PM   #1
Entas is offline Entas  Lithuania
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Default Power Amplifier [150Euros]

Hello,

Checking some Power Amps, the one that I like the t.amp E-400 Thomann Lithuania. Can I find better one than this? Is this thing actually have 190Wrms @ 4ohm?
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Old 21st August 2019, 08:50 PM   #2
Printer2 is offline Printer2  Canada
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At about 7:00 I am wondering on the heat sink area if it actually is Class AB, that is unless it has a fan.
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The T Amp E400 | Audiokarma Home Audio Stereo Discussion Forums

I do not see why it would not do it.
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Old 22nd August 2019, 07:18 AM   #3
Gnobuddy is offline Gnobuddy  Canada
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Quote:
Originally Posted by Printer2 View Post
...unless it has a fan.
Looks like it does use a fan (forced air cooling) - see attached image.

-Gnobuddy
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File Type: gif Fan_Cooling.gif (99.6 KB, 169 views)
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Old 22nd August 2019, 12:25 PM   #4
Entas is offline Entas  Lithuania
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Actually this is a better deal at 159 euros the t.amp E-800 – Thomann Lithuania. Was looking at manual and I found that "Limiter" is an option to limit THD at 5%. However I found that in manual at 50% output power @ 8R it does less then < 0.03%. So how come is that even possible to have 5% THD if at half power is 0.03%?
This one has variable switch between 0.7Vinput and 1.4Vinput. If I have DSP that does 0.9Voutput I have to use 0.7Vinput or 1.4Vinput on amp?

Last edited by Entas; 22nd August 2019 at 12:29 PM.
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Old 22nd August 2019, 09:57 PM   #5
Printer2 is offline Printer2  Canada
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You can overdrive the amp to get more power (area under the curve) and it limits at 5% where you might have been getting 20% without the limiter.
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Old 22nd August 2019, 11:35 PM   #6
Gnobuddy is offline Gnobuddy  Canada
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Quote:
Originally Posted by Entas View Post
So how come is that even possible to have 5% THD if at half power is 0.03%?
All modern solid-state power amps have very, very low distortion over almost the entire range of power output, from zero output up to some maximum. Above that, the amp runs out of capability because the signal starts to clip, and distortion climbs very, very rapidly, until it reaches 100% THD (pure square wave).

It is up to the human operating the amp to keep it from clipping - lower the volume just a little, add a compressor, add a limiter, do something intelligent to prevent the harsh clipping and enormous distortion.
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Originally Posted by Entas View Post
If I have DSP that does 0.9Voutput I have to use 0.7Vinput or 1.4Vinput on amp?
Going by specs, use the 0.7V input; this means you have more than enough gain to get all the way to full power.

But also try the 1.4V input. You will have to turn the volume knob a little higher to get the same volume as before. You may find this actually makes it easier to control the volume, so try it and see.

You will not damage the amp by choosing the wrong input, by the way; just remember to start with the volume at zero, before selecting either input, and then raise the volume smoothly from there until you reach whatever loudness you intend. (And don't let it clip!)

-Gnobuddy
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Old 23rd August 2019, 07:32 AM   #7
Entas is offline Entas  Lithuania
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Thanks for the answers!
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Old 24th August 2019, 02:26 AM   #8
PRR is offline PRR  United States
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Originally Posted by Gnobuddy View Post
...100% THD (pure square wave)....
Just for our information:

A hard square wave seems to be about 42.827627% THD. (Maybe 42.749279%)
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File Type: gif SquareTHD.gif (18.1 KB, 101 views)

Last edited by PRR; 24th August 2019 at 02:29 AM.
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Old 25th August 2019, 04:15 AM   #9
Gnobuddy is offline Gnobuddy  Canada
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Originally Posted by PRR View Post
A hard square wave seems to be about 42.827627% THD. (Maybe 42.749279%)
Here's my reasoning for the 100% THD figure:

1) A sine wave has an RMS value of 1/(square root of 2) of its peak value.

2) A square wave has an RMS value equal to its peak value. (This can be verified graphically or by doing the integral, or here: https://www.google.com/url?sa=t&rct=...Ax0Yvj-X8YrlvI )

3) Since power to a resistive load is proportional to the square of the RMS amplitude, item (2) means that a square wave delivers twice as much power to its load as a same-amplitude sine wave.

4) Item (3) implies that if we run an amplifier up to full sinewave power, and then proceed to overdrive it until it clips all the way into a square, we double the output power (assuming a perfect amp with zero output impedance and a perfectly regulated power supply.)

5) Going from sine to square is done by the addition of harmonics of the sine. Therefore, 50% of the power in the square wave is in the harmonics, and 50% in the fundamental (sine).

6) The definition of total harmonic distortion is (sum of the powers in all harmonics)/(power in the fundamental) x 100.

7) From step 5, this yields 100% THD for the square wave.

Maybe there's a mistake in my reasoning. Or maybe this is a case where the missing 67% THD in your simulation is contained in an infinite number of infinitely smaller and smaller harmonics, which is missed by the numerical simulation, but included in my analytical calculation.

I suspect the latter, since the Fourier series for a square wave has coefficients that fall very slowly with increasing order, going as 1/(2n+1), where n = 0,1,2,3,4.... and so on. In other words, there is still appreciable power contained even in high-order harmonics.

It would be interesting to see the output of an old-fashioned analogue THD meter (which notches out the fundamental frequency, leaving only the harmonics).

There is probably a way to calculate the THD analytically since we know the Fourier coefficients of a square wave ( https://www.stewartcalculus.com/data...rSeries5ET.pdf ). It basically comes down to squaring and summing that 1/(2n+1) series, first including all the sine terms, and second, including all terms but the first.

I'm not sure my brain is awake enough to tackle that tonight, but it might be possible - if we're lucky, it'll turn out to just be a case of summing an infinite geometric or harmonic series.


P.S. In the real world no square-wave is truly square, and we don't care much about the power in the 100th harmonic...so PRR's approach of finding the THD in only the first ten harmonics is probably more realistic in a practical sense.


-Gnobuddy
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Old 26th August 2019, 03:37 AM   #10
PRR is offline PRR  United States
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I didn't go 10 harmonics and I'm sure that makes little difference.

The harmonic level falls with order. Even-order should not appear in a square wave.

The un-normalized numbers:
1 = 17.3
3 = 5.7
5 = 3.5
7 = 2.5

rms(5.7+3.5+2.5) is 7.
Ratio 17.3:7 is 0.41
Adding some more harmonics adds the last couple %.

> clips all the way into a square, we double the output power

Ah. Makes sense on the face of it. I have an early appointment so I'll leave it here.

There ARE alternate ways ("ambiguity") to compute THD. Mostly when you need to cheat.
Total harmonic distortion - Wikipedia
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