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Old 31st March 2012, 05:53 PM   #141
Pemo is offline Pemo  Mexico
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Hi Andrew, this is Pemo again.
Would you please tell me if this C32 + R42 is a Zobel in this design?
Is the Thiele missing?
What is this LM318 for?
Best regards

LM3886 双声道.pdf
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Old 31st March 2012, 06:02 PM   #142
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670 mA into 8 Ohm corresponds to 3,6 W. From that power level upward it is likely that the sound quality will suffer with that inductor. That is more power than you usually need continuously, but with 40 V rails the amp can deliver 20 times as much.

Andrews advice to wind your own is the easiest, cheapest and fastest solution, if you find that the sound quality decreases at higher volume. If you want to replace the inductor with a commercial one, look for ferrite types and make sure the saturation current is 7 A or more.

Luxurious space for the amp. It will need some holes for air circulation, especially if you put the heatsink size in relation to your highish rail voltage. The isolated LM package limits the heatsinking capabilities, too. Should the amp run into thermal protection, consider replacing the LMs with the non-isolated package and reducing the rail voltage.
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Old 31st March 2012, 06:08 PM   #143
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Hi Pemo,

Your answers are here: My "audiophile" LM3886 approach

You can also ask in that thread, whether that seller has permission to sell that design or parts of it.
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Old 31st March 2012, 07:17 PM   #144
Electrons are yellow and more is better!
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Quote:
Originally Posted by AndrewT View Post
About 10Turns of 1.2mm enameled copper wire for the inductor.
Diam. 8 mm
15 turns
1.5 mm copper

= 0.7 uH

Diam. 6.5 mm
13 turns
1.5 mm copper

= 0.45 uH


Diam. 13 mm
15 turns
1.5 mm copper

= 1.48 uH
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Old 31st March 2012, 07:24 PM   #145
AndrewT is offline AndrewT  Scotland
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I am not in agreement that 1u5H is needed.
10T gives ~ from 0u7H to 1uH
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Old 31st March 2012, 08:20 PM   #146
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I have measured it with an expensive LCR bridge (0.1%)

LCR Databridge Model 6451
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Last edited by peranders; 31st March 2012 at 08:23 PM.
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Old 1st April 2012, 01:35 AM   #147
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Quote:
Originally Posted by pacificblue View Post
670 mA into 8 Ohm corresponds to 3,6 W. From that power level upward it is likely that the sound quality will suffer with that inductor. That is more power than you usually need continuously, but with 40 V rails the amp can deliver 20 times as much.

Andrews advice to wind your own is the easiest, cheapest and fastest solution, if you find that the sound quality decreases at higher volume. If you want to replace the inductor with a commercial one, look for ferrite types and make sure the saturation current is 7 A or more.
Hi Pacificblue:

I was checking the electronics component catalog online, the closer match to the 1.5uH inductor is a API Delevan with 5.2 Amps of current saturation, 11 mOhms of DC resistance. They do not show any axial inductors with 7 amps of current saturation, this one seems to be the biggest one in that respect. The part number is API Delevan 2474R-03L, it is available at www.digikey.com . Please let me know if this would work.

If not, I will have to test the amp as it is right now, if the quality of sound is not good, I'll try AndrewT's solution. I just don't feel like playing with copper wire, I'm afraid is not going to come up the right way...

Regarding the heatsinks, do you think those in the picture are not big enough?.

Thank you again.
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Old 1st April 2012, 09:14 AM   #148
AndrewT is offline AndrewT  Scotland
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Quote:
Originally Posted by AndrewT View Post
Think about changing the BOM to
100 10r 600mW 1% metal film resistor
10 47nF 50V, or 63V, or 100V, cheap film capacitor
10 100nF ditto
10 220nF ditto
100gm or 500gm reel of 1.2mm diameter enameled copper wire.
from way back.
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Old 1st April 2012, 12:08 PM   #149
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Quote:
Originally Posted by Aguilabrava View Post
The part number is API Delevan 2474R-03L, it is available at www.digikey.com . Please let me know if this would work.
In theory, yes. In practice you will have to test it, because the high Q factor could lead to ringing.

Quote:
Originally Posted by Aguilabrava View Post
If not, I will have to test the amp as it is right now, if the quality of sound is not good, I'll try AndrewT's solution.
Omitting the resistor and the parallel inductor will not have audible effects. Those components only protect the amplifier from overload, when you connect capacitive loads.
Andrew's solution will not have audible effects either. It is proven practice to do it like that, and you can't make much wrong there.
The solution with the core inductor will have audible effects, when the core runs into saturation, and it could have audible effects due to the Q factor, but that remains to be tested.

Quote:
Originally Posted by Aguilabrava View Post
Regarding the heatsinks, do you think those in the picture are not big enough?
The bigger issue will be the isolated LM package that has a higher thermal resistance than the non-isolated version. Even for 8 Ohm loads your rail voltage is quite high and you can expect ~40 W of worst case power dissipation. Even if the heatsinks are big enough, the amp may not be able to transfer the heat to the heatsinks.

The heatsinks themselves will work well outside of a case, where you can expect 20-25 C ambient temperature. Depending on your listening habits they may or may not work inside of a case. Inside of a case the temperatures will be higher, that is why you should at least make some holes or connect the heatsinks thermally to the case, so that the case itself helps with heatsinking.
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Old 1st April 2012, 02:35 PM   #150
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AndrewT:

I have a question regarding the GRD's in the amplifier. In the PSU, in the DC side I have three connections, +V GRD -V. From what I have been reading, I only connect the +V and -V to each amplifier, then run 2 wires from them to the star point of the ground where they will meet the cable coming from the PSU GRD terminal. Is this correct?. Then, from the Output speaker terminals attached to the chassis, I run one wire from each terminal to that same star spot.

The safety GRD, the one that comes from the wall's mains goes to the chassis at a different location, it doesn't join the rest of grounds that will form the star, right?.

The purple shield wire from the transformer gets attached to the chassis at a different location, no star ground spot, no safety ground from wall's mains.

Please let me know if this is correct.
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