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Old 30th April 2009, 11:19 PM   #1
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Default PA150 (3x LM3886) 2 ohm capability ?

Hello !

I buy PA150 (3x LM3886) but I don't know if this board is able to load 2 ohm ?

If you know what's parts change to make PA150 2 ohm capability please help me !


In the national.com sheet I don't find any spec on PA150 but the PA100 and the BPA200 have 2 ohm capability version.

thanks in advance for your help !


nicK
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Old 1st May 2009, 08:14 AM   #2
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Depends on the supply voltage. According to AN-1192 PA-100 is 2 Ohm stable with up to 28 V rails.

Assume 38 V rails for PA-150. A nominal 2 Ohm speaker can dip down to 1,6 Ohm. 38 V / 1,6 Ohm = 23,75 A. 23,75 A / 3 = 7,92 A, which is just below the 8 A current limit. A bit close for comfort, but PA-150 should therefore be 2 Ohm stable at all supply voltages provided that the heatsink is big enough.

If you assume 38 V rails, the three ICs have a worst case dissipation of ~150 W, requiring a total thermal resistance below 0,033 K/W. Quite unrealistic, when the IC itself already has 1 K/W (T-type) or 2 K/W (TF-type).

Either keep the rails at or below 33 V for PA-150 or build PA-250 or PA-300, so that the 150 W are distributed across 5 or 6 ICs.
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Old 1st May 2009, 12:13 PM   #3
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THANK YOU SO MUCH FOR YOUR ANSWER !!!


nicK
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Old 15th September 2010, 05:33 PM   #4
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I have a spare PA150 board and 2* 31.5VDC / 3.17A power supplies that I'm currently not using... I also have (2) 4ohm woofers so I'm thinking I can build a nice little 150W sub amp for a 2Ohm load.

Q1) Am I correct in thinking that this would be a good match for 2 Ohms? If I understand the AN-1192 notes correctly, in a PA150 config, each chip should see 6Ohms

Q2) Can I create a dual 31.5v supply by joining the (-) of PS1 to the (+) of PS2 [creating my (0) feed], then using the (+) of supply1 and (-) of supply2 for the PA150's +/- (does this present issues with ground? will I fry the PS's by doing this)?

I also have a 36VCT transformer that I guess could work, but I'm guessing the voltage would be a bit low.

Thanks
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Old 15th September 2010, 06:29 PM   #5
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I guess I can't edit my post.

I've discovered that joining the 2 PS's is not as straight forward as I thought and it depends a lot on if, and how they are connected to earth, so I'll leave that alone.

Thanks
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Old 15th September 2010, 08:21 PM   #6
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Quote:
Originally Posted by DarpMalone View Post
I also have a 36VCT transformer that I guess could work, but I'm guessing the voltage would be a bit low.
It would be enough for more than 100 W, provided the transformer has enough power. 100 W means less than 2 dB difference compared to 150 W.

Quote:
Originally Posted by DarpMalone View Post
I guess I can't edit my post.
Only for a limited amount of time after writing it.

Quote:
Originally Posted by DarpMalone View Post
I've discovered that joining the 2 PS's is not as straight forward as I thought and it depends a lot on if, and how they are connected to earth, so I'll leave that alone.
That is right. Usually the negative connection is grounded and you cannot connect two of those SMPS in series without modifying at least one of them.

Anyway 3,17 Apk would not be sufficient for 150 W into 2 Ohm which takes 8,7 Aeff (> 12 Apk) without taking peak current demands into account.
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Old 15th September 2010, 08:39 PM   #7
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Quote:
Originally Posted by pacificblue View Post
It would be enough for more than 100 W, provided the transformer has enough power. 100 W means less than 2 dB difference compared to 150 W.
That's good news. It's a 180VA transformer. So I can still throw this thing together after building a supply.

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Originally Posted by pacificblue View Post
Anyway 3,17 Apk would not be sufficient for 150 W into 2 Ohm which takes 8,7 Aeff (> 12 Apk) without taking peak current demands into account.
Thanks for your reply.

I guess I'm not understanding the concept correctly... If each chip see's 6Ohms, when paralleled, and I'm using 31.5v rails, wouldn't the current demand be much less? (8A from 31.5V rails would be a 500W supply!)

I thought I was beginning to figure this out

Last edited by DarpMalone; 15th September 2010 at 08:43 PM.
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Old 16th September 2010, 12:17 AM   #8
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Rounding to 30V for sake of ease:

30V / 6R = 5A (each IC)
30V * 5A = 150W (still each IC)
150W * 3 = 450W

It doesn't matter from the math perspective whether it's 3 chips seeing 6R or 1 chip seeing 2R. The important part is you spread out the current draw and the heat dissipation.

You weren't wrong with 500W. But that's peak and you don't really need a power supply that large unless you plan on pushing the amp hard.
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Last edited by Redshift187; 16th September 2010 at 12:21 AM.
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Old 16th September 2010, 02:10 AM   #9
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Quote:
Originally Posted by Redshift187 View Post
Rounding to 30V for sake of ease:

30V / 6R = 5A (each IC)
30V * 5A = 150W (still each IC)
150W * 3 = 450W

It doesn't matter from the math perspective whether it's 3 chips seeing 6R or 1 chip seeing 2R. The important part is you spread out the current draw and the heat dissipation.

You weren't wrong with 500W. But that's peak and you don't really need a power supply that large unless you plan on pushing the amp hard.
Hi, again, Redshift187

I'm certainly not trying to argue or challenge, I'm still confused and trying to understand the logic... The 150w should be spread between the 3 IC's, not each (50W x 3 totaling 150W, not 450W), so why are they drawing 3x the total required current?
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Old 16th September 2010, 02:54 AM   #10
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The 150W is a peak power, not sinewave or music power. So the requirement for this current from each IC (5A) is very shortlived.
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