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28th March 2015, 03:03 PM  #1 
diyAudio Member
Join Date: Dec 2012

LM3886 Parallel Output Wattage?
In single ended LM3886 datasheet says
68W4 Ohms28V 38W8 Ohms 28V My question is LM3886 in parallel (2chips) what will be the output wattage? 4 Ohms28V what is the output wattage? 8 Ohms 28V what is the output wattage? 
28th March 2015, 04:58 PM  #2 
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Join Date: Jul 2014

.
I can't say I've thought about. But it seems Mr. Elliott has: Bridging Adapter For Power Amps Simplest Ever Bridging Adapter for Amplifiers . 
28th March 2015, 05:02 PM  #3 
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I would bridge them. If you do, the load must be decreased to 8R minimum to acheive 68W maxmum. The substrates will not take anymore. If you parallel them, it is an unknown as the amplifiers cannot be 100% matched.
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28th March 2015, 05:30 PM  #4 
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working with power is quite confusing for the question you have asked.
Work instead with Voltage. eg PSU is +28Vdc when maximum output is passing and +31Vdc when the output is open. The loss through the amplifier is stated in the datasheet. Let's assume a value of 3Vloss. The maximum unclipped peak voltage (when passing current through the rated load resistance) is therefore 25Vpk (28Vdc3Vloss). The maximum unclipped peak voltage when the load is open is 31Vdc~1Vloss, ~ 30Vpk. Now you have two amplifiers in parallel. The maximum output is still 25Vpk. Does the power output change? No, because the maximum output voltage has not changed. If you need those voltage converted to power use the standard formulae: P=IV=I²R=V²/R (this applies to DC voltages and also to rms values of AC voltages). If you use peak voltage of a sinewave, then the formulae need to be modified for the non rms/DC values and become P=IpkVpk/2=Ipk²R/2=Vpk²/R/2 (this only applies to undistorted sinewave). That 25Vpk is equivalent to 25*25/8/2 = 39W, pretty close to that 38W figure in the earlier post. A bridged pair of amplifiers is capable of delivering double the power into double the load impedance. The total power of the two amplifiers is EXACTLY the same as the output from the bridged pair. For the above case of two amps with 25Vpk output, then they are capable of delivering 39W+39W into 8r0+8r0, i.e 78W into 16r.
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regards Andrew T. Last edited by AndrewT; 28th March 2015 at 05:39 PM. 
28th March 2015, 11:15 PM  #5 
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Join Date: Feb 2011

correct.
if the amplifier can not drive impedance less than 8 ohm than its true. if it can drive a 4 ohm load, then things get a bit more complicated. 68W4 Ohms28V 38W8 Ohms 28V let us assume it stands true. and let us assume you have an 8 ohm load. if you bridge the amplifier, you can not drive a load under 8 ohm. as both amps will see an 4 ohm load. and will deliver the 68 ish watt as they should. in real life something around 120 watt is what you get. just like andrewT wrote. double the power into double the load. if you have an 8 ohm speaker and use LM3886 , you get 38W8 Ohms 28V . if you build it in bridge mode, you get about 120 watt into 8 ohm. if you paralell 2 lm3886 chips into the same load, you get no extra power. but you can use 2 ohm speakers. and there you can get the same close to 120 watt as with bridge config.
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V*A=VA V*A*PowerFactor = WATT 
29th March 2015, 12:02 AM  #6 
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29th March 2015, 01:05 AM  #7 
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Join Date: Dec 2012

Ok thanks guys.
Hmm that means in parallel mode it still delivers 68w in a 4 ohms speaker.which is not enough. I need to build a powered sub amp that delivers 100w or 120 w into 4 ohms speakers. Whats the best method to achieve that? 
29th March 2015, 01:13 AM  #8 
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Join Date: Feb 2011

bridge it, it will give you approx 120 watt into 4 ohm.
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V*A=VA V*A*PowerFactor = WATT 
29th March 2015, 03:25 AM  #9 
expert in tautology
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For a sub, buy a plate amp or a digital amp module?
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29th March 2015, 04:02 AM  #10  
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Join Date: Dec 2012

Quote:
Datasheet shows this in page 6. LM3886 in a twodevice parallel solution driving a 4Ω load will typically provide 110W of output power. Even ebay sells parallel amps using LM3886 mentioning 120W per channel into 4 ohms load. Confusing. 

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