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Old 3rd August 2011, 05:59 PM   #1
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Question Parallel and Bridged question?

I'm designing a 2.1 Gainclone amp using LM3886 chips.

Now, I'm working on an active filter. I will end up with a Left and a Right channel > 200 Hz and a mixed channel < 200 Hz.

The sub will take more power to drive, so I thought I would have 1 x LM3886 for each channel and 2 x LM3886 in bridge or parallel mode for the sub.

Please can someone explain how the chips in work in parallel and bridged......

If you have a single LM3886 with a gain of 20 and 1 volt is applied to the input, you get 20 volts on the output. If you parallel the chips and 1 volt is applied to each input, each chip will have an output of 20 volts, which are connected together and you still have 20 volts???

If your supply voltage is 30v and you speakers are 8 ohms, the maximum current flow will be 30 / 8 = 3.75 amps. Surely no matter how many chips are paralleled 3.75 amps is the max current that will be drawn by the speaker!

In bridged mode, if you applied the same 1v signal to the amps (both with a gain of 20), you end up with +20 volts at one output and -20 on the other output, giving you 40 volts at the speaker terminal to the positive terminal and -20 v to the negative terminal. So you have 40volts at the speaker. Again if the supply voltage was 30v, you could have a maximum of 60 volts at the speaker 60 / 8 = 7.5 amps. Obviously the speaker would be louder than before due to more current and voltage, hence more power.

Surely you are limited to the supply voltage!

Please can someone clarify
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Old 4th August 2011, 01:42 AM   #2
sregor is offline sregor  United States
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Your math is correct. Paralleling doubles the maximum current. Bridging doubles the voltage swing. If you need more current, then parallel chips. A second factor is the power dissipation doubles also. In your bridged example you exceeded the maximum current of the 3886 (7 amps) and are producing over 200 watts, which will burn up your chips (they can't put out more than 60 watts each, and that is into resistive loads). (class ab amps are only about 60 percent efficient at max output, and efficiency gets worse as output goes down. The excess gets dissipated as heat).

Sounds like your trying to decide voltages for supply. If you maximize the voltages for 8 ohm operation, you will not be able to get more power to the sub woofer without bridging it - but if you bridge, you can exceed the power and current capabilities of the chip. Your options is to go bpa (bridge parallel amp) and use 4 chips for the sub. or drop you voltages and use less than max power for the satellites. Or use a higher voltage for the low frequency amp and parallel, or lower voltage and bridge. You are the designer - you decide. Good luck.
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Old 4th August 2011, 05:32 AM   #3
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I recommend you read this first, then come back with your questions.

http://www.national.com/an/AN/AN-1192.pdf

Mike
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Old 4th August 2011, 09:28 AM   #4
AndrewT is offline AndrewT  Scotland
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Quote:
Originally Posted by portreathbeach View Post
The sub will take more power to drive, so I thought I would have 1 x LM3886 for each channel and 2 x LM3886 in bridge or parallel mode for the sub.
this "more power" statement only applies if one chooses a low efficiency bass driver whereas higher efficiency drivers have been chosen for the upper frequencies.

As a first stab, build all the channels with single chipamps and with the same PSU voltages.
Choose drivers that have sensitivities that are very similar, (within +-1dB).
Finding 92dB/W to 94dB/W @ 1m drivers is not too expensive. Drivers from 95dB/W @ 1m start becoming very expensive.
Low sensitivity drivers <88dB/W @ 1m do not suit chipamps. The amps clip and limit and sound terrible when the volume is turned up.
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Old 4th August 2011, 03:50 PM   #5
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Michael, I downloaded that very 'pdf' file after posting my question and understand a bit more now. I've only really ever used the 'overture design guide' for calculations and I think that is where I was getting confused.

For example... If you choose LM3886, 35V, 8ohm load and leave the rest as it is, then...

Voltage peak at load = 31.23V
Current peak in load = 3.9A
1% THD output = 60.94W

How does 31.23 v and 3.9A give 60.94W?

Also, as mentioned in this thread, paralleling 2 chips shares the power, so why does it say in the bottom right of the spreadsheet that parallel mode for the values entered = 118.5 W?


Andrew, when I said "The sub will take more power to drive", what I meant is that I will be mixing L and R channels into it, so assuming dB/W @ 1m is the same for the full range speakers as the sub (where there will only be one), would this not need to produce roughly twice as much power to sound the same volume as the other speakers.

Obviously, as discussed in my 2.1 Gainclone thread, the L and R will be summed and filtered to the subwoofer Gainclone amp. With a 1v signal on L and R input, the L and R speakers would see 20V (assuming each amp has a gain of 20). Now the active filter sums the L and R inputs and puts 2 into the LM3886. With a gain of 20, it needs to put out 40V, it will clip obviously, so will I need to bridge the subwoofer channel to be able to get more power from it?

I suppose if the main speakers are 8ohms and the subwoofer is 4ohms, then a gain of 20 probably wouldn't be required for the subwoofer as it can produce more power due to the lower impedance. Then I could parallel 2 chips for the sub channel. Any thoughts.
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Old 4th August 2011, 05:29 PM   #6
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The figures quoted are the peak values the amp will supply to the load under various load conditions, and not necessarily at the same time, and the power quote is for nominal RMS rating, not peak, into 8 ohms. Remember that speaker loads are not purely resistive and have inductive and capacitive components which cause the amp to experience a wide variation in loading depending on frequency. The parallel 118.5 watt rating is for 4 ohm load, and in my experience is probably somewhat optimistic.
Andrew, your turn.

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Old 5th August 2011, 02:35 AM   #7
sregor is offline sregor  United States
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Quote:
Originally Posted by portreathbeach View Post
Voltage peak at load = 31.23V
Current peak in load = 3.9A
1% THD output = 60.94W
How does 31.23 v and 3.9A give 60.94W?
Power ratings are based on average power, not peak. For a sine wave, average value is .707 (square root 2 divided by 2) times peak, and current also, so endup up with peak voltage times peak current divided by 2 for average power.

Quote:
Originally Posted by portreathbeach View Post
I suppose if the main speakers are 8ohms and the subwoofer is 4ohms, then a gain of 20 probably wouldn't be required for the subwoofer as it can produce more power due to the lower impedance. Then I could parallel 2 chips for the sub channel. Any thoughts.
One factor you are neglecting is speaker efficiency. Tweeters tend to be more efficient than woofers, midranges are more of a mixed bag. Nearly every passive speaker system I've seen uses some form of attenuation to match levels for midrange and tweeter. One of the big advantages (IMO of course) of active speakers. Basic fact is that most of your amplifier power is dissipated as heat by the speaker, not sound output and has to be adjusted for actively or passively. Lower impedence woofers will increase power, but if they aren't efficient, they will need it. Your gains have to be adjusted for the efficiency of the speakers, and normal practice is before the amplifier circuit. Good luck.
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