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Old 12th May 2015, 02:06 PM   #1
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Default Input to Parallel power amps-Basic theoretical question by a newbie

I dont know if I am going against any basic law of electronics here, if so please point out.
If you think I need to learn some specific fundamentals first, please point me to links on the internet. It will be a nice help. Thank you.
I was reading about BPA 300 amp on a website. This basic question popped in my head.
The diagram I have attached says everything about whats happening. Please see it.
My question is : At position A and B what can be done to increase (x/2) watts back to x watts. Perhaps an additional amp stage ? If it can be done then we dont need to bridge amplifiers isnt it ?

This is quite a simple idea. The fact that I dont see anybody try this means I must be missing some basic things. Please make me understand.
Thank you.
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File Type: jpg fig 1.jpg (29.9 KB, 59 views)

Last edited by Prashanthb; 12th May 2015 at 02:08 PM.
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Old 12th May 2015, 02:33 PM   #2
johnr66 is offline johnr66  United States
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Think of this as voltage levels at the input. The CD player will provide a signal with some source impedance. The amplifier inputs will have input impedance. The amplifier's input impedance will typically be an order of magnitude higher than the source impedance so as not to load down the input signal. Think of it as a resistor divider with the top resistor as the source impedance and the bottom the input impedance of the amp. The center node would be the voltage of the signal when connected to the amp(s). As you add more amplifiers, the input impedance drops, but not significant enough to be an issue if the input impedance is at least 10x the source impedance.

Beyond this, I'm not sure what you are trying to accomplish. Are you paralleling or bridging the amps.
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Old 12th May 2015, 02:53 PM   #3
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I am paralleling the amps.

My point is : By paralleling (load is not changed, 4 ohms in my case) we do not increase the output power. Higher power is realised only when i change to a 2ohms speaker(which I dont want to do).
Now...If we use 2 amps in parallel, the signal strength thru each amp will be halved. If we use 5 amps in parallel, then each amp gets 1/5 th the original signal strength.

My question is : Can I put another opamp(s) at positions A and B to increase the signal strength before it is fed into the power amps ?......So that if single LM3886 gets me 68W of power @ 4 ohms, by paralleling and increasing input signal strength can I get >68W into 4 ohms.

Last edited by Prashanthb; 12th May 2015 at 03:07 PM.
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Old 12th May 2015, 03:14 PM   #4
johnr66 is offline johnr66  United States
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Quote:
Originally Posted by Prashanthb View Post
If we use 2 amps in parallel, the signal strength thru each amp will be halved. If we use 5 amps in parallel, then each amp gets 1/5 th the original signal strength.

My question is : Can I put another opamp(s) at positions A and B to increase the signal strength before it is fed into the power amps ?......So that if single LM3886 gets me 68W of power @ 4 ohms, by paralleling and increasing input signal strength can I get >68W into 4 ohms.
Splitting the source signal among two or more amplifier inputs does not halve (or worse) the signal for each amp as previously explained if the source impedance is much less than the input impedance. What is the source impedance? What is the input impedance?

Let's look at it another way. Let's say you have a power outlet with two connectors. You plug in a 60 watt lamp, turn it on and measure the supply voltage. It reads 230 volts AC. Next, you connect a second 60 watt lamp, turn it on and measure the voltage. It is still about 230 volts AC. Why is that? It is because the source impedance is so low that voltage to the loads is barely affected even when the additional load impedance is connected.

Last edited by johnr66; 12th May 2015 at 03:16 PM.
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Old 12th May 2015, 03:30 PM   #5
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If the first lamp is rated 230V/0.260A(60W) then by adding another lamp(230v/0.260A)) in parallel we will be drawing 120watts from the mains.
i.e 230v/0.521A from the mains.

What I want to achieve is this : (Please see attached image; compare with first image)
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Old 12th May 2015, 03:43 PM   #6
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ASSUME(!) the CD player outputs 2Vac / 0.5A.

If it is followed by 2 parallel power amps, then the input signal into AMP1 is 2Vac/0.25A and AMP2 is 2Vac/0.25A. Is there anyway where I can make this 2Vac/0.5A for AMP1 and 2Vac/0.5A for AMP2 ?
Does a pre-amp increase current ?
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Old 12th May 2015, 03:43 PM   #7
rayma is offline rayma  United States
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Quote:
Originally Posted by Prashanthb View Post
If the first lamp is rated 230V/0.260A(60W) then by adding another lamp(230v/0.260A)) in parallel we will be drawing 120watts from the mains. i.e 230v/0.521A from the mains.
What I want to achieve is this : (Please see attached image; compare with first image)
Watts in this case matter only at the output/speaker interface. Before that, just voltage matters. Read up on bridged amplifiers.
Bridging two amplifiers allows you to double the voltage at the speaker. In this case, each amplifier would have to be able to drive 2 Ohms,
(since your speaker is 4 Ohms), and the amplifier probably can't do that. You could bridge these amps if the speaker were 8 Ohms.

Last edited by rayma; 12th May 2015 at 03:59 PM.
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Old 12th May 2015, 03:48 PM   #8
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You have supplied the answer. By connecting 2 lamps you doubled the power and by connecting 2 speakers or 1 with half the impedance you will do the same.
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Old 12th May 2015, 05:10 PM   #9
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So can 1 CD player (assume 2Vac output) directly feed 20 LM3886 amps each powering 4 ohms speakers...and I get an output of 1360 W (68Wx20) ???
I think the final output will still be 68W, and not 1360W.
What will the final output be in watts ?

Last edited by Prashanthb; 12th May 2015 at 05:26 PM.
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Old 12th May 2015, 05:28 PM   #10
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Yes, 68x20 is correct.
A few things to study.
Voltage, current, power and gain.
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