how to wire multiple amp chips?

[big314mp]

I'm building relatively large GC amp, and had a question regarding how to wire multiple chips together.

I'm building the speakers as well, so would I be able to get away with wiring the sub amp chips into the main amp? Basically, instead of having the signal from the source go to the left and right channels, as well as the sub, would I be able to just add some extra power to the left and right channels, wire both channels' outputs to the sub, and forget about a separate sub amp.

Advantage = simpler wiring (I think)

What are the disadvantages?

Thanks!

[Nuuk]

Your question isn't clear! Could you supply a circuit diagram of what you want to build, and supply details of the other equipment involved?

[AndrewT]

Like Nuuk, I'm a bit lost.
You need three channels of amplification to drive three speakers with Left, Right and Sub-bass signals.

[big314mp]

Sorry about the ambiguity.

The amp will be using one LM4780 per channel, and and a third to drive the sub. I was wondering whether I could have 2 LM4780's per channel and drive the sub off of the L/R outputs.

The source is my PC's headphone jack, so there isn't any specific sub output.

[big314mp]

and the other pic. This is the one that I was going to build, but then I had the above idea. Each one will have separate PSUs per channel, so the above schematic would use 2 PSU's, whereas this one would use three.

[DigitalJunkie]

This is/was fairly common with car amplifiers,do a search for "tri-mode"
Also,one channel will need to be inverted.

[big314mp]

thanks a lot for the help!

I'm still not quite sure I understand how the wiring would work. If one of the channels is inverted, then wouldn't it "cancel out" the other channel at the sub?

How would this work especially if the Left and Right channels have different signal outputs (i.e. the drummer is on the left side of the band and the bassist is on the other, wouldn't the two channels be sending completely different information to the sub, and since they are inverted , wouldn't this create some strange effects in the sound)?

If the channels themselves are bridged relative to each other would it be impossible to bridge the individual amps on the LM4780?

I'll go through some of the tutorials on the other threads and see if it makes more sense after that.

[relder]

Consider the case where you feed a mono signal to that setup. The subwoofer terminals would see that same mono signal voltage on both terminals. I.e. there wouldn't be a difference in voltage, 0V, and therefore no sound. You'd have to invert one to get any sound.

Rob

[big314mp]

sorry for my apparent n00bness, but can't you just wire both (L/R) outputs to the sub input, and then wire the other terminal of the sub (the output if you will) to ground? Or back to the respective channels if the chips in each channel are bridged? Or would that create cross talk problems.

Maybe I should drop this line of reasoning and just ask how to wire up a sub amp using LM4780s, if only L/R channels are available from the source.

Thanks a lot!

[sangram]

There are a ton of posts on this topic, a search will throw up quite a few.

To summarise, you can increase power from smaller amps:

1. Into a low impedance load: Connect in parallel mode, a small resistor in series with each amplifier to allow amps to share current equally.

2. Into a high impedance load: Connect in bridge mode. One channel needs to be inverted wrt the other. This can be done by changing the topology (run in inverted mode), or actively inverting the signal through an external inverter or using a single resistor method as detailed on Rod Elliot's site.

3. Into the same impedance load, but with higher power: A combination of all the above.

Read the datasheet for the 4780. The basics are all well-covered in there.

For your case, you will need the third channel. The moment you connect the two channels outputs together, you've lost the stereo information, and even your left and right speakers will be putting out mono.

Also the 4780 may not function correctly due to the very low impedance load presented.