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Can I bridge the two chanells of LM4809 phone Amp;

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aetosa

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Can I bridge the two chanells of LM4809 headphone Amp;

I want to make a headphone Amplifier with the LM4809.
But, I want to combine the two channels in ONE, so that I will take the double power in one channel.
"Bridge" is how this is called?
How can I do it?
In the image bellow, you can see the typical application from page 2 of the datasheet.


**broken link removed**
 
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Not directly because you would need access to the non-inverting input to one of the amplifiers and it's internal to the silcon. You can do it by inverting the signal to one channel with an added cicuit. All you need is a unity gain inverter, one channel connects as normal and the other is fed through the inverter. The loudspeaker than connects across the two outputs.

I would question the wisdom of doing it though, these are only rated at 0.1W. It would be simpler to abandon the LM4809 and use a higher powered amplifier. Even the humble LM386 will produce several times more output power for about the same number of extra components you would need.

Brian.
 

betwixt said:
Not directly because you would need access to the non-inverting input to one of the amplifiers and it's internal to the silcon. You can do it by inverting the signal to one channel with an added cicuit. All you need is a unity gain inverter, one channel connects as normal and the other is fed through the inverter. The loudspeaker than connects across the two outputs.

I would question the wisdom of doing it though, these are only rated at 0.1W. It would be simpler to abandon the LM4809 and use a higher powered amplifier. Even the humble LM386 will produce several times more output power for about the same number of extra components you would need.

Brian.
Click to expand...

I only need 0.2 Watts, it's more than enough.
The two channels of LM4809 give about 0.2 Watts.
Why wouldn't it be wise to bridge them?

(now I'm thinking of it, the addition of a unity gain inverter circuit would make my circuit much more complicated and of bigger size than I had planned)

My application is a portable amplifier, that is battery operated, so I'm looking for the best efficiency at 200 milliWatts of power output.
 
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That was my point. For the cost and complexity of adding an inversion stage you could build a more powerful amplifier anyway.

Brian.
 

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