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CMOY diode question

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Old 03-15-11, 10:46 AM Thread Starter   #1
cmoybutts
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CMOY diode question


I'm working on a A47 headphone amp and I'm putting in a diode based Battery / wall power switch like the "clever method" here:

http://tangentsoft.net/audio/cmoy-tu...ks.html#dcjack

This will work as long as one voltage is at least one voltage drop higher than the other, right? That is, it doesn't matter if the battery is higher voltage or the wall DC, as long as there is a difference.

It makes sense in my head but I just want to make sure I'm not overlooking anything since I'm new to these types of electronics. I want to run two 9v batteries in series ( ~18v) with a 15.3v wall wart I stole off a broken printer.

Thanks!
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Old 03-15-11, 03:47 PM   #2
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Just thought I'd let you know that you check out/post this over at headfi.org. There is a lot of people with tons of knowledge on the subject.

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Old 03-15-11, 04:46 PM   #3
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Quote:
Originally Posted by cmoybutts View Post
I'm working on a A47 headphone amp and I'm putting in a diode based Battery / wall power switch like the "clever method" here:

http://tangentsoft.net/audio/cmoy-tu...ks.html#dcjack

This will work as long as one voltage is at least one voltage drop higher than the other, right? That is, it doesn't matter if the battery is higher voltage or the wall DC, as long as there is a difference.

It makes sense in my head but I just want to make sure I'm not overlooking anything since I'm new to these types of electronics. I want to run two 9v batteries in series ( ~18v) with a 15.3v wall wart I stole off a broken printer.

Thanks!
Sort of.

The Wall voltage always will have to be one voltage drop higher than the battery, else when it's plugged into the wall the voltage drop will mean that the battery would still be used (since it has the higher voltage than the wall)

If the battery was one voltage drop higher than the wall, it would mean that the wall would deliver two voltage drops lower than the batteries and therefore never be used.

In your example, lets say you have a 0.7 voltage drop.

Your batteries - 18V (19V fresh) minus the voltage drop to give the cathode 17.3V (18.3V Fresh).

Your wall - 15.3V minus the voltage drop to give the cathode 14.6V.

Therefore, the wall would never be used, since it has not enough voltage to "overcome" the batteries at the cathode.

Hope this helps.

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