Simple LED lamp with battery backup

Thread Starter

vonstroodl

Joined Aug 17, 2006
7
Howdy,
I'm making a simple lamp using 3 white LEDs. I've set up a simple circuit that works well enough and has the schematic

See in "basicLamp.png"

However, I would like for it to operate off my AC/DC adapter regularily, and only use battery power when the power from the wall adapter is removed. Would a simple circuit using some diodes work?

See "possibleLamp.png"

Any other suggestions would be much appreciated. I was thinking of using a P-Channel JFET or a depletion MOSFET instead. Would that route be more recommended?
 

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beenthere

Joined Apr 20, 2004
15,819
Hi,

Your diodes should work. Just use something like 1N4000's to handle the current - about 220 milliamps.

Another possibility is a SPDT relay with the coil tied to the adapter. It would pick up the adapter 6 volts when that was on, and drop to the battery when AC power was off. With a SPST relay, you have a power failure lamp that only illuminates when power is lost.
 

Mazaag

Joined Oct 23, 2004
255
I just wanted to ask a question about the second schematic you posted.

So how exactly does this work? when the 6v DC are coming in , the diode above the R4 is reverse biased ? and hence acts as an open circuit ? My question is why does it become reverse biased ? would the voltages at both the anode and cathode of the diode be the same? or is there a voltage drop across R4 ? ( and if so , where does the current flow to generate this voltage drop ? )

thanks
 

Thread Starter

vonstroodl

Joined Aug 17, 2006
7
From what I know, the circuit should work as annotated below.
See "explanation.png"

But I think you might be right, without current flow, voltage would not drop over R4.

The following is some speculation: If the forward voltage of the diodes are 0.3 volts, as soon as the DC passes through the its diode, it dropes to 5.7 volts. Therefore the voltage on the cathode side of the battery's diode (above R4) is less than the annode. This would make that diode forward biased, and some current would leak through from the battery.

Would it be better if I had two diodes in series, instead of one diode and R4 in series?
 

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nomurphy

Joined Aug 8, 2005
567
Your circuit is fine, it's a classic wired-OR for power supplies and battery back-up. But, R4 is pointless and can be removed.

If the forward voltage (>3V ?) of the white LEDs is getting close to your supply voltage (such as when the battery is weak), you could use Schottky diodes such as the B340A (3A, 40V) for less drop and better supply margin.
 

ragnorkano

Joined Aug 19, 2006
6
what about using a power jack for your adaptor. It would let you use your adaptor power when pluged in and not use any battery until it was unpluged.
 

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Thread Starter

vonstroodl

Joined Aug 17, 2006
7
Oh brilliant, I didn't know power jacks were packaged like that. This is exactly what I needed. Thanks ragnorkano, and everyone else that pitched in. You're an awful helpful bunch
 

moni_ul

Joined Mar 22, 2009
1
Howdy,
I'm making a simple lamp using 3 white LEDs. I've set up a simple circuit that works well enough and has the schematic below:


However, I would like for it to operate off my AC/DC adapter regularily, and only use battery power when the power from the wall adapter is removed. Would a simple circuit using some diodes work?


Any other suggestions would be much appreciated. I was thinking of using a P-Channel JFET or a depletion MOSFET instead. Would that route be more recommended?


psl give me the schematic this email <SNIP>
 
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Thread Starter

vonstroodl

Joined Aug 17, 2006
7
Hello muni_ol. I uploaded the images in this thread that were originally linked. Ragnorkano had the best solution for this problem though if this is what you're trying to make
 
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