led rechargeable 12v flashlight

Shorting one of the diodes should increase your charge voltage by .3 to .5 volts.

Is it my imagination, or did the OP who doesnt have any experience in or understanding of electronics just try to school the people who are trying to help him?

Go ahead . .  bite my hand . . .

Granted 4 diodes, a resistor and LED pack hardly count as "electronics" but hey . . .

The point stands - if the only output anywhere, at all, is 3.4v then it is rated for the LEDs, and not to charge the batteries.

I didnt make it this way, that's just the way it is and you can't get blood from a turnip.

If this were Judge Judy, she's would slam the gavvel down and declair that the defendent hasnt proven his case the the flashlight will recharge the batteries by design.

If the diodes are meant to be a cheap way to drop voltage, then shunting one in order to increase the voltage could destroy the LEDs. (unless there is a regulator with a constant voltage ic)

Either 2 people are reading the voltage from the wrong point or this is a basic flashlight that delivers 3.4v to the LEDs by design and there's nothing more to go on about.

Hang out at Dealextreme and look at LED arrays and voltage regulators for them.

Here we have:

No model #

No schematic

No parts list

Your battery recharging specs are pretty spot on, the IC parts I used from MAXIM for rechargers spec in that range.

Well yes it does! A long winded way to say you're right, and not a criticicm of your analysis, just over stating the obvious.

And what's a FARN? Wasnt that on Farscape? Must be an Aussie thing :D

Hi all,

I didnt' see the an answer to the OP's question about why dropping a diode from the circuit will increase the voltage out.  It's fairly simple, if diodes are understood.  Remember, they are semi-conductors.  That is to say that they "sort-of" conduct.  Not perfectly or pretty well, like metals (gold, silver, copper, etc....) but a lot better than insulators (glass, plastic, wood, etc...).

Diodes "sort-of" conduct when connected tin a "forward" manner, and "sort-of" insulate when connected in a "reverse" manner.  How much this "sort-of" amounts to varies according to what the diode is made of and how it's desiged.  A strange thing about them is that they conduct at a particular voltage.  Above that magic voltage line and they conduct pretty much like a short circuit, but be below the line they appear as an open.  Usually, "general purpose" diodes, 1n400x series diodes for example, have a forward voltage of 0.7volts  and a reverse, or "inverse" rating of some much higher voltage, depending on the diode.

So how does 3 diodes in series make a difference?  Suppose a circuit with a 2 volt battery powering a 0.7volt diode in series with and some sort of "load" (a light bulb or resistor or whatever).  Of the 2 volts presented on the circuit, the diode will take 0.7 volts of "drop", leaving 1.3 volts for the load.  Add another diode in series and it will take another 0.7 volts, leaving the load with only 0.6 volts.

So, if one had a 12 volt power source, but needed a 3 volt power supply, one could connect 10 of the 0.7 diodes in series to drop 7 volts (0.7 x 10 = 7) leaving 3 for the load.

That's all theory, even if it's got practical value, but safety and good design means that there are other things to consider.  What if the circuit shorts?  What if the requirements of the load changes (like batteries that are charging or discharging) and changes the overall resistance of the circuit?  So a resisitor is sometimes put in to compensate for this or to limit the current.

Why use diodes and not resistors?  Resistors have a fixed resistance (unless they are defective).  Diodes have a fixed voltage (unless they are broken).  Ohms law says that Volts = Amps x Resistance.  Using a resistor, with it's fixed resistance, means that changing the overall voltage on a circuit will change the voltage drop across the resistor.  Using a diode, with it's fixed voltage (drop), means that changing the overall voltage on a circuit will not change the overall voltage drop across the diode, unless or until the min or max  voltage threshold is crossed, in which case the diode will either stop conducting because the voltage is too low or the the diode will open or short because the over voltage blew it to smithereens...

Hope this helps,

Alan

In a real world application I have yet to find a diode that dropped .7 volts.