led rechargeable 12v flashlight
Printed From: the12volt.com
Forum Name: Lights, Neon, LEDs, HIDs
Forum Discription: Under Car Lighting, Strobe Lights, Fog Lights, Headlights, HIDs, DRL, Tail Lights, Brake Lights, Dashboard Lights, WigWag, etc.
URL: https://www.the12volt.com/installbay/forum_posts.asp?tid=120721
Printed Date: July 14, 2025 at 4:25 AM
Topic: led rechargeable 12v flashlight
Posted By: kazoocruiser
Subject: led rechargeable 12v flashlight
Date Posted: March 14, 2010 at 3:18 PM
I tried to post a response to another thread and that thread was closed, so sorry if this was answered somewhere else in the 38 pages of electrical threads. . .
Here is the issue. . .
I have an LED flashlight that plugs into the lighter socket, and is supposed to provide a bright white output. The thing is from China (Iron Forge) and seems to be a good idea. . .
After a week of use, I thought maybe the batteries were the weak link, so I took it into Batteries Plus and paid 12 bucks for another battery pack. . .3, 2/3 1.2 (aaa?) VDC batteries. They went weak.
So I opened up the thing, and started checking. . .and the batteries are only getting about 3.4 volts supplied to them through the conversion process from the 13VDC lighter socket. One resister and four diodes.
The help at the Batteries Plus store said I really need about 4.2 volts to charge the batteries properly. . .so can someone tell me either:
What wall charger / transformer I can use with this unit "as is" to obtain that level after the internal stepdown process, (I have an adapter for the plug I can use) or. .
What resistor is needed to replace the one inside the internal converter inside the flashlight.
Replies:
Posted By: rkanauer
Date Posted: March 14, 2010 at 3:28 PM
I assume the diodes and resistor are all in series. Jump out one of the diodes.
-------------
RARIHTS
Posted By: kazoocruiser
Date Posted: March 15, 2010 at 6:53 PM
Thank you for telling me what to do. You are correct, all the diodes are in series.
And since I am electrodense, can you take a minute and explain how the circuit I am looking at works? Four diodes, and a resistor.
Why would the designers / builders of this device do what has been done in this case?
Sorry if the question is clumsy. . .
Thanks in advance.
Posted By: anonymous1
Date Posted: March 15, 2010 at 8:02 PM
The 3.4v is the standard supply voltage for the LEDs. Who says this unit is supposed to recharge the batteries from the 12v outlet? Is that specifically stated in the documentation? I doubt it, I buy a buttload of cheap crap from China. Another thing is - rechargeable batteries ARE NOT 1.5v, they are rated at about 1.25v. Go to dealnews.com or any of a myriad of discount type sites and search for battery chargers, they average $10 and come with 4 batteries. I have 3 chargers here in my office that stay loaded up with batteries, at least 4 at a time. Finally, when you buy AAA or AA batteries, try to find the highest rating in ma that you can. If you can find AA batts @ 2800ma, get them.
Posted By: oldspark
Date Posted: March 16, 2010 at 6:26 AM
Are you sure isn't a battery torch that can also be plugged into the cig socket - as opposed to a rechargeable torch?
Have you seen those "battery-less" dynamo rechargeable torches? THEY have batteries, and their hand-pored dynamo does NOT recharge the batteries.
They should be banned, but too many people buy them, pull out the batteries and dump the rest. It's cheaper than buying the batteries.
Posted By: i am an idiot
Date Posted: March 16, 2010 at 7:51 PM
Did you let the batteries charge? If you did not let it charge, when you read the voltage, you were reading the voltage of the battery at that time. You need to let it charge for a while and check the voltage again.
Posted By: anonymous1
Date Posted: March 16, 2010 at 8:22 PM
"the batteries are only getting about 3.4 volts supplied to them through the conversion process from the 13VDC lighter socket. One resister and four diodes."
Posted By: anonymous1
Date Posted: March 16, 2010 at 8:25 PM
OH! And I completely forgot, I bought a THOR-X today in great condition for $35 from a guy who lives near. I plan to do a HID conversion on it.
Posted By: i am an idiot
Date Posted: March 16, 2010 at 9:19 PM
With no batteries in the unit, what is the voltage on the charge terminals?
Posted By: kazoocruiser
Date Posted: March 17, 2010 at 6:31 PM
Thanks for the responses. . .
Just for clarification. . .the batteries, like the battery in a car, has to have a higher supply voltage in order to maintain peak output.
A vehicle with a 12.6 battery normally receives 13.6 - 14.2 from the alternator (rectified output) which obviously is higher voltage than the rated 12.6. A battery with 12.4 is partially discharged. An alternator that puts out only 12.6 would be defective.
So a series of (3) 1.2 VDC batteries at 3.4 volts is ALSO partially discharged. The 13 VDC from the cigarette lighter socket is being CUT to 3.4 via the bitty circuit inside the flashlight. So it is being undercharged ALL THE TIME. Since no one has explained to me what the removal of one diode will do by bypassing it, I am going to take a leap of faith and simply wire a bridge across it and then measure the output voltage going INTO the batteries. . .I suspect the simple answer from rkanauer the rookie is what I need to do, but not WHY I need to do this.
"tree swing"
https://www.businessballs.com/treeswing.htm
Posted By: i am an idiot
Date Posted: March 17, 2010 at 8:41 PM
Shorting one of the diodes should increase your charge voltage by .3 to .5 volts.
Posted By: anonymous1
Date Posted: March 18, 2010 at 8:10 PM
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 . . .
Posted By: oldspark
Date Posted: March 18, 2010 at 8:49 PM
Yes - but that's battery stuff - not electronics.
But I like it. It's nice to know another agrees with me. Again.
It's also good to see their reasoning or knowledge or what they have learnt (maybe since since our info? IE - we said rechargeable = 1.2V so OP correctly deduces >3.6V needed).
I like that. Firstly it shows that people on this site are not a FARN STOOPID as the MORONs I have dealt with elsewhere  (Yes, forgive my language, but I am so animated on such morons that I actually included an emoticon!)
Secondly, I just like it... (Apart from showing smarts or education, it could show the misunderstanding or divergence etc.)
Then again, I didn't take it as dictatorial or arrogance.
And whilst I didn't like the " Since no one has explained to me...", the point is the later " I am going to take a leap of faith..." - I take that as a compliment...
Besides, that was my dislike... which is funny because OP explained himself (>3.6V etc) and merely expects the same, and I am one that usually does (try to) explain - and hence one reason for my l o n g replies - yet I react to someone wanting the same from others.... LOL! (But I don't answer anywhere near the number of posts nor provide the same "detail" as others do, etc etc.)
Alas, the fun of different audiences, customs and expressions.
Don't worry Kazoo, I'll help you fight these critics....
Then again, I do try to help most people... especially good ones like these critical experts....
FYI - I think NiMH cells charge at 1.4–1.6 V/cell, hence 4.2V-4.8V for 3; a full cell is about 1.25 => 3.75V/3. But I'm unsure of float voltages etc.
And NiCd may be similar but current limited. A full NiCd is ~1.3-1.4V/cell => 3.9-4.2V/3.
A diode is supposed to usually have a 0.6-0.7V drop across it, but this can vary... (with load (as that Idiot said) and type).
Posted By: anonymous1
Date Posted: March 19, 2010 at 8:00 PM
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.
Posted By: oldspark
Date Posted: March 20, 2010 at 7:58 AM
Hmmm - that sounds like reply #5....
Posted By: anonymous1
Date Posted: March 21, 2010 at 2:00 AM
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
Posted By: oldspark
Date Posted: March 21, 2010 at 4:00 AM
Oops - post #5 = reply #4....
But I agree with what you said - assuming it is a rechargeable (battery) torch....
Farn is an Aussie abbreviation for the Aussie workd "farken".
Farscape is a merkan SciFi series that was filmed in Oz. Its stars later joined the SG1 team.
Posted By: saw830
Date Posted: April 27, 2010 at 9:00 PM
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
Posted By: i am an idiot
Date Posted: April 27, 2010 at 9:07 PM
i am an idiot wrote:
Shorting one of the diodes should increase your charge voltage by .3 to .5 volts.
In a real world application I have yet to find a diode that dropped .7 volts.
Posted By: oldspark
Date Posted: April 27, 2010 at 9:41 PM
Tried alternator diodes? From memory the external tests showed 1.4xxV which means 0.7+ V each (2 in series) except after jump starting whereby the "scale went off".
Interestingly my Jap alternators show ~0.6V for the same - ie, ~0.3 per diode. (That's probably 4/7ths of the reason they do NOT blow under the same circumstances!)
But like transistors - also a "diode FWD voltage drop of 0.7V", 0.6V seems the norm for older components (Germanium excluded).
I've lost track of modern manufacture & compounds. (Though I understand Schottky diodes are still ~0.6. I couldn't understand why many insisted on them for normal 12V auto use? CDI etc yes, but alternators?)
Semiconductors & "sort off" conduct? Yeah - sort off.
I like the description that they "can act as conductors or insulators" and anywhere in between - ie, they can semi conduct. Or they are semi-conductors in that they can conduct, but not necessarily.
Idiot & I probably refer to them as semi-insulators.
But yeah - big difference between actual voltages etc. Not only manufacturing variations, but use, size & currents involved - eg signal versus power diodes.
And a "diode model" (circuit) is like a battery - an ideal diode (for voltage drop & uni-direction) and a series resistor.
Try explaining those variances (and what is allowable) to those wanting to use series LEDs... It's similar to parallel batteries!
Posted By: saw830
Date Posted: April 27, 2010 at 10:07 PM
My experience has been that "most" diodes commonly used for DIY stuff are "power" or "rectifier" diodes, generally made of silicon (SI). I don't recall every having a silicon one of these that was not 0.7 fV, but then I don't consider myself an expert on the matter. Due to the "semi-conductiveness" or diodes, saying that a diode has is "0.7fV" may be a nominal rating, a bit like saying that a car battery is "12 volts". :) From Wikipedia, complete with a simple and enlightening voltage/current graph, comes: The current–voltage curve is exponential. In a normal silicon diode at rated currents, the arbitrary “cut-in” voltage is defined as 0.6 to 0.7 volts. The value is different for other diode types — Schottky diodes can be rated as low as 0.2 V and red or blue light-emitting diodes (LEDs) can have values of 1.4 V and 4.0 V respectively. At higher currents the forward voltage drop of the diode increases. A drop of 1 V to 1.5 V is typical at full rated current for power diodes. https://en.wikipedia.org/wiki/Diode#Current.E2.80.93voltage_characteristic
Posted By: oldspark
Date Posted: April 27, 2010 at 10:52 PM
Damned Kiwis! Always smarter than us mainlanders.
(But our Mainland dairy products are nicer!) (LOL!)
And imagine that, someone that researches!
No Saw, I'm not having a go. Well - not at you!
My compliments on doing what I just joked about elsewhere - that people do NOT search etc.
Also in part that there is no "fixed" parameter... hence exponential and other behaviours.
EG - I recently saw some refer to "shonky figures" because gains etc were quoted at a certain frequency (1kHz) and they varied elsewhere. It's as if the variables are not understood. (To me it's like saying that "statistics lie". But stats are merely data, so data lies? Isn't it usually their interpretation - whether deliberately misrepresented, or by the non-enlightened?)
FYI the Schottky's - I expected the lower voltage units, but they weren't. So why pay heaps more for the same voltage drop when speed is NOT an issue? I think it was merely a "Schottky is better" attitude.
But is this thread being hijacked... I lost track...
|