7135 specsheets a max of 6v input, but efficiency at those voltages looks crappy. I'm expecting the 123s to drop to maybe 2.5 volts under a 1.2-1.4a draw (shown in the battery shootout), but even after this how well will the board run? Will it overheat/what kind of runtimes am I looking at?
7135 board on 6v, 2x123 input?
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jasonck08
Flashlight Enthusiast
It's my understanding that these drivers run best at 3.6v to 4.5v. They should be most efficient when the battery voltage is closest to the VF of the LED. Correct me if I'm wrong but I believe any voltage higher than the VF of the LED is just basically burnt off as heat waste, and thats why efficiency drops so much with higher voltages...
Packhorse
Flashlight Enthusiast
Exactly.
As far as run time its a little difficult to calculate since you have not told us any specs of the batteries or what LED you are using.
Even given this info you can only guestimate.
What will happen is the LED will run at full power until the Vf matches the Vout of the batteries. Then it will slowly dim down. This dimming may take many hours.
Lets assume you use 1400ma cells. Run 2 amc7135 chips and you get 700ma. That should gie you 2 hours run time ( on paper). But the battery probably will not deliver the full capacity on a short run time like that. So maybe 1.5 hours. But then as the battery voltage drops to below the Vf of the LED the LED will draw less current meaning run time is increased but at lower output. How long is a product of the Vf of the LED and the discharge curve of the battery and more.
As far as run time its a little difficult to calculate since you have not told us any specs of the batteries or what LED you are using.
Even given this info you can only guestimate.
What will happen is the LED will run at full power until the Vf matches the Vout of the batteries. Then it will slowly dim down. This dimming may take many hours.
Lets assume you use 1400ma cells. Run 2 amc7135 chips and you get 700ma. That should gie you 2 hours run time ( on paper). But the battery probably will not deliver the full capacity on a short run time like that. So maybe 1.5 hours. But then as the battery voltage drops to below the Vf of the LED the LED will draw less current meaning run time is increased but at lower output. How long is a product of the Vf of the LED and the discharge curve of the battery and more.
Lighthouse one
Flashlight Enthusiast
I have a home made light with the DX driver that uses 2 7135 chips...750 ma. It runs great on an 18650 or 2 123 primaries. I don't do timed runs, but I don't notice any overheat problem with the 2 123 primaries. I don't believe it's quite as inefficient as some have thought. Unless the light has very poor heatsinking- should be no problem. Remember- you can't just turn this on and let it sit for an hour...you must be holding it or using shorter run times like is normal.
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Packhorse
Flashlight Enthusiast
As I said before its not something that can really be accurately calculated.
IF the battery can deliver full voltage and they are 1400ma batteries you will get 1 hour run time.
Remember with these drivers more voltage will not give you more run time.
eg a 4v 1400ma battery will run the light for the same time and brightness as a 6v 1400ma battery. The 2v difference will simply cause the driver to heat up more as this needs to be dissipated.
So, after a little more research on 7135's, I've discovered something..
Earlier I saw that cr123's drop to around 2.5v under a 1.4A load, x2 gives you around 5v input. I've read that the reverse polarity diode on the 7135 boards drop voltage another ~.6V, wouldn't this give you an input of 4.4v after losses? Am I missing something?
Earlier I saw that cr123's drop to around 2.5v under a 1.4A load, x2 gives you around 5v input. I've read that the reverse polarity diode on the 7135 boards drop voltage another ~.6V, wouldn't this give you an input of 4.4v after losses? Am I missing something?
Linger
Flashlight Enthusiast
re: missing something: There's a small drop across the 7135 chip, .0x of a volt (iirc .02 but confirm that)
Packhorse
Flashlight Enthusiast
No there is a huge drop across the 7135 chip. If its a 4.4v input and the LED has a Vf of 3.4v at the driven current the drop will be 1 volt. This is how 7135s work.
Its only when the Vin drops below the Vf that you will get a 0.0X volt drop.
Doh!Nut
Newly Enlightened
Wouldnt this be a good candidate for a Keenan board?
I was thinking of making one up in this configuration
I was thinking of making one up in this configuration
all right! Post #2 for me. Newbie here, so please bare with me. I looked up the information on this 7135 chip and it looks to me like there is a 200 mv "dropout voltage" (which is probably why these sorts of chips are called low drop out regulators). If the input (battery) voltage falls below the load voltage plus this dropout voltage, then the chip isn't in complete regulation anymore. I think this is what linger might be thinking of.
Al Combs
Enlightened
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Was it this spreadsheet you were talking about? Excellent source of info about the AMC7135 if you haven't seen it. At 6 volts he was getting 61.8 % efficiency. But he was testing the board with an SSC P4 that had a Vf of 3.7 volts. An XP-G has a Vf about a half a volt lower than a P4 so it would be even worse in the efficiency department using the 7135. Here is the pdf of the AMC7135. The specs say the minimum dropout voltage of the chip is 120 millivolts. But as Packhorse pointed out, at 6 volts it's going to be a lot higher than 0.12 volts. The reverse polarity protection diode on the DX boards is wired in series with the VDD pin of the chip. The chip's minimum operating voltage is 2.7 volts. So your ≈ 0.6 volts or whatever the diode drop is relates to how low the battery voltage can go before the chip turns off. It basically has no effect on the voltage going to the LED. If you really want to use the AMC7135 with a pair of CR123A's, you would be better off leaving the reverse polarity protection diodes on the board. There would be less likelihood of driving one of the batteries into reverse polarity.
It would work great with a single 18650 powering an XP-G. I have a pair of 1,400 ma AMC7135's wired in parallel driving a P7 with an IMR26650 in a 2C MagLite. The boards don't even get warm to the touch. With proper potting you could get rid of the heat a pair of CR123A's would create. But if you really want to use CR123A's for something other than backup in emergencies only, you should look for a good buck switching regulator instead.
It would work great with a single 18650 powering an XP-G. I have a pair of 1,400 ma AMC7135's wired in parallel driving a P7 with an IMR26650 in a 2C MagLite. The boards don't even get warm to the touch. With proper potting you could get rid of the heat a pair of CR123A's would create. But if you really want to use CR123A's for something other than backup in emergencies only, you should look for a good buck switching regulator instead.
kosPap
Flashlight Enthusiast
well i would liek to input some more on this....
few days ago I tested a ShiningBeam MC-E/P7 driver with 2xCR123....this driver is an 8xAMC7135 application...
Current draw was 2.85A and working voltage of the batteriers 4.05....No this is atxing for the batts to say the least...but works nevertheless
few days ago I tested a ShiningBeam MC-E/P7 driver with 2xCR123....this driver is an 8xAMC7135 application...
Current draw was 2.85A and working voltage of the batteriers 4.05....No this is atxing for the batts to say the least...but works nevertheless
TorchBoy
Flashlight Enthusiast
FYI, it's "Kennan". Or was Kennan when they were still sold under that name. There are Kennan-like drivers around which do pretty well on 4.8 V and up.
Sorry, I don't strip with newbies. 😗
:welcome: DelDotB.
alpg88
Flashlight Enthusiast
- Joined
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- 5,833
IT will work, but it will heat up a lot. it will need heat sinking, even at 4.8v (4xnimh) without heatsink it heats up to the point individual chips shut off on internal thermal protection. if your 7135 does not heat up from 6v, your entire system has too much resistance. some 7135 drivers have reverse polarity protection diode that drops too much voltage, even on fresh 1 li ion they do not put out rated current, with 8 chips, they max out at 2A, i had a bunch of those, it drove me nuts before i figured the problem.
