Help! 1.4A driver only delivering 700ma...is it me?

archer6817j

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Hi all,

I purchased several of these drivers to build 1x18650 lights.

I hooked one up on the bench today and at 4.2V I only get about 700ma instead of 1.4A. The input voltage is 2.8-6v. If I crank up the power supply to 6 volts I can get (near) the full output. I forget the exact number. However, et occurs to me that this relationship (input volts/output amps) might be very logical.

I'm relatively new to this, so should I expect that you have to drive it at 6v to get the 1.4A? I had assumed that any voltage in the stated input range would deliver the full current.
 

VidPro

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nope and yup, and . . .
Buck : to reduce the voltage so the current to the LED doesnt go over
Linear : hard to make without some voltage drops involved
2.8v still operates :) if only you had a 2V led to drive with it

usually something like this would be ok, and would dwindle down only after the battery was discharged somewhat, even this should read higher than that with the full 4.2v, better ask ShiningBeam because they would know best. did you test a second one yet?

is possible but hard to tell that reverse polarity diode could be bypassed , and it could work to lower voltages, then you just dont accidentally reverse it and burn out the chips.

It is not easy to make either boost OR buck curcuits to work with changing voltages of battery that are VERY similar to the voltage the led will be driven to.
Boost wants the battery voltage to be lower (when bat is high), buck wants the voltage to be higher (when Bat is low)

DC-DC "wizz" "flex" , Boost&Buck is the only choice that can keep fully regulated at both higher and lower than the led voltages.
 
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HKJ

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I hooked one up on the bench today and at 4.2V I only get about 700ma instead of 1.4A. The input voltage is 2.8-6v. If I crank up the power supply to 6 volts I can get (near) the full output. I forget the exact number. However, et occurs to me that this relationship (input volts/output amps) might be very logical.

This type of drivers needs about 0.2 volt to work, you problem might be to high Vf in the led or maybe to thin wires?


is possible but hard to tell that reverse polarity diode could be bypassed , and it could work to lower voltages, then you just dont accidentally reverse it and burn out the chips.

The protection diode has nothing to do with voltage drop in this circuit, i.e. shorting the diode will only remove the protection, not change the working voltage in any way!
 

archer6817j

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With regard to the small wires, the driver had leads attached that seem really tiny to me. The lead wires I attached to the battery pads were 20awg and the (long) leads from my meter and some alligator leads are all about 3 feet long (16awg) so maybe that adds up to a lot of resistance. I'll try replacing the stock driver leads and make up some shortys to use on my meter.

How can I check the vf of the LED itself?

I also bought the other shining beam driver that goes up to 2.5 amps and I had about the same results...seeing about half the rated amperage on my meter with the same bench setup.
 

VidPro

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The protection diode has nothing to do with voltage drop in this circuit,

so this driver is the usual AMC7135 type of driver? i thought it might be, but that doesnt sound like what he is testing, so i wasnt sure.
the pic shows 3 of the chip things and a controller? so is it a different chip to get to 1.4A with 3 or is that 2 and the mosfet?
http://www.shiningbeam.com/servlet/the-133/3-dsh-Mode-Regulated-Circuit-Board/Detail

archer if you can post a close scan of the chip, it would be helpfull.
The "other" one i got has 8 of them on it, but it is old.
this one the positioning of the diodes Looks different in the curcuit, and are only leetel glass diodes.
soooo, this ShiningBeam one looks more like it is before the curcuit itself, I couldn't even trace/understand it with a good picture anyway, i would just disable it and see what happened.
 
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archer6817j

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Oh so I forgot, is it correct that I should be able to get (close) to 1.4A with a 4.2v battery assuming the LED vf is something reasonable like 3.2-3.4?
 

VidPro

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Oh so I forgot, is it correct that I should be able to get (close) to 1.4A with a 4.2v battery assuming the LED vf is something reasonable like 3.2-3.4?

yes , the leds VF goes up remember over the Speced 350ma , when driven at 1.4A, most of the white leds will be more like 3.6+ even ~3.9 for some of the older ones.
so
~4V battery under load, ~.2V loss, ~3.8V max to the led <--- or somewhere around there.
 

HKJ

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I also bought the other shining beam driver that goes up to 2.5 amps and I had about the same results...seeing about half the rated amperage on my meter with the same bench setup.

Your meter might also be part of the problem, it will also add a small voltage drop.

If you get the same result with 1.4 and 2.5 A drivers, it is definitely somewhere in your connections. If you have two meters, use the extra meter to measure voltage to find where the problem is located.

How can I check the vf of the LED itself?

Use your meter to measure the voltage over the led. You can also use it to measure battery voltage and voltage drop in wires.

Oh so I forgot, is it correct that I should be able to get (close) to 1.4A with a 4.2v battery assuming the LED vf is something reasonable like 3.2-3.4?

Yes, each chip is usual 350 mA.
 

LilKevin715

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Is the 700ma reading at the start or middle/end of your test? AMC7135 linear regulators dump off the excess input voltage as heat, so it could be 2 of the 4 chips thermally shutting down to prevent damage. For example if your LED Vf is 3.6v and your Vin is 4.2v, then the excess heat would be 4.2v-3.6v x 1.4A = 0.84watts. Its not a lot of heat initially but after a long run-time the chips can get hot enough to shutdown to prevent damage.

It would be helpful if we knew what LED you are using to test with. For AMC7135 drivers its best to have Vin of 0.12v above the Vf of the LED to have the highest efficiency and lower heat generated from the chips.

If you are using the stock teflon wires already soldered on the emitter side of the driver then do yourself a favor and desolder those and use 20-24 guage wire instead. I think the stock emitter wires are somewhere around 28 guage, too thin and they will have too much resistance to them.
 

Justin Case

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That's a multimode AMC7135-based driver. AMC chips need about 0.12V above Vf (excluding any reverse polarity protection diodes) to run in full regulation. If you have the driver running on a bench power supply, then I doubt you have trouble with power source voltage drop. This assumes that your bench supply wires are of sufficient gauge, which seems to be the case (20 ga and 16 ga wires). Even factoring in any contact resistance from the bench supply wires to the driver (if you are using probes to touch the driver's center pad and outer ground ring to power the driver -- it sounds like you are not) and additional voltage drop from the reverse polarity protection, you shouldn't need 6V to get 1.4A drive current.

If your Vf is around 3.5V, then 4.5V should be plenty to reach regulation (3.5V Vf, 0.5V protection diode, 0.12V AMC7135 overhead, and let's say 200 mohms contact resistance for your bench supply connection to the driver).

The 200 mohms is most likely way more than in your case. It sounds like you soldered some 20 ga wires to the driver center Batt+ pad and outer ring Batt-. Then used alligator clips on your bench power supply's 16 ga wires to hook up to the driver. The voltage drop with that setup is most likely low.

So 4.2V could be enough to run in full regulation.

If the LED Vf is lower than the assumed 3.5V, then the voltage to reach regulation is that much lower.

Two suggestions:

1. Make sure you are running in max mode when making your measurements. If you are in a lower mode, you'll get a lower tail current draw. Typically, these drivers use a sequential on-off power cycling method to change modes. So, it can be quite easy to unintentionally enter the wrong mode during testing.

2. Make sure that all of the AMC7135 chips are properly connected in parallel on the driver board (this shouldn't be an issue with your SB drivers, but some boards from DX a few years ago didn't have all of the chips connected and the end user had to make those connections).
 
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HKJ

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If your Vf is around 3.5V, then 4.5V should be plenty to reach regulation (3.5V Vf, 0.5V protection diode, 0.12V AMC7135 overhead, and let's say 200 mohms contact resistance for your bench supply connection to the driver).

Again: There is no voltage drop from a protection diode on a typical AMC7135 driver!

1. Make sure you are running in max mode when making your measurements. If you are in a lower mode, you'll get a lower tail current draw. Typically, these drivers use a sequential on-off power cycling method to change modes. So, it can be quite easy to unintentionally enter the wrong mode during testing.

This could be the reason for the problem.
 

Justin Case

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With regard to the small wires, the driver had leads attached that seem really tiny to me. The lead wires I attached to the battery pads were 20awg and the (long) leads from my meter and some alligator leads are all about 3 feet long (16awg) so maybe that adds up to a lot of resistance. I'll try replacing the stock driver leads and make up some shortys to use on my meter.

How can I check the vf of the LED itself?

I also bought the other shining beam driver that goes up to 2.5 amps and I had about the same results...seeing about half the rated amperage on my meter with the same bench setup.

The stock LED hookup wires are typically junk. Using good quality 24 gauge stranded wire is a good idea. But I doubt that is the cause of your difficulties.

Put a DMM across the LED to measure Vf.

When you say you saw "about half the rated amperage on my meter", are you talking about a separate DMM hooked into your circuit, or the current meter built into your bench supply?
 

Black Rose

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Those particular drivers lock the mode into memory after about 2 seconds of being on.

After 2 seconds, you need to interrupt the power twice (double tap) to get it to jump to the next mode.

I use these drivers in several of my lights and P60 drop-ins, and they do put out 1.4A.
 
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archer6817j

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My readings are at the beginning of the test. I tested with a star board XR-E mounted to an aluminum slug. Oh, I also tried all three modes and they read about half.

Also, as I increase the voltage on the power supply the amperage also increases. I was anticipating a constant output amperage. I'll put new (larger) wires on the board and recheck my setup. I assumed these are good drivers so I've got to be doing something funky.

As far as checking the led vf, I just attach to the led leads while under power?

Thanks for all the help! This forum rocks.
 

HKJ

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As far as checking the led vf, I just attach to the led leads while under power?

Yes.
But try moving the probes around and find how much voltage is lost in the different parts of the circuit.
I.e. measure voltage from power supply, measure voltage for each wire, measure voltage across driver, measure voltage across led.

If you have a decent DMM, change to AC voltage and check output from power supply, this must be very close to zero (if not, you meter can not be used for this test), then check across the led. If the led reads much more AC than the power supply, the drive is using using pwm to reduce the brightness.
This test cannot prove that there is no pwm (Except if you have a very good meter), but if it shows pwm you know the reason for the reduced brightness.
 

HKJ

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Why not? Isn't the diode between the power supply and the AMC7135's Vin pin?

Yes it is, but the led is from battery + to the led terminal on the AMC7135, i.e. the led current does not pass through the protection diode.
The idea behind this is that the led is also a diode, i.e. it will protect the led terminal on AMC7135.
 

bshanahan14rulz

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But then, what would prevent reverse voltage across Gnd and Vin pin on AMC7135? The LED would prevent reverse voltage across Vin and output pin, right?

And while we're on the subject, I was hoping to run a load with a Vf of ~6V. If I supplied 8.4V to the load and used diodes in series to drop the supply to the chip to closer to 6V, would those diodes interfere with its current-limiting functionality? Like, would the chip think that they are part of the load and adjust for it?

Edit: ok, now I understand what you are getting at, the diode doesn't drop voltage before it gets to the LED, and that I definitely agree with. Still would like your input on my idea, though.
 

HKJ

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But then, what would prevent reverse voltage across Gnd and Vin pin on AMC7135? The LED would prevent reverse voltage across Vin and output pin, right?

The led would prevent reverse voltage on output, the diode would prevent reverse voltage on Vin.

And while we're on the subject, I was hoping to run a load with a Vf of ~6V. If I supplied 8.4V to the load and used diodes in series to drop the supply to the chip to closer to 6V, would those diodes interfere with its current-limiting functionality? Like, would the chip think that they are part of the load and adjust for it?

You have to keep the voltage between ground and Vin or output below 6 volt, output preferable much lower due to reduce heat. Some circuit connects extra leds in series with the AMC7135 circuit, i.e. these led would both carry the supply current and the led current from the AMC7135.
 

bao123

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archer6817j said:
Hi all,

I purchased several of these drivers to build 1x18650 lights.

I hooked one up on the bench today and at 4.2V I only get about 700ma instead of 1.4A. The input voltage is 2.8-6v. If I crank up the power supply to 6 volts I can get (near) the full output. I forget the exact number. However, et occurs to me that this relationship (input volts/output amps) might be very logical.

I'm relatively new to this, so should I expect that you have to drive it at 6v to get the 1.4A? I had assumed that any voltage in the stated input range would deliver the full current.

Hi,

The AMC7135 chips function by supplying a fixed voltage output, 3.3V or something like that, and limiting the current of the output. I suspect that as the voltage input increases the output voltage also increases slightly.

SO, the reason you are getting only 700ma at 4.2V input voltage, is because the LED will only take 700ma at the voltage supplied by that input.

In other words, your XRE chip is in a higher Vf bin. For it to function at it's rated current, it needs a higher voltage, one that is only supplied as the input voltage approaches 6V.

If this is the case, no matter how many drivers you parallel to drive the LED, it will only consume 700ma at 4.2V driver input.

You can test this out since you have several drivers by running a pair in parallel.

XRE's do have higher Vfs. What this statement means is that on average, a higher voltage is needed to drive the LEDs to a given current than say XPG chips. However, due to variations that occur in manufacturing, individuals of a class of chips(XRE, XPE, XPG, etc) fall within a range of Vfs. If the reseller doesn't specify Vf bin, you really have know way of knowing what you get until you test it.
 
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