why there are two resistor on the flashlight drive?

[xchcui]

Hi
i know that,usually,to run a led we limit the current with only one resistor.
why in the following image there are two resistors?
this driver running xr-e q5 cree led with 18650 li-ion battery,3 mode.
does the CAB1Q working as pwm,only for controling the brightness of the led without regulation,or it is also work as current regulation?
in other words how does this driver work?

thanks in advance.

 

[NCF8710]

The device marked 471 is a 470 ohm resistor. The tan device at the top of the image is a capacitor.

 

[xchcui]

thanks NCF8710:thumbsup:

now its more reasonable.so how is it exactly work?
does the value of the resistor determine the value of the current?like simple circuit with led and resistor in series?and while i change to lower modes does the function of the CAB1Q is only to change duty cycle?(in the way that it create permanent duty cycle for med mode,permanent duty cycle for low mode?also even the battery voltage drop?)

 

[xchcui]

:candle: :candle:

 

[xchcui]

The device marked 471 is a 470 ohm resistor..
.

why does the resistance so high?
if the battery voltage is 4v and the vf is about 3v so the voltage on the resistor is about 1v.
1v/470ohm=0.002A.
i will very appreciate if someone can explain me that.
thanks in advance.

 

[yellow]

als You already found out:
there is only one resistor, the driver and a capacitor.
* the driver is programmed to give the levels it is supposed to,
* the resistor limits the max current of the driver (... usually. Something like ...)

the way HOW it works ---> You have to get datasheet for that part and read.

PS: and that whole "system" does not have anything to do with what You mean with:

usually,to run a led we limit the current with only one resistor

 

[xchcui]

...* the resistor limits the max current of the driver (... usually. Something like ...)

Hi yellow.
so the calculation that i did(that used,usually,when choosing a resistor value to determine the current in a circuit with led+resistor in series)is not right for that driver circuit?because i,still,dont understand why it is so high?(other than the fact you mentioned).

...You have to get datasheet for that part and read.

believe me,i looked for the datasheet for that part,but without luck.:shrug:

 

[Everett]

The current through the LED does not run through that 470 Ohm resistor. You can follow the traces and see that current flows from the + connection into the LED, then from the LED into the black IC. The IC has two legs connected to the ground ring to complete the current path. The capacitor is presumably to power the IC for a short time during power cycles so that it can change modes. The resistor is either providing power to the IC or it could be configuring it to operate in a specific way. Based on the small size of the IC, it is likely direct-driving the LED through an internal MOSFET (which it can PWM to dim the LED). You could adjust the drive current down by putting a series resistor in line with the LED, but you probably cannot increase the drive current.

 

[xchcui]

Hi Everett.
thank you very much for your reply.

...The capacitor is presumably to power the IC for a short time during power cycles so that it can change modes...

isn't it possible that the capacitor function is:to supply current to the led with the battery in order to prevent voltage drop of the battery and to prolong working time and battery life?

..The current through the LED does not run through that 470 Ohm resistor...

so what is determine the max current to the led?only the IC?

. Based on the small size of the IC, it is likely direct-driving the LED through an internal MOSFET(which it can PWM to dim the LED)...

is it mean that the current on the high mode is the max. with 100%dutycycle,and at the lower modes the dutycycle is being reduced to dim the led?and when the battery voltage decrease, Is the max.current on the high mode stay the same as when full battery or decrease too?(and ,as result, also the currents on the lowers modes will be lower ?)

...You could adjust the drive current down by putting a series resistor in line with the LED...

only to be sure,you don't mean instead of the 470 ohm resistor,do you?
where can i solder it on the driver board that seen in the image?
and how do i calculate the value of the resistor if the current that i measure in the tail cap on high mode is,lets say, 1400mA?(i know how to choose a resistor when there is only led+resistor in series,but here there are combination of IC(pwm),capacitor,unfamiliar 470 ohm resistor so i am not sure how to choose resistor for determine different max current.:thinking:

 

[FoxyRick]

xchcui, first understand that the part number on the integrated circuit is probably a manufacturer's own code - it is not a general part number and thus we cannot really identify the integrated circuit. All we an do is guess at its function.

isn't it possible that the capacitor function is:to supply current to the led with the battery in order to prevent voltage drop of the battery and to prolong working time and battery life?

Definitely not. Where would the capacitor get its own power from? The cell powers everything and Everett is most likely correct that the capacitor is there to power the integrated circuit (only) during the brief moment that the switch is off while changing modes.

so what is determine the max current to the led?only the IC?

Yes, only the IC. I think that the 470 ohm resistor might be a reference for the integrated circuit, to determine how it steps down the drive in the modes. But that is only a guess. If I had the circuit here I could test things, as could many others here. Without it, we are guessing.

Do you have access to a multimeter and oscilloscope?

 

[NCF8710]

thanks NCF8710:thumbsup:

now its more reasonable.so how is it exactly work?
does the value of the resistor determine the value of the current?like simple circuit with led and resistor in series?and while i change to lower modes does the function of the CAB1Q is only to change duty cycle?(in the way that it create permanent duty cycle for med mode,permanent duty cycle for low mode?also even the battery voltage drop?)

I am not familiar with this IC and not able to locate any data on it. As others have speculated, it is probably a PWM chip. I would guess that the 470 ohm resistor is used to program the PWM function. Its value is so high that it cannot be used to directly control the LED current. In fact, if this resistor were placed directly across a fully charged Li Ion cell (4.2V) it would draw less than 9 mA and dissipate less than 38 mW. The capacitor is probably used as a bypass capacitor to filter the voltage transients caused by the PWM operation. It is too large to incorporate on a monolithic chip and is therefore an external component.

 

[xchcui]

...Definitely not. Where would the capacitor get its own power from? ...

i thought,maybe,the capacitor gets its power from the cell,between the off/on pulses of the pwm.at "off"times the capacitor is being charged by the battery and at "on" times it discharge its power to the led together with the battery.in that way keeping the battery stay longer and not drop too much.but i understand that this is not the situation.

xchcui, first understand that the part number on the integrated circuit is probably a manufacturer's own code - it is not a general part number and thus we cannot really identify the integrated circuit. All we an do is guess at its function.

i understand.but from your experience if you measure the tailcap with dmm and you read on the highmode-1400mA(medium -700mA,low-300mA)on this flashlight that consider cheap and power by ~4.2v,is it mean that the highest peak current is 1400mA with 100% duty cycle or the peak current is higher,with lower dutycycle?(like 2800mA with 50%dutycycle?)ofcourse i mean what is more reasonable to assume for this particular flashlight?)

if i want to adjust and decrease the current to 350mA on highmode,how do i calculate what value of resistor to choose?and where can i solder it on the driver?

 

[xchcui]

:help:
Anyone can help me about my last post?
i thank you about your replies but i got all the time partial answers.
i'm sure that in this respected forum there are alot of experts that can easily answer my questions.
thanks in advance.

 

[argleargle]

if i want to adjust and decrease the current to 350mA on highmode,how do i calculate what value of resistor to choose?and where can i solder it on the driver?

I think you'll have to know exactly how that IC works to make that call.

 

[yellow]

without the exact info on the IC, nothing can be calculated,
and even when knowing the IC, it could be one that can be programmed (think ATiny) and so without the programming, again nothing can be done.


But You can risk and do the following: add another resistor to the one that is present.
As You want to reduce the current ...
* if the current measured gets higher, You have to use a higher value for the (single) resistor You solder in finally
* current lower --> solder in a resistor with lower number than the original 471

Start with ... 100 ohms. Thats 1/4 more/less and everything should survive a short burst ...
(or with any "normal" resistor You have, when at least has 100 ohms. You bend the legs and hold them on both sides of the 471 while its powered)


PS: with a small soldering iron (small tip) its not too complicated to solder these parts. They simply are small ...

 

[argleargle]

...and that resistor may well be limiting current from the battery simply to power the IC. We don't know yet. Messing with that resistor may destroy the IC.

 

[yellow]

... and risk killing it, correct

I still think its 99 % that the resistor does "make" the drive current.
If You can live with damaging Your light engine, try; if not, get another driver

 

[yellow]

PS: if there were a resistor where the FULL drive current goes "through", it were placed at one of the 2 free spaces on the right side.
The ones that connect "-" with "led -" ("L-)

(strange printing! imho that kind of setup is really out dated; if I read that correct)
:thinking:

 

[VidPro]

If you want to reduce the current some, using a cheap resister method, on the output to the led, try about 1 ohm and 1/2ohm.

Use an online LED calculator like this http://www.superbrightleds.com/led_info.php
Figure both the high(est) voltage and lowest voltage of the power source (battery).
Toss in some relevent voltage for the LED , like around 3.7V.
Get some close enough resistance ammount from the calculator there, remembering that resistance is everywhere already, springs and wires add in small resistances already.

Put a bit of resistance IN the curcuit, between the board and the led, and check it with a meter or your eye, to see where your at. The resistance will need to handle the power shown in the calculator, so it usually needs to be a 1W resister, with no cooling available.

I have some of the same junk here, and the high voltage of the li-ion or even 3X ni-mhy drives the led hard. Small ammounts of resistance will stop the insanity, drive the led more normal, and increase the runtime.
They do this because it is cheap, it is very bright (overdriven), and it works for some time then destroys the led. (for the price of them you could just buy more and destroy them too

The "better" thing to do might be to put one of them 7135 (regulator) driver things in, the "glorified resister" or replace the non-driver that you have with a real driver.

I think everyone called it pretty well, its a "Direct Drive" , with PWM, I wouldnt try messing with the setup there, because there isnt anything to "change".
Add some resistance to the main power, to reduce it some, or replace it with a real driver. (because a real driver wouldn't cost any more than stacking pwm to a real driver)

I would/have tossed a resister in, to make the LED more long term, but I would want to switch the resistance Out again (switch or button), for both higher drive currents, and when the battery voltage is lower. Without a switch , I would probably want that glorified resister thing, or driver. Each has its pros and cons.

Use any one of the LED calculators, and change up the battery (input) voltage, and current, after making selections, and you will see about what happens (in math).
Long term dwindling outputs have some really good advantages, but many people would not like them. If you get a ~350ma drive at say some ~3.9v of the battery ~1/2ohm (0.5ohm), the rest of it will work close enough. Cheap easy, get it done, and "fix" what is junk, to usable junk and its good to go.

It Looks like there is even space on the board to tack in 2X 1ohm (1/2ohm when parelleled) tiny resisters, and move one wire over to L-. (to small for me to do though) adding small resistance to the - there. like yellow said

 

[xchcui]

But You can risk and do the following: add another resistor to the one that is present.
As You want to reduce the current ...
* if the current measured gets higher, You have to use a higher value for the (single) resistor You solder in finally
* current lower --> solder in a resistor with lower number than the original 471...

it looks to me the ideal thing to do,but you said,also,it may risk the IC,so i prefer no to take the risk,at least not now.

Put a bit of resistance IN the curcuit, between the board and the led, and check it with a meter or your eye, to see where your at.

VidPro suggestion seems to be safer,and i want to try it,but i didn't understand completely how to connect the resistor?

It Looks like there is even space on the board to tack in 2X 1ohm (1/2ohm when parelleled) tiny resisters, and move one wire over to L-. (to small for me to do though) adding small resistance to the - there. like yellow said

from where and to where should i move one wire?do you mean to move the blue wire that already solder in the bottom L- to the upper L- empty space?
does this changes is going to bypass the IC and all the components in the circuit,and i will ,practical,get a led in series with resistor circuit with only one mode?
i don't have problem with that,but i don't want to make any mistakes.