Agony of Confusion About Batteries and Current

HKJ · 2019-01-29T07:21:46.000Z

danallen:

transitioning into wrap up mode for this thread

Just a few really simple questions.

1. What is the effect of adding 7135 resistors to a driver. My impression is each resistor provide an additional step between off and brightest, because. according to my unreliable knowledge, the resistors a are in parallel with one another and there ends my theory.

The purpose is to limit the current and in the case of 7135 also to stabilize the current. Without you would get a brighter light, but probably not for long (The led would die due to over current).

danallen:

If I say I have a “7amp flashlight” , what am I saying? I know what amps are, but is that statement usually going to reference a certain type of light, such as a single cell single led light? It seems as if there are specs besides the current that are assumed and I missed the memo covering them.

Current is not saying everything, you need both voltage and current to get power. When talking about flashlight the voltage is often assumed to be the voltage drop for one led, at the specified current.
A freshly charged LiIon battery has more voltage than a led needs, a depleted battery less, this means a flashlight usual needs something to limit the current from the freshly charged battery, to avoid blowing the led (like 7135), but when the battery gets discharge the brightness will drop.

danallen · 2019-01-29T07:29:34.000Z

HKJ:

Current is not saying everything, you need both voltage and current to get power. When talking about flashlight the voltage is often assumed to be the voltage drop for one led, at the specified current.

That is what I was looking for, thank you.

With regard to adding a 7135 I understand going from zero to 1 resistor. What is the point of adding 5 or 8 instead?

HKJ · 2019-01-29T07:34:03.000Z

danallen:

HKJ:

Current is not saying everything, you need both voltage and current to get power. When talking about flashlight the voltage is often assumed to be the voltage drop for one led, at the specified current.

That is what I was looking for, thank you.

With regard to adding a 7135 I understand going from zero to 1 resistor. What is the point of adding 5 or 8 instead?

7135 is not a resistor, it is a current regulator, each one will pass 350mA, using two means 700mA, using 3 means 1050mA, etc.
With resistors it is often a question about getting the correct values and power ratings, it is easier to parallel a couple of common values, than try to get a special value in a high enough power rating.

ggf31416 · 2019-01-29T10:17:30.000Z

Read Driver - Flashlight Wiki for the basics.

LEDs are most efficient at lower currents, so using only as few 7135s as needed is better.

FET+N drivers(often FET + 1 like in the BLF A6) have 2 separate channels that run at different rates using PWM. For currents below N * 350mA they will use only the 7135 and keep the FET channel at 0%. For higher currents they will use both, but then current is not longer regulated.

Search for FET and FET+1 in the search dialog.

A more complex driver often will have a better microcontroller (i.e an attiny25 or attiny85 instead of an attiny13) with higher storage capacity, so a more complex firmware will fit in it.

Also check out FW3A, a TLF/BLF EDC flashlight - SST-20 available, coupon codes public - #5622 by ToyKeeper

As how to make a triple led flashlight, I never done it, but i’m sure it’s well covered in previous threads.

danallen · 2019-01-29T15:52:09.000Z

HKJ:

danallen:

HKJ:

Current is not saying everything, you need both voltage and current to get power. When talking about flashlight the voltage is often assumed to be the voltage drop for one led, at the specified current.

That is what I was looking for, thank you.

With regard to adding a 7135 I understand going from zero to 1 resistor. What is the point of adding 5 or 8 instead?

7135 is not a resistor, it is a current regulator, each one will pass 350mA, using two means 700mA, using 3 means 1050mA, etc.
With resistors it is often a question about getting the correct values and power ratings, it is easier to parallel a couple of common values, than try to get a special value in a high enough power rating.

Oh!!! well that paints a different picture from the wrong one I had in my head. Thank you very much.

danallen · 2019-01-29T15:56:15.000Z

ggf31416:

Read Driver - Flashlight Wiki for the basics.

LEDs are most efficient at lower currents, so using only as few 7135s as needed is better.

FET+N drivers(often FET + 1 like in the BLF A6) have 2 separate channels that run at different rates using PWM. For currents below N * 350mA they will use only the 7135 and keep the FET channel at 0%. For higher currents they will use both, but then current is not longer regulated.

Search for FET and FET+1 in the search dialog.

A more complex driver often will have a better microcontroller (i.e an attiny25 or attiny85 instead of an attiny13) with higher storage capacity, so a more complex firmware will fit in it.

Also check out FW3A, a TLF/BLF EDC flashlight - SST-20 available, coupon codes public - #5622 by ToyKeeper

As how to make a triple led flashlight, I never done it, but i’m sure it’s well covered in previous threads.

The information in this thread is immensely helpful. Now I am starting to feel like I am amongst the information needed to start having a clue what’s going on the number of people who have contributed directly to my questions is a lot, it is unbelievable the level of support. I am actually moved deeply.

HKJ · 2019-01-29T15:59:48.000Z

danallen:

HKJ:

With resistors it is often a question about getting the correct values and power ratings, it is easier to parallel a couple of common values, than try to get a special value in a high enough power rating.

Oh!!! well that paints a different picture from the wrong one I had in my head. Thank you very much.

With resistors there are some other points:

Resistor have many other application than limiting the current to the led, i.e. when there are multiple resistors it can also be for other reasons. The current limiting resistors will often be the largest ones.

Capacitors may look like resistors (They usual have another color and no letters/numbers on them), but have other functions.

When modding a driver to higher output power, it is often easier to solder a resistor “piggy back” on the current limiting resistor, than it is to replace the original resistor.

danallen · 2019-01-29T16:35:34.000Z

HKJ:

danallen:

HKJ:

With resistors it is often a question about getting the correct values and power ratings, it is easier to parallel a couple of common values, than try to get a special value in a high enough power rating.

Oh!!! well that paints a different picture from the wrong one I had in my head. Thank you very much.

With resistors there are some other points:

Resistor have many other application than limiting the current to the led, i.e. when there are multiple resistors it can also be for other reasons. The current limiting resistors will often be the largest ones.

Capacitors may look like resistors (They usual have another color and no letters/numbers on them), but have other functions.

When modding a driver to higher output power, it is often easier to solder a resistor “piggy back” on the current limiting resistor, than it is to replace the original resistor.

When it comes to modding a light, I am having trouble identifying a worthwhile mod, because it seems as if everything I might want already is available at the same cost or better than it would cost me. I still want to do some mods, because that has to be best way to learn how the lights work, once you have a reasonable framework for understanding. Most of these components are theoretical for me at this point. In a college lab, there was a poster: One experiment is worth one thousand theories.

danallen · 2019-02-05T22:34:34.000Z

Another example of something I don’t understand. Convoy L2 with XPL HI U6-3A.

It runs great on one 26650 cell, using the short tube (~3.7 volts).

It runs equally great on two 26650 cells in the long tube (~7.4 volts).

How does it run the same on 3.7 volts as 7.4 volts?

BurningPlayd0h · 2019-02-05T22:40:29.000Z

danallen:

I understand that rapid charging of high drain Li ion cells doesn’t save much time because any gain on the rapid charge up to around 4.0-4.1 volts mostly is paid back when the current drops for the rest of the charge cycle, which will run longer after a rapid charge for the first part of the cycle.

I don’t think that’s true at all. I don’t have any charging curves in front of me to look at but I can say that my 1.5A li-ion charger is CONSIDERABLY faster at bringing cells from ~3v to 4.15/4.2v (measured with DMM) than the 0.5A max charger I had before.

I haven’t timed it but the roughly 3x faster it should be sounds about right.

Marc_E · 2019-02-05T23:03:50.000Z

danallen:

Another example of something I don’t understand. Convoy L2 with XPL HI U6-3A.

It runs great on one 26650 cell, using the short tube (~3.7 volts).

It runs equally great on two 26650 cells in the long tube (~7.4 volts).

How does it run the same on 3.7 volts as 7.4 volts?

It doesnt run the same, something will change. What changes depends on how the driver is designed. (The driver is the electronics that sit between the battery and the LED to control things.)

The 3.7v and 7.4v are supplied to the driver, the driver can control the current that it gives to the LED.
Voltage is not power, power is a combination of voltage and current (voltage multiplied by current to be precise, 4 amps at 3.7 volts has the same power as 2 amps at 7.4 volts. 4x3.7 = 2x7.4)

If the driver is designed to supply the same power to the LED regardless of the voltage it is being fed then the LED will act the same with 2 batteries as it does with 1, but 2 batteries will give twice the runtime at the same power as 1 battery. (ie 2 batteries last twice as long as 1).
If the driver is designed to give more power to the LED when the voltage it is being fed increases then the LED will be brighter with 2 batteries than with 1 for the same runtime.

In other words, with 2 batteries you would either get more runtime or more brightness than with 1 battery, and sometimes a bit of both.

danallen · 2019-02-05T23:34:38.000Z

Marc E:

danallen:

Another example of something I don’t understand. Convoy L2 with XPL HI U6-3A.

It runs great on one 26650 cell, using the short tube (~3.7 volts).

It runs equally great on two 26650 cells in the long tube (~7.4 volts).

How does it run the same on 3.7 volts as 7.4 volts?

It doesnt run the same, something will change. What changes depends on how the driver is designed. (The driver is the electronics that sit between the battery and the LED to control things.)

The 3.7v and 7.4v are supplied to the driver, the driver can control the current that it gives to the LED.
Voltage is not power, power is a combination of voltage and current (voltage multiplied by current to be precise, 4 amps at 3.7 volts has the same power as 2 amps at 7.4 volts. 4x3.7 = 2x7.4)

If the driver is designed to supply the same power to the LED regardless of the voltage it is being fed then the LED will act the same with 2 batteries as it does with 1, but 2 batteries will give twice the runtime at the same power as 1 battery. (ie 2 batteries last twice as long as 1).
If the driver is designed to give more power to the LED when the voltage it is being fed increases then the LED will be brighter with 2 batteries than with 1 for the same runtime.

In other words, with 2 batteries you would either get more runtime or more brightness than with 1 battery, and sometimes a bit of both.

With this light, the runtime doubles with 2 cells vs. 1 cell, led power appears the same, might be slightly different if precise measurements are taken.

How does a driver output the same power to the led over such a wide range of input voltages?

ggf31416 · 2019-02-06T00:27:23.000Z

The buck drivers have a small ‘transformer’ that reduces the voltage from whatever the input is to the 3.x V needed by the led. As for how it works, an electrical engineering background is needed to really understand them and I don’t have such background.

Edit: Read Switched-mode power supply - Wikipedia and Buck converter - Wikipedia . Don’t expect that it will be easy to understand.

BlueSwordM · 2019-02-05T23:53:08.000Z

It’s using a buck driver, a driver that steps down the high DC voltage to a usable lower DC voltage.

Let’s start with an example: with a 2S battery pack, you get a maximum battery voltage of 8,4V.

However, in this case, you have a 3V LED driven at 3A. You would burn it in direct drive since the delta in voltage is massive!

So you use a buck driver to step down the voltage. As the battery voltage goes down, to sustain the same power level, the current goes up.

teacher · 2019-02-06T00:31:34.000Z

Nice explanation Blue, easy to understand.

LouieAtienza · 2019-02-06T01:26:14.000Z

danallen:

With this light, the runtime doubles with 2 cells vs. 1 cell, led power appears the same, might be slightly different if precise measurements are taken.

How does a driver output the same power to the led over such a wide range of input voltages?

The driver (in your case) does not output power to the LED, it supplies a_ constant current_ to the LED, and the voltage will vary with a specific range to maintain the current. Power is the energy used by the LED (over time), and the wattage (power) rating of an LED does not necessarily determine its brightness. This is because certain LEDs are more or less efficient than others.

When running two (identical) cells in series the voltage doubles but the capacity rating stays the same. When running two cells in parallel the voltage remains the same but the capacity doubles. Running two cells in your light, in your case, would not seem to make sense, since the driver does not seem to take advantage of the extra voltage.

ggf31416 · 2019-02-06T02:58:57.000Z

LouieAtienza:

The driver (in your case) does not output power to the LED, it supplies a_ constant current_ to the LED, and the voltage will vary with a specific range to maintain the current. Power is the energy used by the LED (over time), and the wattage (power) rating of an LED does not necessarily determine its brightness. This is because certain LEDs are more or less efficient than others.

When running two (identical) cells in series the voltage doubles but the capacity rating stays the same. When running two cells in parallel the voltage remains the same but the capacity doubles. Running two cells in your light, in your case, would not seem to make sense, since the driver does not seem to take advantage of the extra voltage.

I believe you are mistaken. Current to a given LED is mainly function of the voltage (temperature has some effect but it’s less important). If you keep the current to the LED constant the voltage will remain constant as well. A regulated buck driver will keep the output current constant while varying the input current to adjust for the input voltage. In an ideal buck driver (disregarding losses) if the input voltage is doubled because of two cells, the input current will be halved thus getting about double the runtime.

Most linear drivers (which can’t use the extra voltage) will just overheat if you attempt to use two cells to drive a 3V LED.

danallen · 2019-02-09T18:47:42.000Z

ggf31416:

The buck drivers have a small ‘transformer’ that reduces the voltage from whatever the input is to the 3.x V needed by the led. As for how it works, an electrical engineering background is needed to really understand them and I don’t have such background.

Edit: Read Switched-mode power supply - Wikipedia and Buck converter - Wikipedia . Don’t expect that it will be easy to understand.

Thank you, that article is exactly what I needed. For the flashlight to produce the same output across a range of input voltages that varies from one cell to two cells in series, it uses a buck driver that varies the duty cycle with the input voltage, and has enough inductance or capacitance to handle an input voltage twice as high as needed to run the LED. When only one cell is used and its voltage is low enough, the switch controlling the duty cycle can stay closed, because bucking is not needed at that level. I don’t know the details of how to make the switching vary the duty cycle based on input voltage, but I sure do want to know.

BlueSwordM:

It’s using a buck driver, a driver that steps down the high DC voltage to a usable lower DC voltage.

Let’s start with an example: with a 2S battery pack, you get a maximum battery voltage of 8,4V.

However, in this case, you have a 3V LED driven at 3A. You would burn it in direct drive since the delta in voltage is massive!

So you use a buck driver to step down the voltage. As the battery voltage goes down, to sustain the same power level, the current goes up.

The part of this I don’t understand is how the driver can handle the voltage from two batteries in series and also handle running only one battery.

LouieAtienza:

danallen:

With this light, the runtime doubles with 2 cells vs. 1 cell, led power appears the same, might be slightly different if precise measurements are taken.

How does a driver output the same power to the led over such a wide range of input voltages? That is the part I don’t understand.

The driver (in your case) does not output power to the LED, it supplies a_ constant current_ to the LED, and the voltage will vary with a specific range to maintain the current. Power is the energy used by the LED (over time), and the wattage (power) rating of an LED does not necessarily determine its brightness. This is because certain LEDs are more or less efficient than others.

When running two (identical) cells in series the voltage doubles but the capacity rating stays the same. When running two cells in parallel the voltage remains the same but the capacity doubles. Running two cells in your light, in your case, would not seem to make sense, since the driver does not seem to take advantage of the extra voltage.

The advantage of two cells is a result of less current out of the batteries when the voltage is higher. The two cells in series make the voltage into the driver higher than when there is only one battery. The output voltage and current both are the same whether there are two batteries or one. Lower current from the batteries means longer run time. To make the work, the driver has to use transistors to route the electricity into inductors or capacitors and control the current from the batteries. It is the combination of power switching and energy storage of the inductor or capacitor that makes this work efficiently. I want to find out the details of how these components perform.

BurningPlayd0h · 2019-02-10T16:21:05.000Z

danallen:

The part of this I don’t understand is how the driver can handle the voltage from two batteries in series and also handle running only one battery.

Because the buck driver only steps down the voltage once it passes a certain point. So for one cell it just takes the voltage, while for two it steps it down.

ggf31416 · 2019-02-10T19:38:20.000Z

I think this article would be interesting: A matter of light, Part 2 - Buck whenever possible

Edit: Here are the other parts of the series:

A matter of light, Part 1 - The ABC’s of LEDs
A matter of light, Part 3 - When to boost and buck-boost
A matter of light, Part 4 - PWM dimming

Some of the talk is specific to AC current, but I think most of it is still applicable to flashlights.