Tailcap current vs emitter current?

gcbryan

Flashlight Enthusiast
Joined
Oct 19, 2009
Messages
2,473
Location
Seattle,WA
I'm sure this has been asked before but I couldn't find it. Is there some rule of thumb as to related tailcap current draw to emitter current draw?

One is easy to measure and one is not.

With newly a new charged 18650 battery measuring 4.2 V I meansured the tailcap current for my XR-E R2 drop-in and it's in the 1A range and using the same battery when I measure my XP-G R5 drop-in I get 1.7A.

Can I assume that the emitter using new batteries is seeing close to 1A with the XR-E and close to 1.5 A with the XP-G at least when the batteries are new?

Also, the the reason it's more accurate to measure the actual emitter because using the tailcap method you can't tell if the driver has a boost circuit to it? Is that not reflected in a tailcap measurement?

Just trying to learn here and I just got a multi-meter than can read higher DC current (old smaller one only went to 800mA).
 
Last edited:

Russel

Enlightened
Joined
Jan 31, 2009
Messages
583
Location
California
The emitter current also depends upon the voltage.

Let's say you have 1 amp at the emitter at 2 volts, for example:

1A * 2V (example emitter voltage) = 2W

So, at the tail cap:

2W / 4.2V = 0.48 Amps

Assuming the circuit is 90% efficient:

0.48A / .9 = 0.53 Amps at the tail cap. (4.2 volts)


You have to take voltage and current into account:

4.2V * 1 = 4.2 Watts

4.2V * 1.7 = 7.14 Watts

What is the emitter voltage when operating?

[Edit]

I just looked up the specifications for a Cree XR-E emitter. It looks like the forward voltage runs around 3.5 volts at 700mA and 3.7 at 1000mA so your estimate may not be as far off as I implied.

[End edit]
 
Last edited:

Ajay

Enlightened
Joined
Dec 12, 2008
Messages
284
Location
Queens,NY
Yep I think you're on the right track but a few things:

Some circuits can pull 1.7amps at the tail and give less than 1amp at the LED, they are just not very efficient. You would have to research the circuit itself to know. I love the AMC7135 circuits because they are pretty efficient/cheap and well used here at CPF. SKU S004659 at KD for instance.

The current reading on the tail varies depending on the number of cells or the voltage of the combined cells. Rough example: 2 lithium cells total = 8.4v to a XRE driven at 1amp @ LED might read about 0.5 amp at the tail assuming 100% efficiency. That's because the current draw on the cells will decrease as the number of cells increase...given the LED and circuit remains the same.

Boost or buck circuit doesn't say anything about the tail current or vice versa to my knowledge. It's just a way to drop or increase the voltage to the LED.

Veteran CPFers please correct me if I am wrong.
 

Russel

Enlightened
Joined
Jan 31, 2009
Messages
583
Location
California
OK, let's estimate:


XR-E

4.2V * 1A = 4.2 Watts

4.2W / 3.5V = 1.2A


XP-G

4.2V * 1.7A = 7.14 Watts

7.14 / 3.3V = 2.16A
 

gcbryan

Flashlight Enthusiast
Joined
Oct 19, 2009
Messages
2,473
Location
Seattle,WA
OK, from those numbers it looks like both emitters are being fully driven (driven quite hard).

How does the driver fit into all this.

How can I tell from the tailcap numbers whether I have a buck/boost driver or just a buck driver.

Once the batteries are somewhat used (as in the ones in there now) they are measuring at 3.6V. The tailcap current when using those batteries (for the XP-G) is something like .9V at the tailcap.

Does this mean there is no boost circuit in this light? That's what I'm assuming but I just want to make sure I understand that aspect as well.
 

mdocod

Flashaholic
Joined
Nov 9, 2005
Messages
7,544
Location
COLORado spRINGs
Both boost and buck circuits should draw more current as the voltage input drops, however, the difference is that, a buck circuit will only draw more current as the input voltage drops down to the Vf of the LED combined with whatever over-head voltage the regulator requires to stay in regulation, after that point, lower voltage will translate to lower current consumption. Boost circuits will often have more difficult to identify behavior because many of them will operate in something close to "direct drive" when the input voltage is above the Vf of the LED (like a fully charged 18650 with low resistance). As the input voltage drops down to the Vf of the LED, the current will also drop, until the input voltage drops below that point, at which point the boost circuit should attempt to pull more current from the cell to maintain current to the emitter.

Modules that are rated for say, 3-8.4V or 4-18V, or 3-12V etc etc, are generally buck regulated. Modules that are rated for 3.0-4.2V or 3.7-6V (or something like that) are often semi-direct drive, or they might contain a buck style setup that has minimal over-head.... Modules that are rated like 1-4V or something like that are often going to have a true boost circuit in them.

Eric
 

gcbryan

Flashlight Enthusiast
Joined
Oct 19, 2009
Messages
2,473
Location
Seattle,WA
So unless the circuit is designed for alkaline batteries it's not generally going to have a boost component?

If the light is designed to use 18650 or in some cases 18650 and CR123 and if it states "constant current" wouldn't it have to have a boost component as well?
 

Lynx_Arc

Flashaholic
Joined
Oct 1, 2004
Messages
11,096
Location
Tulsa,OK
Both boost and buck circuits should draw more current as the voltage input drops, however, the difference is that, a buck circuit will only draw more current as the input voltage drops down to the Vf of the LED combined with whatever over-head voltage the regulator requires to stay in regulation, after that point, lower voltage will translate to lower current consumption. Boost circuits will often have more difficult to identify behavior because many of them will operate in something close to "direct drive" when the input voltage is above the Vf of the LED (like a fully charged 18650 with low resistance). As the input voltage drops down to the Vf of the LED, the current will also drop, until the input voltage drops below that point, at which point the boost circuit should attempt to pull more current from the cell to maintain current to the emitter.

Modules that are rated for say, 3-8.4V or 4-18V, or 3-12V etc etc, are generally buck regulated. Modules that are rated for 3.0-4.2V or 3.7-6V (or something like that) are often semi-direct drive, or they might contain a buck style setup that has minimal over-head.... Modules that are rated like 1-4V or something like that are often going to have a true boost circuit in them.

Eric
actually not all boost circuits are current regulated, these will drop in current as the battery voltage drops with depletion, they will not go up in current to compensate as they are not designed to regulate at all but simply boost output. If you have a light that maintains output at a constant level some current regulated boost circuits will "fall off the cliff" or suddenly just black out when the battery is almost depleted while others will drop out of regulation and just fade dimmer till nothing is left. cheaper lights with boost circuits start out bright and just keep dimming till the circuit hits its lower limits.... them *poof* it goes out. My roomy has an older luxeon 2AA magled that just quits when the batteries hit about 1v and I can take the batteries out of it and run my cheap 2AA dorcy for hours on the power left as lower levels while it slowly dims to where you can only see the led glowing if you look at it directly.
 

bshanahan14rulz

Flashlight Enthusiast
Joined
Jan 29, 2009
Messages
2,789
Location
Tennessee
Best way to do this is convert all volts or amps readings to watts. Use the amps reading that you measured and also the voltage of the battery (best to use voltage while in operation, but standing voltage can serve as decent approximation).
Power(Watts)=Current(Amps)*Voltage(Volts)
So with a tailcap reading, you end up with how many watts the flashlight as a whole uses. You can then easily apply various driver efficiency percentages to it (you know, if you want to see what the power to the LED is if your driver were 75% efficient, 85% efficient, etc.) and find out how many Watts to the emitter you are seeing.

From there, it's a little harder to turn watts to amps and volts. Basically, divide watts by whatever the measured Vf was at the same time you measured the current at the tailcap, or pick a Vf value from the IxV graph on the datasheet that you think the LED would be dropping and use that. Voltage doesn't drift as much as current on these curves, so it would be easier to guesstimate which voltage to use instead of which current to use.

So, summary:
turn tailcap amps into power
apply driver efficiency percentage
divide by Vf of LED
 

Der Wichtel

Enlightened
Joined
Mar 12, 2007
Messages
829
Location
Germany
Best way to do this is convert all volts or amps readings to watts. Use the amps reading that you measured and also the voltage of the battery (best to use voltage while in operation, but standing voltage can serve as decent approximation).
Power(Watts)=Current(Amps)*Voltage(Volts)
So with a tailcap reading, you end up with how many watts the flashlight as a whole uses. You can then easily apply various driver efficiency percentages to it (you know, if you want to see what the power to the LED is if your driver were 75% efficient, 85% efficient, etc.) and find out how many Watts to the emitter you are seeing.

From there, it's a little harder to turn watts to amps and volts. Basically, divide watts by whatever the measured Vf was at the same time you measured the current at the tailcap, or pick a Vf value from the IxV graph on the datasheet that you think the LED would be dropping and use that. Voltage doesn't drift as much as current on these curves, so it would be easier to guesstimate which voltage to use instead of which current to use.

So, summary:
turn tailcap amps into power
apply driver efficiency percentage
divide by Vf of LED

But only if the light is using a buck or boost driver.

with linear regulators such as amc7135 or direct driven leds tailcap current = led current
 

hoongern

Enlightened
Joined
Apr 19, 2009
Messages
435
Location
Cambridge, MA & Malaysia
But only if the light is using a buck or boost driver.

with linear regulators such as amc7135 or direct driven leds tailcap current = led current

The general formula will work for any driver/regulator/DD. Maybe to make it clearer, the electrical formulas dictate that

[power at the LED] = [power from batteries] - [any losses due to resistance / driver efficiency]

and power is calculated with P=V*I

So, ignoring efficiency losses,

LED_Current = battery_voltage*battery_current / LED_voltage

from that, you can see that with a buck/boost driver, your LED voltage and battery voltage differ, hence their currents also differ.

With a DD setup, battery voltage = LED voltage, hence LED current = battery current.
 

Der Wichtel

Enlightened
Joined
Mar 12, 2007
Messages
829
Location
Germany
just wanted to point out that with non switching regulators there is no need to do the calculation.

Just tailcap current = LED current
 

Russel

Enlightened
Joined
Jan 31, 2009
Messages
583
Location
California
What if it is a linear regulator (non-switching) dropping the battery voltage significantly for the LED?
 

Der Wichtel

Enlightened
Joined
Mar 12, 2007
Messages
829
Location
Germany
Still the same. A linear current regulator is nothing else than a current controled resistor.

So the resistor will burn the excess voltage to heat.

And the current through the resistor is the same as the current through the LED or through every part of the circuit. So it does not matter where you are measuring current. Always the same in linear regulators
 

Russel

Enlightened
Joined
Jan 31, 2009
Messages
583
Location
California
OK, that makes sense.

I've been spending too much time lately learning about switching mode regulation.

[Edit]

That being the case, you could figure the efficiency of a linear regulator by calculating the power difference due to voltage drop. For example:

Linear regulator:

4.2V 1.5A input

3.5V output (1.5A)

4.2V * 1.5A = 6.3W
3.5V * 1.5V = 5.25W
(5.25W / 6.3W)*100 = 83%

So 1.05 watts is dissipated in the regulator as heat.

[End edit]
 
Last edited:

bshanahan14rulz

Flashlight Enthusiast
Joined
Jan 29, 2009
Messages
2,789
Location
Tennessee
Yes, logic sometimes fails me (or perhaps it is me who is failing in logic:poof:).

Totally makes sense that with a linear regulator setup, current is current, no need to even convert to power.
 

gcbryan

Flashlight Enthusiast
Joined
Oct 19, 2009
Messages
2,473
Location
Seattle,WA
On DX when you see info for the driver for a flashlight and it says 4.2V max and it only takes (1) 18650 would you assume a linear regulator and if it says 8.4V and also takes 2 CR123's would you assume it's using a buck regulator?

So, in the case of a light that can only use (1) 18650 and that has a linear regulator does that mean that regulation is only until the voltage falls close to the vf of the emitter (let's say 3.6V)?

In that case, which is a very common case, the light spends almost all of it's life unregulated right? Since 18650's fall to that range pretty quickly.

Therefore is the only reason for that type of regulation just to make sure that the emitter doesn't burn out right after a 18650 comes of the charger and otherwise it's essentially direct drive?

If you wanted constant current regulation in a light using only (1) 18650 would that be accomplished with a buck/boost driver with constant current regulation?

What I'm asking here is that the correct way to describe it (constant current regulation)?
 

TorchBoy

Flashlight Enthusiast
Joined
Jan 15, 2007
Messages
4,486
Location
New Zealand
1. On DX when you see info for the driver for a flashlight and it says 4.2V max and it only takes (1) 18650 would you assume a linear regulator and if it says 8.4V and also takes 2 CR123's would you assume it's using a buck regulator?

2. So, in the case of a light that can only use (1) 18650 and that has a linear regulator does that mean that regulation is only until the voltage falls close to the vf of the emitter (let's say 3.6V)?

3. In that case, which is a very common case, the light spends almost all of it's life unregulated right? Since 18650's fall to that range pretty quickly.

4. Therefore is the only reason for that type of regulation just to make sure that the emitter doesn't burn out right after a 18650 comes of the charger and otherwise it's essentially direct drive?

5. If you wanted constant current regulation in a light using only (1) 18650 would that be accomplished with a buck/boost driver with constant current regulation?

6. What I'm asking here is that the correct way to describe it (constant current regulation)?
1. I'm not sure it's safe to assume anything with DX. :sick2: The most common linear regulators they use are the AMC7135 which on paper have a maximum input of 6 V. At least some of the drivers they claim have a maximum 4.2 V input are actually boost drivers which would be well into direct drive territory at 4.2 V.

2. Yes. With the AMC7135 there's a 0.12 V overhead.

3. When it actually falls out of regulation depends on how well the battery holds its voltage under load while discharging and what the LED's Vf is at the current it's being run.

4. Yes, that's one reason, but I wouldn't say the only reason.

5. To guarantee it, yes, but bear in mind that buck boost drivers typically have lower efficiency that buck drivers or linear regulators (when linear regs are run in a suitable situation - they can be very inefficient if run in poorly suited situations). Also, there aren't many buck/boost drivers around.

6. Sounds OK. Constant current regulation is what the ideal aim is.
 

gcbryan

Flashlight Enthusiast
Joined
Oct 19, 2009
Messages
2,473
Location
Seattle,WA
I guess a follow up question is...why are so few DX lights constant current? I understand the obvious if it's related to cost but is it?

For example most any "dedicated thrower" would be most useful with a constant current driver.

A general purpose light may or may not need it depending on what the user was more concerned with...constant brightness or runtime.

The throwers don't seem to have constant current drivers however.
 
Top