M60 + ?ohms = M60LL (Multi Level Malkoff's?)

[ElectronGuru]

In the spirit of solve for X, Malkoff M60 + ?ohms = M60LL, where:

M60 - The current draw is 780ma at 6 volts
M60LL - The current draw is 170ma at 6 volts


Which resister (how many ohms) is needed to turn an M60 into an M60LL, such that we get high output or high runtime in the same module/light - and easily switch between them?


Here's the switch itself and a [Pre Malkoff] discussion thread:

http://theledguy.chainreactionweb.com/product_info.php?products_id=641

http://candlepowerforums.com/vb/showthread.php?t=83844


Thanks!

 

[gilrain]

Well, Ohm's Law is: V = IR. We can solve that for resistance by dividing both sides by I.

V/I = R

6v/0.17a = 35.3 ohms

So, you'd want, roughly, the 30 ohm version of the switch.

Edit: Or actually, the resistance of the original M60 needs to be taken into account. It's about 7.7 ohms. We want to add the difference between them, using the switch. 35.3 - 7.7 = 27.6, so you'd still want to use the 30 ohm version.

 

[LED_astray]

I don't think it works this way. (But note I don't have a Malkoff or that 2 level switch, this is just a theory.) I believe the Malkoff drop-ins are regulated. The switch on low seems to add a series resister to drop the current in a direct drive circuit. If so, the switch will just lower the input voltage to the regulator until the regulator falls out of regulation. The other side effect is that it will lower your battery life because you would be dissipating power in the series resister and therefore drawing more power from the battery (since the regulator keeps the power to the LED constant, or tries to.)

 

[gilrain]

That is entirely probable. I don't have any idea how the light itself works, but if it's regulated like that then there's certainly something happening behind the scenes that'd make the basic equations pretty worthless.

Is it fair to say, then, that the switch mentioned isn't useful for any regulated light? Just direct drive?

 

[gswitter]

I don't think it works this way. (But note I don't have a Malkoff or that 2 level switch, this is just a theory.) I believe the Malkoff drop-ins are regulated. The switch on low seems to add a series resister to drop the current in a direct drive circuit. If so, the switch will just lower the input voltage to the regulator until the regulator falls out of regulation. The other side effect is that it will lower your battery life because you would be dissipating power in the series resister and therefore drawing more power from the battery (since the regulator keeps the power to the LED constant, or tries to.)

Yup. If you want two levels of output from an M60, the 2-stage switch for Surefire Classic series (or the McE2S, if you're using an E-series host) can work. But, you'll need to do the math and account for your battery voltage and resistor value, and keep in mind that using the low output can decrease your runtime.

Is it fair to say, then, that the switch mentioned isn't useful for any regulated light? Just direct drive?

Provided you know the specs of the regulation circuit (specifically, below what voltage it drops out of regulation) and you do the math, sure, it can be useful in a regulated light. You can turn a single-stage, regulated light into a two-stage light with a regulated high and a DD low.

The M60 is a barn burner and I love it, but sometimes I really prefer a temporary low output mode, so I use a 60 ohm 2-stage switch with primary CR123As.

 

[ElectronGuru]

Never mind:

http://www.lighthound.com/index.asp?PageAction=VIEWPROD&ProdID=3514

 

[Bullzeyebill]

Never mind:

http://www.lighthound.com/index.asp?PageAction=VIEWPROD&ProdID=3514

Probably a better solution since your M60 will run in regulation at the lower levels, something that the resistor setup you proposed will not do. I have one of these UNIQ tailcaps, and it works ok at 6 volts or more.

Bill

 

[ElectronGuru]

I've read since that the Uniq's have a slow drain when off. Any issues with that on your end?

I'm now waiting for AW to blow this market away...

 

[Bullzeyebill]

I've read since that the Uniq's have a slow drain when off. Any issues with that on your end?

I'm now waiting for AW to blow this market away...

Not sure about drain, though it probably exists. When I don't use it I turn the tailcap out from contact with body, sort of lock it out. AW has sort of entered this market when he introduced the soft start, 3 level tailcap switch mod that works with C series SF lights. Orginally intended for incan lights, AW's soft start switch also works with regulated LED lights. Like incan apps you need at least 6 volts to run an LED light properly.

Bill

 

[smopoim86]

Is there a pwm option for the e series. I'm using the vme adapternand woul like a low as well. Something other than a resistorex tail would be great.

Daniel

 

[Justin Case]

The Malkoff web site says that the M60 drops out of regulation below 3.8V. Let's assume that the drop-in uses a Cree Q5 and is driven at 1000mA at full power (the 3.8V regulation figure above probably is based on the Q5's Vf at 1000mA drive current plus some amount of voltage headroom required by the buck driver).

To get to the M60LL's output of 80 lumens, the Q5 has to be driven at about 350mA, which corresponds to a Vf of about 3.2V.

Let's assume that you drive the M60 with a single Li-ion cell, e.g., a 17670 in a 6P host. Then, you need the right resistor to force the M60 to drop out of regulation and into direct drive.

Let's say that the 17670 voltage under load is 3.8V, which is the limit to run the M60 in regulation. You want to drop the input to 3.2V and 350mA.

That's about a 1.7 ohm resistor.

You could try a similar calculation for a 2x16340 battery configuration. Now, you need to drop the voltage from a nominal 7.4V or 7.6V to 3.2V and 350mA. You need about a 12 ohm resistor now.

 

[Kestrel]

There is nothing at all wrong with using resistors to obtain lower output from the Malkoff M60. My experience with the M60 & the McC2S on 2xCR123:
30 ohms: -> ~50 lumens
60 ohms: -> ~20 lumens
120 ohms: -> ~10 lumens

But, you'll need to do the math and account for your battery voltage and resistor value, and keep in mind that using the low output can decrease your runtime.
Provided you know the specs of the regulation circuit (specifically, below what voltage it drops out of regulation) and you do the math, sure, it can be useful in a regulated light. You can turn a single-stage, regulated light into a two-stage light with a regulated high and a DD low.

I'm pretty sure that running the current through the resistor can in most cases dramatically increase runtime:

Example 1:
SF L1 Luxeon, 1xCR123: 22 lumens for ~1 hr -> 2 lumens for ~50 hrs (adding 10 ohms in series via the two-level tailcap)
SF L1 Cree, 1xCR123: 65 lumens for ~1.5 hrs -> 10 lumens for ~16 hrs (+10 ohms in series)
SF L2 Luxeon, 2xCR123: 100 lumens for ~1 hr -> 15 lumens for ~15 hrs (+10 ohms in series, dropping it from buck-regulated to DD) Edit: buck-regulated is probably incorrect, see BB's post below.

Example 2:
While it's been way too long since my EE-for-non-EE-Engineers class, from the following link:
http://www.doctronics.co.uk/resistor.htm

What is the power output of a resistor when the voltage across it is 6 V, and the current flowing through it is 100 mA?
P = VI = 6x100mA = 600 mW = 0.6 W
Therefore, 0.6 W of heat are generated in this resistor
​

Running the M60 'full bore' on 2xCR123, you are consuming ~4.5 watts. Running the M60 on a resistor which yields 100mA (sort of like the current for a theoretical M60LLL, ~40 lumens on ~100 mA), the resistor is dissipating 0.6 Watts of power. Not sure what power an M60 running at 40 lumens would dissipate at the emitter, but my guess is you're looking like something around one watt for the total circuit. Much lower than 4.5 watts that the M60 circuit (i.e. just the LED) is consuming running 'full bore'. Therefore, runtime should be extended considerably, albeit with considerably less efficiency than a 'true' low-output Malkoff.

I understand that these numbers are very rough and that I could easily be off by perhaps a factor of 2, I am just using the example calculation from the web link and attempting to apply it to this discussion.

 

[Bullzeyebill]

Good info here. Actually the L2, KL4, KL5, U2 use boost constant current drivers. L2 was always sort of "overdriven". This for the Lux V LEDs.

Bill

 

[Kestrel]

Actually the L2, KL4, KL5, U2 use boost constant current drivers. L2 was always sort of "overdriven". This for the Lux V LEDs.

Thanks for the correction, Bill. I guess I don't understand the LuxV L2 very well, I only started learning the 'nuts&bolts' with the current generation of Crees.

Edit: At some point, I also hope to have 250 & 500 ohm McC2S boards. I will also have two different M60 setups, 2x17500 and 1x18650. I am looking forward to seeing the difference between the two, although I feel that the McC2S configuration could be more efficient from 1x18650 than from 2x17500. I definitely have no interest in having a resistor inline with the boost-regulated M30.

I am unsure about doing runtime tests, as the current should be decreased to below the point where I can rely on the AW protection circuit protecting the cell(s) from over-discharge.