rizky_p
Flashlight Enthusiast
Has anyone measured a total resistances on Standard Mag 2D?
Thanks.
Thanks.
Resistance on Standard mag 2D.
- Thread starter rizky_p
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Mr Happy
Flashlight Enthusiast
I have, but like a fool I don't remember where I wrote it down.
I was using some cheap 3AA-1D adapters with Eneloops, and from memory the results were something like this:
Internal resistance of cells 6 x 50 m = 300 m (m = milliohm)
Resistance of battery holders = 2 x 150 m = 300 m
Combined resistance of Mag springs, switch, etc. = 400 m
Total circuit resistance = 300 m + 300 m + 400 m = 1 ohm.
I was using some cheap 3AA-1D adapters with Eneloops, and from memory the results were something like this:
Internal resistance of cells 6 x 50 m = 300 m (m = milliohm)
Resistance of battery holders = 2 x 150 m = 300 m
Combined resistance of Mag springs, switch, etc. = 400 m
Total circuit resistance = 300 m + 300 m + 400 m = 1 ohm.
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You may want to specify what components you will be using in your standard 2D Mag. Obviously, the switch & spring are stock, but as you can see there are variable amounts of resistance in battery holder/pack, bulb holder. Then there are resistance variations if you use a KIU with various ways of connecting bulb holder wires, or AW's D Driver, etc.
Mr Happy
Flashlight Enthusiast
With an incan bulb, it means three things:
1) The extra resistance reduces the current, so the batteries run down slower
2) The reduced current means the bulb runs dimmer
3) A bit of power is wasted in overcoming the circuit resistance
Now in certain cases, if you are overdriving the bulb right to the limit, the circuit resistance might be all that saves you from an instaflash; so reducing the resistance isn't always a good thing...
With an incan bulb, it means three things:
1) The extra resistance reduces the current, so the batteries run down slower
2) The reduced current means the bulb runs dimmer
3) A bit of power is wasted in overcoming the circuit resistance
Now in certain cases, if you are overdriving the bulb right to the limit, the circuit resistance might be all that saves you from an instaflash; so reducing the resistance isn't always a good thing...
Will all this result in an explosion? Just making sure, about to build myself a hotwire 4D Mag with KaiDomain 3AA to D adapters and Duracell 2650.
Mr Happy
Flashlight Enthusiast
No explosion, but it was the KaiDomain 3AA to D adapters that had the 150 milliohm resistance in my post above. So four of them is going to add up to 600 milliohms. They are not really suitable for high current hotwires due to their high resistance.
What does this mean? Well, for each 1 amp drawn by the bulb, you are going to lose 0.6 V in the adapters. If you tried to use a 4 amp bulb in the hotwire, you would lose 2.4 V. (A 12 V 50 W halogen would be a 4 amp bulb.)
Now consider the cell resistance. Eneloops are quite good at 50 milliohms, but put 12 of them in series and you get 600 milliohms. That's another 2.4 V down on a 4 amp bulb. Other cells could be 100 milliohms or greater, which could put you 4.8 V down. Let's assume 4.8 V worst case with the Duracells.
Now suppose the stock springs and switch without resistance mods are 400 milliohms. That loses you 1.6 V.
So add all this up:
12 AA cells nominally will give you about 12 x 1.3 V open circuit voltage = 15.6 V.
For a 4 amp hotwire, you subtract 2.4 V + 4.8 V + 1.6 V = 8.8 V.
Therefore your final voltage at the bulb terminals will be 15.6 - 8.8 = 6.8 V at the bulb. Not so much...
Now there are lots of ifs and buts and let's assumes in this, so it's just an illustration. But what it means is you will need to measure and calculate and test with the specific modifications, battery choice, battery holder and bulb that you choose to find what works out nicely.
What does this mean? Well, for each 1 amp drawn by the bulb, you are going to lose 0.6 V in the adapters. If you tried to use a 4 amp bulb in the hotwire, you would lose 2.4 V. (A 12 V 50 W halogen would be a 4 amp bulb.)
Now consider the cell resistance. Eneloops are quite good at 50 milliohms, but put 12 of them in series and you get 600 milliohms. That's another 2.4 V down on a 4 amp bulb. Other cells could be 100 milliohms or greater, which could put you 4.8 V down. Let's assume 4.8 V worst case with the Duracells.
Now suppose the stock springs and switch without resistance mods are 400 milliohms. That loses you 1.6 V.
So add all this up:
12 AA cells nominally will give you about 12 x 1.3 V open circuit voltage = 15.6 V.
For a 4 amp hotwire, you subtract 2.4 V + 4.8 V + 1.6 V = 8.8 V.
Therefore your final voltage at the bulb terminals will be 15.6 - 8.8 = 6.8 V at the bulb. Not so much...
Now there are lots of ifs and buts and let's assumes in this, so it's just an illustration. But what it means is you will need to measure and calculate and test with the specific modifications, battery choice, battery holder and bulb that you choose to find what works out nicely.
diff_lock2
Newly Enlightened
Mr happy, are you saying that lowering resistance will decrease run time?
And if that is true, is it due to the bulb drawing more power from the relatively higher voltage?
And if that is true, is it due to the bulb drawing more power from the relatively higher voltage?
Mr Happy
Flashlight Enthusiast
With an incan bulb, then yes, lowering resistance will decrease the run time.
It's not only the bulb we have to look at but the total resistance of the circuit. A lower circuit resistance means more current drawn from the batteries and thus a faster battery drain. In the opposite case, if we add resistance we reduce the current and thus lengthen the run time.
(If we had a regulated LED driver this would be different, but with an unregulated incan it is as I describe.)
rizky_p
Flashlight Enthusiast
Hi Lux, everything are standard as fas as the Mag itself is concerns, I am running a G4 bulb on a PR to bi-pin adapter. So everything can be considered standard. I am curious how much resistances between the positive contact point(bottom of the switch assembly) and the tailspring (Battery setup is not considered)?
Thanks.
rizky_p
Flashlight Enthusiast
Mr Happy
Flashlight Enthusiast
The 400 milliohms might not be exact, I am going from memory there, but it's in the ballpark. There are various steel parts in the current path of a Mag and I think these account for most of the resistance. I believe the main contributors are the tail spring and the long thin spring in the bulb tower. Secondary contributors are the steel strip on the outside of the bulb focus slider and the switch contacts.
You can bypass many of the steel parts with copper wire to reduce the resistance.
rizky_p
Flashlight Enthusiast
Yes i just did, Dont have a decent DMM to measure total resistance to confirm reduction after the mod. :mecry:
Mr Happy
Flashlight Enthusiast
Actually I have found it very difficult to get accurate resistance measurements in the fractional ohm range with a DMM. The most accurate solution I've found, if you can set it up, is to pass a known current like 1 amp through the part to be measured and then measure the voltage drop with the millivolt setting on the meter. If you have two meters you can measure the current and voltage simultaneously.
rizky_p
Flashlight Enthusiast
Nice idea, in fact i have two DMMs i might try it. I assume some load is required?
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so now we know that stock mag (switch/body/spring) has high resistance.
but how much is the total resistence of a modded mag?
for exapmle with tail spring fixed, KIU socket...and for batteries adapter a Fivemega one...:thinking:
I think that this is the best work we can have...right??
but how much is the total resistence of a modded mag?
for exapmle with tail spring fixed, KIU socket...and for batteries adapter a Fivemega one...:thinking:
I think that this is the best work we can have...right??
Mr Happy
Flashlight Enthusiast
Yes, I typically just use a bulb of some sort in the circuit to limit the current to an amp or two.
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