Explain voltage to me

I am a machinist with mechanical inclination. Electrical stuff just makes my head hurt so pardon my ignorance.

I took out my volt meter and measured the voltage on some CR123 batteries. They came in at 3.28v or 6.54v when running 2 in, say, my Fenix TK12. Now, my TK12 is reported to take an 18650 battery. I went on Battery Junction to price out a charger and a couple of these 18650 batteries when it dawned on me...they are only 3.7 volts.

This got me to thinking, how can my light still be as bright and last longer on lower voltage and lower Mah (2600 vs 2800 total on 2 CR123's)? I mean, I would imagine that if the CR123's dropped down to 3.7v total, I would have a pretty dim light...


Right?

I dunno.

Voltage is like the pressure from the source of your garden hose.
Resistance is how skinny your hose is. The more resistance, the harder it is to get water through it.
Amperage is the water flow itself.

Have you ever used a garden hose attachment? Or simply used your finger to make the water shoot further?
What you have done is acted as a voltage converter. Instead of a bunch of water dribbling out the hose, you now have a lesser amount of water spewing out a lot faster. I.e. you are drawing more amperage to boost your effective voltage.
Ok, my example is starting to fall apart but you get the idea.

Most good flashlights are voltage regulated. Many power emitters have about a 3 volt Vf or forward voltage. They then use a resistor to limit the current. However, they can accept a range of input voltage and then increase or decrease it in order to get the right forward voltage.

If you have a 1.5 volt battery, in order to get 200 lumens out of your XP-G, you will need to transform it to about 3 volts. The price of doing so is that instead of sucking 1 Amp, the transformation now means you suck TWO amps from the battery at 1.5 volts to equal 1 Amp at 3.0 Volts.

Fortunately, NiMH batteries are pretty low resistance and can support a decent amperage pull. An alkaline will suck though because they aren't made for such high current demands.

Lithium Ion batteries are no problem at all. The XP-G only needs 1 amp of current at about 3 volts. Some push it to 1.4 or so. But even partially depleted CR123s are going to be well over 3 volts (with 2 of em). They are also capable of cramming out about 3+ Amps each so no worries there.

This a huge subject with diferent explanations, depending on what light you are looking on.

But first: A LiIon battery is not 3.7 volt, the voltage depends on charge state and how much current the light is using, just off the charger it will be 4.2 volt. With a heavy load and just before it is empty it might be down to 2.5 volt.

A CR123 battery is about 3 volt when full, but under load the voltage drops down to 2.5 volt or lower.
Using two CR123 batteries in series will give double the voltage, but the same mAh, i.e. 2 x 3volt/1400mAh will be 6volt/1400mAh for series connection (or 3volt/2800mAh for parallel connection).

I have not tested the TK12, but it probably can keep full brightness down to about 3.8 volt, the the brightness will gradually go down.
The TK12 has a smart circuit (Called a buck regulator) that will take any voltage above 3.8 volt and reduce it to 3.8 volt, with a very small power loss, i.e. it will consume less current at 6 volt than at 3.8 volt.
This makes it difficult to directly use the mAh to estimate runtime.


With the TK12 and 2xCR123 you have 5volt/1400mAh with lower current draw or 1x18650 3.7volt/2600mAh with higher current draw. Both the above voltages are a estimated average.

TwinBlade said:

... how can my light still be as bright and last longer on lower voltage and lower Mah (2600 vs 2800 total on 2 CR123's)? I mean, I would imagine that if the CR123's dropped down to 3.7v total, I would have a pretty dim light...


Right?

I dunno.

Click to expand...

First of all... Panasonic USA CR123 cells (for example) are only ~1500 mah (at 3 volts). With two primaries in series capacity does NOT increase, only the total Voltage does.

Watt density looks like this for the 2 primaries:
6V x 1.5Ah = 9 Watts

Watt density looks like this for the 18650:
3.7V x 2.6Ah = 9.92 Watts

In regards to your brightness question, Cree LEDs only need ~3.6V to obtain full brightness. Any higher voltage beyond that is not needed, and is "bucked" down to ~3.6V by the driver circuit in the light. Thus the LED relies on the delivered current to obtain brightness... not voltage.

kramer5150 said:

First of all... Panasonic USA CR123 cells (for example) are only ~1500 mah (at 3 volts). With two primaries in series capacity does NOT increase, only the total Voltage does.

Watt density looks like this for the 2 primaries:
6V x 1.5Ah = 9 Watts

Watt density looks like this for the 18650:
3.7V x 2.6Ah = 9.92 Watts

In regards to your brightness question, Cree LEDs only need ~3.6V to obtain full brightness. Any higher voltage beyond that is not needed, and is "bucked" down to ~3.6V by the driver circuit in the light. Thus the LED relies on the delivered current to obtain brightness... not voltage.

Click to expand...

Ahhh...many thanks. I get it now. Between this explanation and the others, I am up to speed on this.

Basically, depending on how batteries are wired up (series/parallel) you can either achieve double capacity output, or double voltage output, not both. I also did not realize that Cree LED's require that little power to hit maximum levels.

Really amazing the technology behind all this. Thumbs up to all of you for helping me out.:thumbsup::thumbsup::thumbsup:

Dont forget that V x Ah = Watt-hour (Wh)
If i remember right from of top of my head

joe1512 said:

Voltage is like the pressure from the source of your garden hose.
Resistance is how skinny your hose is. The more resistance, the harder it is to get water through it.
Amperage is the water flow itself.

Click to expand...

Very good simple explanation +1:thumbsup:
Then to add, the same would apply to air as well!

Great explanation with the water! I've always liked explaining these concepts. I'd like to go a bit further...

With electricity, in the end what you are looking for is energy. Imagine each water drop is an energy unit. So to power up your lights you need energy, the bigger the light, more energy, more water drops....

So if you need more energy that's the same as needing more water drops. In this case you could do two things: use a wider hose or pipe (more amperage) or increase the water pressure (more voltage). With both alternatives you get more water drops (energy units), but using different methods and equipment.

This is the principal that's used to carry power in big power grids... in order to move lots of energy without having to use massive cables, they increase the voltage to several thousand volts, results in huge amounts of energy with a much smaller diameter cable than using 110V.

In the end, what are those energy units? Watts!

Voltage * Amperage = Watts

So you can balance V and A to get same amounts of power!

For example, here in Chile, we use 220V, so the mains breaker at most homes is just 15 amps. At the same time, a hotwire Mag moves 10-12 amps. That's more than what a house consumes! But not really more energy, as the Mag is running at around 24V. So really:

220V * 15A = 3300 W
24V * 12A 288 W

Hope you enjoy it, cheers!

Robinson

I've had the same question for a long time, so can I throw in another example or two for dissection?

I have a malkoff m60. What is the benefit of running it off of a 17670 vs 2 primary cr123s or vice versa?

Also I have a LF d26 3 mode drop in that is 3.6-6v...
When using the low mode will it be the same brightness regardless of battery used? It will be if it's buck regulated right?


Thanks for the vastly informative thread

Since the LED is essentially a diode it can only accept voltage one way. Why is it Vf or forward voltage?

Forward voltage.:thumbsup:

Actually, some diodes can take a voltage in the reverse direction without damage, look up zener diodes.

I'm lost.

The big 18650 battery has more surface area on its electrodes than the little CR123 cells have, so the chemicals inside the 18650 can crank out more electrons than the little CR123 cells can. When the light is busy running at full-tilt, the 18650's ability to sustain a more vigorous chemical reaction means it's able to maintain a higher electrical pressure (i.e. voltage) on each electron that goes out the door. That higher pressure, in turn, means more electrons are getting pushed into the circuit to light the LED. On the other hand, even when you put two CR123 cells end-to-end so they double the voltage, they still have a physically smaller supply of electrons to apply all that pressure to, and with more energy than stuff to carry the energy, the CR123 cells just end up getting hot. So the LED shines brighter with the 18650 cell, because it can supply electrons at a higher rate than the CR123s can, without having to work so hard that it ends up wasting energy as heat instead of doing productive work.

TwinBlade said:

I took out my volt meter and measured the voltage on some CR123 batteries. They came in at 3.28v or 6.54v when running 2 in, say, my Fenix TK12. Now, my TK12 is reported to take an 18650 battery. I went on Battery Junction to price out a charger and a couple of these 18650 batteries when it dawned on me...they are only 3.7 volts.

Click to expand...

One thing that I come across a LOT with peoples understanding of batteries is the voltage - people often think that whatever the voltage specified on the battery is exactly what they will get from it, this isn't even close to being true.

For some reason the non-rechargeable batteries seem to be the worst at providing what they say - 1.5V alkalines will NOT supply 1.5V under any decent load and 3V CR123a cells will NOT provide 3V under higher loads. In fact with a high drain bulb you can get fresh 123 cells to drop to 2V almost immediately.

Rechargeable batteries seem to hold up to their rated voltage much better. 1.2V NiCd & NiMH seem to be able to supply 1.2V at a fairly high current. 3.7V Li-ion cells are even better - they can supply over 3.7V at fairly high currents and not drop below 3.7V until they are half flat.

I have often had people ask me about whether a device like a digital camera will work OK from NiMH cells because they are only 1.2V whereas alkalines are 1.5V. But if the camera wont be happy with 1.2V then it is going to stop working on alkalines before they are half used up - that would make for a pretty battery hungry device IMO.

Generally a light that takes 2 x CR123a cells will be getting more voltage from its batteries from those 2 cells than it would from 1 x 18650 cell - but maybe not by as much as what you would think. If it is a regulated LED light then the lower voltage may not matter much and the maximum output may be close to the same, but with MUCH better run time.

Thanks for the links. Looks like I have some readi

red02 said:

Thanks for the links. Looks like I have some readi

Click to expand...

...ng to to? :thinking:

:lolsign:

kramer5150 said:

Watt density looks like this for the 2 primaries:
6V x 1.5Ah = 9 Watts

.

Click to expand...

NO !

6 volts x 1.5 amps = 9 watts - a measure of instantaneous power.

6 volts x 1.5 amphours = 9 watthours - a measure of energy - power for a period of time.

Some really good basic info here, fellas. Well done.

obi