For fun, I am increasing the life of little laser pointers for cats.

I noticed that 3 button batteries, (either alkaline or silver-oxide,
357, or zinc-air hearing aid, 675) which produce 3x1.5 3x1.55 3x1.4
volts respectively, are not nearly as bright as 2 x C cells producing
only 2x1.6 volts initially.

Why is this?

The 3 button batteries are all over 4 volts although their capacities
are around 105 165 540 mAH respectively.

The C batteries are a little over 3 volts for 2 of them but their mAH
are quite high relatively, 8350 mAH. Alkaline batteries for the C.

There are discharge characteristics of batteries that I am woefully
ignorant of, well, maybe not woefully. Got me curious.

For a little test, I noticed that 2 C cells enable the laser pointer
to draw 25 mA but 3 C cell will draw 27 mA. For those 2 extra mA
coming from using 4.5 volts instead of 3 volts (nominally), the laser
is noticeably brighter. I assume the nM of the laser is possibly being
changed or just the power throughput or both. I read that as the laser
drops from 660-670 nM to 650 nM the light gets brighter.

My long-winded question is that I thought a device draws only the
current it needs. But if there is more reserve capacity as with the C
over the button batteries, the device draws more current or the
batteries allow more current to be drawn or what?

Wuold the true be same for D batteries over C also?
I noticed that the lumens stay the same with Maglites using C or D
batteries but the PBC or goes up slightly using D over C batteries.
So again, bigger batteries allow more current to be used or what?

Thanks for any input into this burning question. I cross-posted this
since it's hard for me to post and this is the sort of question that
can be answered by those who either tinker or design or repair.

The small button batteries under the load of the laser diode, are most
likely having a greater voltage drop, than the higher capacity C batteries.
The small button batteries, cannot have the capacity of the C batteries.

The breakover voltage where any LED device (laser diode) will function is
very tight. A small difference of the supply voltage in this region will
normally make a large difference in the output of the LED. Take care to not
exceed any LED's maximum voltage rating. They can be easily damaged.

--

Greetings,

Jerry Greenberg GLG Technologies GLG
=========================================
WebPage http://www.zoom-one.com
Electronics http://www.zoom-one.com/electron.htm
=========================================


"One Two" wrote in message
om...
For fun, I am increasing the life of little laser pointers for cats.

I noticed that 3 button batteries, (either alkaline or silver-oxide,
357, or zinc-air hearing aid, 675) which produce 3x1.5 3x1.55 3x1.4
volts respectively, are not nearly as bright as 2 x C cells producing
only 2x1.6 volts initially.

Why is this?

The 3 button batteries are all over 4 volts although their capacities
are around 105 165 540 mAH respectively.

The C batteries are a little over 3 volts for 2 of them but their mAH
are quite high relatively, 8350 mAH. Alkaline batteries for the C.

There are discharge characteristics of batteries that I am woefully
ignorant of, well, maybe not woefully. Got me curious.

For a little test, I noticed that 2 C cells enable the laser pointer
to draw 25 mA but 3 C cell will draw 27 mA. For those 2 extra mA
coming from using 4.5 volts instead of 3 volts (nominally), the laser
is noticeably brighter. I assume the nM of the laser is possibly being
changed or just the power throughput or both. I read that as the laser
drops from 660-670 nM to 650 nM the light gets brighter.

My long-winded question is that I thought a device draws only the
current it needs. But if there is more reserve capacity as with the C
over the button batteries, the device draws more current or the
batteries allow more current to be drawn or what?

Wuold the true be same for D batteries over C also?
I noticed that the lumens stay the same with Maglites using C or D
batteries but the PBC or goes up slightly using D over C batteries.
So again, bigger batteries allow more current to be used or what?

Thanks for any input into this burning question. I cross-posted this
since it's hard for me to post and this is the sort of question that
can be answered by those who either tinker or design or repair.

"Jerry G." writes:

The small button batteries under the load of the laser diode, are most
likely having a greater voltage drop, than the higher capacity C batteries.
The small button batteries, cannot have the capacity of the C batteries.

The breakover voltage where any LED device (laser diode) will function is
very tight. A small difference of the supply voltage in this region will
normally make a large difference in the output of the LED. Take care to not
exceed any LED's maximum voltage rating. They can be easily damaged.

Both LEDs and laser diodes operate on current. So, the resistance
of the battery in conjunction with the internal resistor of the laser pointer
control the current through the laser diode. Higher quality laser pointers
include an active current regulator which provides the current needed to
maintain the same light output via monitoring it in a feedback loop.
The pointers you find in Dollar stores only use a 50 ohm resistor. So,
the resistance of the batteries becomes significant.

Laser diodes are also exceedingly sensitive to overcurrent. Those
cheap pointers already tend to be unreliable and have limited life.
Increasing the brightness will shorten it further - possibly to 0.

--- sam | Sci.Electronics.Repair FAQ Home Page: http://www.repairfaq.org/
Repair | Main Table of Contents: http://www.repairfaq.org/REPAIR/
+Lasers | Sam's Laser FAQ: http://www.repairfaq.org/sam/lasersam.htm
| Mirror Site Info: http://www.repairfaq.org/REPAIR/F_mirror.html

Important: The email address in this message header may no longer work. To
contact me, please use the feedback form on the S.E.R FAQ Web sites.

You are right about these pointers. I have several cheap ones. After a lot
of use, they get weak, and become useless. Infact, the batteries are
lasting longer than the laser itself!

--

Greetings,

Jerry Greenberg GLG Technologies GLG
=========================================
WebPage http://www.zoom-one.com
Electronics http://www.zoom-one.com/electron.htm
=========================================


"Sam Goldwasser" wrote in message
...
"Jerry G." writes:

The small button batteries under the load of the laser diode, are most
likely having a greater voltage drop, than the higher capacity C
batteries.
The small button batteries, cannot have the capacity of the C batteries.

The breakover voltage where any LED device (laser diode) will function is
very tight. A small difference of the supply voltage in this region will
normally make a large difference in the output of the LED. Take care to
not
exceed any LED's maximum voltage rating. They can be easily damaged.

Both LEDs and laser diodes operate on current. So, the resistance
of the battery in conjunction with the internal resistor of the laser
pointer
control the current through the laser diode. Higher quality laser pointers
include an active current regulator which provides the current needed to
maintain the same light output via monitoring it in a feedback loop.
The pointers you find in Dollar stores only use a 50 ohm resistor. So,
the resistance of the batteries becomes significant.

Laser diodes are also exceedingly sensitive to overcurrent. Those
cheap pointers already tend to be unreliable and have limited life.
Increasing the brightness will shorten it further - possibly to 0.

--- sam | Sci.Electronics.Repair FAQ Home Page: http://www.repairfaq.org/
Repair | Main Table of Contents: http://www.repairfaq.org/REPAIR/
+Lasers | Sam's Laser FAQ: http://www.repairfaq.org/sam/lasersam.htm
| Mirror Site Info: http://www.repairfaq.org/REPAIR/F_mirror.html

Important: The email address in this message header may no longer work. To
contact me, please use the feedback form on the S.E.R FAQ Web sites.

"Jerry G." wrote in message ...
You are right about these pointers. I have several cheap ones. After a lot
of use, they get weak, and become useless. Infact, the batteries are
lasting longer than the laser itself!

--

Greetings,

Jerry Greenberg GLG Technologies GLG
=========================================
WebPage http://www.zoom-one.com
Electronics http://www.zoom-one.com/electron.htm
=========================================


"Sam Goldwasser" wrote in message
...
"Jerry G." writes:

The small button batteries under the load of the laser diode, are most
likely having a greater voltage drop, than the higher capacity C
batteries.
The small button batteries, cannot have the capacity of the C batteries.

The breakover voltage where any LED device (laser diode) will function is
very tight. A small difference of the supply voltage in this region will
normally make a large difference in the output of the LED. Take care to
not
exceed any LED's maximum voltage rating. They can be easily damaged.

Both LEDs and laser diodes operate on current. So, the resistance
of the battery in conjunction with the internal resistor of the laser
pointer
control the current through the laser diode. Higher quality laser pointers
include an active current regulator which provides the current needed to
maintain the same light output via monitoring it in a feedback loop.
The pointers you find in Dollar stores only use a 50 ohm resistor. So,
the resistance of the batteries becomes significant.

Laser diodes are also exceedingly sensitive to overcurrent. Those
cheap pointers already tend to be unreliable and have limited life.
Increasing the brightness will shorten it further - possibly to 0.

--- sam | Sci.Electronics.Repair FAQ Home Page: http://www.repairfaq.org/
Repair | Main Table of Contents: http://www.repairfaq.org/REPAIR/
+Lasers | Sam's Laser FAQ: http://www.repairfaq.org/sam/lasersam.htm
| Mirror Site Info: http://www.repairfaq.org/REPAIR/F_mirror.html

Important: The email address in this message header may no longer work. To
contact me, please use the feedback form on the S.E.R FAQ Web sites.

I have several laser pointers dating back as far as 1994, and have
never had one die, just need new batteries. They were gifts from
various functions I attended, but were not real cheap or real
expensive. The batteries can test ok for voltage using a simple
voltmeter, but you must test them under load so that you get a true
indication of their capacity. I typically use a 470 ohm resistor
across the battery, that puts a reasonable load on the battery and you
can sort thru a lot of ?? batteries in a short time t find whoich ones
are truly "good".

H. R. (Bob) Hofmann