Basic Circuit Theory

[Matthew&Wendy]

I am trying to built an Infrared Illuminator. I would like to use standard
AA type batteries. These batteries are state that they are 1.3 volts. The IR
leds in the catalog are rated at 1.2 volts. How could I built a circuit
using two or four batteries and some leds? I do not know what happens to the
voltage and resistance as the leds are placed in series or parallel. Thanks
for the assistance.
Matthew

 

[Activ8]

I am trying to built an Infrared Illuminator. I would like to use standard
AA type batteries. These batteries are state that they are 1.3 volts. The IR
leds in the catalog are rated at 1.2 volts. How could I built a circuit
using two or four batteries and some leds? I do not know what happens to the
voltage and resistance as the leds are placed in series or parallel. Thanks
for the assistance.
Matthew

Question. Would you like to spend $0 on a trip to the library to
learn ohms law and read "fun with electricity" (for a start) or
spend a pile of time energy and $ on leds and stuff?

The answer to your question is ... with your hands or someone
else's.

It's either gotta be 2 or 4 batts and a specific "some" leds.

Try this. 2 AAs at 1.5 V each fresh. and 1 150 ohm resistor and 1
cheap led, which counts as "some leds".

But wouldn't it be more fun to just hook up stuff and watch it
smoke?

 

[Andrew Holme]

I am trying to built an Infrared Illuminator. I would like to use standard
AA type batteries. These batteries are state that they are 1.3 volts. The IR
leds in the catalog are rated at 1.2 volts. How could I built a circuit
using two or four batteries and some leds? I do not know what happens to the
voltage and resistance as the leds are placed in series or parallel. Thanks
for the assistance.

You have to put something in series with the LEDs to control the
current. The simplest solution is to use two batteries and a resistor
for each LED e.g.

Voltage drop across resistor = 2.6 - 1.2 = 1.4V
Current = 1.4/R

A more efficient way is to use four batteries and groups of three LEDs
e.g.

Voltage drop across resistor = 5.2 - 3.6 = 1.6V
Current = 1.6/R

You need some headroom because the battery voltage will fall as they
discharge and the LED drop will not be exactly 1.2V.

 

[Dale Benjamin]

Put your cells in parallel to make a battery, hook that to a LED, or several
in parallel. The difference between 1.2 and 1.3 volts is irrelevant here,
as in many aspects of electronics. Any difference less than 1 in 10 is
irrelevant. But sometimes it really does matter. Consult any good school.

 

[Jim Thompson]

Put your cells in parallel to make a battery, hook that to a LED, or several
in parallel. The difference between 1.2 and 1.3 volts is irrelevant here,
as in many aspects of electronics. Any difference less than 1 in 10 is
irrelevant. But sometimes it really does matter. Consult any good school.

[snip]

Hopefully not the one you went to ;-)

...Jim Thompson

 

[Dale Benjamin]

Hopefully, no.

Put your cells in parallel to make a battery, hook that to a LED, or several
in parallel. The difference between 1.2 and 1.3 volts is irrelevant here,
as in many aspects of electronics. Any difference less than 1 in 10 is
irrelevant. But sometimes it really does matter. Consult any good

school.
[snip]

Hopefully not the one you went to ;-)

...Jim Thompson