Re: Running LED off 9.3 Volts AC

On Thu, 22 Nov 2007 06:00:21 -0800 (PST), "Dave.H"
<the1930s@xxxxxxxxxxxxxx> wrote:

I have an upcoming project involving running three LEDs off a single 9
volt battery, what resistors would I need, and would I also need a
capacitor, as in this circuit?


The specifications for the LED's I am planning to use, (the same one's
I used for the radio dial) are

5mm (T1 1/2) round with flange
16,000mcd (typical) output
Water Clear Lens
High Brightness
620 - 625nm Wavelength (Orange-Red)
15o Viewing Angle
Forward voltage, Vf = 2.0V (min), 2.3V (max)
Forward current, If = 20mA

---
Teach a man to fish...

View in Courier:

If you connect the LEDs in series, like this,:

.. +------[R]--[LED>]--[LED>]--[LED>]--+
.. |+ |
..[BATTERY] |
.. | |
.. +-----------------------------------+

then the current in the circuit will be everywhere the same and in
order to determine the resistance you must know how much current you
want in the LEDs, the total voltage they will drop, and the battery
voltage.

Since you said the battery will put out 9 volts and the forward
current will be 20 milliamperes, all we need to do is find out how
much voltage the LEDs will drop.

To be safe, we'll use the minimum Vf of 2.0 volts each and, since
they're in series and the voltages will add, there'll be 6V across
the LEDs with 20mA through them.

Now, since the battery will be supplying 9V and the LEDs only want
6V, we have to get rid of that 3V difference by dropping it across
the resistor.

Invoking Ohm's law, we can say:


E Vbat - Vled 3V
R = --- = ------------- = ------- = 150 ohms
I I 0.02A


The power the resistor must dissipate will be:


P = IE = 0.02A * 3V = 0.06 watts = 60 milliwatts,


so a standard 150 ohm 5% 1/4 watt resistor will be fine.

How long will the battery last?

An alkaline 9V battery has a capacity of about 580
milliampere-hours, so:

C 580mA * hr
T = --- = ------------ = 29 hours
I 20mA


You could also hook up the LEDs like this, in parallel:


.. +---------+------+------+
.. | | | |
.. |+ [R] [R] [R]
..[BATTERY] | | |
.. | [LED] [LED] [LED]
.. | |A |A |A
.. +---------+------+------+

In this case, though, since each LED wants to see 2V and the battery
is putting out 9V, the extra 7V would have to be dropped across each
resistor, wasting a lot of power.

Also, since the LEDs don't share a common current the battery would
have to supply each of the LEDs with a separate 20mA, reducing the
battery's life to about 1/3 of what it would have been with the LEDs
in series.


--
JF
.

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