# Running LED off 9.3 Volts AC

J

#### John Popelish

Jan 1, 1970
0
Dave.H said:
6.75 volts. I must have had the multimeter on AC mode before.
6.75 divided by 390 = 0.0173076923

17 mA. much better.

D

#### Dave.H

Jan 1, 1970
0
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

I built this circuit, works excellent, substituted the red LED's for
some 6500 mcd orange ones, the switch is a push button 250 volt
"lampswitch" this can only be installed on the positive battery lead
not the negative (has only two terminals), this won't drain the
battery when the unit is off will it? If so, I'll just install
another switch on the negative side.

J

Jan 1, 1970
0
E

#### Ecnerwal

Jan 1, 1970
0
Dave.H said:
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

Two simple options:

9V battery, 3x 2V (min), 3V to be dropped across a resistor X at 20mA -
V=IR, R=V/I R=3/0.020, R=150 ohms.

9V battery, 3x 2.3V (max) 9-6.9=2.1V headroom - you could use an LM317
set up as a current source. That would allow driving reliably with 20mA
(actual voltage not that big of an issue for LEDs, and not a very
reliable number). The LM317 acts as whatever size resistor is needed to
run 20mA (when it's set up to act as a current source). There are also
newer devices that do that as a unit, with no external parts.

R

#### redbelly

Jan 1, 1970
0
Teach a man to fish...

.... and he will pay you for one day.

Give a man fish, and he will pay you for a lifetime.

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