Running LED off 9.3 Volts AC

[Dave.H]

I also have another similar circuit running on 12 volts. Is it OK to
use the same resistor setup? I've read somewhere that you have to use
560 k.

 

[Dave.H]

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

 

[John Fields]

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.

 

[John Popelish]

I the 390 ohm I used was measuring on my multimeter as 390 k, bought
another, also read as 390 k. What's going on here? The colour code
is Orange White Brown Brown tolerance band, tolerance is supposed to
be 1%, I'm thinking my multimeter is at fault, even though it's only a
couple of weeks old.

1% resistors generally have 3 bands for significant digits
and 1 band of multiplier, plus a brown tolerance band
meaning 1%. The standard 1% resistor that is closest to 390
and starts with a 3 is 392.
http://www.logwell.com/tech/components/resistor_values.html

That is color coded orange white red black brown.
or 392*10^0, 1%

I have no good idea what the value of a resistor would be
with a color code of orange white brown brown.

http://www.elexp.com/t_resist.htm

 

[John Popelish]

The resistors run pretty warm, and the LED is also slightly warm. Is
this normal. The LED I have installed now is a 16, 000 millicandela
red 5mm unit. This lights the dial much better than the 8,000 mcd
unit.

Yes, that is normal for this light output and supply
voltage. The resistor power is about (9.3-2.7)^2/390 watts
= .11 watt, which is warm to the touch. If your 9.3 volt AC
supply is actually a bit higher, the power goes up pretty
fast (square of voltage).

 

[John Popelish]

I also have another similar circuit running on 12 volts. Is it OK to
use the same resistor setup? I've read somewhere that you have to use
560 k.

If you meant 560 ohms, I agree. The resistor has to be
proportional to the excess voltage, to keep the LEDs at the
same current.

 

[John Popelish]

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

If the sum of all the LED forward voltage drops is
significantly less than 9 volts, just divide the remaining
voltage (that must be used up by the resistor) by the
current you want the series string of LEDs to pass.

For instance, if the LED forward voltage is 2 volts, 3 of
them in series will need 6 volts. So the resistor must burn
the remaining 3 volts (9-6=3). If you want those LEDs to
see a current of 10 mA (0.01A), you would need 3/0.01=300
ohms in series. No other parts are needed.

By the way, it is generally a good idea to derate the LED
current by about half, if you want a long life with a steady
light output. Same goes for wattage of resistors, versus
their rated wattage.

 

[Dave.H]

If you meant 560 ohms, I agree. The resistor has to be
proportional to the excess voltage, to keep the LEDs at the
same current.

I did mean 560 ohm. I have salvaged some more resistors from some old
CFL ballasts, these test OK, if I add them to the two resistors
installed that give about 565 ohms, which I'm assuming is OK.

 

[Don Klipstein]

The resistors run pretty warm, and the LED is also slightly warm. Is
this normal. The LED I have installed now is a 16, 000 millicandela
red 5mm unit. This lights the dial much better than the 8,000 mcd
unit.

I have been out of this thread for a few days, so I forget how much
current the LED has. I am going on 9.3 volts AC and a 390 ohm resistor.

But rectified filtered 9.3 VAC is about 12 volts DC after diode drops.
With an LED dropping 2 volts (10 volts across the resistor) and a 390 ohm
resistor, this means 25.5 mA through the LED and .26 watt dissipated in
the resistor.
A usual 5 mm (T1-3/4) LED does usually get a little noticeably warm at
25 mA, and a half watt resistor will get fairly warm at .25 watt.

If this 9.3 VAC is unfiltered fullwave rectified, then the average
voltage across the resistor will be roughly 6 volts and the RMS voltage
across the resistor will be a bit under 7 volts (roughly).
A 390 ohm resistor here will pass about 15 mA average current. My
experience is that most LEDs that drop ~2V do not noticeably warm up at 15
mA, but if you are paying attention for that you may notice a temperature
rise.

The LED will also get warmer if you use no rectification and the LED
experiences reverse breakdown every other half cycle.

The LED will also warm up if heat conducts to it from a very nearby
resistor.

7 volts RMS across a 390 ohm resistor will cause about 1/8 watt to be
dissipated in the resistor. A 1/2 watt one gets just a little warm with
1/8 watt, and a 1/4 watt one usually gets fairly warm, but you may not
notice much heat since your fingertip could significantly heatsink the
resistor if you touch the resistor.

Check the voltage - it may be higher than nominal.

- Don Klipstein ([email protected])

 

[Dave.H]

I have soldered 7 resistors in series, giving a total of 552 ohms,
(although it still reads as 552 k) I am guessing this is OK to run
now, as it's only 8 ohms short of 560. Will eventually get a single
560 ohm resistor. The LED current is 20 mA, the voltage source is 12
volts, I've checked this with my multimeter.

 

[John Popelish]

I have soldered 7 resistors in series, giving a total of 552 ohms,
(although it still reads as 552 k) I am guessing this is OK to run
now, as it's only 8 ohms short of 560. Will eventually get a single
560 ohm resistor. The LED current is 20 mA, the voltage source is 12
volts, I've checked this with my multimeter.

Sounds perfectly functional.

 

[Dave.H]

Except there 26 mA across the LED. I want 20 mA.

 

[Dave.H]

How would a 1 watt 390 ohm resistor go. I have one.

 

[John Popelish]

How would a 1 watt 390 ohm resistor go. I have one.

Will work where any lower wattage is needed.

 

[John Popelish]

Except there 26 mA across the LED. I want 20 mA.

Then you need more resistance.

 

[Dave.H]

Then you need more resistance.
Solved that. Put two 390 ohm resistors in series one .25 watt, and one
1 watt, as that's all I had., 2.1 volts 19.12 mA across the LED, LED
is designed for 2.0 volts, 20 mA, figured the slight over voltage
can't really hurt it too much, I'm happy with the result.

 

[John Popelish]

Solved that. Put two 390 ohm resistors in series one .25 watt, and one
1 watt, as that's all I had., 2.1 volts 19.12 mA across the LED, LED
is designed for 2.0 volts, 20 mA, figured the slight over voltage
can't really hurt it too much, I'm happy with the result.

That assumption might bite you. The current through an LED
changes dramatically with very tiny voltage changes.
Something like a 0.06 volts change will swing the current
through about a 10 times multiple. Measure the voltage
across one of the resistors and divide that voltage by the
resistance to find the actual current.

 

[Dave.H]

0.455 divided by 390 = 0.00116666667

 

[John Popelish]

0.455 divided by 390 = 0.00116666667

You measure 0.455 volts across a 390 ohm resistor in series
with the LEDs? That can't be right.

 

[Dave.H]

6.75 volts. I must have had the multimeter on AC mode before.
6.75 divided by 390 = 0.0173076923