Running LED off 9.3 Volts AC

D

Dave.H

Guest
I need to replace an incandescent dial lamp in a modern radio, and I
would love to replace it with a high brightness LED, to save having to
replace it in the near future. Problem is, the source voltage is 9.3
volts AC, not suitable for running an LED off directly. What value
resistor would I need for this?

The specifications of the LED I had in mind are as follows:

# Emitted Colour: Sunset Red
# Forward Voltage = 2.0V Typical @ IF = 20mA
# Reverse Voltage = 5V
# Maximum Voltage = 2.6V @ IF = 20mA
# Luminous Intensity = 6,000mcd Minimum, 8,000mcd Typical
# Peak Emission Wavelength = 640nm
# Power Dissipation = 80mW
# Continuous Forward Current = 50mA
 
On Nov 19, 8:27 pm, "Dave.H" <the19...@googlemail.com> wrote:
I need to replace an incandescent dial lamp in a modern radio, and I
would love to replace it with a high brightness LED, to save having to
replace it in the near future. Problem is, the source voltage is 9.3
volts AC, not suitable for running an LED off directly. What value
resistor would I need for this?

The specifications of the LED I had in mind are as follows:

# Emitted Colour: Sunset Red
# Forward Voltage = 2.0V Typical @ IF = 20mA
# Reverse Voltage = 5V
# Maximum Voltage = 2.6V @ IF = 20mA
# Luminous Intensity = 6,000mcd Minimum, 8,000mcd Typical
# Peak Emission Wavelength = 640nm
# Power Dissipation = 80mW
# Continuous Forward Current = 50mA
Click to expand...
You've just about got it already, Dave. Put the LED in series with a
1N4001 diode (blocks 50VDC, will work just fine), and a series
resistor.

With a series diode, the LED will only be on half the time, so you can
probably bump up the LED current to 30mA or so. Here's the drill:

9.3VAC will give about 13V peak. Subtracting out the 2V for the LED
and the 0.7V for the diode, that leaves you with about 10V. Using
Ohms Law:

R = 10V / .03A = 333 ohms.

Use a 330 ohm 1/2 watt resistor, and you'll be just fine.

Good luck
Chris
 
Dave.H wrote:
I need to replace an incandescent dial lamp in a modern radio, and I
would love to replace it with a high brightness LED, to save having to
replace it in the near future. Problem is, the source voltage is 9.3
volts AC, not suitable for running an LED off directly. What value
resistor would I need for this?

The specifications of the LED I had in mind are as follows:

# Emitted Colour: Sunset Red
# Forward Voltage = 2.0V Typical @ IF = 20mA
# Reverse Voltage = 5V
# Maximum Voltage = 2.6V @ IF = 20mA
# Luminous Intensity = 6,000mcd Minimum, 8,000mcd Typical
# Peak Emission Wavelength = 640nm
# Power Dissipation = 80mW
# Continuous Forward Current = 50mA
Click to expand...
AC --->|---+---[510R]---+
D1 |+ |
[100uF] [LED]
| |
AC --------+------------+

The 510 ohm resistor will limit the current to
about 20 mA. The cap will prevent flicker.
Use a 16 volt or higher cap. Any standard
diode will be fine.

Ed
 
In <59c42359-62a9-43d6-9263-8ba92f3b0869@i12g2000prf.googlegroups.com>,
Dave.H wrote:

I need to replace an incandescent dial lamp in a modern radio, and I
would love to replace it with a high brightness LED, to save having to
replace it in the near future. Problem is, the source voltage is 9.3
volts AC, not suitable for running an LED off directly. What value
resistor would I need for this?

The specifications of the LED I had in mind are as follows:

# Emitted Colour: Sunset Red
# Forward Voltage = 2.0V Typical @ IF = 20mA
# Reverse Voltage = 5V
# Maximum Voltage = 2.6V @ IF = 20mA
# Luminous Intensity = 6,000mcd Minimum, 8,000mcd Typical
# Peak Emission Wavelength = 640nm
# Power Dissipation = 80mW
# Continuous Forward Current = 50mA
Click to expand...
If you need just one LED and a fairly pure shade of red (like 632.8 nm
He-Ne laser or half a fine red hair more-pure-red) is OK and light output
maybe a bit on the low side is OK, do this:

1. Get a Radio Shack 276-307 red LED. Use some fine sandpaper on it to
make it more-diffusing. If you need really wide spread light dispersion
pattern, then sand it with sandpaper until it is shortened by about 1-1.25
millimeters (.04-.05 inch), and make it shaped to be slightly less pointy
in shape than it was.

2. Get or make just about any full wave bridge rectifier. I like to get
plenty of 1N4000 series diodes, either of type 1N4004 (400 volt) or 1N4007
(1000 volt) - cost increase with increasing voltage rating is small enough
to get only higher voltage ones for all needs as far as I see things!

3. Feed the LED with the bridge rectifier. The LED drops about 2 volts,
and the bridge rectifier at 20 mA or whatever drops about 1.3 volts. For
a rough-and-dirty first order approximation, subtract 3.3 volts from 9.3
to get 6 volts across the dropping resistor. To get 20 mA, nearest
very-common value is 220 ohms. I see RMS current being about 18 mA and
average current being about 16 mA as guesstimates for now. This makes me
suspect 270 ohms is good for 20 mA average LED current, and 220 ohms
should make the average LED current about 24-25 mA, which the LED should
take quite well.

-----------------------------------------------------

For a brighter and slightly orangish red LED: Get them from Digi-Key!
I really like Avago HLMP-ED33-SVOOO, DigiKey catalog number 516-1390-ND.

Do same as above, get about 3 times as much light as with the Rat
Shack LED and the color is a slightly orangish shade of red. In some
situations it will appear plainly red, and this color is more of a
slightly orangish shade of red rather than a reddish shade of orange.

Cost is 80 cents apiece for 1-9 pieces, $4.76 per 10 pieces for 10-90
pieces, plus shipping (just a few dollars for UPS ground), plus a $5
handling charge if your order before tax and shipping is under $25 (IIRC).

Both of these LEDs have MCD under 6,000, but at 20 mA average current
the Rat Shack one produces about half a lumen of red light, and the Avago
one above produces about 1.6 lumens of red light.

5V 60 mA incandescent indicator lamps rated to last 25,000 hours produce
about .63 lumen, and 5V .115 amp incandescent indicator lamps rated to
last 40,000 hours produce about 1.9 lumens, without any color filtering.

Keep in mind that with an incandescent, filtering it to red or even
orange-red will block roughly half to 2/3 of the luminous output. Even if
this is a low wattage low current long life incandescent and the color
temperature is close to 2000 K, filtering it to red will block at least
half the luminous output and filtering it to a borderline red-orange
close to that of red-orange-glowing clear neon sign tubing will remove
about 40% of the luminous output.

The Rat Shack one has peak wavelength around 650-660 nm and dominant
wavelength (color specification that approximately means "hue") of about
637-640 nm. This is half a fine red hair more pure a red than a 632.8
nm He-Ne laser. The Avago one has peak wavelength in the mid-upper 630's
of nm and dominant wavelength in the mid 620's of nm. This is more orange
than a He-Ne laser, but still basically red, slightly less orangish than
most automotive taillights, less orangish than clear red-orange neon
tuning, and slightly less orangish than a display of "pure red stimulus"
on most computer monitors and TV sets.

- Don Klipstein (don@misty.com)
 
"Dave.H" <the1930s@googlemail.com> wrote in message
news:78efb23a-4306-45db-9d45-8d21a1caca67@s6g2000prc.googlegroups.com...
AC --->|---+---[510R]---+
D1 |+ |
[100uF] [LED]
| |
AC --------+------------+

The 510 ohm resistor will limit the current to
about 20 mA. The cap will prevent flicker.
Use a 16 volt or higher cap. Any standard
diode will be fine.

Ed

I'm interested in doing this but I'm having trouble following your
diagram.
Click to expand...
Be sure to view it in a fixed-width typeface such as Courier so things will
line up.

The AC goes through a diode and then across a 100-uF capacitor. From the
positive side of the capacitor there's a 510-ohm resistor to the anode of
the LED, and then the cathode of the LED goes to the negative side of the
capacitor.
 
---+
D1 |+ |
[100uF] [LED]
| |
AC --------+------------+

The 510 ohm resistor will limit the current to
about 20 mA. The cap will prevent flicker.
Use a 16 volt or higher cap. Any standard
diode will be fine.

Ed
Click to expand...
I'm interested in doing this but I'm having trouble following your
diagram.
 
Thanks, that made it a lot easier to understand. Just thinking how
I'm going to mount it, there's heaps of room in the radio, so I'm
thinking a second PC board, to make it neat and tidy. Thanks for all
your help, appreciate it.
 
Dave.H wrote:
I need to replace an incandescent dial lamp in a modern radio, and I
would love to replace it with a high brightness LED, to save having to
replace it in the near future. Problem is, the source voltage is 9.3
volts AC, not suitable for running an LED off directly. What value
resistor would I need for this?

The specifications of the LED I had in mind are as follows:

# Emitted Colour: Sunset Red
# Forward Voltage = 2.0V Typical @ IF = 20mA
# Reverse Voltage = 5V
# Maximum Voltage = 2.6V @ IF = 20mA
# Luminous Intensity = 6,000mcd Minimum, 8,000mcd Typical
# Peak Emission Wavelength = 640nm
# Power Dissipation = 80mW
# Continuous Forward Current = 50mA
Click to expand...
If you can use these LEDs in pairs, paralleled, one goingh
each way, one will run on one half cycle of the AC, and one
on the other. A single resistor in series with the pair
will limit the current from the AC source. I would run them
at a peak current around 20 mA, for long life.

9.3 volts AC hits peaks of about +-13.2 volts. Since each
way, there will be one LED using up voltage of about 2.6
volts (in each direction, the resistor must use up the rest,
while letting through 20 mA. (13.2-2.6)/0.02=530. a 470
ohm 1/4 watt would be close enough. If you could use 4
LEDs, put another pair in series with the first, and maybe
lower the resistance a bit, maybe to 390 ohm.

Since each of the LEDs is hitting a peak of only about 20 mA
and conducting only half the time, if this isn't bright
enough, you could go as low as half these values to double
the light output (that is a just noticeable increase), but
use 1/2 watt resistors.
 
I have two resistors, a 390 ohm, and 120 ohm, do I connect these in
seris or parallel?
 
Dave.H wrote:
I have two resistors, a 390 ohm, and 120 ohm, do I connect these in
seris or parallel?
Click to expand...
If you are going to try driving a paralleled (one reversed)
pair, I would try, first with the 390 ohm resistor in series
with the 120 ohm in series with the pair.

AC-----390--120----+----+
| |
- v LEDs
9.3AC ^ -
| |
AC-----------------+----+


If you go for 4 LEDs (2 inverse paralleled pairs in series)
then just use the 390 ohm in series with them.

AC-----390---------+----+
| |
- v LEDs
^ -
| |
9.4VAC +----+
| |
- v LEDs
^ -
| |
AC-----------------+----+
 
On Nov 20, 6:41 pm, "Dave.H" <the19...@googlemail.com> wrote:
I have two resistors, a 390 ohm, and 120 ohm, do I connect these in
seris or parallel?
Click to expand...
Series (view in fixed font or cut&paste to Notepad):

| 1N400X
| ___ ___
| .----->|--|___|--|___|--o----.
| | 390 120 | |
| | | |
| o | |
| / \ LED V ~ - ~
| ( ~ ) 9.3VAC - ~ ^ ~
| \_/ | |
| o | |
| | | |
| '-----------------------o----'
|
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)

Good Luck
Chris
 
Chris wrote:
On Nov 20, 6:41 pm, "Dave.H" <the19...@googlemail.com> wrote:
I have two resistors, a 390 ohm, and 120 ohm, do I connect these in
seris or parallel?

Series (view in fixed font or cut&paste to Notepad):

| 1N400X
| ___ ___
| .----->|--|___|--|___|--o----.
| | 390 120 | |
| | | |
| o | |
| / \ LED V ~ - ~
| ( ~ ) 9.3VAC - ~ ^ ~
| \_/ | |
| o | |
| | | |
| '-----------------------o----'
|
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)

Good Luck
Chris
Click to expand...
Why add the 1N400X?
It will just prevent one of the LEDs from lighting.
 
Thanks, I have just finished building and installing it in the radio,
works really good at illuminating the dial. Thanks for all your help,
I appreciate it.
 
On Nov 19, 10:06 pm, Chris <cfoley1...@yahoo.com> wrote:
On Nov 19, 8:27 pm, "Dave.H" <the19...@googlemail.com> wrote:



I need to replace an incandescent dial lamp in a modern radio, and I
would love to replace it with a high brightness LED, to save having to
replace it in the near future. Problem is, the source voltage is 9.3
volts AC, not suitable for running an LED off directly. What value
resistor would I need for this?

The specifications of the LED I had in mind are as follows:

# Emitted Colour: Sunset Red
# Forward Voltage = 2.0V Typical @ IF = 20mA
# Reverse Voltage = 5V
# Maximum Voltage = 2.6V @ IF = 20mA
# Luminous Intensity = 6,000mcd Minimum, 8,000mcd Typical
# Peak Emission Wavelength = 640nm
# Power Dissipation = 80mW
# Continuous Forward Current = 50mA

You've just about got it already, Dave. Put the LED in series with a
1N4001 diode (blocks 50VDC, will work just fine), and a series
resistor.
Click to expand...
One thing I wonder about, when you're in the reverse-bias portion of
the line cycle. The LED spec says max reverse bias is 5V (see OP's
specs). What, if anything, insures that the 1n4001 takes up most of
the voltage in reverse mode?

(Of course, adding a cap as Ed/ehsjr suggested would make my question
a moot point.)

Mark

With a series diode, the LED will only be on half the time, so you can
probably bump up the LED current to 30mA or so. Here's the drill:

9.3VAC will give about 13V peak. Subtracting out the 2V for the LED
and the 0.7V for the diode, that leaves you with about 10V. Using
Ohms Law:

R = 10V / .03A = 333 ohms.

Use a 330 ohm 1/2 watt resistor, and you'll be just fine.

Good luck
Chris
Click to expand...
 
redbelly wrote:

One thing I wonder about, when you're in the reverse-bias portion of
the line cycle. The LED spec says max reverse bias is 5V (see OP's
specs). What, if anything, insures that the 1n4001 takes up most of
the voltage in reverse mode?

(Of course, adding a cap as Ed/ehsjr suggested would make my question
a moot point.)
Click to expand...
I agree. LEDs generally have very low reverse leakage
current, so I am reluctant to ever use a rectifier diode in
series and assume that its leakage will be a small fraction
of the LED's leakage current. That is why I recommended the
pair of LEDs in inverse parallel. Each protects the the
from excessive reverse voltage, while spreading the light
out over two sources.
 
On Nov 20, 10:35 pm, John Popelish <jpopel...@rica.net> wrote:
Chris wrote:
On Nov 20, 6:41 pm, "Dave.H" <the19...@googlemail.com> wrote:
I have two resistors, a 390 ohm, and 120 ohm, do I connect these in
seris or parallel?

Series (view in fixed font or cut&paste to Notepad):

| 1N400X
| ___ ___
| .----->|--|___|--|___|--o----.
| | 390 120 | |
| | | |
| o | |
| / \ LED V ~ - ~
| ( ~ ) 9.3VAC - ~ ^ ~
| \_/ | |
| o | |
| | | |
| '-----------------------o----'
|
(created by AACircuit v1.28.6 beta 04/19/05www.tech-chat.de)

Good Luck
Chris

Why add the 1N400X?
It will just prevent one of the LEDs from lighting.- Hide quoted text -

- Show quoted text -
Click to expand...
Oops -- thanks for the spot, Mr. Popelish. Less haste next time.

Cheers
Chris
 
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.
 
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.
 
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