I've got 8 or so white LEDs, but no specs on them. They look average size
with no particular markings. I want to make a small light using a 12V power
source. The online calculators I see ask for their voltage and current
ratings.

How do I determine these numbers?

"garlicman" None wrote in message
. ..
I've got 8 or so white LEDs, but no specs on them. They look average size
with no particular markings. I want to make a small light using a 12V
power source. The online calculators I see ask for their voltage and
current ratings.

How do I determine these numbers?

If you have a variable DC supply (0 to 10 volts or so) and a multimeter,
just connect them with a series resistor of 150 ohms or so and slowly crank
up the DC voltage. Stop when the brightness is adequate or no longer much
increasing and measure the current and voltage drop.

"Charles" wrote in message
. ..

"garlicman" None wrote in message
. ..
I've got 8 or so white LEDs, but no specs on them. They look average size
with no particular markings. I want to make a small light using a 12V
power source. The online calculators I see ask for their voltage and
current ratings.

How do I determine these numbers?

If you have a variable DC supply (0 to 10 volts or so) and a multimeter,
just connect them with a series resistor of 150 ohms or so and slowly
crank up the DC voltage. Stop when the brightness is adequate or no
longer much increasing and measure the current and voltage drop.

Thanks, I have a 0 to 15VDC variable power supply. I just have a cheap
multimeter that doesn't measure DC amps. Time to upgrade.

"garlicman" None wrote in message
. ..

"Charles" wrote in message
. ..

"garlicman" None wrote in message
. ..
I've got 8 or so white LEDs, but no specs on them. They look average
size with no particular markings. I want to make a small light using a
12V power source. The online calculators I see ask for their voltage and
current ratings.

How do I determine these numbers?

If you have a variable DC supply (0 to 10 volts or so) and a multimeter,
just connect them with a series resistor of 150 ohms or so and slowly
crank up the DC voltage. Stop when the brightness is adequate or no
longer much increasing and measure the current and voltage drop.

Thanks, I have a 0 to 15VDC variable power supply. I just have a cheap
multimeter that doesn't measure DC amps. Time to upgrade.

You can measure the voltage drop across the 150 ohm resistor and calculate
the current.

On Mon, 27 Aug 2007 11:33:06 -0700, "garlicman" None wrote:

I've got 8 or so white LEDs, but no specs on them. They look average size
with no particular markings. I want to make a small light using a 12V power
source. The online calculators I see ask for their voltage and current
ratings.

How do I determine these numbers?

---
Empirically.

Download a few data sheets for common white LEDs and make a note of
their typical forward voltage drops with nominal current through
them.

You'll probably find something like a 3.5V drop with 20mA through
the LED, but don't trust me... Get the data sheets and find out for
yourself.

Since you have a variable power supply, you'll need to put a
resistor in series with the LED so that you can measure the voltage
drop across the resistor and, using Ohm's law, determine the current
through the resistor.

Then, since the resistor and LED will be in series, the current
through the resistor will be the same as the current through the
LED, so measuring the voltage drop across the series resistance will
allow you to calculate that common current.

For example, using a 100 ohm resistor will result in your setup
looking like this: (View in Courier)



+-------+ +-----+
| +|-----+------|+ |
| | | | |
| | [100R] | |
| | | | |
| | +------|- |
| | |A +-----+
| | [LED] METER
| | |
| -|-----+
+-------+
SUPPLY

Now, in order to determine when 20mA is in the resistor, invoke
Ohm's law:


E = IR = 0.02A * 100R = 2V


So, crank up your supply until you read 2 volts across the resistor,
and then you'll know that there's 20mA in both the resistor and the
LED.

Next, in order to determine the voltage dropped across the LED,
leave the supply where it is and reconnect the voltmeter like this:

+-------+
| +|-----+
| | |
| | [100R]
| | | +-----+
| | +------|+ |
| | |A | |
| | [LED] | |
| | | | |
| -|-----+------|- |
+-------+ +-----+
SUPPLY METER


VOILA!

You now have all the information you need to plug into those online
calculators.

With just a little more effort you can blow off those online
calculators and figure it all out for yourself.

Interested?


--
JF

John Fields wrote:
On Mon, 27 Aug 2007 11:33:06 -0700, "garlicman" None wrote:


I've got 8 or so white LEDs, but no specs on them. They look average size
with no particular markings. I want to make a small light using a 12V power
source. The online calculators I see ask for their voltage and current
ratings.

How do I determine these numbers?


---
Empirically.

Download a few data sheets for common white LEDs and make a note of
their typical forward voltage drops with nominal current through
them.

You'll probably find something like a 3.5V drop with 20mA through
the LED, but don't trust me... Get the data sheets and find out for
yourself.

Since you have a variable power supply, you'll need to put a
resistor in series with the LED so that you can measure the voltage
drop across the resistor and, using Ohm's law, determine the current
through the resistor.

Then, since the resistor and LED will be in series, the current
through the resistor will be the same as the current through the
LED, so measuring the voltage drop across the series resistance will
allow you to calculate that common current.

For example, using a 100 ohm resistor will result in your setup
looking like this: (View in Courier)



+-------+ +-----+
| +|-----+------|+ |
| | | | |
| | [100R] | |
| | | | |
| | +------|- |
| | |A +-----+
| | [LED] METER
| | |
| -|-----+
+-------+
SUPPLY

Now, in order to determine when 20mA is in the resistor, invoke
Ohm's law:


E = IR = 0.02A * 100R = 2V


So, crank up your supply until you read 2 volts across the resistor,
and then you'll know that there's 20mA in both the resistor and the
LED.

Next, in order to determine the voltage dropped across the LED,
leave the supply where it is and reconnect the voltmeter like this:

+-------+
| +|-----+
| | |
| | [100R]
| | | +-----+
| | +------|+ |
| | |A | |
| | [LED] | |
| | | | |
| -|-----+------|- |
+-------+ +-----+
SUPPLY METER


VOILA!

You now have all the information you need to plug into those online
calculators.

With just a little more effort you can blow off those online
calculators and figure it all out for yourself.

Interested?


Couldnt have said it better myself..

On Thu, 30 Aug 2007 19:12:03 -0500, Skenny
wrote:

John Fields wrote:
On Mon, 27 Aug 2007 11:33:06 -0700, "garlicman" None wrote:


I've got 8 or so white LEDs, but no specs on them. They look average size
with no particular markings. I want to make a small light using a 12V power
source. The online calculators I see ask for their voltage and current
ratings.

How do I determine these numbers?


---
Empirically.

Download a few data sheets for common white LEDs and make a note of
their typical forward voltage drops with nominal current through
them.

You'll probably find something like a 3.5V drop with 20mA through
the LED, but don't trust me... Get the data sheets and find out for
yourself.

Since you have a variable power supply, you'll need to put a
resistor in series with the LED so that you can measure the voltage
drop across the resistor and, using Ohm's law, determine the current
through the resistor.

Then, since the resistor and LED will be in series, the current
through the resistor will be the same as the current through the
LED, so measuring the voltage drop across the series resistance will
allow you to calculate that common current.

For example, using a 100 ohm resistor will result in your setup
looking like this: (View in Courier)



+-------+ +-----+
| +|-----+------|+ |
| | | | |
| | [100R] | |
| | | | |
| | +------|- |
| | |A +-----+
| | [LED] METER
| | |
| -|-----+
+-------+
SUPPLY

Now, in order to determine when 20mA is in the resistor, invoke
Ohm's law:


E = IR = 0.02A * 100R = 2V


So, crank up your supply until you read 2 volts across the resistor,
and then you'll know that there's 20mA in both the resistor and the
LED.

Next, in order to determine the voltage dropped across the LED,
leave the supply where it is and reconnect the voltmeter like this:

+-------+
| +|-----+
| | |
| | [100R]
| | | +-----+
| | +------|+ |
| | |A | |
| | [LED] | |
| | | | |
| -|-----+------|- |
+-------+ +-----+
SUPPLY METER


VOILA!

You now have all the information you need to plug into those online
calculators.

With just a little more effort you can blow off those online
calculators and figure it all out for yourself.

Interested?


Couldnt have said it better myself..

---
Obviously.


--
JF

John Fields wrote:
On Thu, 30 Aug 2007 19:12:03 -0500, Skenny
wrote:


John Fields wrote:

On Mon, 27 Aug 2007 11:33:06 -0700, "garlicman" None wrote:



I've got 8 or so white LEDs, but no specs on them. They look average size
with no particular markings. I want to make a small light using a 12V power
source. The online calculators I see ask for their voltage and current
ratings.

How do I determine these numbers?


---
Empirically.

Download a few data sheets for common white LEDs and make a note of
their typical forward voltage drops with nominal current through
them.

You'll probably find something like a 3.5V drop with 20mA through
the LED, but don't trust me... Get the data sheets and find out for
yourself.

Since you have a variable power supply, you'll need to put a
resistor in series with the LED so that you can measure the voltage
drop across the resistor and, using Ohm's law, determine the current
through the resistor.

Then, since the resistor and LED will be in series, the current
through the resistor will be the same as the current through the
LED, so measuring the voltage drop across the series resistance will
allow you to calculate that common current.

For example, using a 100 ohm resistor will result in your setup
looking like this: (View in Courier)



+-------+ +-----+
| +|-----+------|+ |
| | | | |
| | [100R] | |
| | | | |
| | +------|- |
| | |A +-----+
| | [LED] METER
| | |
| -|-----+
+-------+
SUPPLY

Now, in order to determine when 20mA is in the resistor, invoke
Ohm's law:


E = IR = 0.02A * 100R = 2V


So, crank up your supply until you read 2 volts across the resistor,
and then you'll know that there's 20mA in both the resistor and the
LED.

Next, in order to determine the voltage dropped across the LED,
leave the supply where it is and reconnect the voltmeter like this:

+-------+
| +|-----+
| | |
| | [100R]
| | | +-----+
| | +------|+ |
| | |A | |
| | [LED] | |
| | | | |
| -|-----+------|- |
+-------+ +-----+
SUPPLY METER


VOILA!

You now have all the information you need to plug into those online
calculators.

With just a little more effort you can blow off those online
calculators and figure it all out for yourself.

Interested?



Couldnt have said it better myself..


---
Obviously.


But your ascii drawings really suck.

On Fri, 31 Aug 2007 07:58:03 -0500, Skenny
wrote:

John Fields wrote:
On Thu, 30 Aug 2007 19:12:03 -0500, Skenny
wrote:


John Fields wrote:

On Mon, 27 Aug 2007 11:33:06 -0700, "garlicman" None wrote:



I've got 8 or so white LEDs, but no specs on them. They look average size
with no particular markings. I want to make a small light using a 12V power
source. The online calculators I see ask for their voltage and current
ratings.

How do I determine these numbers?


---
Empirically.

Download a few data sheets for common white LEDs and make a note of
their typical forward voltage drops with nominal current through
them.

You'll probably find something like a 3.5V drop with 20mA through
the LED, but don't trust me... Get the data sheets and find out for
yourself.

Since you have a variable power supply, you'll need to put a
resistor in series with the LED so that you can measure the voltage
drop across the resistor and, using Ohm's law, determine the current
through the resistor.

Then, since the resistor and LED will be in series, the current
through the resistor will be the same as the current through the
LED, so measuring the voltage drop across the series resistance will
allow you to calculate that common current.

For example, using a 100 ohm resistor will result in your setup
looking like this: (View in Courier)



+-------+ +-----+
| +|-----+------|+ |
| | | | |
| | [100R] | |
| | | | |
| | +------|- |
| | |A +-----+
| | [LED] METER
| | |
| -|-----+
+-------+
SUPPLY

Now, in order to determine when 20mA is in the resistor, invoke
Ohm's law:


E = IR = 0.02A * 100R = 2V


So, crank up your supply until you read 2 volts across the resistor,
and then you'll know that there's 20mA in both the resistor and the
LED.

Next, in order to determine the voltage dropped across the LED,
leave the supply where it is and reconnect the voltmeter like this:

+-------+
| +|-----+
| | |
| | [100R]
| | | +-----+
| | +------|+ |
| | |A | |
| | [LED] | |
| | | | |
| -|-----+------|- |
+-------+ +-----+
SUPPLY METER


VOILA!

You now have all the information you need to plug into those online
calculators.

With just a little more effort you can blow off those online
calculators and figure it all out for yourself.

Interested?



Couldnt have said it better myself..


---
Obviously.


But your ascii drawings really suck.

---
Yup. Here are some particularly bad ones you can View in Courier:



+V
| IN OUT
+------+ | |
|K | O-+ O-+
[DIODE] [COIL]- -|-|- - - - - -| |
| | O-- | | O-- | |
+------+ | K1A | |K1B |
+V | +------|-[A]--+ |
| ___ +----+ C | |
+-O O--+---|S Q|-------+------[1000]--B NPN | |
| S1 | +-|R | | E | |
| | | +----+ A--|---+ | | |
| | +---------Y OR | | GND | |
| | B--|-+ | | |
| [10K] | | | +V | |
| | | | | | | |
| GND | | | +------+ | |
| | | | |K | O-+ O-+
| | | | [DIODE] [COIL]- -|-|- - - - - -| |
| | | | | | O-- | | O-- | |
| | | | +------+ | K2A | |K2B |
| | | | | +------|-[b]--+ |
| ___ +----+ | | | C | |
+--O O-+---|S Q|-------|-+-|--[1000]--B NPN | |
| S2 | +-|R | | | | E | |
| | | +----+ A--+ | | | | |
| | +---------Y OR | | | GND | |
| | B--|-|-+ | |
| [10K] | | | +V | |
| | | | | | | |
| GND | | | +------+ | |
| | | | |K | O-+ |
| | | | [DIODE] [COIL]- -| |
| | | | | | O-- | |
| | | | +------+ | K3A |
| | | | | +--------------------+
| ___ +----+ | | | C
+--O O-+---|S Q|-------|-|-+--[1000]--B
S3 | +-|R | | | E
| | +----+ A--+ | |
| +---------Y OR | GND
| B----+
[10K]
|
GND






VCC------+----+
| | V+
| [RT] |
| | [10K]
[R7] +----|-\ |
| | | --+--180s
+----|----|+/
| | V+
| | |
[R6] | [10K]
| +----|-\ |
| | | --+--150s
+----|----|+/
| | V+
| | |
[R5] | [10K]
| +----|-\ |
| | | --+--120s
+----|----|+/
| | V+
| | |
[R4] | [10K]
| +----|-\ |
| | | --+--90s
+----|----|+/
| | V+
| | |
[R3] | [10K]
| +----|-\ |
| | | --+--60s
+----|----|+/
| | V+
| | |
[R2] | [10K]
| +----|-\ |
| | | --+--30s
+----|----|+/
| |
| +----D
[R1] | G--+
| [CT] S |
| | | |
GND------+----+----+ |
_ |
R/T---------------------+





+12V---------+--------------+-----+-------+--------------+
| | | | |
| | | | [510]
[10K] [1M] | | |A
| | | | [LED]
| +--[10M]--+ |R2 | |
|R1 | | [10K]---|-\ E
[10K]-+-|+\ | | | -+-[10K]--B 2N3906
| | ----+-----|------|+/ | C
IN--[100K]---|------|-/U1A | | U1B| |+ |
| LM393 |+ | | [22µF] |
| [10µF] | | |C2 |
| | | | | |
GND----------+--------------+-----+-------+---+----------+





Vcc
|
+-----+---+ U1A
| | | +-------+
| | | |_ |
+-[500K] +-O|R |
| | |
Vin-+------|------|T |
| | | |
| +------|RC |
| | | |
| [0.01µF] | |
| | | |
| +------|C |
| | |_ _|
| GND +-O|T Q|--+
| | +-------+ |
| +-------------+-[0.1µF]---+----------------+
| | |
| Vcc [10K] |
| | | A--+
| +----+---+ GND +--Y U2A
| | | | | B--+
| +-[500k]| U1B | |
| | | +-------+ +--B |
| | | |_ _| U2B Y--+--OUT
| | +-O|R Q|----+ Vcc +--A
| | |_ | | | |
+------|------|T | [0.1µF] [0.1µF] |
| | | | | |
+------|RC | +---+ +---+ |
| | | | | | | |
[0.01µf] | | [10K] | | [1MR] |
| | | | | | | Y
+-----+------|C | GND | | GND U2D
| | | | | A B
GND +-O|T Q|--+ | +---A | |
| +-------+ | | U2C Y-+---+
+-------------+ +------B





+2V
| Vcc
[10.0K] |
| [10K]
+----|+\ |
| | --+----+
x---+--[10K]--------|-/ |
| | Vcc | +-------+
| [10.0K] | | 0V---|b |
| | [10K] +--B | _ |
+--[10k]--------|+\ | and Y-------|a/b y|---y
| | | --+-------A | |
| +----|-/ +----|a |
| | | +-------+
| GND | S1
| |
| |
| +-------+ +-------+
| 1V--|1 | +----------------+ |
| +------+ | 1-x²|--|1-x² sqrt(1-x²)|--+
+--|x x²|--|x² | +----------------+
+------+ +-------+





Vcc-+---------+-----+-------+-------------+----+-------+
| | | | | | |
[10K] [10K] [1M] +---+---+ [10K][1M] +---+---+
| | | |_ Vcc | | | |_ Vcc |
+-[0.1µF]-+-----|--O|T OUT|-[0.1µF]-+----|--O|T OUT|--+
| | | | | | | |
+-----------+ +---|TH 7555| +---|TH 7555| |
| | | |_ _| | |_ _| |
| O | +--O|D R|O-+V +--O|D R|O-|-+V
| O | +| | GND | | | GND | |
| | [2µF] +---+---+ [0.1µF]+---+---+ |
| | | | | | |
GND-+-----------|---+-------+------------------+-------+ |
| | |
| +--A |
| | Y----+ +------------------------------+
| +--B | |
[0.1µF] | A B
| +-----+ Y +V
| | |A | |
| [1M] [1N4148] | +-------+
| | | | |K | O--C
+-----+-----+ A--+ [DIODE] [COIL]- - - -|
| +--Y | | O- |
| | B--+ +-------+ |
| | | | +------NO
| +--A | C
| Y--+------[1000]--B NPN
+------B E
|
GND


Just makes ya wanna throw up, huh?


--
JF