LED help

G

garlicman

Guest
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
news:EKydnZJHraZwhU7bnZ2dnUVZ_rWtnZ2d@comcast.com...
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?
Click to expand...
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" <charlesschuler@comcast.net> wrote in message
news:v5qdnVea_89Pxk7bnZ2dnUVZ_vamnZ2d@comcast.com...
"garlicman" <None> wrote in message
news:EKydnZJHraZwhU7bnZ2dnUVZ_rWtnZ2d@comcast.com...
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.
Click to expand...
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
news:MNidnQW35O8LKUnbnZ2dnUVZ_r2nnZ2d@comcast.com...
"Charles" <charlesschuler@comcast.net> wrote in message
news:v5qdnVea_89Pxk7bnZ2dnUVZ_vamnZ2d@comcast.com...

"garlicman" <None> wrote in message
news:EKydnZJHraZwhU7bnZ2dnUVZ_rWtnZ2d@comcast.com...
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.
Click to expand...
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?
Click to expand...
---
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..
Click to expand...
 
On Thu, 30 Aug 2007 19:12:03 -0500, Skenny
<Skenny@highstream.invalid> 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..
Click to expand...
---
Obviously.


--
JF
 
John Fields wrote:
On Thu, 30 Aug 2007 19:12:03 -0500, Skenny
Skenny@highstream.invalid> 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.
Click to expand...
 
On Fri, 31 Aug 2007 07:58:03 -0500, Skenny
<Skenny@highstream.invalid> wrote:

John Fields wrote:
On Thu, 30 Aug 2007 19:12:03 -0500, Skenny
Skenny@highstream.invalid> 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.
Click to expand...
---
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 |
| | | | | +------|---+ |
| ___ +----+ | | | 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 |+ | | [22ľF] |
| [10ľF] | | |C2 |
| | | | | |
GND>----------+--------------+-----+-------+---+----------+





Vcc
|
+-----+---+ U1A
| | | +-------+
| | | |_ |
+->[500K] +-O|R |
| | |
Vin>-+------|------|T |
| | | |
| +------|RC |
| | | |
| [0.01ľF] | |
| | | |
| +------|C |
| | |_ _|
| GND +-O|T Q|--+
| | +-------+ |
| +-------------+-[0.1ľF]---+----------------+
| | |
| Vcc [10K] |
| | | A--+
| +----+---+ GND +--Y U2A
| | | | | B--+
| +->[500k]| U1B | |
| | | +-------+ +--B |
| | | |_ _| U2B Y--+-->OUT
| | +-O|R Q|----+ Vcc +--A
| | |_ | | | |
+------|------|T | [0.1ľF] [0.1ľF] |
| | | | | |
+------|RC | +---+ +---+ |
| | | | | | | |
[0.01ľf] | | [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-x˛|--|1-x˛ sqrt(1-x˛)|--+
+--|x x˛|--|x˛ | +----------------+
+------+ +-------+





Vcc>-+---------+-----+-------+-------------+----+-------+
| | | | | | |
[10K] [10K] [1M] +---+---+ [10K][1M] +---+---+
| | | |_ Vcc | | | |_ Vcc |
+-[0.1ľF]-+-----|--O|T OUT|-[0.1ľF]-+----|--O|T OUT|--+
| | | | | | | |
+-----------+ +---|TH 7555| +---|TH 7555| |
| | | |_ _| | |_ _| |
| O | +--O|D R|O-+V +--O|D R|O-|-+V
| O | +| | GND | | | GND | |
| | [2ľF] +---+---+ [0.1ľF]+---+---+ |
| | | | | | |
GND>-+-----------|---+-------+------------------+-------+ |
| | |
| +--A |
| | Y----+ +------------------------------+
| +--B | |
[0.1ľF] | 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
 
John Fields wrote:
On Fri, 31 Aug 2007 07:58:03 -0500, Skenny
Skenny@highstream.invalid> wrote:


John Fields wrote:

On Thu, 30 Aug 2007 19:12:03 -0500, Skenny
Skenny@highstream.invalid> 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 |
| | | | | +------|---+ |
| ___ +----+ | | | 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 |+ | | [22ľF] |
| [10ľF] | | |C2 |
| | | | | |
GND>----------+--------------+-----+-------+---+----------+





Vcc
|
+-----+---+ U1A
| | | +-------+
| | | |_ |
+->[500K] +-O|R |
| | |
Vin>-+------|------|T |
| | | |
| +------|RC |
| | | |
| [0.01ľF] | |
| | | |
| +------|C |
| | |_ _|
| GND +-O|T Q|--+
| | +-------+ |
| +-------------+-[0.1ľF]---+----------------+
| | |
| Vcc [10K] |
| | | A--+
| +----+---+ GND +--Y U2A
| | | | | B--+
| +->[500k]| U1B | |
| | | +-------+ +--B |
| | | |_ _| U2B Y--+-->OUT
| | +-O|R Q|----+ Vcc +--A
| | |_ | | | |
+------|------|T | [0.1ľF] [0.1ľF] |
| | | | | |
+------|RC | +---+ +---+ |
| | | | | | | |
[0.01ľf] | | [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-x˛|--|1-x˛ sqrt(1-x˛)|--+
+--|x x˛|--|x˛ | +----------------+
+------+ +-------+





Vcc>-+---------+-----+-------+-------------+----+-------+
| | | | | | |
[10K] [10K] [1M] +---+---+ [10K][1M] +---+---+
| | | |_ Vcc | | | |_ Vcc |
+-[0.1ľF]-+-----|--O|T OUT|-[0.1ľF]-+----|--O|T OUT|--+
| | | | | | | |
+-----------+ +---|TH 7555| +---|TH 7555| |
| | | |_ _| | |_ _| |
| O | +--O|D R|O-+V +--O|D R|O-|-+V
| O | +| | GND | | | GND | |
| | [2ľF] +---+---+ [0.1ľF]+---+---+ |
| | | | | | |
GND>-+-----------|---+-------+------------------+-------+ |
| | |
| +--A |
| | Y----+ +------------------------------+
| +--B | |
[0.1ľF] | 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?


Well, my terms wouldnt be so nice.
Click to expand...
 
On Fri, 31 Aug 2007 10:24:14 -0500, Skenny
<Skenny@highstream.invalid> wrote:

John Fields wrote:
On Fri, 31 Aug 2007 07:58:03 -0500, Skenny
Skenny@highstream.invalid> wrote:


John Fields wrote:

On Thu, 30 Aug 2007 19:12:03 -0500, Skenny
Skenny@highstream.invalid> 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 |
| | | | | +------|---+ |
| ___ +----+ | | | 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 |+ | | [22ľF] |
| [10ľF] | | |C2 |
| | | | | |
GND>----------+--------------+-----+-------+---+----------+





Vcc
|
+-----+---+ U1A
| | | +-------+
| | | |_ |
+->[500K] +-O|R |
| | |
Vin>-+------|------|T |
| | | |
| +------|RC |
| | | |
| [0.01ľF] | |
| | | |
| +------|C |
| | |_ _|
| GND +-O|T Q|--+
| | +-------+ |
| +-------------+-[0.1ľF]---+----------------+
| | |
| Vcc [10K] |
| | | A--+
| +----+---+ GND +--Y U2A
| | | | | B--+
| +->[500k]| U1B | |
| | | +-------+ +--B |
| | | |_ _| U2B Y--+-->OUT
| | +-O|R Q|----+ Vcc +--A
| | |_ | | | |
+------|------|T | [0.1ľF] [0.1ľF] |
| | | | | |
+------|RC | +---+ +---+ |
| | | | | | | |
[0.01ľf] | | [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-x˛|--|1-x˛ sqrt(1-x˛)|--+
+--|x x˛|--|x˛ | +----------------+
+------+ +-------+





Vcc>-+---------+-----+-------+-------------+----+-------+
| | | | | | |
[10K] [10K] [1M] +---+---+ [10K][1M] +---+---+
| | | |_ Vcc | | | |_ Vcc |
+-[0.1ľF]-+-----|--O|T OUT|-[0.1ľF]-+----|--O|T OUT|--+
| | | | | | | |
+-----------+ +---|TH 7555| +---|TH 7555| |
| | | |_ _| | |_ _| |
| O | +--O|D R|O-+V +--O|D R|O-|-+V
| O | +| | GND | | | GND | |
| | [2ľF] +---+---+ [0.1ľF]+---+---+ |
| | | | | | |
GND>-+-----------|---+-------+------------------+-------+ |
| | |
| +--A |
| | Y----+ +------------------------------+
| +--B | |
[0.1ľF] | 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?


Well, my terms wouldnt be so nice.
Click to expand...

---
Blow me.


--
JF
 
John Fields wrote:
On Fri, 31 Aug 2007 10:24:14 -0500, Skenny
Skenny@highstream.invalid> wrote:


John Fields wrote:

On Fri, 31 Aug 2007 07:58:03 -0500, Skenny
Skenny@highstream.invalid> wrote:



John Fields wrote:


On Thu, 30 Aug 2007 19:12:03 -0500, Skenny
Skenny@highstream.invalid> 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 |
| | | | | +------|---+ |
| ___ +----+ | | | 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 |+ | | [22ľF] |
| [10ľF] | | |C2 |
| | | | | |
GND>----------+--------------+-----+-------+---+----------+





Vcc
|
+-----+---+ U1A
| | | +-------+
| | | |_ |
+->[500K] +-O|R |
| | |
Vin>-+------|------|T |
| | | |
| +------|RC |
| | | |
| [0.01ľF] | |
| | | |
| +------|C |
| | |_ _|
| GND +-O|T Q|--+
| | +-------+ |
| +-------------+-[0.1ľF]---+----------------+
| | |
| Vcc [10K] |
| | | A--+
| +----+---+ GND +--Y U2A
| | | | | B--+
| +->[500k]| U1B | |
| | | +-------+ +--B |
| | | |_ _| U2B Y--+-->OUT
| | +-O|R Q|----+ Vcc +--A
| | |_ | | | |
+------|------|T | [0.1ľF] [0.1ľF] |
| | | | | |
+------|RC | +---+ +---+ |
| | | | | | | |
[0.01ľf] | | [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-x˛|--|1-x˛ sqrt(1-x˛)|--+
+--|x x˛|--|x˛ | +----------------+
+------+ +-------+





Vcc>-+---------+-----+-------+-------------+----+-------+
| | | | | | |
[10K] [10K] [1M] +---+---+ [10K][1M] +---+---+
| | | |_ Vcc | | | |_ Vcc |
+-[0.1ľF]-+-----|--O|T OUT|-[0.1ľF]-+----|--O|T OUT|--+
| | | | | | | |
+-----------+ +---|TH 7555| +---|TH 7555| |
| | | |_ _| | |_ _| |
| O | +--O|D R|O-+V +--O|D R|O-|-+V
| O | +| | GND | | | GND | |
| | [2ľF] +---+---+ [0.1ľF]+---+---+ |
| | | | | | |
GND>-+-----------|---+-------+------------------+-------+ |
| | |
| +--A |
| | Y----+ +------------------------------+
| +--B | |
[0.1ľF] | 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?



Well, my terms wouldnt be so nice.


---
Blow me.


HUH HUH HUH... poser..
Click to expand...
 
On Fri, 31 Aug 2007 11:02:02 -0500, Skenny
<Skenny@highstream.invalid> wrote:


HUH HUH HUH... poser..
Click to expand...
---
LOL, that from the idiot with the hammer and sharp chisel...


--
JF
 
On Fri, 31 Aug 2007 12:09:22 -0500, John Fields
<jfields@austininstruments.com> wrote:

On Fri, 31 Aug 2007 11:02:02 -0500, Skenny
Skenny@highstream.invalid> wrote:


HUH HUH HUH... poser..

---
LOL, that from the idiot with the hammer and sharp chisel...
Click to expand...

---
Oops...

I meant: "LOL, that from the poser who said: 'Couldn't have said it
better myself..' when he couldn't have said it at all!"


--
JF
 
John Fields wrote:
On Fri, 31 Aug 2007 11:02:02 -0500, Skenny
Skenny@highstream.invalid> wrote:



HUH HUH HUH... poser..


---
LOL, that from the idiot with the hammer and sharp chisel...


At least my chisel is sharp unlike yours, oh thats not a chisel you have
Click to expand...
there, its a used comdom. Why are you hanging on to that?
Waiting for that next "special guy"?
 
John Fields wrote:
On Fri, 31 Aug 2007 12:09:22 -0500, John Fields
jfields@austininstruments.com> wrote:


On Fri, 31 Aug 2007 11:02:02 -0500, Skenny
Skenny@highstream.invalid> wrote:



HUH HUH HUH... poser..

---
LOL, that from the idiot with the hammer and sharp chisel...



---
Oops...

I meant: "LOL, that from the poser who said: 'Couldn't have said it
better myself..' when he couldn't have said it at all!"


Well actually, I could have said it better, I just didnt want to hurt a
Click to expand...
special persons feelings.
 
Skenny wrote:
But your ascii drawings really suck.

John Fields wrote:
Yup. Here are some particularly bad ones you can View in Courier:
Click to expand...
....and the WORST thing about ASCII schematics
is that *persistance* thing:
http://groups.google.com/group/sci.electronics/browse_frm/thread/86bb4239891d93d5/4523c80acf6aa356?q=R1+R2+Q2&fwc=1
 
On Fri, 31 Aug 2007 12:42:14 -0500, Skenny
<Skenny@highstream.invalid> wrote:

John Fields wrote:
On Fri, 31 Aug 2007 11:02:02 -0500, Skenny
Skenny@highstream.invalid> wrote:



HUH HUH HUH... poser..


---
LOL, that from the idiot with the hammer and sharp chisel...


At least my chisel is sharp unlike yours, oh thats not a chisel you have
there, its a used comdom.
Click to expand...
---
"Comdom"???
---

Why are you hanging on to that?
Waiting for that next "special guy"?
Click to expand...
---
Yup, and I figured from the tone and timbre of your previous posts
that you wouldn't mind wrapping your lips around something that had
previously been up someone's ass.


--
JF
 
On Fri, 31 Aug 2007 12:44:13 -0500, Skenny
<Skenny@highstream.invalid> wrote:

John Fields wrote:
On Fri, 31 Aug 2007 12:09:22 -0500, John Fields
jfields@austininstruments.com> wrote:


On Fri, 31 Aug 2007 11:02:02 -0500, Skenny
Skenny@highstream.invalid> wrote:



HUH HUH HUH... poser..

---
LOL, that from the idiot with the hammer and sharp chisel...



---
Oops...

I meant: "LOL, that from the poser who said: 'Couldn't have said it
better myself..' when he couldn't have said it at all!"


Well actually, I could have said it better,
Click to expand...
---
Easy to say, now that you've been shown the trick.
---

I just didnt want to hurt a special persons feelings.
Click to expand...
---
Geez, you don't have to worry about that, just post your "better
said" version and let's see what you've got going on.

BTW, if you want to be considered serial, you might want to learn
how to punctualize more better than you does now.


--
JF
 
John Fields wrote:
On Fri, 31 Aug 2007 12:42:14 -0500, Skenny
Skenny@highstream.invalid> wrote:


John Fields wrote:

On Fri, 31 Aug 2007 11:02:02 -0500, Skenny
Skenny@highstream.invalid> wrote:




HUH HUH HUH... poser..


---
LOL, that from the idiot with the hammer and sharp chisel...



At least my chisel is sharp unlike yours, oh thats not a chisel you have
there, its a used comdom.


---
"Comdom"???
---


Why are you hanging on to that?
Waiting for that next "special guy"?


---
Yup, and I figured from the tone and timbre of your previous posts
that you wouldn't mind wrapping your lips around something that had
previously been up someone's ass.


That explains why you are such a sore asshole.
Click to expand...
 
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