Help on small LED circuits

[chrisp9au (Chris)]

Time to admit to the depth of my ignorance, again!

I need some advice on resistors and circuits. I suspect Bojan will be one of the first to reply.

Using a 12v gel battery, I have successfully rigged up my dob mount to provide 12v power to the fan at the back of my mirror, and also to my David Ek DSC box, with switches on each. I've also provided for spare 12v and 5v outlets, anticipating power requirements for an Air Cable to provide bluetooth comms from the DSCs to my laptop. Up until now everything works, and I haven't smelt any smoke!

Of course, I now want to complicate things! I would like to add LEDs to indicate when my switches are on/off.

My research suggests that most small red LEDs work on 2 volts and 20 milliamps.
Am I right in thinking that I need 560 ohm resistors on the positive side of the LED?
I understand these are identified by green, blue and brown bands?

I've checked the small supply of resistors that I have against an online chart, but can't understand which end of the resistor to start at!
When I tested what I thought was a 560ohm, 4 band, green, blue, brown or gold, gold, I got smoke!

I guess I'll need a trip to Jaycar in Bendigo, or is there a reliable online source?

2nd question, my battery reads 12V7.5Ah/20HRS, what does the 20 hours signify?

Any and all advice would be appreciated!

Cheers
Chris

[kinetic (Steve)]

Sorry, I'm not Bojan

http://led.linear1.org/1led.wiz

Use 1.5v for a LED...colours vary in voltage but
1.5v is typical.
20mA is very bright too...10mA is better in a dark dome
FWIW

Steve

[RB (Andrew)]

I've checked the small supply of resistors that I have against an online chart, but can't understand which end of the resistor to start at!

Chris the Gold band signifies the resistors tolerance and is at the end of the number chain, so hence you read the values starting from the opposite end.
Green, blue, brown = 560 ohm

[chrisp9au (Chris)]

Thanks guys,

1.5v, 10ma = 1.2k ohms, brown-red-red-gold, I'm learning!

Cheers
Chris

[RB (Andrew)]

Originally Posted by chrisp9au

2nd question, my battery reads 12V7.5Ah/20HRS, what does the 20 hours signify?

Any and all advice would be appreciated!

Cheers
Chris

On your second question, I found this from a pdf, hope it helps.

Typical ratings for the discharge of nickel-cadmium batteries areover 10 hours and for lead-acid batteries are over 20 hours. Asa result, a rating of 4 or 6.5 ampere-hours for a battery does notmean you can draw 4 or 6.5 amperes from the battery for onehour. The actual amount at the one-hour rate is typicallyone-half the “rated” capacity of the battery, or less. To determinethe actual battery rating, you must take the nominal capacity(4 or 6.5 ampere-hours) and divide it by 10 or 20, depending onthe battery type.In the case of the Model 934 Nickel-Cadmium Battery, you use afigure of 0.4, and for the Model 946 Lead-Acid Battery you willuse 0.325. Customarily, currents below 1.0 ampere are expressedin milliamperes (mA). This means you can expect to dischargethe nickel-cadmium battery at 400 mA for 10 hours. You wouldexpect to discharge the lead-acid battery at 325 mA for 20 hours.Battery manufacturers provide curves in their literatureshowing the discharge rates in percentages of “C,” which standsfor rated capacity in ampere-hours (Ah). If you discharge thebattery at rates greater than the 20-hour rate, you can drawmore current, but for a considerably shorter time; that is, thevalue of time multiplied by current will amount to much lessthan “C.” For more than 20 hours, you can draw less current for alonger time.For times beyond 20 hours, the product of current and time willamount to more than “C,” at least, up to a point. The value of“C,” or more likely a percentage of it, is also used to describe theproper charging current for a battery.

[chrisp9au (Chris)]

Thanks Andrew!

Chris

[kustard (Simon)]

In general, you can use the following formula to work out what resistor you need for your LED.

R = (V - Vf) / I

Where:
V = supply voltage (ie 12V)
Vf = forward voltage of the LED (usually 1.5V to 2V for a red LED)
I = desired current through the LED.

For your 12V @ 10mA you'd end up with:

R = (12 - 1.5) / 0.01
R = 10.5 / 0.01
R = 1050 ohms (1K rounded down or 1K2 rounded up)

What you'll find is that Steve is quite correct, and you might even find that 10mA is too much depending on where you are placing the LEDs and their direction (right into your eyes for example).

What you can do is invest in a 5Kohm potentiometer and wire that up with your LED then at night adjust the pot until you get an acceptable brightness. Measure the resistance across the pot and find the nearest standard value resistor....

...or along with the 1.2K (1K2) resistor grab a 1K5 and a 1K8 as well and try those as well.

[GrampianStars (Rob)]

12V 7.5Ah/ 20HRS = continuous use of 10 - 20 hours of battery @ 12V 7.5A/h

[chrisp9au (Chris)]

Thanks everyone, knew I would get lots of good advice.
The LEDs are going to be down low on the dob, around knee height, nowhere near eye level, so I can't forsee a problem there.

Cheers
Chris

[mithrandir (Andrew)]

I've just stuck a red flashing LED across the input to the power distribution box I'm building, after the circuit breaker. It proves the battery is connected the right way round and the circuit breaker has not tripped. These LEDs are rated 3-10v and 60mA at 25% duty cycle.

I've stuck a 1k1 resistor in series and with the 12.5V from a lead acid battery it appears to use about 3mA. That gives 12.5/0.003 ~4k2 total.

1100/4200*12.5 is 3.2V across the resistor and (4200-1100)/4200*12.5 or 9.3V across the LED.

Maybe I could use 2k1 which would make it 2.3mA and 7.5V across the LED.

Is my math right?

[rally]

For an extra dollar or so you can buy "12v" LEDs already mounted in a bezel for easy mounting.
The resistor must actually be inbuilt - so you only have polarity to worry about and that is easy - mine came with a black a red leads.