a couple last questions and then i think i will have all my questions answered.

the main power source on the input side is 12-12.6V, now once the contact is made to close the circuit, is there any 'burst' power that i need to worry about?

type of resistors - ceramic ok? something better? should the resistor W be twice what is needed for acceptable headroom?

thanks all

 

^^^ any input in regards to any 'burst' V i need to deal w/?

thanks in advance,
bob

 

Hello,

Resistors are available in carbon film and metal film.
http://www.piclist.com/images/www/hobby_elec/e_resistor.htm

This is a link from the EDUCYPEDIA on resistors:
http://www.educypedia.be/electronics/resistors.htm

For this application it does not care much.
The power needed for R2 is (U*U)/R = (12*12)/470 = 0.31 Watt.
A 1 Watt resistor will be good.

Bertus

 

thanks for the info & links

 

There actually a few more types of resistors.
In the early days, carbon resistors were most frequently used. These have a cylindrical brown body, and powdered carbon inside. This type of resistor usually does not age well, even when not in use. The resistance usually increases over time.

Carbon film resistors generally have a light tan body and are shaped like a dumbbell. This type resistor is very stable compared to the old carbon type, and doesn't change over time.

Metal film resistors are more stable than carbon film resistors, and generate less noise; so are popular with audiophiles and others who want a minimum of added noise in their circuits. For your application, a carbon film resistor is fine. These resistors frequently have a light blue colored body. They are usually much more expensive than carbon film resistors.

There are also wire wound resistors that generally have low resistance and are used in high power applications. Many of these wire-wound resistors are inductive, but some are specifically made to be non-inductive for applications like current sensing in power supplies.

 

another noob question - somebody (can't remember) brought it up before about V drop from the diode. i read the datasheet and it says 1V drop, so if i have 12V going, should i be doing my math for 11V and not 12V?

also, does the location of the diode and resistor matter? should the resistor be after the diode instead of the resistor first like it currently is?

 

Hello,

You can calculate the resistor with the following formula:
Rcurrent_limiting = (Vsupply - Vdrop_led) / Iled

Here is an overview of leds and the (avarage) voltage drops:
http://www.oksolar.com/led/led_color_chart.htm

This is a link from the EDUCYPEDIA on leds:
http://www.educypedia.be/electronics/composemiconductorsled.htm

Bertus