>>>>>> I would like to add lights to a loco using a dcc chip.
>>>>>> What can I use? and how do you do it.
[quoted text clipped - 32 lines]
> Regards,
> Greg.P.
"> As I understand it, and as I have discovered, LEDs like all diodes have
> a rated voltage, and will blow if you expose them to much more than that
> voltage, which for most LEDs is around 1.5 to 2.5 volts. That is, the LED
> will block up to about the rated voltage. Since directional lighting
> circuits generally rely on the current blocking function of the LEDs,
> exceeding the rated voltage is a bad idea.
No you are confusing 2 different parameters, all diodes have a reverse
voltage rating, and a forward voltage voltage drop.
The reverse voltage is how high a voltage it will withstand when NOT
conducting. This is important for rectifier diodes, but not relevant to
LED's unless you are running them from an ac supply or, as you say, reverse
biasing them for directional lighting. LED's only light up when they are
conducting.
LED's do not have a rated votlage for operating at. The forward voltage drop
is just that, how many volts dissapear in the diode. For a rectifier diode
it might be 0.7V or even less, LED's vary with colour any may be anywhere
from 1V to 3V. It also varies with current. Obviously the applied voltage
must be greater than the voltage drop for the led to light.
LED's are when forward biased (ie lit) have effectively a low resistance and
will try to take as much current as they can. So unless there is something
limiting it they will take as much as the psu will give until they destroy
themselves.
> Also AIUI, the current capacity of an LED depends on its size.
Correct
<At any
> given voltage, the LED will absorb a certain current. At higher voltages,
> higher currents are possible, so again, it's a bad idea to exceed an LED's
> rated (or working) voltage.
No things don't work like that. If you have a simple circuit of a psu, a
resistor and then an LED to ground what you will see is this.
If you measure the voltage between ground and the junction of the led and
resistor you will *always* see just the forward voltage of the led
regardless of the current through the led . (The forward voltage will change
slightly with current but not by a lot).
The resistor is limiting the current through the LED. Lower the resistor
value and the current will go up, the LED will get brighter, but the voltage
across the LED will remain much the same. So you see you cannot, as such
exceed the forward voltage of the LED it is just a parameter the the led
has. Waht you must not do is put too much current through the led and exceed
its forward *current* rating.
The situation is different if you reverse bias the LED, (not lit). Since
virtually no current is flowing the voltage seen by the led is not limited
by the resistor so if the psu voltage exceeeds the *reverse* breakdown
voltage then you may damage the led.
> Further AIUI the brightness of an LED depends on its composition, which
> also determines its colour. IOW, the amber LED will have a lower
> brightness regardless, simply because less of the current is converted in
> photons than in a green or red LED.
More or less correct.
> For the above reasons, I understand voltage to be the key rating for LEDs
> used in model railroading. A lighting engineer would of course be as
> concerned with current rating.
Voltage is of little consequence in the application of led's, current is the
key!
You only need consider voltage this way:
your supply must be greater than Vf (forward voltage) in order to get it to
light.
If you are reverse biasing it (directional or ac supply) Vr (reverse
voltage) must be greater than the supply
Brightness is dependant on current so the series resistor must limit the
current to a safe value, less than Ifmax.
Vf is of some use when you are calculating the value of the series resistor
if you wish the set a particular current.
Say you have a 12V, and an led with a 2V Vf. Subtract Vf from the psu
voltage (12-2=10V). That will give you how many volts need to be dropped
across the resistor. Then use Ohms Law to work out the resistor value for
the current that you require. Let's say 10mA (0.01A) , so R=V/I 10/0.01 =
1000ohms or, 1kohm.
Hope this helps
Regards
Jeff
Jeff - 26 Nov 2007 09:54 GMT
> No you are confusing 2 different parameters, all diodes have a reverse
> voltage rating, and a forward voltage voltage drop.
[quoted text clipped - 4 lines]
> reverse biasing them for directional lighting. LED's only light up when
> they are conducting.
I forgot to add that the Reverse breakdown voltage Vr for LED's is greater
than the Forward Voltage drop, usually in excess of 5V. If you do wish to
reverse bias them, for directional lighting for example, then be careful not
to exceed this value. If you do have a problem, the solution it to add a
rectifier type diode in series with the LED. Rectifier diodes have Vr
anywhere up to 1000's of volts!. Don't forget to allow for the additional
forward voltage drop of the rectifier diode when calculating resistor
values, this can be in the range of about 0.5 to 1V.
If you are running leds from a pulsed supply, (eg DCC) then they may not
glow as brightly as you expect because of the pulsed nature of the supply.
The max forward current allowed is usually higher for short pulses, so that
may be taken into account.
Another way around it may be to add a series diode as above, and a large
capacitor across the diode to smooth out the voltage seen by the LED,. This
will also help if the led appears to flicker because of the pulsed supply.
Regards
Jeff
Wolf Kirchmeir - 26 Nov 2007 14:09 GMT
> "> As I understand it, and as I have discovered, LEDs like all diodes have
[snip discourse on LEDs]
> Hope this helps
It does indeed. Thanks.
> Regards
> Jeff