@itsnotworkingatall If you place the LED in series with the resistor and connect the transistorās emitter to GND, the LED will light up at lower voltages (approx 0.6). In the schematic drawn it will only light up at 0.6 + LED-voltage which could be as high as 3.6 Volt. By changing the position of the LED you will see it produce light at lower input voltages. The LED current will remain approximately the same as in the original schematic.
actually I have been wondering this to . guess we could try to look it up somewhere 
. dam didnāt realize there would be so many answers to the same question , but I think you guys cover it well . now to figure out which answer applies to my particular situation .
and it certainly has, I was thinking just the same, as I wanted more flashing lights on the front bottom panels of my rig. now whereās that soldering iron. Hmm what power resistor though? or is that like how long is a piece of string?
A quarter watt until proven otherwise.
Power is voltage times current.
In a 12V system like we deal with here, you can safely pass up to 20 milliamps in a quarter watt resistor.
So if youāre not wasting those precious milliamps, you should be fine with quarter watt resistors.
I added this circuit too my CGS joystick and it is working well.
When the Red LED lights up when its 0 too +5v And the blue one lights up with 0 too -5v
Strictly speaking the maximum potential difference is 24V. Iād have to wonder why anybody would pass that kind of juice between the positive and negative rails, though.
Ok, it is a schematic diagram, so in principle it does not matter at all, but ā¦ there are conventions. To draw LED 1 in this way, with ground pointing to the top of the schematic is really odd. This makes the schematic more difficult to read. I just noticed that the LED was not connected to +Vdd, but to ground. Iād advice to follow convention and always draw GND (and -Vcc) at or towards the bottom side of the schematic and +Vdd at or towards the top. B.t.w. you could omit one of the resistors R2 or R5 and connect the LEDs anti parallel to one another. This is also better for LEDs with low reverse voltages. The anti parallel LED will prevent the voltage from getting too high.
Whoops about having ground pointing up. Lol that is pretty funny the led offset subcircuit is the only section with ground on the top. Also thanks for telling me about getting rid of one of the resistors no need for that too be there if it is not needed. Someone also told me about Bicolor leds Which is a great idea too.
I have a bunch of those bicolor leds, but those wouldnāt fit your purpose. They have 3 leads and their cathodes (or anodes, Iām not sure at the moment) are connected together, so you canāt use them in an anti parallel fashion. But there may be other types of packaging.
There are 2-lead bicolors which are essentially antiparallel red and green.
sorry what I should have said was what impedance, or does that depend on the circuit or led?
Basic rules:
- the smaller the resistor the brighter the LED
- too small a resistor and you blow something up
Super simple conservative rule of thumb:
Start with a standard resistor value close to 200 times the voltage you are driving the LED with.
E.g. 12V x 200 = 2400, use a 2k4 resistor
If that is not bright enough use a smaller resistor.
That rule of thumb gives you a current through the LED which should be safe in most cases.
If you are driving the LED with an external transistor and want a brighter LED, you can lower the resistor significantly (like divide it by 4), but if you are driving multiple LEDs directly from a microprocessor such as the Arduino, you should probably do some more reading before lowering the resistor values.
Iāve not hear of this rule of thumb before. I prefer to calculate the resistor needed, which is not very complicated.
R-series = (U-power - U-led) / I-led
U-led is the forward voltage of the LED, which depending on the color of the led can be 2.x to 3.x Volts.
U-power is the DC powering voltage of the circuit the LED is connected to ( Iām assuming +V -> series resistor -> LED -> GND).
I-Led is the max forward current the LED supports. Something like 10 - 20 mA.
So at 12 Volts a LED with 3 Volts forward voltage allowing a current of 10 mA needs a serial resistor of
(12 - 3) / 0.01 = 900 Ohm
for a current of 20 mA this will be
(12 - 3) / 0.02 = 450 Ohm
The power the resistor consumes P = i^2 * R = 0.09 W in the first and 0.18 W in the second example.
So for the first example a 1/8 watt resistor will do fine, for the second strictly speaking youād need a 1/4 watt resistor (or allow a 1/8 watt resistor to go hot).
In my experience modern LEDs are so bright, that a single signalling LED can light up a room, which is too bright to my taste, so I go for much lower currents (thus higher resistor values, like 5 to 10 kOhm).
Yeah, I just made it up.
Exactly why using the maximum forward supported current is not the best starting point and why I chose a value which would give less than 5mA.
indeed! I just copy and past that when I need it haha
yes I also do exactly the same schem for example on VCA which did not have led indicator