Can I drive a 5v LED without a current resistor?

[Bruce]

For a standard LED, use the series limiting resistor. If you're short on space, you
can find LEDs with built-in current limiting resistors.

[sayzer]

If your program logic allows you to do, you can pulse it or pwm it at or below its operating voltage.

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[wellyboot]

I assume a 5v LED would have a built in resistor as standard LEDs are around 2v.

Just need to check the datasheet to make sure that the current flow of the LED is within the PIC limits (normally 25mA)

[Bruce]

If your program logic allows you to do, you can pulse it or pwm it at or below its operating voltage.

Even with a very short duty-cycle, you're still over-driving the PIC output pin, and the LED
during peak-pulse times. I for sure don't recommend you do this without the series resistor
when directly driving the LED with a PIC output pin.

With a transistor driving the LED, you may be able to get away with high peak-pulsed
current levels, for short durations, but I would never try this directly driving the LED from
the PIC output pin.

I assume a 5v LED would have a built in resistor as standard LEDs are around 2v.

I wouldn't assume the LED had a built-in resistor. I would check the data sheet for the LED
first. If you have an LED with a forward voltage drop of 2V, and a MAX forward current of
10mA, then just subtract 2V from 5V for the working voltage, then /10mA to find the series
resistor value.

I.E. 5V - 2V = 3V / 10mA = 300-ohm series resistor you'll need to operate the LED at 5V
with 10mA current through the LED.

What if you need to operate this same LED with a 12V supply? Same thing applies. 12V -2V
= 10V. 10V / 10mA = 1K. So you just pop in a 1K resistor. Most LEDs don't care what the
voltage is, you just have to limit current through the LED at whatever voltage you're using.

Now, if you want to use PWM to dim or brighten the LED, your peak-pulse current levels will
not exceed the LED max DC current levels during the peak-pulse periods.

If you drive the LED with a 50-50 PWM duty cycle, your time average current will be
50% of 10mA, but your peak-pulse-current will still be the full 10mA.

If you exceed the MAX drive levels for the PIC or LED, even for brief periods, you can
pretty much count on failure at some point.

Seems pretty silly when you could have prevented total failure, or degredation over a
period of time with a 1-cent resistor...;o}

[Melanie]

I have a design in which a PIC drives two seven-segment displays directly. No segment is ON more than 1mS in any 15mS period (allowing all 14 segments and the Decimal Point to be cycled across 15mS). I dispensed with the Common Anode Resistor in each display saving two Resistors on the Basis that having the LED ON for only 6% of the time was an acceptable risk (with the advantage of a brighter display) and the product was unlikely to fail within the warranty period. It wasn't the cost, but the labour in their insertion that was the consideration in a cheap $20 product. 25,000 sales across six years with zero returns - I can live with that.

[ardhuru]

I dispensed with the Common Anode Resistor in each display saving two Resistors on the Basis that having the LED ON for only 6% of the time was an acceptable risk

Same technique used in this very versatile project; http://members.cox.net/berniekm/super.html

I have built this, and been using it for a couple of years with no problems either to the pic, or the display.

Regards,

Anand Dhuru

[wellyboot]

I dispensed with the Common Anode Resistor in each display saving two Resistors

wouldnt having the resistor on the common anode cause you to get different brightness for different digits?
ie 8 would be dimmer than 1 as the current would be split between 7 segments to display 8, and only 2 segments to display 1.

i normally put the resistors in series with each segment, and use a transistor on the common pin