LED. Question.

mike30

Hi there can any one help with this query. I am of limited electrical expertise but I would like to fit an led into my charging
system of my portable electric screw driver, I apologise for what I suspect is a very basic procedure.

1. Charging system plug : Input 240 V ac.
Output 3.3 V. 200 ma.

2. Voltage measured at the screwdriver charging points : 3.34 V ac. with no polarity ?

3. Screw Driver data plate : Black & Decker Type 1 : 2.4 V ac.
1300 rpm.

The insertion of a suitable 'led' is so that I can see at a glance if the charger is working as the charging point is in a
very darkened area. I know move the charger but I just want to complete this very small project.


regards,
mike............

kwinn

You will need a diode, a current limiting resistor, and of course the led. The value/rating of the resistor and diode depend on whether you can connect to the 3.34VAC or the 240VAC. Connecting to the 3.34VAC would be safer and simpler if it is at all possible.

Duane C. Johnson

Hi Mike;

mike30 wrote: »

Hi kwinn thanks for your reply to me query, I am afraid that for me there is still some confusion on my part. You advised a diode but I thought with my very limited knowledge that the led fitted into the charging 2.4V circuit was a diode. And why do I require a current limiting resistor
if the voltage is very small. Will you please take a look at my very simple diagram and comment as you see fit please.

Since your charging power supply is only 3.34VAC you don't need the diode, but you do need a current limiting resistor.
The peak voltage would be:
3.34VAC * 1.414 = 4.72V peak
Most, if not all, LEDs have reverse breakdown voltages higher than this, so no harm no foul.
Start with a ballast resistor of about 100Ω or so for about 10mA average LED current.

Duane J

tonyp12

>You advised a diode but I thought with my very limited knowledge that the led fitted into the charging 2.4V circuit

If you don't the LED to stay on from the power of the battery, you may need to cut a trace and put a diode there.
Even if you have a low voltage that is close to the LED forward voltage, you still should put a limiting resistor, 120-200ohm is OK.

Mike Green

1) Yes, an LED is a kind of diode, but its design is optimized for producing light efficiently, not at blocking reverse voltages (when it's connected backwards or when, with AC, the voltage reverses). If you're connecting an LED to AC, you normally put a standard silicon diode in series with the LED with direction of the diodes the same (cathode of one connected to anode of the other). That way, the silicon diode blocks the reverse voltage.

2) You need a current limiting resistor because, above the threshold (Vf - Forward Voltage) of the diode, the diode will conduct as much current as can be supplied (until it burns out). See this chart for common Vf values. A bright red LED will typically have a Vf of around 1.7V. If you connect that to a 2.4V power supply, it will promptly burn out. If you put a silicon diode in series, the total Vf will be 1.7V + 0.7V = 2.4V. If you connect that to 3.34VAC, the LED will burn out (first). The silicon diode will probably last longer because of how it's constructed. You would usually pick a resistor value that would limit the current to about 10mA.

Resistor value = Voltage / Current = (Power Supply Voltage - LED Vf - Silicon Diode Vf) / 0.01A [ You could use a lower current value with corresponding less light produced ]

With a 3.34VAC (peak voltage) supply, that would be (3.34V - 1.7V - 0.7V) / 0.01A = 0.96V / 0.01A = 96 Ohms (use 100 Ohms as nearest standard value)

You could actually use a higher value for less light, maybe 150 or 200 Ohms.

Duane's calculation for peak voltage assumes that the 3.34VAC value is RMS voltage (root mean square) which is more commonly used than peak voltage for AC voltage ratings.

mike30

Thanks guys for your much appreciated help in what should and probably is a vey simple job, but as they say there are no
silly questions only silly answers.

thanks much appreciated,
Mike