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Topic: Checking my resistor values (Read 595 times) previous topic - next topic

Grumpy_Mike

So you will not be driving it with such a short duty cycle then so stick to the maximum continuous current of 100mA. Then the resistor calculation will produce a sensible resistor value. Use the next highest standard value from your calculation.

kenishi86


However, the whole concept of driving the LED so hard is wrong, what are you doing with the LED that needs driving it for so short a duty cycle and as hard as this?

It'll be mounted on the wall, above the door, across the room for remote controlling of my TV, AC, and lights. All via IR.

Grumpy_Mike

As well as what every one else says apart from the resistor power ratings stuff:-
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but since people say datasheets tend to be wrong, I'm aiming at 900mA.

Data sheets are not wrong but you need to understand them. That 1A value is given under absolute maximum but here is a note:-
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Note 1: Pulse width ?100?s, repetitive frequency = 100 Hz

This means to achieve this the LED needs to be on less that 0.1mS and off for at least 10mS, that is a 100 : 1 duty cycle. Can you ensure this happens? This duty cycle means you can also reduce the power ratings of the resistor to 1/100th of the constant current values.

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Since the arduino pin supplies 20mA, that means I'll need a resistance of around 172ohms. Correct?

No it will only supply 20mA if you let it. You only need to use a 1K resistor or so, it is not critical.

To get 900mA you need at 5V to have a series resistor of:-
Supply - Vce - diode forward drop = voltage across the resistor
5 - 1.4 - 1.3 = 2.3V across the resistor
So the resistor needs to be:-
2.3 / 0.9 = 2.5 Ohms

That is a silly value and points you to the fact that you are doing it wrong. In fact with large currents down an LED you need a constant current supply to drive it.

However, the whole concept of driving the LED so hard is wrong, what are you doing with the LED that needs driving it for so short a duty cycle and as hard as this?

tack

Drive voltage via an Arduino pin will be 5V

A Darlington will typically have 1.4v drop across the Base/Emitter junction.

The LED forward voltage is required to complete any calculation.

tack

You'll also need the forward volt drop of the IR LED.

Also, you'll not want to control that 0.9A just by calculating Base current for gain. You want to fully saturate the transistor, to use a a switch, and then use an appropriate resistor in the CE leg.

Since you want 0.9A you'll also have to consider the power being dissipated in that resistor, as it's likely to be well above a standard 1/4 or 1/2W resistor, probably more like a couple of Watts.

If you try using the transistor outside of saturation then you'll be effectively trying to dissipate that power int he transistor which means it will be getting very hot and probably dying quite quickly.

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