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 Author Topic: Checking my resistor values  (Read 529 times) 0 Members and 1 Guest are viewing this topic.
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 « on: June 30, 2013, 10:04:00 am » Bigger Smaller Reset

I really don't know much on electronics and most of what I know to date is self taught. I'm have a difficult time understanding transistors so I'm trying to check that I have the right resistors for my circuit.

I have a Fairchild TIL-122 Darlington NPN transistor that I'm trying to use to drive a Toshiba TLN-108(F) IR LED. The LED can burst at up to 1A so I'm wanting to do that as close as possible, but since people say datasheets tend to be wrong, I'm aiming at 900mA.

The TIL122 has a current gain ratio of 1000. So because I want about 900mA off the collector, that means I need .9mA off the base. Since the arduino pin supplies 20mA, that means I'll need a resistance of around 172ohms. Correct? This will pull the 900mA I need on the collector to power the LED.

What I'm confused about is what resistor I need in series off the collector. Can someone explain how to figure this out?
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 « Reply #1 on: June 30, 2013, 10:23:55 am » Bigger Smaller Reset

The open loop gain of the transistor is only a nominal value.

You might want to consider driving the transistor hard on and limit current through the LED via either a control resistor or a constant current driver.

If using a control resistor, then, based on a supply voltage of 12 volts, assuming a voltage "loss" across the transistor of 0.7 volts and an LED drive voltage of 2 volts, then you need to drop 12-(0.7 + 2) volts across the control resistor viz 9.3 volts.  With a current flow of 900mA the resistor value is 10.3ohms.   Say 10ohms

Power rating will be 8.6watts, so say 15watts

To ensure the transistor is turned hard on I'd suggest sinking at least 10mA into the base, so try using a 390ohm drive resistor

Note also that the transistor will be dissipating 0.9 x 0.7 volts = 0.63 watts so will need a heat sink.

You'd be better using a mosfet which has virtually 0 volts drop so the control resistor will need raising to around 110ohms.  The trigger resistor can be quite a high value, say 4k7
 « Last Edit: June 30, 2013, 10:51:27 am by jackrae » Logged

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 « Reply #2 on: June 30, 2013, 10:29:27 am » Bigger Smaller Reset

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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 « Reply #3 on: June 30, 2013, 10:32:04 am » Bigger Smaller Reset

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.
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 « Reply #4 on: June 30, 2013, 10:55:20 am » Bigger Smaller Reset

As well as what every one else says apart from the resistor power ratings stuff:-
Quote
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:-
Quote
Note 1: Pulse width ≦100μs, 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.

Quote
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?
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 « Reply #5 on: June 30, 2013, 11:36:21 am » Bigger Smaller Reset

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.
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 « Reply #6 on: June 30, 2013, 01:12:28 pm » Bigger Smaller Reset

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.
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