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[Help] Using one BJT(BC637) as switch with Arduino digital pin.

Former Member

Hello, i don't have one reasonable experience with electronics component, and i would like to someone with more experience than me to check if my calculation done right.
First i will control one backlight led that the recommended current draw is 22.7 mA.

Looking up the DC Current Gain on the BC637 datasheet and using the formula Ib = Ic/b i calculated:

Ib = 22.7/25

Ib = 0.908 mA

And arduino logic level is 5V to HIGH and 0V to LOW so i should add a current limiting resistor in series on the BC637 base pin.

So using the formula Rb= (Vcc - Vbe)/Ib i calculated

Rb = (5V - 1V)/0.000908A

Rb is aprox. 4405 ohm

 

So i'm planning to use one 4.7k ohm resistor on Rb  so the backlight should be always less brighter (since less current will flow).

Its my calculation done right? It should work the way i planned?

 

Btw here the simplified schematic

Schematic

  • Your close, it is ok to slightly over drive the base as long as you don't exceed its current rating , I cant see the MAX base current in the specs

    Base − Emitter On Voltage is about 1V max (Typically assumed to be a diode forward voltage of 0.6V so this needs to be subtracted from your formula, you used 1V so that's ok

     

    I would not assume you will get the full 5V out of the CPU but it will be close

     

    So just to be sure the transistor is saturated (Fully turned on so less power dissipated in the transistor, I would reduce the base resistor to 3K3 or 2K8 ish. Basically your using it as a switch, not in its linear region so we want to ensure it is fully on or fully off.

     

    You will also need a current limiting resistor in series with the LED

     

    So 5V - Backlight Volts at 22mA / 22mA to give you series resistor value. There is also probably a certain amount of volts dropped across the Collector / Emitter so if you want it perfect then this should also be factored into your equation

     

    Hope this helps

  • The current gain of a transistor is an instable parameter. It will be hard to regulate the current with the transistor alone.

    it's safer to put a resistor in series with the LED that limits the current under the maximum,

    and then fully saturate the transistor as explained by

  • A BJT is a current amplifier but it's not sensible to work with it directly like that. As says, the current gain is very variable. It varies from device to device and it is dependent on other things as well.

     

    The part you've chosen is an old part [the ON website says it's just about to go obsolete]. It was designed for 'high current' use - high current being up to 1A. The 25 figure for the gain is a minimum figure and it will only decline to that at the maximum current of 1A. If you have a look at Figure 2 in this datasheet it shows how the typical gain varies with collector current.

     

    http://www.onsemi.com/pub_link/Collateral/BC635-D.PDF

     

    Further to what said above, Figure 4 shows that if you use it to switch 22mA, the voltage at the base will be about 0.7V and the saturation voltage [collector to emitter] around 30mV.

     

    Since you are only switching 22mA, you might consider using a BC547B or 2N3904, or something cheap and cheerful like that [if you haven't already bought the transistors].

     

    Another alternative is to simply drive the LED from the Arduino. Although the output drivers on the Atmel part are only characterised up to 20mA, the datasheet does say it's acceptable to drive higher currents as long as you don't do it on too many of the pins at the same time [the absolute max for a pin is 40mA]. Note that, at 22mA, you'd get a voltage drop across the lower output FET in the driver which you'd need to take into account with your LED current-limiting resistor calculation.