Calculating Base Resistor for BC549 NPN Transistor Driving Paired LEDs with PIC18F46K80

Hi Guys,

New to this forum, hope this is a good place to come for help with this sort of stuff.

I am a pretty involved EE hobbyist and and am designing a unit to interface with automotive electronics over a CANBUS. I am learning massive amounts as I go and really enjoying this challenge.

On the particular MCU I have chosen (PIC18F46K80) I have decided I want to have a series of warning LED's, as well as the ability to drive a standard automotive relay (in the future anyhow)

LED's I have are grouped in pairs, and draw around 25ma each at 2.2v. I am currently running each one with a 150R resistor.

Max current sourced/sunk by the PIC on some pins is 25mA, 8mA or 2mA, so by putting them into pairs (parallel) I am saving I/O pins but increasing current draw, meaning I cannot supply directly from the MCU anymore.

So I am going to use a basic NPN transistor as a switch, *which is a first for me.*

The unit I have on hand is a BC549 and has the following characteristics:
Ic (max) 100mA (suitable for the LED's)
Hfe min 420
VCEsat 200mV at IC = 100 mA; IB = 5 mA
VBEsat 900mV at IC = 100 mA; IB = 5 mA

What I am struggling to understand is how to equate the base resistor value. I have read probably 20 different web theories and still do not fully understand, and have come up with a few different values as a result.

Both the LED and the IC (connected to base) operate off the same VDD of +5v, and the pin I want to control the transistor from has a Ic max of 2mA, but I do also have a pin with Ic max 8mA also.

Can anyone point me to some clear-cut theory or help me to understand this a little better?

One website I found most helpful was http://electronicsclub.info/transistorcircuits.htm
But I am obviously missing something here.

Thanks,
Hayden

http://www.nxp.com/documents/data_sheet/BC549_550.pdf

I assume your circuit is a common-emitter amplifier.

The datasheet parameters you cited tell us that to achieve a Vce of 600 mV (design to the max value, not typical) at 100 mA the base must be driven with 5 mA. We say that a "forced beta" of 100mA/5mA = 20 is used.

Looking at page 4 of the datasheet (the link is above) we see that the transistor starts to run out of gain well before 100 mA is reached. If you can drive your LEDs at 10 mA page 3 of the datasheet has a direct datapoint you can use. It shows Vce of 250 mV max at 10 mA for 0.5 mA of base current. For current between 10 mA and 100 mA we can interpolate and use a forced beta of 20.

http://ww1.microchip.com/downloads/en/DeviceDoc/39977f.pdf

To drive 25 mA with a forced beta of 20 the base current is 1.25 mA. Looking at page 556 of the PIC datasheet we see that PORT D-G can drive 2 mA with a voltage of Vdd - 0.7 volts when Vdd is 5.5 volts.

At your Vdd of 5.0 volts and 1.25 mA of drive we don't have exact data. But based on the data we have let's calculate the base resistor like so:

The PIC output of 4.3 volts driving a transistor having a Vbe of 0.7 volts. The current to drive is 1.25 mA.

R = (4.3V-0.7V)/1.25mA = *2880 ohms*. I would use a 2.74k or 3.32k ohm base resistor and call it good (and conservative).

The LED series resistor is (5.0V-2.2V)/25mA = 112 ohms.