Could someone please check my understanding?
Given: Vcc = 3.6V on the EXP-MSP4320G2 board
First off, from the article, "When picking a transistor, its datasheet must be analyzed for its characteristic values.".
Any idea why was the PN2369 chosen? I'm not sure what characteristics the author was looking for when he chose that part.
Second, Ra is the current limiting resistor. Formula: Ra =(Vcc - Vf - Vcesat) / If
Vf and If are from LED datasheet, let's call it 2V and 10 mA using the article examples.
Vcesat is from the transistor datasheet. For the PN2369 used, the datasheet says that it's 0.25V max.
Plugging in the numbers, we get Ra = (3.6V - 2.0V - 0.25V) / 0.010 A = 1350 ohm
Is this right -- that is, do I plug in the maximum Vcesat value from the datasheet?
Third, Rb is the base resistor sitting between the transistor base and the MCU GPIO.
From the formula, Rb = [(Vcc - 0.3) - Vbe] / Ib
Where (Vcc - 0.3) comes from the processor datasheet, Voh = high-level output voltage on output ports.
Vbe is given in the transistor datasheet as 0.7 V min, 0.85 V max, so let's call it 0.8 V
Beta is given in the transistor datasheet as 5 (listed as Hfe)
Ib is the current the MCU can drive (source) and the reason we're fussing with the transistors in the first place.
Ib isn't given, but in the linked article and the datasheet (for the msp430g2553), we read that "The maximum total current, I(OHmax) and I(OLmax), for all outputs combined should not exceed ±48 mA to hold the maximum voltage drop specified." Let's assume worst case and have 24 output pins, which would give Ib = 48 mA / 24 output pins = 2 mA
This gives Rb = (3.6 - 0.3 - 0.8) / 0.002 = 1250 ohm
But in the article, a 2K2 resistor is used. Where did I go wrong?
Fourth, if I work backwards and compute Ib = Ic / Beta and use Beta = 5 (from transistor datasheet) I still get Ib = 2 mA and the same result as above.