7seg display transistor and resistor values.

[evol_w10lv]

I want to connect PIC16F690 (http://ww1.microchip.com/downloads/en/DeviceDoc/41262C.pdf) with 7seg display. Have got this model: ELT-511SURWA/S530-A3 (datasheet: http://www.datasheets360.com/pdf/-3210061775093412092)
Forward current-Max: If = 0.025 A
Forward voltage-Typ: Vf = 2 V
R = (5-2)/0.0250 = 120 Ohms (150 Ohms, if If = 0.02 A)
Resistor wattage:
P = I * V = 0.02 * 3 = 0.06 Watts
Is this correct calculation for current limiting resistor value for anodes of the display?

I'm going to multiplex these segmets (only two of three digits will be used). Idea is to use two NPN transistors to sink current from each of the two common cathodes.
8 * 25 = 200 mA current for the collector. It means I need ~20 mA base current, but maximum output current (any I/O pin of PIC) is 0.025 A. So can I use any NPN transistors which are rated for 200mA collector current? Have read before, when comon cathode 7seg display used, we need PNP transistors.

 

[Arouse1973]

You can do it either way. The NPN transistor will drop a larger voltage through it's base emitter connection than the PNP version. So you will need to have different values of LED resistor to maintain the same LED current if you tried both versions. In the following example the NPN transistor will dissipate 5 times the power as apposed to the PNP version. You will also need to increase the base resistor for the PNP to reduce excess power dissipation due to the larger voltage drop across the base resistor.

Thanks
Adam

Example:
LED VF = 1.58 Volts @ 15.21 mA
NPN transistor (common collector configuration) 2N2222
Base resistor 1K
Base current = 100.8 uA
LED resistor 170 Ohms
LED supply +5 Volts
Power dissipated = 12.6 mW

LED VF = 1.58 Volts @ 14.75 mA
PNP transistor (common collector configuration) 2N3799
Base resistor 43K
Base current = 98 uA
LED resistor 220 Ohms
LED supply +5 Volts
Power dissipated = 2.65mW