| ??? 07/20/06 04:24 Modified: 07/20/06 04:24 Read: times |
#120637 - Three currents arround the base... Responding to: ???'s previous message |
Suresh said:
isn't the direction of base current towards the port pin?
thats why i did 430uA + 70uA, which is obviously going to be the sink current. 
From the schematic you can see, that three currents are flowing arround the base of BC556: 1. Current through the 10k resistor, which is mounted between base of BC556 and port output: (5V-0.7V)/10kOhm = 430µA. This current is flowing into the port pin and is the sum of emitter base current and current through upper 10k resistor, which is mounted between +5V and base of BC556. 2. Current through the upper 10k resistor, which is mounted between +5V and base of BC556: 0.7V/10kOhm = 70µA. 3. Emitter base current, which is the difference of current (1) and current (2): 430µA - 70µA = 360µA. This current is flowing internally of BC556 from emitter to base, means is flowing out of the base and is then flowing into the port pin. Kai |
