??? 07/18/06 18:49 Modified: 07/18/06 18:51 Read: times |
#120510 - sure.but why? this is \"grab bag values\" Responding to: ???'s previous message |
I could not sort out the amount of voltage drop accross the 10K connected between B-E of BC 556 (assuming emitter current = 10mA). Also unable to calculate the voltage drops in the 1K2 resistors.
Could you help me sort it out? I would consider all in that schematic "grab bag values" If I have to drive a transistor that need to drive 5 mA, I GRAB a 10k. For anythibg less than, say, 100mA I never calculate, just make sure that the "grab bag value" is enough to saturate the transistor, Kais 'rule' of 1/10 of collector current sounds OK and leave enough headroom that you can ignore saturation voltages and such. So, if the 10k is from the PNP base to the port pin it will probably supply something like 0.4mA (5V - two diode drops across it) which is more than enough (relay current/100 (10 * 10), the grab bag calculation, ignoring two diode drops give 0.5mA but who cares about that, there is plenty of headroom. NOW, IF this is something that is risetime critical or such, it is another story, but driving a relay - come on. Erik. |