| ??? 07/19/06 14:40 Read: times |
#120593 - Some thoughts Responding to: ???'s previous message |
Oleg said:
First of all, this voltage level is not safe for pin itself and may destroy it due over voltage conditions and CMOS "latchup" effect as well Hardly, as long as he limits the base current by his 10k resistor. Also, built-in protection diodes prevent worse. Oleg said:
Secondly, transistor will be kept open by this voltage anyway as long as difference above ~0.7V. Yes, but only if a built-in protetcion diode is forward biased. Oleg said:
Another point is "collector goes to ground". This is not ordinary connection of transistors. During such connection about 2,5V or so drops over collector-emitter when transistor opens. So what? This circuit IS a regular circuit, showing a certain advantage, namely that this circuit will forgive turn-off levels, which are not reaching Vcc. These circuits are widely used when it's needed to provide a high turn-on threshold. Not only with the PNP transistor, but also with NPN: 10k resistor between port output and base, collector to Vcc, relay netween emitter and ground: As drive level a voltage is needed, which is much higher than the forward voltage drop of base emitter junction. This circuit can very often be seen and is totally immune against noise. Kai |
