??? 08/22/09 02:04 Modified: 08/22/09 02:37 Read: times |
#168452 - I would use a PMOSFET Responding to: ???'s previous message |
Hi Asok,
I would use a suited PMOSFET (e.g. IRF7104) instead of BC369, with a 4k7 resistor between source and gate. To this gate connect the drains of two NMOSFETs (e.g. BS170), via 1k resistor each. One of the BS170 is at the left section, the other at the right. Both can turn on the IRF7104, controlled by their corresponding circuitry. Connect the gates of BS170 via 4k7 resistors to the outputs of two additional 74HC04 inverters, with their inputs connected to the corresponding port pins of 8051 micros. On the left, a low state at port output of micro turns on the corresponding BS170 and by this the IRF7104. This will make the +5V supply rise and by this turn on the right 8051 micro. After finishing the reset of this micro, toggle the corresponding port output low, which will turn on the right BS170 and by this the IRF7104. After pushing the off button, toggle the corresponding port output of the right micro high, which will turn off the right BS170 and by this the IRF7104. When the supply voltage of the right 8051 micro falls, the reset chip will reset this micro, resulting in the ports emitting high state. This keeps the right BS170 turned off until the supply voltage is so small, that the right BS170 cannot be turned on again by an undetermined output state of right inverter (74HC04). Connect a pull-down of 4k7 to the output of right inverter (74HC04), a pulldown of 10k to the input of 7805 and two pullups of 10k to the inputs of inverters (74HC04). Eventually, add a LED directly to the output of the right inverter, to force a voltage drop of about 2V. Kai Klaas |
Topic | Author | Date |
PNP SWITCH | 01/01/70 00:00 | |
I would use a PMOSFET | 01/01/70 00:00 | |
MOSFETS | 01/01/70 00:00 | |
MOSFETs help to save supply current | 01/01/70 00:00 | |
stock depleted :-)![]() | 01/01/70 00:00 |