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When power is turned off, VCC and ground are at the same voltage, ...
Ok I'm self taught so bear with me here,
The Positive side of the capacitor goes to ground, but what is happening at the negative side. Doesn't the anode of the diode need to be more positive than the cathode for it to be forward biased?
During the regular operation of my circuit I will need to periodically send pin 4 low. Is it ok fo me to put a OR gate between C1 and R1 so i can turn on the transistor.
...
P.S. MikeMI... like the plane, I fly a 172S (rented) out of 1B9.
Think of it slightly differently. The negative end of the capacitor is clamped one diode drop (≈0.65V) below ground. Ground is always ground
Because you cannot change the voltage on a capacitor in zero time, right after the input to the supply is turned off, C1 (and the filter capacitor in the powersupply) holds Vcc at almost the previous supply voltage. C1, and the supply's filter caps slowly discharge toward ground. When the dust settles several seconds later, C1 and the supply's filter capacitor are both left discharged...
hi MIke,
The circuit description on the posted drawing is for a power ON, RESET pulse.
At power up, the cap charges via the diode ...
... When the cap is charged the Vbe ON disappears and the RESET line goes high.
IMO a 10K from the RESET pin to +V and 1uF cap to 0V would be just as effective and is commonly used.
The duration of the reset pulse would be much shorter that way, 6ms vs about 60ms the way the posted circuit does it. The reason you might want the longer time constant is because the turn-on slew rate of the main power supply might be slower than 6ms, therefore you want the RESET pulse to be there longer than it takes the supply to come up.
Would there be any danger in just increasing the value of the resistor
and/or capacitor to get a longer time constant.
Haxxx.
Would there be any danger in just increasing the value of the resistor
and/or capacitor to get a longer time constant.
Haxxx.