Timeline for Time delayed deactivation of a circuit
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
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| Aug 4, 2017 at 3:18 | comment | added | AnalogKid | True, but the turn-on and turn-off transitions will be much more gradual. The voltage range for a base-emitter junction to go from non-conducting to conducting is much shorter that the equivalent gate voltage range for a MOSFET. | |
| Jul 31, 2017 at 14:14 | comment | added | Jason Han | You're right. If I change the BJT to a mosfet, then C1 and R2 will decide the on time. | |
| Jul 30, 2017 at 2:25 | comment | added | AnalogKid | Yes, but not as long as you might think. The Q1 base-emitter junction prevents the cap from charging above 0.6 V, and the transistor will turn off at around 0.45 V. If you move C1 to the other side of R3, and delete R2, you will bet a much longer ON time for the same sized capacitor. | |
| Jul 30, 2017 at 0:15 | history | answered | Jason Han | CC BY-SA 3.0 |