Timeline for Determining how much load capacitance a 40-series logic IC can safely drive
Current License: CC BY-SA 4.0
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| Oct 10, 2022 at 17:23 | comment | added | jp314 | When you are using the diode to charge the cap, most of the enerfy is dissipated in the IC (you can ignore the ~ 0.3 V across the diode). When discharging, presumably the 1k ohm dominates and that portion of the energy is dissipated in the resistor. | |
| Oct 10, 2022 at 15:07 | comment | added | Peter Cordes | (update to previous comment: If the only resistance anywhere is in the CMOS gate, then yes. Charging from 0 to 5V is exactly like discharging a capacitor from -5V to 0 through a resistor; all the capacitor energy ends up as heat in the resistor.) | |
| Oct 10, 2022 at 13:30 | comment | added | danmcb | that's nice. To avoid an extra IC you could also "diode OR" the output of the first gate and the top of C to do the same thing - one extra diode instead (probably same package if using SMD devices). You can also still have a low value resistor across the second diode which charges C quickly enough (whatever that means). | |
| Oct 10, 2022 at 12:46 | history | edited | jp314 | CC BY-SA 4.0 |
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| Oct 10, 2022 at 4:17 | comment | added | Peter Cordes | That amount of energy has to make it through the CMOS chip into the capacitor, but it's not necessarily dissipated (as heat) inside the chip. Unless the chip is the only resistance anywhere in the loop from power supply to ground, or nearly so. (e.g. capacitor ESR and other factors negligible vs. output resistance.) Then yes, since we know the cap won't charge instantly with infinite current, there was an RC time constant with the CMOS gate being most of the R. And it sounds about right that the integral of voltage drop x current across that R would be the capacitor energy. | |
| Oct 9, 2022 at 18:37 | history | answered | jp314 | CC BY-SA 4.0 |