3 ms·
The PN boundary (assuming that's what you are trying to explain) doesn't work like that. The current at given voltage is determined by the work which has to be
by varjag 8y ago
The PN boundary (assuming that's what you are trying to explain) doesn't work like that.
The current at given voltage is determined by the work which has to be done for switching the load. For signal electronics, we're talking microamperes here, that's not anywhere near "high current". Even the TTY loop I mentioned, a 1920s tech used to drive large relays and solenoids directly is just 40mA.
- thesz 8y agoThe load is either 0V or Vcc. Which is 0 and 5V and 0 and 12V and you can easily guess everything else. Also look here: https://en.wikipedia.org/wiki/Low-power_electronics https://en.wikipedia.org/wiki/Low-power_electronics Energy dissipated by change of voltage in the capacitor is E=CU^2/2. Now I stop.