|
An example of charging/discharging a capacitor inversely.
When the output of the inverter is low (sinking), the voltage above the capacitor (opposite the source) is pulled down toward 0 V. At first glance it would appear that the cap is discharging quickly through the diode and 1k resistor and slowly charging back up via the 100k resistor. That’s what the green waveform is showing isn’t it??
Nope.
In reality it’s exactly the opposite. When the inverter goes low, a path is created for the cap to begin charging rapidly through the 1k. When the inverter goes high, the diode is reverse biased and the now charged capacitor begins to discharge slowly through the 100k.
This is not something that’s easy to see with a scope because the only place you can really measure is on the side of the cap opposite the source which makes it look like it’s charging!!! You could put a bleeder resistor in series with the cap and watch the current, but that’s not easy to do with SMT on a finished PCB. In that case it is something that must be conceptualized instead… a thought experiment if you will.
The simulation is kind enough to show the charging/discharging action of the capacitor with the +/- symbols getting larger and smaller respectively. The concept is backed even further by looking at the simulated current through the cap and the diode. When the voltage above the capacitor goes toward 0 V, the current through the capacitor/diode is almost instantaneously high and then begins to taper back down towards 0 amps as if the capacitor were charging… because it is.
Another way to look at it: when the output of the inverter is high, the voltage on both sides of the cap is equal (for this example), i.e., there’s 0 potential across the capacitor. How can it charge then? It can’t. Why? Because it’s already charged.
|
