Yesterday, I tried to get to grips with how a capacitive power supply works. Real samples are a bit messy, though.
So let me try something else this time, and simplify a bit:
The setup is very similar, but I’m leaving out the zener diode and the rest, and more importantly, I’m going to feed a real 50 Hz sine wave signal into this circuit, using my little sine wave generator.
Here again, because scopes need to measure with a common ground, I’m placing that common ground in between the resistor and the cap, and I’m using the scope’s internal “invert” feature to treat this as if the channel 1 probe were connected the other way around. Here’s what we get with this new setup:
The vertical scale of the resistor was adjusted to display the same amplitude for the resistor as for the capacitor.
- the yellow line is the voltage over the capacitor
- the blue line is the voltage over the resistor
- the red line is the sum of the yellow and blue lines
The scale of the red line is not quite accurate, but its shape is. So the red line is in essence the input signal.
So what’s going on here?
As you can see, all these signals are 50 Hz sine waves. That’s quite remarkable already. Obviously the red line is a 50 Hz sine wave, since that’s what we’ve been feeding in. But so is the voltage over the capacitor, the voltage over the resistor, and hence also the current through this circuit!
What you see, is a set of sine waves which differ only in phase and in amplitude:
- the voltage over the capacitor (yellow) lags the input signal (red): it’s forever trying to catch up
- the current through the capacitor, i.e. the voltage over the resistor (blue), is leading in phase
And something else, as we saw yesterday: the current through the capacitor is related directly to the slope of the capacitor’s voltage change (i.e. its derivative). When the yellow line is steepest, the blue line is at its highest.
Let’s throw one more calculation into the mix: power. Power is input voltage (red) times input current (blue):
Looks very sine wave’ish again! There is some amplitude variation, which can probably be attributed to signal asymmetry or amplitude differences between the different sine waves. The phase of this wave is different from the ones already shown, but note that it’s also twice the frequency, i.e. 100 Hz.
Let’s step back for a moment. With a purely resistive circuit (as used in a resistive transformer-less supply), the current and voltage would be in lock step, i.e. sine waves with exactly the same phase, and the power consumption would be maximal (high current at times of high voltage).
With this capacitive setup, currents get “moved around”. That means power consumption will be less than when voltage and current match up (since this gives the largest possible results).
I’ll include one more screenshot, this time using the same vertical scale for all signals (except power):
This puts things more in perspective: the voltage over the capacitor (yellow) is slightly lagging the input signal (red) now. And the input current (blue) is out of phase w.r.t. the input signal (red). So the power consumption (red x blue) is substantially lower than with a resistive circuit: when the current is maximal, the input voltage is only a fraction of its maximum range. IOW, we’re drawing current when it “costs” little.
This is why a capacitively coupled supply is cheaper: the electricity company charges us for real power (i.e. V x I). We’re being charged for what happens inside a pure resistor.
But we’re doing a lot more: we’re taking charge out of the AC mains line on one half of the cycle and pushing it back on the other half. It might seem as if that doesn’t use up energy, but it does: current through a wire causes resistive losses (in the form of heat), and “returning” that energy one half cycle later causes those losses again! So the electricity company sees its electricity turn into waste heat, and they can’t charge us for it.
Here’s a thought experiment: suppose you had a huge capacitor, and hooked it up directly to AC mains, next to the electricity meter. According to what we’ve seen, it’ll track the 230V cycles, with current exactly 90% out of phase with AC mains voltage. Huge currents would flow at the time of zero volts. You’d be charged relatively little for the large out of phase current that flows (not quite zero because a practical capacitor has some internal losses that look like a resistive component from the outside). But your garden would be nicely heated by that current in the resistance of wires between the electricity company and your meter…
You can read more about “real”, “reactive”, and “apparent” power on Wikipedia.
(with a tip of the hat to Martyn for helping me understand this stuff a little better)