(Thanks for your overwhelming understanding w.r.t. not making the Low-power supply available as kit!)
The concept of the transformer-less capacitive power supply still puzzles me – intuitively I still don’t get it:
(the above image was copied from this excellent site with a web-calculcator for it all)
So what exactly is going on here? Tomorrow, I’ll simplify that circuit in an attempt to really get it, but for now let me show you what I see on the oscilloscope. What I did was feed a ≈ 100 Vpp signal into the above circuit, which is essentially the same as in the Low-power Supply.
To interpret the graph, you need some info:
- the scope ground was connected between the capacitor and the resistor
- the yellow trace is the voltage over the resistor
- the blue trace is the voltage over the capacitor
- the yellow trace is inverted, i.e. negative voltages at the top, positive at the bottom
- zero is the middle of the screen for both signals
Here’s the scope capture:
Ignore the fact that these waves are hideously complex, ignore the red lines for now, and also note that watching voltage over a resistor is the same as watching current through that resistor. Voltage and current are always proportional in a resistor, that’s what defines a resistor (Ohm’s law: voltage = current x resistance).
So what are we seeing here?
Well… when the yellow line is high, the blue line rises sharply. When the yellow line is zero, the blue line is flat.
That makes a lot of sense: the resistor is charging the capacitor. And similarly for negative values, it’s discharging the cap (and then charging it negatively).
So ignoring the zener and the rest of the righthand side of the circuit, this is really all that’s going on: when the input voltage is highly positive, current flows in one direction, charging the cap and dropping to zero, and when the input voltage is highly negative, the whole process unfolds in the opposite direction.
One more piece of the puzzle: current is “charge per time unit” (in units: Amperes is Coulombs per second).
In other words, the capacitor accumulates the charge pushed into it, in either polarity. And while it does so, the resistor “takes the heat”, so to speak: it limits the current by creating a voltage drop over itself.
Please let this sink in, dear reader. It’s essential to get a solid intuitive grasp on what’s going on.
Note also that it really makes no difference at all how complex the input signal is. The resistor and capacitor work in tandem, sharing the task of dealing with that signal. In other words: the capacitor is always trying to catch up!
This is where it gets interesting.
You may or may not have given up in high school when it came to advanced maths / calculus – derivatives and integrals, in particular. If so, then get ready to finally get to grips with these incredible concepts.
Let me explain the red lines in the above image – they are generated by the built-in math functions of this scope. One red line is the integral of the value measured across the resistor, and the other red line is the derivative of the voltage across the capacitor. But here’s the big surprise:
- the integral of the voltage over the resistor is the same as the voltage over the capacitor!
- the derivative of the voltage over the capacitor is the same as the voltage over the resistor!
What’s the point? Well, this means that I didn’t have to measure both signals to see what’s going on. I could have omitted the blue trace, because it can be calculated from the yellow trace (and vice versa). Even though these traces have completely different shapes, they are in fact totally inter-locked and inter-related.
As I said before, the voltage over a resistor is proportional to the current through it. So the derivative of the voltage over the cap is the same as the current through it (the current flowing through the resistor and the capacitor is always the same, since they are connected in series).
The derivative is the rate of change, i.e. the slope of the graph. Integrating the current (i.e. the derivative) is like adding “all the little currents together over time. A capacitor is no more and no less than a “current integrator”.
And that’s exactly the same as saying that a capacitor accumulates charge. It’s like a tiny rechargeable battery, it takes the current pushed through it and it stores that current (as charge). As the charge accumulates, the voltage rises. Loosely speaking, this is the same as saying that it pushes back harder and harder against the incoming current. At some point it pushes back so hard, that no more current comes in. At that same point, there will be zero volts over the resistor, and the voltage over the cap stays constant. Check the graph to see where that happens.
One last observation is that the blue line is a lot smoother than the yellow line. That’s not surprising: when you integrate (accumulate) a jittery signal, things tend to smooth out. That’s why capacitors are also a fundamental component in filters, i.e. circuits which let some frequencies through more and others less. That schematic we’ve been looking at here is also known as an RC circuit – if you ignore the zener and the rest. One way to look at an RC filter is to see the capacitor as the sluggish part, and the resistor as taking up the slack. So with any input signal, the voltage over the cap is related to the low frequencies, while the resistor follows more the high frequencies.
Did this explain how a capacitive power supply works? Probably not. But first we need to get to grips with what a capacitor does, and hopefully this little experiment helped you get some intuition for what’s going on.
I’ll try to take this further tomorrow, by simplifying things a bit. Stay tuned!
