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I am evaluating a new controller and see a circuit I do not quite understand. I've built the circuit and I see the following (see image), but not understand why. Can someone please give me the name of this circuit so I can look it up?

Bottom line, I do not understand how the cap and resistor in parallel = DC.

enter image description here

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  • \$\begingroup\$ Such supplies are lethally dangerous BUT as lethally dangerous supplies go, that example is probably the very best implemented of its sort that I've ever seen. The person who designed that supply DESIGNED it, they knew what they were doing, how it worked, what they wanted and how to achieve it. It's 'astoundingly good'. The 14 VDC label is misleading as that point sees half wave positive DC of about 30V and negative half cycles of about -0.6V. | What is it out of? That was 2+ years ago - how did they work out? \$\endgroup\$ – Russell McMahon Oct 5 '16 at 12:26
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It is a capacitive power supply with a half-wave rectifier. Most of the grid voltage is dropped on the capacitor due to its impedance. The rest is rectified and regulated using a Zener diode.

You must be careful when using this kind of power supply because it is not isolating the grid from low voltage side. Also remember that special (X-rated) capacitor is needed in order to provide minimum security.

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  • \$\begingroup\$ @becjaslAs Szymon notes - the capacitor is "special" - and they have noted the actual capacitor to use it's listed here at Digikey. 630V Polyester metallised. \$\endgroup\$ – Russell McMahon Oct 5 '16 at 12:31
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...Otherwise known as a reactive dropper.

Cheap, but dangerous. If the coupling cap goes short your equipment catches fire!

I was once involved in investigating an insurance claim due to a house fire caused by a PIR security lamp with such a circuit. The company that produced the lamp were sued for a lot of money!

Ways to make is safer are to use 'fusible' resistors and a clamping diode as well as the correct type of coupling capacitor. That way if the cap fails, the current rises, flows through the clamp diode and safely burns out the resistor.

I will reiterate the need for 'self healing' Class X, or even better Class Y capacitors mentioned in other answers. Even better, unless cost is really the driver, scrap this circuit and use a proper off line SMPS.


To answer your later question, "How the cap + resistor in parallel = DC" it does not.

A capacitor is like a frequency dependent resistor. As you probably understand two resistors in series across a voltage will produce a 'potted' down lower voltage at their centre.

This is just the same, but instead you get a lower AC voltage. The resistor in parallel with the cap is there to discharge the cap for safety reasons, it plays no part in the circuit operation. The other 'resistor' is the remains of the circuit, diodes etc.

You could do this with two resistors, rather than a 'resistor' and a capacitor, but you'd need a physically large resistors due to the power dissipated.

The cap solves this problem because it's reactance dissipates zero power (actually it has parasitic resistances which do dissipate a little power).

Think of it this way; A capacitors' reactance (AC resistance) is Zc=1/(2*PI*f*C) at 50Hz or 60Hz a 1nF capacitor will be about 3MOhm, so you'd get very little current through it at 120Vac.

This one is 1.2uF, so is about 2k2 Ohm at 60hz and it has 105V across it. If it were a resistor it would dissipate (V^2/R) = 5W, yet as a capacitor it dissipates virtually zero.

So we just get a smaller 'potted' AC voltage which we then rectify with diodes (D1, D2) (to remove the negative part of the cycle). Now we have (very lumpy) DC which can smooth with a capacitor and regulate. The last component is a 5V linear regulator. Z1 is there to protect the regulator from voltages spikes which make it through the coupling cap.

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