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I have some trouble figuring out a power supply. I understand most parts of it, but have some doubts. Is D2 and D4 both zenerdiodes to get the wanted voltage at the output? And how does the switching of the transistors for the relay function?

Thanks in advance!

Power supply

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    \$\begingroup\$ Why is it powered from live and neutral? \$\endgroup\$ – Jason Han Aug 15 '17 at 7:26
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    \$\begingroup\$ @JasonHan This circuit could for example be used to switch a mains load (like a light) through the relay based on input from a sensor (like a motion sensor, which needs 5 V, output of IC1). To keep it small and cheap it is directly mains powered so without using a transformer. Automatic garden (flood) lights with movement sensor usually have this kind of circuit. \$\endgroup\$ – Bimpelrekkie Aug 15 '17 at 7:48
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Since this circuit is powered directly from AC-mains through a capacitive dropper circuit (C1) there needs to be some form of voltage clamping for the low voltage supply across C2, C3 and C4. Without that IC1 would be destroyed when too little current is drawn as that 24 V could become much more than 24 V.

So D2 and D4 must be zener diodes otherwise they would serve no purpose.

The relay is engaged by closing Q2 via the Signal node from the right. Note how relay coil and Collector-Emitter of Q2 are in series and that combination is across the 24 V supply rail.

Q1 is an oddity, it will only close when Q2 is open so when the relay is not engaged. Then (Q1 closed, relay off) Q shorts D4 meaning that only D2 will clamp the supply. So in this situation, the supply voltage will be somewhat lower as there is now only one zender diode (D2) clamping it.

A possible reason for this configuration is to save some power. When the relay is on then it will draw some current so less current can flow through the zener diodes (D2 and D4).

When the relay is off it does not draw current so perhaps that would cause the supply voltage to rise too much, this is prevented by using only one zener diode (D2, and shorting D4).

For this to work properly I would expect that D2 and D4 have very different zener voltages, for example D2 = 20 V and D4 = 4 V for example.

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  • \$\begingroup\$ Ah, nice explanation. "An oddity" describes my feelings towards this circuit pretty well. \$\endgroup\$ – Marcus Müller Aug 15 '17 at 7:51
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    \$\begingroup\$ @MarcusMüller Indeed, to me it looks like Q1 was added to fix a problem the designer had later on (supply increasing too much). I would simply use a 24 V zener (instead of D2, D4), remove Q1, R4, and dimension C1 and relay such that the supply would drop to 20 V for example when the relay is on. Then there is still pleny of margin for IC1 to supply 5 V at its output (I'm assuming only a few mA are needed there). \$\endgroup\$ – Bimpelrekkie Aug 15 '17 at 7:55
  • \$\begingroup\$ Thanks alot! This helped me clear my mind. Yeah was thinking about doing some similar changes and yes the powerconsumption isn't that high. \$\endgroup\$ – TheOne Aug 15 '17 at 8:03
  • \$\begingroup\$ I really wonder where this is from. There's so much that says "uhhh and what if". For example, I can't see anything that would inhibit the relay from starting to oscillate if whatever load's attached to L' draws enough current to cause a spike through C1... \$\endgroup\$ – Marcus Müller Aug 15 '17 at 8:06
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    \$\begingroup\$ @DiBosco Thanks, I was not thinking that :-) I was thinking you were paying attention and spotted the typo which I put in there to test if anyone was paying attention ;-) Only joking, it was a typo. \$\endgroup\$ – Bimpelrekkie Aug 15 '17 at 8:45
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Might just be an interesting way of controlling Q1 to switch on the relay only when the input voltage is in a defined range:

When the upper, possibly Zener, diode breaks down, Q1 can get a reasonable I_ce, which in turn might activate the relay. When the lower breaks down, the current would flow through that instead of Q1, and the relay would turn off.

I use "might", "would" because it's actually impossible to know for sure without knowing the diodes and transistors involved.

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