🚀 go-pugleaf

In article <1992Aug15.082131.29146@discus.technion.ac.il> s2859479@techst02.technion.ac.il (Yaron Keren) writes:
>I need to supply 5-8V DC at current of 0.7mA avg, 2mA max from the line
>(220V here) voltage. Since the current is very low, I thought to use a
>diode-resistor-zener+capacitor as a power supply instead of a tranformer.
>What are the disadventages, dangers, etc. inbvolved?
>Is there other simple way to do this?
>
>Please email and I'll summarize if there are enough responses.
>
>Yaron
>-- 
>Yaron Keren                                 s2859479@techst02.technion.ac.il


Assuming that you can stand to have whatever you're powering directly
connected to one side of the powerline, you could use a line-clocked
charge pump circuit:

At double your load's maximum current of 2mA, it would consume 4 / 60
milliCoulombs of charge every line cycle (I'm assuming the US mains
frequency of 60Hz here).  Moving that amount of charge by changing
the voltage on a coupling cap by 220 * SQRT(8), or 622 V would require
a capacitor no larger than about 0.1uF.

Your load-voltage variation of 3V is roughly 200X smaller this peak-to-peak
line voltage, so you'd need an output filter cap of at least 20uF.

So, you could try this circuit:

                    D2
 O-------||------*-->|-*------->
         C1      |     |+        8V, 2mA out
LINE            ---   --- C2
 IN           D1 ^    ---
                 |     |
 O--*FFF*--------*-----*------->
     Fuse

C1 -   0.1uF, 400V (Minimum) non-electrolytic

C2 -   100 to 200uF, 16V electrolytic

D1 -   8.7V (or so) Zener Diode.  It's important that this be a single-
       junction part, which will conduct in the "forward" direction,
       as well as in the `breakdown' direction.  If not, reverse-parallel
       it with a small-signal silicon diode.

D2 -   General purpose small-signal silicon diode. It'll only see 10V reverse.


Brian Rhodefer