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'PIC PSU - Multiple Zener Diodes'
2000\05\19@024213 by Werner Soekoe

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Hi Everyone

There are a few circuits on the Internet, including a Microchip App Note,
which uses a 5.1V Zener Diode (amongst other components) to convert mains to
5 volt for pic use. I need to build a PIC Circuit which will draw a total of
500mA, and I need a transformerless PSU to convert 230VAC to 5VDC. I
thoought of using a Zener Diode based PSU, but I have the following
questions:

1. Suppose you use a 1watt Zener, does that mean that the PSU can deliver
200mA at 5V?

2. Suppose you were to connect 5 identical 5.1V 1W Zener diodes in parallel,
would the PSU be able to deliver 1000mA?

3. If the PIC uses a 5V Zener PSU, and draws 100mA, how much current is
drawn from the AC Mains Supply???

Thanks!

Regards,
Werner Soekoe
Information Systems Manager
Free State Legislature
spam_OUTwernersTakeThisOuTspamfsl.gov.za
Tel. (051) 407-1109
Fax. (051) 407-1137
Cell. 082 376 8383
Country Code. +27

2000\05\19@044801 by Vasile Surducan

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On 19 May 00, at 8:40, Werner Soekoe wrote:

> Hi Everyone
>
> There are a few circuits on the Internet, including a Microchip App Note,
> which uses a 5.1V Zener Diode (amongst other components) to convert mains to
> 5 volt for pic use. I need to build a PIC Circuit which will draw a total of

> 500mA, and I need a transformerless PSU to convert 230VAC to 5VDC.

  NOT RECOMMENDED ! For such high currents you will don't find
a good  HV capacitor or you'll find in a hard way...
I
> thoought of using a Zener Diode based PSU, but I have the following
> questions:
>
> 1. Suppose you use a 1watt Zener, does that mean that the PSU can deliver
> 200mA at 5V?
   for a short time...and hot to ...ouch! on zener
   The output power depends of capacitor value ( Xc ). Only if load
has a high value all current will flow to zenner and for a 1W/5V and
200mA zener, will blow up quickly your diode... max current for this
type would not exceed 100mA...
>
> 2. Suppose you were to connect 5 identical 5.1V 1W Zener diodes in parallel,
> would the PSU be able to deliver 1000mA?
   untill one of diodes will crush ... ( but you'll need a lot of zenners
for select 5 identical ) 10W zenner diodes has allready invented...
>
> 3. If the PIC uses a 5V Zener PSU, and draws 100mA, how much current is
> drawn from the AC Mains Supply???
  like OHM law said: if you're men with me, i would be men with
you... exactly the sum between current flow through load and
zenner...and all power disipation will drop across capacitor ( who will
becomes hot after a while and in a best way will be interrupted or
short-circuited and adio PIC! )
  So, use a transformer.

*********************************************
Surducan Vasile, engineer
mail: .....vasileKILLspamspam@spam@l30.itim-cj.ro
URL: http://www.geocities.com/vsurducan
*********************************************

2000\05\19@093949 by Russell McMahon

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Werner,

>There are a few circuits on the Internet, including a Microchip App Note,
>which uses a 5.1V Zener Diode (amongst other components) to convert mains
to
>5 volt for pic use. I need to build a PIC Circuit which will draw a total
of
>500mA, and I need a transformerless PSU to convert 230VAC to 5VDC. I
>thoought of using a Zener Diode based PSU, but I have the following
>questions:
>
>1. Suppose you use a 1watt Zener, does that mean that the PSU can deliver
>200mA at 5V?


No!
The zener is a so called "shunt regulator".
A series element (usually a resistor) passes current from the high voltage
source to the load and the zener "shunts" part of this to ground.
The idea is that the series resistor is desiggned in conjunction with the
maximum load to provide a load voltage slightly above that required and the
zener then "shunts" excess current to ground until the load voltage drops to
the design voltage.

The zener power dissipation is the product of its rated voltage and the
current which it  shunts to ground.
The actual  zener dissipation will depend on how much excess current you
design in that must be shunted to ground.

If you are using a 230VAC supply this will rectify to about 320 VDC.
If you drop this witha resistor to 5v then (320-5) = 315 v will drop across
the resistor.
At 500mA the power dissipatioj in the resistor will be 315 * 0.500A = 157
watts !!!!
(!!)

This is not a practical way of getting 500mA at 5V from the mains!
A 3 VA transformer is small, cheap and much much safer.

An alternative method is to use a capacitor to drop the voltage. At 500 mA
this is also not really practical but let's see what is needed.
Effective dropping resistance R = V/I = about 300/0.5A = 600 ohms.
The impedance of a capacitor is Z = 1/(2*Pi*F*C) or
C = 1/(2*Pi*F*Z)
This gives C = 5uF approximately.
This will provide this current for one half cycle so you would need a 10uF
capacitor or higher.
This would need to be a non-polar (ie NOT an eletrolytic) and should be a
mains X-rated part. This would be large and expensive and the end result
would be potentially lethal.

A small transformer is a much more practical solution.

There are other potential transformerless solutions but at this current
rating they are liable to be unnatractive for anything except high volume
and/or specialist applications.




regards,


     Russell McMahon
_____________________________

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2000\05\20@031013 by William Chops Westfield

face picon face
>> I need to build a PIC Circuit which will draw a total of 500mA,
>> and I need a transformerless PSU to convert 230VAC to 5VDC.
>
> At 500mA the power dissipatioj in the resistor will be 315 *
> 0.500A = 157 watts !!!!  (!!)
> This is not a practical way of getting 500mA at 5V from the mains!

What he said.

> A 3 VA transformer is small, cheap and much much safer.

You didn't specify why you wanted a "transformerless" design.  Perhaps
you'd be more interested in some of the somewhat new switching power
supply chips designed to operate directly from rectified AC wall power.
(ie Motorola/ONSemi MC333622)  By operating the transformer at high
frequencies, the physical size of the iron shrinks considerably - I
seem to recall an press release describing a 5V/1A supply that fit in
the space of a standard 3-prong plug...

BillW

2000\05\20@103234 by Dan Mulally

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A series capacitor can be used along with a shunt zener (rated at the
desired voltage plus 0.6V). This is then followed with a series diode (like
a 1N4004) and a shunt filter capacitor. It looks something like a voltage
doubler circuit with the first diode replaced with a zener. For a positive
supply the cathode of the zener is connected to the series cap and the anode
of the 2nd diode is connected to this same point. The output voltage is
taken across the 2nd filter cap (100uF or so). The series cap must be rated
for the AC peak voltage used. The circuit uses energy from both positive and
negative cycles and since the capcitor is reative, little energy is
dissipated in it. It's value depends on the current requirements and line
frequency. I've used a 0.1 uF at 240V RMS 60Hz to supply about 15 mA or so.
Be careful: there is no isolation from the AC line as would be with a
transformer!

Also, its a good idea to have a resistor in series with the 1st cap to limit
current on startup. Maybe 20 Ohms or so.
Dan Mulally

{Original Message removed}

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