On 2017/08/01 3:47 AM, wrote:
On Tuesday, August 1, 2017 at 6:27:17 AM UTC, Phil Allison wrote:
Dave M wrote:

---------------------------


Are you trying to make a DC power source or an AC source?


** The OP is making a PSU for a relay the runs from the AC supply with no isolation using a series capacitor.

There is virtually zero chance he is using an AC relay cos they come in 120VAC and 240VAc versions.


Is the load the relay coil or a load that is being switched by the relay?

** The OP has told us: " ... the load is a relay ".

Obvious since capacitor fed PSUs have very limited current - but enough for many relays.


What kind of instrument are you using to measure the current, and how are
you hooking it into the circuit?

** We need to know what the circuit is first, many possible variations on the theme exist.

Is the relay a DC or AC relay?


** Forget the AC case.



If it's a DC relay coil, then just measure the coil resistance, divide that
value into the voltage across the coil and you'll know what the current
should be.

** Well, that will give you the average value.


If it's an AC relay coil,


** You are hooked on that wild card.

BTW:

The OP is clearly a novice and I hope he is aware how dangerous transformerless PSUs are to work on and takes all the precautions needed.

If this is a task set by his electronics instructor, big bad on him or her.


.... Phil

You seem to understand my problem very well. The psu is DC 12volts and the relay is also DC 12v. I am using a 66uF cap a bridge rectifier, a 12v zener diode and a filter cap.I have a 220 ohm resistor in series with the 66uF cap and a 1 meg bleeder resistor across the 66uF cap. In fact the circuit is working as it should but I want to know the amount of power being drawn by the circuit when the relay is off and the power drawn when it is on so that l can compare it to the a similar one using a transformer. I don't know how to post an image of the circuit here else l would have done that so please forgive me. Thanks to you all.


Just to get this straight - you are sourcing 12VDC to the 12VDC relay by
using a bridge rectifier, filter cap, bleeder resistor, series resistor
and a zener. How is the relay hooked to this circuit? Is it simply wired
across the 12VDC power so it is always energized when the PSU is on, or
is there some sort of switch? Something like below (in simplest form)?

+ -\-{--| (12VDC relay)
1 } - (EMF diode)
2 { ^
- ---}--|

(lousy ASCII drawing)

What I am trying to figure out is why you need a bridge rectifier,
zener, bleeder, and the 66ufd cap considering that your power supply is
already 12VDC - which matches your 12VDC relay. Or are you making it
12VDC by having something around 10VAC connected to your bridge
rectifier, then off to the 66ufd cap via the 220R series resistor on the
positive line. The 66ufd cap has a bleeder resistor which may be
pointless if it is connected directly to the relay.

Put a back EMF diode (1N400X) on the relay coil to protect any solid
state devices (diodes, etc.) from reverse discharge when the relay is
de-energized.

Is your purpose to have a time delay element for the relay? In other
words, when the relay is powered up, do you want it to stay energized
for a short period of time after power is removed via the 66ufd cap and
resistor? Or is the relay simply switched in/out of the circuit?

John :-#)#

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