Please I have to make a transformerless power supply using a capacitor. I h ave made the circuit and its working well, my problem is how to measure the current when its on load and the no-load current. When I use the normal wa y of measuring current the meter is not stable at all. Please how do l do it. Also for academic purpose how do I calculate the cur rent drawn by the load (the load is a relay) What are the advantages of using the transformerless PSU(using a capacitor) thanks
A few questions to you first, to get an idea as to what you want to do, and what you've done so far.
Are you trying to make a DC power source or an AC source? Is the load the relay coil or a load that is being switched by the relay? What kind of instrument are you using to measure the current, and how are you hooking it into the circuit? Is the relay a DC or AC relay? DC and AC relays are built differently, so they have different characteristics. 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. If it's an AC relay coil, then you really need the Mfr's specs to know how much current the relay coil will draw. It's not just the coil's DC resistance that determines the current; also the coil inductance.
Be careful of blanket statements. The boiler in our house uses 24VC relays for various functions. As happens, the 24VAC unregulated, unrectified control voltage switches various 120V pumps, fans and valves. As is the case with many US heating systems.
erless PSUs are to work on and takes all the precautions needed.
r.
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 c ap and a 1 meg bleeder resistor across the 66uF cap. In fact the circuit i s working as it should but I want to know the amount of power being drawn b y the circuit when the relay is off and the power drawn when it is on so th at 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 ple ase forgive me. Thanks to you all.
So you have a 12v AC load, and a 68uF capacitor feeding it what? 50Hz at 240V it would be in Australia...
That's 5 amps peak, maybe 3A RMS. I don't think your Zener will survive very long dissipating 36W (3*12).
Try again with a 1uF capacitor for maybe 60mA. If your relay needs that much, otherwise smaller. Smaller again if it's 60Hz, but double it for 120VAC in.
the relay is also DC 12v. I am using a 66uF cap a bridge rectifier, a 12v z ener diode and a filter cap.I have a 220 ohm resistor in series with the 66 uF cap and a 1 meg bleeder resistor across the 66uF cap. In fact the circu it is working as it should but I want to know the amount of power being dra wn by the circuit when the relay is off and the power drawn when it is on s o that l can compare it to the a similar one using a transformer. I don't k now how to post an image of the circuit here else l would have done that so please forgive me. Thanks to you all.
** The OP is not making any sense.
66uF implies amps of current and 220 ohms in series implies hundreds of watts of dissipation.
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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(Please post followups or tech inquiries to the USENET newsgroup)
John's Jukes Ltd.
I do hope it is a class X1 or X2 type - rated for mains AC voltage.
It's impedance at 50Hz is 4680 ohms so the average *rectified* current flow is 40mA. Average, full wave, rectified sine wave current = 0.63 times the peak value. Look it up.
The 12V relay must operate reliably at that current, so coil resistance needs to be not more than 300 ohms.
If the relay coil is switched off, there is Zero current flow.
The * HUGE * advantage of using a cap to drop the AC supply voltage to suit the relay is that is dissipates NO power.
Thanks a lot Phil you were really helpful. All l needed you have provided I also had a lot more info from hackabay.com. But, but l am surprised the high level of respect for members in this group has gone down. Any way all is not lost yet people like Phil are always there to help.
On Wednesday, August 2, 2017 at 10:00:17 AM UTC-4, snipped-for-privacy@yahoo.com wrote: But, but l am surprised the high level of respect for members in this group has gone down. Any way all is not lost yet people like Phil are always there to help.
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