Transformerless PSU using a capacitor

[fynnashba@yahoo.com]

Please I have to make a transformerless power supply using a capacitor. I have 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 way 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 current drawn by the load (the load is a relay)
What are the advantages of using the transformerless PSU(using a capacitor)
thanks

 

[Phil Allison]

fynn...@yahoo.com wrote:

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

Please I have to make a transformerless power supply using a capacitor.
I have made the circuit and its working well, my problem is how to
measure the current when its on load and the no-load current.

** Our problem is we cannot see you damn circuit !!

Is there one diode or a bridge ?

Is the rely coil switched on and off or shorted to disable ?



..... Phil

 

[Tim R]

Seriously? bringing your class homework here?

 

[Dave M]

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.

Cheers,
Dave M


fynnashba@yahoo.com wrote:

Please I have to make a transformerless power supply using a
capacitor. I have 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 way 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 current drawn by the load (the load is a relay)
What are the advantages of using the transformerless PSU(using a
capacitor)
thanks

 

[Phil Allison]

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

 

The big problem is there is no isolation.

 

[pfjw@aol.com]

On Tuesday, August 1, 2017 at 2:27:17 AM UTC-4, Phil Allison wrote:

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

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.

Similarly on the hot-tub. Also quite common.

Peter Wieck
Melrose Park, PA

 

[fynnashba@yahoo.com]

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.

 

[Clifford Heath]

On 01/08/17 20:47, fynnashba@yahoo.com 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.

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.

Clifford Heath.

 

[Phil Allison]

fynn...@yahoo.com wrote:

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

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.

** Fantastic.

Cos it is bloody obvious YOU do not understand it one tiny bit.

Have you realty built this thing ??

I doubt that VERY much.

Maybe you simmed it.

FFS - you have NOT mentioned the AC supply voltage so far.

Slipped your mind ?




..... Phil

 

[Phil Allison]

Clifford Heath wrote:

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

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.


So you have a 12v AC load, and a 68uF capacitor feeding
it what? 50Hz at 240V it would be in Australia...

** The OP is not making any sense.

66uF implies amps of current and 220 ohms in series implies
hundreds of watts of dissipation.

The fool probably means 0.68uF.

Wish he would get one tiny fact RIGHT !!!



..... Phil

 

[Clifford Heath]

On 02/08/17 00:13, Phil Allison wrote:

Clifford Heath wrote:

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


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.


So you have a 12v AC load, and a 68uF capacitor feeding
it what? 50Hz at 240V it would be in Australia...

** The OP is not making any sense.

66uF implies amps of current and 220 ohms in series implies
hundreds of watts of dissipation.

Ugh, yeah, I chose to ignore the 220R. He's probably trying
to limit inrush current when switched at the wrong point.

The fool probably means 0.68uF.
Wish he would get one tiny fact RIGHT !!!

Indeed. It's hard to do electronics without knowing arithmetic.

Clifford Heath.

 

[John Robertson]

On 2017/08/01 3:47 AM, fynnashba@yahoo.com 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 :-#)#

--
(Please post followups or tech inquiries to the USENET newsgroup)
John's Jukes Ltd.
MOVED to #7 - 3979 Marine Way, Burnaby, BC, Canada V5J 5E3
(604)872-5757 (Pinballs, Jukes, Video Games)
www.flippers.com
"Old pinballers never die, they just flip out."

 

[John Robertson]

On 2017/08/01 6:50 PM, John Robertson wrote:

On 2017/08/01 3:47 AM, fynnashba@yahoo.com 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 :-#)#

Oh, wait, subject line - transformerless PSU. Sigh, I hope your health
insurance is up to date.

John :-#(#

--
(Please post followups or tech inquiries to the USENET newsgroup)
John's Jukes Ltd.
MOVED to #7 - 3979 Marine Way, Burnaby, BC, Canada V5J 5E3
(604)872-5757 (Pinballs, Jukes, Video Games)
www.flippers.com
"Old pinballers never die, they just flip out."

 

[fynnashba@yahoo.com]

On Tuesday, August 1, 2017 at 2:03:26 PM UTC, Phil Allison wrote:

fynn...@yahoo.com wrote:

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


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.



** Fantastic.

Cos it is bloody obvious YOU do not understand it one tiny bit.

Have you realty built this thing ??

I doubt that VERY much.

Maybe you simmed it.

FFS - you have NOT mentioned the AC supply voltage so far.

Slipped your mind ?




.... Phil

sorry please the cap is 0.66uf and the mains voltage is 220v 50hz.

 

[Phil Allison]

fynn...@yahoo.com wrote:

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

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.




You seem to understand my problem very well.



** Fantastic.

Cos it is bloody obvious YOU do not understand it one tiny bit.

Have you realty built this thing ??

I doubt that VERY much.

Maybe you simmed it.

FFS - you have NOT mentioned the AC supply voltage so far.

Slipped your mind ?



sorry please the cap is 0.66uf and the mains voltage is 220v 50hz.

** The cap is 0.68uF - put your glasses on.

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.

A resistor would dissipate nearly 10 watts.



..... Phil

 

[fynnashba@yahoo.com]

On Wednesday, August 2, 2017 at 1:22:32 PM UTC, Phil Allison wrote:

fynn...@yahoo.com wrote:

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



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.




You seem to understand my problem very well.



** Fantastic.

Cos it is bloody obvious YOU do not understand it one tiny bit.

Have you realty built this thing ??

I doubt that VERY much.

Maybe you simmed it.

FFS - you have NOT mentioned the AC supply voltage so far.

Slipped your mind ?



sorry please the cap is 0.66uf and the mains voltage is 220v 50hz.



** The cap is 0.68uF - put your glasses on.

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.

A resistor would dissipate nearly 10 watts.



.... Phil

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.

 

[John-Del]

On Wednesday, August 2, 2017 at 10:00:17 AM UTC-4, fynn...@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.

LOL...

 

[pfjw@aol.com]

On Wednesday, August 2, 2017 at 12:19:19 PM UTC-4, John-Del wrote:

On Wednesday, August 2, 2017 at 10:00:17 AM UTC-4, fynn...@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.

LOL...

Now, wait just a darned minute. Phil is quite reasonable when he takes his meds!

Peter Wieck
Melrose Park, PA

 

[rickman]

pfjw@aol.com wrote on 8/2/2017 3:43 PM:

On Wednesday, August 2, 2017 at 12:19:19 PM UTC-4, John-Del wrote:
On Wednesday, August 2, 2017 at 10:00:17 AM UTC-4, fynn...@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.

LOL...

Now, wait just a darned minute. Phil is quite reasonable when he takes his meds!

And which holiday would that be?

--

Rick C