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

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

Seriously? bringing your class homework here?

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


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

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.

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

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.

On 01/08/17 20:47, 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.

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

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

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.

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 :-#)#

--
(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."

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


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."

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

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

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

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

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, 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

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, 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

On Wednesday, August 2, 2017 at 3:56:46 PM UTC-4, rickman wrote:

And which holiday would that be?

The one celebrating the day he took his meds... .

Peter Wieck
Melrose Park, PA

wrote on 8/2/2017 4:03 PM:
On Wednesday, August 2, 2017 at 3:56:46 PM UTC-4, rickman wrote:

And which holiday would that be?

The one celebrating the day he took his meds... .

Yeah, I think you are right, that would be a holiday.

--

Rick C

I want to know what is wrong with using a five buck transformer rather than creating a shock hazard.

Such a scheme beats on the cap and it may go bad. In fact it may short out and teach you why they do not do this. Hopefully your fire insurance is paid up.

This is worse than those little transformerless cellphone chargers. They should not be allowed to sell those things.

There is a right way and a wrong way, and a capacitively coupled doodad like this is the wrong way.

On Wednesday, August 2, 2017 at 6:05:49 PM UTC-4, wrote:
I want to know what is wrong with using a five buck transformer rather than creating a shock hazard.

Such a scheme beats on the cap and it may go bad. In fact it may short out and teach you why they do not do this. Hopefully your fire insurance is paid up.

This is worse than those little transformerless cellphone chargers. They should not be allowed to sell those things.

There is a right way and a wrong way, and a capacitively coupled doodad like this is the wrong way.



Yes, but I believe it was only an exercise in a class. But you can use this technique on the secondary of a transformer instead of using a dropping resistor, although I can't recall seeing it done.

The only time I used it was very recently on an old AC/DC radio from the 30s that had about 80V worth of filaments but needed to run on 120V. The radio originally used a resistive AC cord which consisted of two normal conductors and one length of asbestos yarn wrapped in a spiral of nichrome wire that ran the entire length of the AC cord and connected to one side of the AC plug. This long wire resistor provided the necessary drop for the filament requirement. The cord was shot so I put in a two wire cord and used something like 6uf worth of film caps out of Panasonic plasma sustain to get down to the proper voltage for the tube's series string. It's fused, but it's a temporary solution just to get the radio operational. If I ever restore the radio I'll probably locate a replacement cord as there's no room for a dropping resistor of the necessary wattage in the case (miniature mantle radio).

wrote:

I want to know what is wrong with using a five buck transformer
rather than creating a shock hazard.


** No shock hazard if built sensibly.

Smoke detectors use dropping caps all the time.


Such a scheme beats on the cap and it may go bad.

** They sometimes go open.

In fact it may short out and teach you why they do not do this.


** A fuse or fusible resistor in series with the incoming AC is wise.


..... Phil

Yeah, another thing that raises the cost of fire insurance.

I remember when GM used resistive wire as a ballast for the coil, which was shorted out during cranking by the extra terminal on the solenoid. We have advanced past that and resistive power cords. WE now know how to control power pretty accurately.

If you really must restore some old thing to its original, unsafe and inefficient state, by all means do so. But don't come crying to me if it burns your house down. Mainly, don't fall asleep with the thing running.

Sorry, I tend to err on the side of safety. Things need to be isolated and that means a transformer. A capacitive coupled device ? I would not have it in the house.

I have had dangerous things, for one guns. Then my buddy brings over this oxygen hydrogen separator that runs on 12 volts. And he didn't realize that it was better to separate those two gases, because together they are most dangerous.

He wanted to use it to boost his car, but we found out that the thing was pulling like 11 amps at 12 volts and the load on the alternator would negate most of the gain.

It is like what I have been saying, fossil fuel beats about anything, that is why it sells. When something better comes along, the oil companies won't be oil companies, they will be selling whatever it is.

But anyway, dropping voltage/current with a capacitor is not the way to do it. There are times when you just cough up the money for a transformer.

wrote:

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

Sorry, I tend to err on the side of safety.
Things need to be isolated and that means a transformer.
A capacitive coupled device ? I would not have it in the house.


** Your toaster isn't isolated, nor is your hair dryer and a host of everyday appliances.

The only appliances that NEED isolation are electronic ones that have external connections to other devices. The rest can safely rely on being fully insulated or by safety grounding any external metalwork.

FYI: transformers do not automatically provide *safety isolation* unless that are designed to meet class 2 requirements. Very few off the shelf ones are.

Your constant posting of rambling drivel is getting worse - jurb.

You have way too many loose screws.



..... Phil

In article ,
says...

This is worse than those little transformerless cellphone chargers. They should not be allowed to sell those things.

Really? All the ones I have taken apart (and not just for cellphones)
are switch-mode with tiny transformers. Admittedly I don't know how they
make them miniature secure isolation devices...

Mike.

Mike Coon wrote:

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

says...

This is worse than those little transformerless cellphone chargers.
They should not be allowed to sell those things.


Really?



** The term "transformerless" sometimes only means there is no iron transformer - just small ferrite cored one instead.

Nowadays,they can be made as SMD devices and be incredibly small for DC powers of 1 watt or so.



All the ones I have taken apart (and not just for cellphones)
are switch-mode with tiny transformers. Admittedly I don't know how they
make them miniature secure isolation devices...


** Enough insulation is there to tolerate the AC supply voltage and not fail from overloading.

The latest thing is a "piezo transformer" but I have no evidence that are being used in cell phone chargers.


..... Phil

In article ,
says...

The latest thing is a "piezo transformer" but I have no evidence that
are being used in cell phone chargers.

I hadn't heard of that, but I suppose it is a no moving part (in usual
interpretation of that phrase) motor+dynamo combination; brilliant!

Mike.

On Thursday, August 3, 2017 at 4:54:43 AM UTC, Phil Allison wrote:
wrote:

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

Sorry, I tend to err on the side of safety.
Things need to be isolated and that means a transformer.
A capacitive coupled device ? I would not have it in the house.


** Your toaster isn't isolated, nor is your hair dryer and a host of everyday appliances.

The only appliances that NEED isolation are electronic ones that have external connections to other devices. The rest can safely rely on being fully insulated or by safety grounding any external metalwork.

FYI: transformers do not automatically provide *safety isolation* unless that are designed to meet class 2 requirements. Very few off the shelf ones are.

Your constant posting of rambling drivel is getting worse - jurb.

You have way too many loose screws.



.... Phil

thanks to you all for that fruitful discussion.
As indicated by Phil its not all appliances that need to be duly isolated. A device like an automatic dawn to dusk light switch can be safe without an isolation transformer besides during the day the transformer's primary will still take some power and warm up.
I have also seen a Chines made portable radio set without transformer. It was boldly written on it "transformerless AC radio" apparently the manufacturer was boasting. A guy opened it to find out what was inside and showed us this cap drop power supply so he warned the friend against possible electric shock when he touch any metal part. the friend replied there were no metal parts so its safe. Not long before that he got zapped when he forgot that the antenna was made of metal. l am going to let him read all your post.
Thanks once again.
i ve asked him to read all your post

On Thursday, August 3, 2017 at 12:16:59 AM UTC-4, wrote:


If you really must restore some old thing to its original, unsafe and inefficient state, by all means do so. But don't come crying to me if it burns your house down. Mainly, don't fall asleep with the thing running.


If you're referring to the resistance line cord in my example, there aren't many options even if you don't care about originality (I don't). That radio in my example has 5 tubes, two IF XFRs, a gang tuner and a dynamic speaker all crammed into a 4X10 chassis pan. The plastic cabinet covers this like a glove. There's simply no room to mount a 25W resistor or mitigate the heat that it produces.

Still, in the trade, the resistive AC line cord was known as a "curtain burners" because people would often coil them up in a neat little ball instead of spreading them out as the owner's manual directed.

I do restore an occasional antique radio for my customers, and even after adding fuses and polarized line cords to them, I instruct them on the dangers of line connected metal chassis and never plug in the radio if one of the plastic knobs falls off. I also instruct them to leave all old radios unplugged when not in use and to not leave them running unattended.

The sovereign remedy for a curtain-burner is a unique plug to an external box that contains the resistor and heat-sink assembly. A nice, fat DALE aluminum resistor of the appropriate rating does well.

In this way, the radio does not get plugged into the wall by accident (nor anything else), and the external box can have a fuse, switch and other niceties as well as serve multiple radios.

Safety is a good thing, and despite the fact that our parents survived all sorts of deadly things such as a lack of seat belts, hard dashboards, Open-mesh electric fans with metal blades, and much more, there is no reason to lean into a punch.

Peter Wieck
Melrose Park, PA

On Thursday, August 3, 2017 at 8:39:43 AM UTC-4, wrote:
The sovereign remedy for a curtain-burner is a unique plug to an external box that contains the resistor and heat-sink assembly. A nice, fat DALE aluminum resistor of the appropriate rating does well.

In this way, the radio does not get plugged into the wall by accident (nor anything else), and the external box can have a fuse, switch and other niceties as well as serve multiple radios.

True, but that doesn't preclude the possibility that someone in the future will find the radio, cut off it's unique plug, and replace it with a standard AC cord. The radio also loses some of it's charm having to be connected to a life support box instead of being independent.

BTW, does anyone else remember the Sanyo/Fisher receivers of the 80s that used LV AC taps off the transformer to power the matching Sanyo/Fisher tuner, CD, and cassette deck? I saw a bunch that had the unique plugs cut off and replaced with regular AC plugs. Feeding 120V into a tuner that worked on 16VAC left a big black smudge inside the cabinet.

The 1930s radio in question was only powered up with capacitive voltage dropping as a triage to assess the rest of the radio. I haven't yet decided whether to restore it. If so, I may keep the capacitors and add a diode, fuse, and zener crowbar to eliminate any possibility of a shorted cap running amok. These things can be safely installed below the deck and produce no heat.

Generally, if I make any sort of functional change to any given radio - dropping resistor, dropping capacitor, whatever, a 3 x 5 card will go under the chassis with such notes on it. I am not above using an etcher as well to make sure that something really important is not lost over time.

Stuff "below the chassis" has its significant advantages but unless explained becomes a mystery to the next user.

Peter Wieck
Melrose Park, PA

On Wed, 2 Aug 2017 21:16:53 -0700 (PDT), wrote:

Yeah, another thing that raises the cost of fire insurance.

I remember when GM used resistive wire as a ballast for the coil, which was shorted out during cranking by the extra terminal on the solenoid. We have advanced past that and resistive power cords. WE now know how to control power pretty accurately.

If you really must restore some old thing to its original, unsafe and inefficient state, by all means do so. But don't come crying to me if it burns your house down. Mainly, don't fall asleep with the thing running.

Sorry, I tend to err on the side of safety. Things need to be isolated and that means a transformer. A capacitive coupled device ? I would not have it in the house.

I have had dangerous things, for one guns. Then my buddy brings over this oxygen hydrogen separator that runs on 12 volts. And he didn't realize that it was better to separate those two gases, because together they are most dangerous.

He wanted to use it to boost his car, but we found out that the thing was pulling like 11 amps at 12 volts and the load on the alternator would negate most of the gain.

It is like what I have been saying, fossil fuel beats about anything, that is why it sells. When something better comes along, the oil companies won't be oil companies, they will be selling whatever it is.

But anyway, dropping voltage/current with a capacitor is not the way to do it. There are times when you just cough up the money for a transformer.
Actually, the load on the alternator must be MORE than any gain,
otherwise you would have a perpetual motion device.
Eric

On Thursday, August 3, 2017 at 12:51:32 PM UTC-4, wrote:

I have had dangerous things, for one guns. Then my buddy brings over this oxygen hydrogen separator that runs on 12 volts. And he didn't realize that it was better to separate those two gases, because together they are most dangerous.

He wanted to use it to boost his car, but we found out that the thing was pulling like 11 amps at 12 volts and the load on the alternator would negate most of the gain.





Actually, the load on the alternator must be MORE than any gain,
otherwise you would have a perpetual motion device.
Eric




The HHO guys say you can get "free" energy out of tap water by building a hydrogen generator and feeding the resulting hydrogen into the intake of the engine giving it more power and using less gasoline. What the poster was saying was any gain (if any) from any hydrogen produced and burned would be negated by electrical toll it took to create it.

Even if the hydrogen scheme produced a net positive, it still wouldn't be perpetual motion (or breaking the conservation of energy law) any more than a gas engine is because the water is an expendable fuel.

On Thu, 3 Aug 2017 13:07:55 -0700 (PDT), John-Del
wrote:

On Thursday, August 3, 2017 at 12:51:32 PM UTC-4, wrote:

I have had dangerous things, for one guns. Then my buddy brings over this oxygen hydrogen separator that runs on 12 volts. And he didn't realize that it was better to separate those two gases, because together they are most dangerous.

He wanted to use it to boost his car, but we found out that the thing was pulling like 11 amps at 12 volts and the load on the alternator would negate most of the gain.





Actually, the load on the alternator must be MORE than any gain,
otherwise you would have a perpetual motion device.
Eric




The HHO guys say you can get "free" energy out of tap water by building a hydrogen generator and feeding the resulting hydrogen into the intake of the engine giving it more power and using less gasoline. What the poster was saying was any gain (if any) from any hydrogen produced and burned would be negated by electrical toll it took to create it.

Even if the hydrogen scheme produced a net positive, it still wouldn't be perpetual motion (or breaking the conservation of energy law) any more than a gas engine is because the water is an expendable fuel.
What you are saying is that it would be not be breaking any
conservation of energy law if you get more power from an engine by
feeding it hydrogen or even HHO that the power required from the
engine to break the water molecule apart into HHO. I don't get it.
Please explain.
Eric

John-Del wrote on 8/3/2017 4:07 PM:
On Thursday, August 3, 2017 at 12:51:32 PM UTC-4, wrote:

I have had dangerous things, for one guns. Then my buddy brings over this oxygen hydrogen separator that runs on 12 volts. And he didn't realize that it was better to separate those two gases, because together they are most dangerous.

He wanted to use it to boost his car, but we found out that the thing was pulling like 11 amps at 12 volts and the load on the alternator would negate most of the gain.





Actually, the load on the alternator must be MORE than any gain,
otherwise you would have a perpetual motion device.
Eric




The HHO guys say you can get "free" energy out of tap water by building a hydrogen generator and feeding the resulting hydrogen into the intake of the engine giving it more power and using less gasoline. What the poster was saying was any gain (if any) from any hydrogen produced and burned would be negated by electrical toll it took to create it.

Even if the hydrogen scheme produced a net positive, it still wouldn't be perpetual motion (or breaking the conservation of energy law) any more than a gas engine is because the water is an expendable fuel.

When you burn the hydrogen and oxygen you produce more water that could be
fed back into the tank along with the water produced from burning the
hydrocarbons resulting in *MORE* fuel than you started with. But the
problem is that the energy produced by separating the oxygen and the
hydrogen is at *least* as much as what you get from burning the hydrogen.
Taking into account all the losses and you get a net loss of energy by some
80% or more.

Eric is right. Water is not a fuel. The fact that in any practical machine
you just refill it is because of the cost rather than the issue of it
providing any energy.

--

Rick C

rickman wrote on 8/3/2017 5:23 PM:
John-Del wrote on 8/3/2017 4:07 PM:
On Thursday, August 3, 2017 at 12:51:32 PM UTC-4, wrote:

I have had dangerous things, for one guns. Then my buddy brings over
this oxygen hydrogen separator that runs on 12 volts. And he didn't
realize that it was better to separate those two gases, because together
they are most dangerous.

He wanted to use it to boost his car, but we found out that the thing
was pulling like 11 amps at 12 volts and the load on the alternator
would negate most of the gain.





Actually, the load on the alternator must be MORE than any gain,
otherwise you would have a perpetual motion device.
Eric




The HHO guys say you can get "free" energy out of tap water by building a
hydrogen generator and feeding the resulting hydrogen into the intake of
the engine giving it more power and using less gasoline. What the poster
was saying was any gain (if any) from any hydrogen produced and burned
would be negated by electrical toll it took to create it.

Even if the hydrogen scheme produced a net positive, it still wouldn't be
perpetual motion (or breaking the conservation of energy law) any more
than a gas engine is because the water is an expendable fuel.

When you burn the hydrogen and oxygen you produce more water that could be
fed back into the tank along with the water produced from burning the
hydrocarbons resulting in *MORE* fuel than you started with. But the
problem is that the energy produced by separating the oxygen and the
hydrogen is at *least* as much as what you get from burning the hydrogen.
Taking into account all the losses and you get a net loss of energy by some
80% or more.

Err, that should read, "energy *consumed* by separating the oxygen and the
hydrogen"

--

Rick C