Line voltage to 5VDC supply, transformerless

Hey, all, I tried to build a circuit that I found online and had a
problem with it. I e-mailed the creator but didn't get a reply so
here I am.

What I have is a 120VAC device that switches on and off through
existing circuitry; it is a fully functioning device and circuit.
Periodically the existing circuitry will power up the device for a
period of time. I want a circuit that will detect when the current
device cycles from off to on to off and, upon returning to the off
state, will enable a second device for a configurable period of time.
That is, I have a circuit (I did not create or build) that controls a
120VAC load. I want to introduce another 120VAC load that will run
for a set amount of time each time the original load turns on and then
off.

My goal is to use a 555 timer in mono-stable configuration controlling
a 5 VDC relay to switch the 120VAC line voltage to my second load. I
found this circuit:

http://ourworld.compuserve.com/homepages/Bill_Bowden/page10.htm
Look at the third circuit under AC Line powered LEDs.

I bought all the components specified except the 270 ohm 2 watt
resistor. I had to buy two 560 ohm 1 watt resistors and run them in
parallel (measured 273 ohms). The thing works..but the resistors get
hot as sin in about 30 seconds and the whole time the voltage is
creaping up. When I say hot I mean they are painful to touch and give
off an odor. I added two more resitors for a total of four, two in
series, two series in parallel and still they get hot (painful). With
the four in parallel/series I measured a little over the 560 ohms the
capacitors were designed to be. Still, that's over twice the
resistance the orginal circuit calls for and I thought that would
reduce the current and therefore the heat. Apparently I am wrong.

I did some searching in here before making this post and the best I
found as an alternative is on Microchip's site:

http://www.microchip.com/stellent/idcplg?IdcService=SS_GET_PAGE&nodeId=1824&appnote=en011121

It's TB008.

I can do that (see Figure 4 in that PDF) but I only have 1.5 uF
capacitors in that voltage range (specially purchased for this
project) and, mostly, I hate to abandon an idea that fails without
understanding the failure. Plus, the Microchip shematic doesn't
indicate the wattage of the resistors (that I noticed) whereas the
first circuit I listed does...so I'm wondering if I can get away with
1/4 watt resistors or not. I bet I make popcorn. And, as I get
nervous as hell handling 120VAC in these kinds of situations
(electronics where wires are bare and alligator clips are in use), I
thought I'd see what experience might be out there to learn from.

I could get transformers but that will make my project much more bulky
and I'd like some finesse here.

Thank you for your time and help.

--HC

 

"HC"

** Yep - you are.

Did you use a film cap for the 1.5 uF or an electro ??

** READ THIS URGENTLY !!

http://sound.westhost.com/articles/power-supplies2.htm#s8

Scroos doen to the secion headed

" 7 Cheap Death ( or How Not to Design a Power Supply )
--------------------------------------------------------------------

The utter moron who created and published that circuit ought to be taken out
and shot.




...... Phil

 

Hey, Phil, thanks for the reply. I used a "xicon radial metallized
polyester film capacitor" according to the label on the baggy from
Mouser. I took care to get a non-polarized cap. I would think an
electrolytic would have gone of like a grenade.

I looked at the site you referenced and I see the point there, I
believe: tying the ground to neutral at the device is silly
(dangerous). So, if I use the schematic they mention and if the
device was installed by a trained professional so that the polarity
was properly connected (hot to the positive side, neutral to the
negative side, earth left connected to the chassis for its intended
purpose (to trip a circuit breaker if a short-circuit happens), then
this could work. I think.

My problem is that I need three low voltage DC sources from 120VAC:
one to power the 555 timer circuit all the time, one to take the
120VAC feed to the load and make it an input-level voltage for the 555
timer circuit mentioned, and one to run a relay with a 5 VDC coil
(since I already bought some).

So, if I build the circuit in section 6 of your referenced web page,
hardwired it myself to be sure the polarity was right, then couldn't
that work?

Thank you for your time and information. The site you linked to maded
sense and I see the danger inherent in the Microchip design.

--HC

 

"HC"


** You got some pathological aversion to using a small transformer or wall
wart ? ??

Please explain.



BTW

Under NO circumstances would I ever suggest that any *novice* like you
build an electronic circuit that operates from the AC supply without one.




...... Phil

 

We had a discussion about this a while ago, and I have used similar
circuits for some applications where the entire circuit is contained in an
enclosure and the interface to the outside world is an optoisolator or
relay. The Microchip circuits are particularly bad because of the incorrect
use of a fuse in the neutral leg.

As Phil pointed out, you probably used a polarized electrolytic capacitor,
which will break down during reverse voltage excursions and cause excessive
current flow. You were probably lucky that it did not explode. The 270 ohms
limited the current to about 1/2 amp which would be about 70 watts in the
resistors. They would get seriously hot almost instantly.

I have had problems with the series resistor getting much hotter than
expected, even with properly sized metal film capacitors, and it may have
been due to high frequency harmonics on the AC line (possibly if used on an
inverter). The best solution is really a small wall-wart supply that is
relatively fail-safe and isolated, and can be had very cheap (or free).

Otherwise, you can use a power resistor to drop the line voltage. You can
use a series diode to eliminate the reverse voltage of the applied AC, or a
full wave bridge for better efficiency. You can get 30 mA out if you use a
resistor of R=120/.03 = 4000 ohms, which will dissipate about 3.6 watts.
That is a lot of heat, and even a 10 watt resistor will get too hot to
touch.

There is a linear regulator that will accept rectified line-voltage up to
450 VDC, and will put out 3 mA continuous or 30 mA pulses.
http://www.supertex.com/pdf/datasheets/LR645.pdf

It should be possible to make a very small line operated switching supply
for 1 watt or less, but I have not found any commercially available. The
difficult part would be making such a small transformer with the number of
turns necessary. About the smallest line operated switchers I have come
across are about 5 watts, and they are about 2" x 3", like these:

http://www.mouser.com/Search/ProductDetail.aspx?qs=pqZ7J9Gt/mrLjF0GsFyr0A==
http://www.mouser.com/Search/ProductDetail.aspx?qs=hGGJxEs%2bIdQh/65zCOYtlA==
http://www.mouser.com/Search/ProductDetail.aspx?qs=pqZ7J9Gt/mq5Y0PEQSR1UA==

Paul

 

No, nothing pathological. It's just that since I need one to drive
the logic circuit and maybe also the relay, and another to transform
the original load supply voltage, that would mean I would need TWO
transformers and, if there is a way to elegantly do it without that
overhead, I'd like to do it. It's a terrible waste to put two
southern-engineered wall-warts into play (there won't be any
electrical outlets, this will all be hard-wired) when there might be a
better, more economical solution.

It's interesting that you use the word pathological because it
describes a mentally disturbed condition. I say it's interesting
because you have a mentally disturbed condition, Phil. Listen, you
chastised me for a fair question once before and I didn't get all
nutty on you when you replied to me today because you seemed to have
some useful information. I'm willing to let bygones be
bygones...chalk it up to you bein' on the rag at the time or
whatever. But this is twice, Nut Job. Listen to me carefully, I'm
sorry you have a small dick, or you're fat, or you have beady eyes, or
AIDS or whatever is wrong with you. For whatever you have that you
think is wrong with you, I'm sorry, but I didn't make that happen, so
don't take it out on me. I come here seeking information, prostrating
myself before those who know more. I don't need you telling me with
callouts that I'm a *novice*. I know I don't know it all. Hence the
post here. It's called LOGIC, I highly suggest you look into it.
Listen. I read your posts from years back, prior to when you first
roasted my ass here...you used to be a nice guy and handle questions
with some respect and knowledge and care. I don't know what happened
to you...your wife left you, died of cancer, maybe. Something bad
happened, or maybe your mind just rotted...some kind of disease.
Maybe you woke up one morning and decided you like other men. I don't
know...but your latter posts are full of hate and self-loathing thinly
disguised as sarcasm to those of us online who seek knowledge from
those we believe know more than us. To attack us for not knowing is
pathetic. Get a cup of coffee. Find a comfy spot in the house (tell
your mother to get her fat ass out of the way if she's already in that
spot) and take some time to think about yourself and your goals and
your life and get your head right with yourself. Really, man,
straighten your life out. Don't waste our time and don't take it out
on us.

And Merry Christmas!

--HC

 

"Halfwit Cunthead "

No, nothing pathological.


** Oh I am absolutely sure it is !!!

Cos YOU re FUCKING NUT CASE

Electrocute yourself - ASAP

****.


..... Phil



BTW:

You cannot legally hard wire home made devices to the AC supply.

Fuckwit.

 

LOL! You are so weak! Really, laughing out loud! Oh, and they say
entertainment bargains don't exist anymore! Ha! This is good stuff!

My sides hurt or I'd write more....

--HC

 

Hey, Paul, thank you for your reply. I did not use a polarized
capacitor unless I'm a complete doofus (I tried to be careful about
choosing a non-polarized capacitor and got, according to the baggie
before me, a "xicon radial metalized polyester film capacitor").
That's from Mouser's baggie description. FWIW. I believe that is not
a polarized capacitor.

I would gladly go to a small wall-wart but here's the problem I see:
I have 120VAC supply that needs to supply my watch-dog 555 timer
circuit, my 5VDC relay, and also a separate 120VAC circuit that is the
trigger. It goes like this: a 120VAC circuit comes to life
periodically. It is energized for a while and then shuts off. When
it shuts off I want my device to come to life and run for a while,
powering another 120VAC device. So, I thought a 555 mono would do for
the "come to life and run for a while" thing, and it would power the
coil on a 5 VDC relay which would switch the 120VAC to my newly added
device. But that means I need a way to feed the 555 a trigger that
follows the 120VAC feed to the first device...and that means I would
need two transformers: one for the power to the logic circuit/relay
which needs constant power and one to feed the 555 a state signal (off-
on-off). I could do it, and I suppose the space is available
but...the environment is such that less space is better and the
environment is slightly corrosive so it'd be great if the whole thing
was small and could easily be encapsulated in epoxy or some other
protection. :-/

Thank you for your time.

--HC

 

How much power are you interested in pulling from this circuit?

You can analyze the circuit by using the 'phasor' method, where the
impedance of the cap is Z = 1/(2*PI*f*C)

Using Zout = sqrt(Zc^2+R^2), the current will be

I = 120/sqrt(Z^2 + R^2) ~= 67mA

The power dissipated in the resistor is therefore 1.23W. So, the
resistors will get hot, but should not fail.

The other thing to note is that the zener is passing any current your
circuit isn't using. If you have a 1k load, then the 5mA through that
will lower the current through the zener to about 60mA. Thus, the
power dissipated by the thing will be

60mA * 5.6V / 2 = 168mW

So, you need a 1/4W zener.

Since the power only flows into your load 1/2 of the time, however,
you can only use 1/2 the maximum current above, a whopping 33mA.

You should also use a fuse with this circuit to prevent a fire if the
cap fails closed. Use a 'slo blo' fuse, because the initial blast of
current to charge up the cap can be something like 120/270 = 444mA. A
slow blow 250mA fuse should be fine.

The maximum voltage across the cap is about 170V in both directions,
so you need either a single non-polar cap that is designed to
withstand that voltage (and more), or two polar caps that are arranged
in series, facing opposite ways, and that can withstand twice that
voltage (initally; it'll eventually calm down to 170V due to leakage).

Note that a little transformer will be safer, more efficient, and much
cooler than the circuit above. You can get fairly small transformers.
To get equivalent current to the circuit above, you would need about a
1/2VA transformer. The page below has a 1/2VA transformer that is less
than an inch tall. See the 161C10 here:

http://www.hammondmfg.com/pdf/5c0011-12.pdf

It is less than 1 in on a side.

You can get 100mA at 5V out of it if you wire it in parallel, rather
than the 33mA you'll get from the circuit you picked.

Regards,
Bob Monsen

 

they will get hot there is no way to cure that that doesn't
compromise the circuit, if it produces too much heat you could try
reducing the capacitor.

the circuit shown is unsafe in that it can present a high voltage on
the plug after it's unplugged.

use a transformer instead

you could use a switched mode-powersupply but they are complex to design
and build so you'd be buying one pre-built, possibly you could
scrounge one from an old phone charger etc...

bye.

 

Hey, Bob, thank you for the reply. Okay, I got a reply from the
creator of that schematic I referenced and between what he said and
what you say I understand this a lot better. He suggested using
capacitive reactance so I looked up a calculator for that and tried
some things (I did a lot of reading between yesterday and today) and
here's what I came up with: 120 VAC / 1,768 ohms capacitive reactance
(1.5 uF @ 60 Hz) = 68 mA. You took the time to post the formula so I
did it that way, too: 1/ (2 * pi * 60 * 0.0000015) = 1,767.67 Okay,
now divide the voltage by the capacitive reactance squared multiplied
by the resistance (is that the restance of that resistor in series
with the capacitor, the 270 ohm one in the circuit I referenced?):
120 / sqrt(1768 ^ 2 + 270 ^ 2) = 120 / 1788.5 = 67 mA. And that gets
us to the dissipation of 1.23 watts (0.067 ^2 * 270) 1.21 (I rounded
my numbers off as I went along).

Okay, so now I know a lot more than I did when I started. The relays
I bought consume 66.5 mA (measured) which means this circuit won't be
able to supply the power I need for my logic circuit and the relay.
So, a transformer is a guarantee now. Thank you for the reference to
the PDF, that thing should work fine.

However, I still need a way to trigger my 555 from the 120 VAC circuit
I'm trying to monitor. So, using the information above, could I do
this: put two 1.5 uF caps in series for (1.5 * 1.5 / (1.5 + 1.5)) =
0.75 uF capacitance, with a capacitive reactance of 1/ (2 * pi * 60 *
0.00000075) = 3,535 ohms. 120 / sqrt (3535 ^2 + 270 ^2) = 120 / 3545
= 34 mA which gives me 0.034 ^2 * 270 = 312 mW, about one fourth the
power I was dissipating before, well within the energy dissipating
rating of two parallel 560 ohm, 1 Watt resistors.

I could then feed that into my 555 trigger and power the 555 and relay
from the transformer. The Zeners I have are 1 Watt, so I should be
plenty safe there (34 mA * 5.6 / 2 = 95.2 mW).

I can put a fuse on it as you suggest, no sweat.

Am I doing this right? I could even use some kind of an optocoupler
between the 120 VAC bastardized "transformer/trigger" circuit and the
555 logic circuit so they wouldn't have to share ground and there'd be
no way the 120 VAC could come to full line-level through the 555 logic
circuit. Sound right?

Thank you for your time.

--HC

 

Hey, Jasen, thank you for your reply. I have gotten a lot more
information now and have found that I will have to go with a
transformer for my logic circuit and the relay. However, I still need
a way to trigger the 555 when the 120 VAC monitored circuit cycles
from off to on to off. I put up another post detailing my intent.
Basically, if I put two capacitors in series (I already bought the 1.5
uF caps) I can drop the capacitance and reduce the current, put a fuse
on this thing, and trigger my 555 circuit with it. This thing will be
hard-wired but your point is well taken about the cap retaining
voltage. A few schematics I looked at had a 1 meg resistor in
parallel with the cap to discharge it and I will definitely do that.

Bob had replied with a link to a tiny standard transformer that will
do what I need. The relay I have chosen draws 66.5 mA (measured), and
the 555 circuit draws something like 2 mA (measured) so the
transformer that Bob suggested, with a capacity of 100 mA, should work
fine.

Yeah, the switch-mode would be a pill to design and my circuit is so
silly-simple that the power supply would be about 95% of the project
effort.

Thanks again.

--HC

 

Lot of work. Programmable time delay relays with 120V AC supply,
control and load circuits are easily available sub $100.

However if you *must* build it, consider rolling your own optoisolators
with neon bulbs and phototransistors. You *could* also get there with
simple resistive droppers into optoisolators (two in parallel with the
source side LEDs in antiparallel so they conduct on alternate half
cycles for a fairly steady oytput or a shunt diode for reverse
protection accross the LED if you dont mind a pulsed output. Direct line
derived low voltage supplies are over-complicated for simple on-off
sensing.

A commercial product would probably be non isolated with direct line
sensing via a resistive dropper but the hobbyist without a dedicated
isolation transformer fed workbench is well advised to stick to projects
with integral isolation and with the non-isolated parts of the circuit
kept as simple as possible. The experianced skip-rat would scavenge an
old mobile phone charger for the isolated low voltage supply and look in
larger switched mode power supplies for suitable optoisolators or roll
their own as above.

 

You need a one-shot timer which can be done via a 555.

The remote relay or signal will simply hold the Reset line of
the timer low while the first device is on!..
When the first device turns off, it'll release the hold on the
reset line..

Look up one-shot timer..


http://webpages.charter.net/jamie_5"

 

I have been using direct line operated supplies for years. In most cases I
use a .047uF 600 volt cap as a current limiter. In order to keep the
current requirements lower I use a low current 24 volt relay. Tell me again
what you want to achieve via e-mail and I can give you a circuit and relay
that works. I must apologize for the response you got from the news group.
.

 

A single wall wart can do it.

Connect a 5.5 V regulated wall wart like Allelectronics CAT#
DCTX 532 - to the AC input to the existing device at points
A and B below.

---------- ---------
|Existing |---A---|Existing |
|Switching | |Device |
|Circuit |---B---| |
---------- ---------

The circuit below will give you an adjustable delay to the second
device you want to add - maybe long enough. You didn't say how long
the second device should run, so that's why there is a maybe.

The 5.5V in the diagram below comes from the 5.5V wall wart plugged
in to points A and B above.

+5.5 ---+----+-[D2]-[16R]-+----+----+
| | | | |
[Rly1] [D1] [Rly2] [C1] [Rx]
| |a | | |
Gnd ----+----+------------+----+----+


AC----o o--------o o-----AC to second device
| | | |
|< | <
Rly1 N/C Rly2 N/O

AC------------------------AC to second device

Use a 1 Farad super cap for C1 (CAT# CBC17), a 16 ohm 5 watt
resistor, 1N400x diodes and 5 volt relays with contacts rated
to switch power on/off to the second device. The maximum time
you can get the second device to run depends upon the coil
resistance of Rly2 - the higher the resistance, the greater
the time. If you don't get enough time you can add another super
cap in parallel with C1. If it runs too long, you can reduce the
time by adding a resistor at Rx.

When the existing device turns on, the 5.5 volt supply turns
on and energizes Rly1. The normally closed points on Rly1
open, preventing AC from reaching the second device you are
adding. Shortly after Rly1 energizes, Rly2 energizes. When
power drops to the existing device, it also drops to the
5.5V supply, and Rly1 drops out, and its points close completing
the path for AC to reach the added device. Rly2 stays energized
while the super cap discharges through it, until the cap voltage
drops low enough for the relay to drop out. As long as Rly2
is energized, the second device gets AC through the closed
point on Rly1 and the normally open point on Rly2. D2 prevents
the cap from discharging through Rly1. It depends on the
individual relay, of course, but in general terms Rly2 will
remain transfered for a bit over 2 time constants, and with
a 1 Farad super cap, the time constant will be about equal to
the relay coil resistance. Figuring a 130 ohm relay coil, the
second device would be on for about 4-5 minutes. If you add a
resistor (Rx), that will reduce the time constant. You need
to compute the power dissipation for Rx if you add it.
P = I^2R, I = 5/R

Parts are available from Allelectronics:
http://www.allelectronics.com/

Ed