Appliance LED indicator

Hello,


Sometimes it would be nice to have an LED indicator for some common
home appliances so it's possible to know when they are running
or not. An LED indicator is easy to connect to appliances that
draw low currents by connecting a series resistor of the right
value and another resistor to limit LED current and an anti-
parallel diode to protect the LED from reverse voltages, but
what about for the higher current appliances that draw 5,10, or
even more current when operating? Sometimes it's possible to
open the appliance and add something to it to allow it to
light up an LED when it's running, but usually
it's not very desirable to modify the appliance directly.
Also, it would be nice sometimes to have a stand alone indicator
that could be used with any appliance.

The Series Resistor...
The series resistor idea doesnt work very well with high current
appliances because the size of the resistor becomes too large
and it gets too hot while the thing is running. It also could
drop an awful lot of current with some appliances, so
another idea is to use what is commonly called a
'current transformer'.

The Current Transformer...
Believe it or not, we all have several current transformers
in-the-making lying all around the house doing nothing
All we have to do is take one of those dc (or ac) wall
wart adapters and convert it. This means whereever they
sell wall warts they also sell current transformers!

Converting a standard wall wart into a current transformer...
To convert, follow these steps:
1. Break apart the plastic package so you can get to the
transformer itself.
2. Remove the 'primary' winding. This is the one with the
very fine wire. Sometimes a hacksaw helps, but be careful
not to damage the plastic bobbin the wire is wound onto.
3. Create a new 'primary' winding by wrapping 1 turn of
approximately #16 gauge wire around the core. This winding
should go through one side, around the back, and up though
the other side.
4. Connect a 10 ohm 10 watt resistor to the original secondary.
Connect the primary in series with an appliance that draws
approximately 10 amps. Measure the ac voltage across the
10 ohm resistor. Compute the turns ratio by dividing 100
by this voltage: Tr=100/vac
Compute the output current by io=vac/10.
Compute the excess current ie by subtracting the LED current
from io: ie=io-0.020 (20ma LED).
Compute a current diverging resistor value Rcd by taking the ie
value at the LEDs nominal voltage. Lets say the LED is 20ma at
2 volts, then Rcd=2volts/ie. Connect this resistor in PARALLEL
to the LED. The power rating for this resistor can be computed
from the LED voltage squared divided by Rcd, or P=(2^2)/Rcd.
5. Connect a 1N4002 (or similar) diode in anti-parallel to
the LED.
Remove the 10 ohm resistor, then
connect the combination of LED, resistor, and diode across the
wall warts original secondary winding.
6. Connect the primary again in series with an appliance that
draws 10 amps and check the current though the LED. It should
be around 20 milliamps.

Extra:
In step 4 above we computed the excess current by subtracting the
LED current from the output current. If the output current
happens to be LESS then the required LED current, then we need
to add more turns to the primary and then repeat step 4.
If it's say only 10ma and we need 20ma we simply double the
number of turns (add one more turn).


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

Part 2: Measuring appliance current using the current transformer.

If instead we wish to use our current transformer to measure the
current draw for an appliance all we have to do is skip all the
steps that mention the LED. Instead, we keep the 10 ohm resistor
connected and measure the ac voltage across it while the device
is running. If at 10 amps we measure 2 volts ac, then our scale
is 1 volt per 5 amps, or 0.2 volt per amp. If we switched to
using a 5 ohm resistor instead we would get 0.1 volt per amp.


Benefits of using the current transformer:
1. Dont have to use power resistors in series which might get hot.
2. The output is isolated from the line voltage reducing the
risk of shock. This also makes it easier to use with some
computer monitoring equipment.
3. The equivalent series impedance is very low so it doesnt
drop the voltage getting to the appliance which could
hamper the proper operation of some common appliances.
4. It's very portable...you can use the same one on several
appliances by simply inserting it into one wire of an
extension cord. Plug the cord into the wall and the device
into the socket and you can easily monitor current and/or
device running status.
5. By connecting several appliances to the same transformer you
can monitor the current to all of the appliances at once.
The total output will be the sum of that which each appliance
would produce on its own. This is especially nice when you
need to know if any of a group of devices switch 'on'...if
any one switches on the transformer produces an output.
6. The output can be half wave or full wave rectified producing
a dc output when an ac appliance is running.


Take care,
Al

Mr. Al,
Greetings on the day.

I once did something similar with an environmental test chamber that had failed.

Hooked up a number of neon lamps (the control ckts were all 110v AC) and used them to watch the heaters and other circuits operate. Also I would hang Neon's on liquid freon selonoid valves so I had a visual on when they were being triggered.

So on some consumer devices this might be fun or interesting.

Until you have to hire an outside service dude. Who won't know or like what you have done to a stock unit. They also don't have time to fiddle with user installed electronics no matter how simple. Should the repair dude choose to stay and fix it, expect to pay a preimium for the effort.

Couple of things to consider.

Dirty Power.
The signal that comes out of the outlets is far from clean. Spikes and sags can destroy semiconductors, and LED's are semiconductors. That's why you rarely see just an LED and a large resistor hooked across a 120 ckit.

Cracking wall wart units is a good first start. Using the transformer to reduce the voltage is the better choice. Even better if there are some MOV's to clip spikes.

Another thing is mechanical issues.
Most consumer goods a lot of effort goes into making the item producable and after market items don't slide in well. No holes for wires, no strain releif, sharp edges to cut insulation, the works. So by adding circuits and sub systems you loose some minor advantage of MTBF (mean time between failures)

The last major issue is type accecptance.
In radios it's FCC testing that thumbs up or down any given radio product.

In consumer goods it's people like Underwriters Lab (UL). If a modified appliance lights up and burns down your home, you might just be stuck without recorse. Unlikely, sure but an expensive possibility.

What your outlining might be a fun project, it just may not be a worth while project. It's like buying some kids hot rod and having to figure out what the last owner did to it. When you find the power antenna hooked into the break light.

Have a good one
Jack Crow in Iraq

ps we need an Icon for geting blitzed with high voltage. /ubbthreads/images/graemlins/dedhorse.gif

Hello Jack,

It's kind of funny, your ideas sound like they support the
idea of an external current monitor vs an internal one
but then you knock the idea at the end?

Anyway, if it wasnt worth the effort i wouldnt have done it.

Also, i have one now and it works fine and i can now
measure current to an appliance by using a voltmeter
and dont require a high current ac ammeter.
Plus, it doesnt interfere with the operation of the
appliance.

Also, it saves on having to use a current shunt and
amplifier, which requires a separate power supply.
The current transformer doesnt require extra power to
run it.

If you are worried about the LED getting harmed,
you can always add protection to it (if you are in
fact using an LED, note my original post didnt assume
this).

BTW, it's just a current transformer, and these are used
all the time in industry for monitoring various processes
where the ac current has to be measured.

Take care,
Al

Al,

I may have mis understood the nature of the project. Just what were you trying to do. Some energy efficency project I gather. Perhpas some current draw limit to avoide tripping a demand meter and the resultant utility charges?

As for measuring current draw, mostly I have used an amp clamp and been happy at the chance. Haven't had much need to hand fab current xfm's. Most of what I used to work on would blow 75 amp fuses if a compressor had a flash over.

As for the LED assumption, yeah got bit. Lot of LED fans here and got suckered in. Guilty as charged.

It's first thing in the AM here and waking up ain't as easy as it was when I was 30.

Hope all is well.
Keep it warm
Jack Crow in Iraq

Hi there Jack,

It's early here now too, and im having trouble waking up
today myself so i know the feeling

I see what you are talking about (current clamp on) and
i like that idea a lot. The only thing i dont like
is the price of clamp on's But whats more is, if
you want to monitor several devices without having
to swap meters you can also build one of these current
transformers and connect a cheap analog meter for
each device you want to monitor.

If you wanted too you could build one into a small electrical
box too, with an outlet and extension cord so all you
have to do is plug something into it etc and you are
quickly monitoring the current without having to use
your clamp on.
Some clamp on's require batteries too.

So the motivation is
1. cheap (wall warts can be had at 2 bucks each.
2. no batteries (or other ps needed)

Where i was comming from:
1. Built a 1 milliohm current shunt but it required
a power supply and op amp 200 gain amplifier to drive
a meter (a meter that didnt require batteries for
24/7 monitoring needs). Wanted to get rid of the
extra wall wart for powering the amplifier.
2. Had an old wall wart with shorted primary, so it
wasnt good for anything else anyway so i thought i would
put it to use. After removing the primary it was able
to be used for a current transformer.

Another use that comes to mind:
If you happen to have an electric coffee pot (which i
always need) and it doesnt have an indicator light on it
to tell you when it's done, you can add one of these
current transformer indicators without having to
modify the coffee pot or use up your current meter or
clamp on just for monitoring the coffee
Lets see what the cost is:
Wall wart $2.00
Resistor 0.05
Diode 0.05
LED 0.25
Elect box $0.50
Outlet 0.50
Extension cord $1.00

So for under $5 you can build one
and because of the way it works you never have
to change batteries because there is none.

Take care,
Al

Mr Al,

It's a great method - thanks for posting it! /ubbthreads/images/graemlins/thumbsup.gif

I've used this method to monitor the cycling of two sump pumps and it works well. These are induction motor pumps, and one has a start winding with a centrifugal switch.

The one pump (along with most washers and dryers) draws a huge amount of current while the motor is starting. This might also be the case with a heating type appliance (toaster, space heater, waffle maker, etc) - drawing a lot of power until the heating element got cherry red.

It was necessary for my circuit to withstand these large initial inrush currents while maintaining an even brightness level and not blowing the LED. I, too, wound a 1-3 turn primary from heavy 600V rated insulated wire - being careful not to nick the insulation on the sharp steel laminations of the transformer core. I then connected the "secondary" to a series resistor followed by a 5V zener diode (1 watt) with a normal IN400x in anti-parallel. I chose the series resistor to allow approximately 70+ mA to flow into the zener diode during inrush. I then connected a LED with it's own current limiting resistor to the 5V zener, allowing the LED to run at 20-30mA. This circuit allows for the LED to illuminate at a fairly even level for a large range of appliance current. I, too, didn't want to go overboard, but the addition of a second resistor and a zener diode seemed like a good compromise.

This unit is installed in a small grounded metal box with a single grounded UL outlet recepticle and a short three wire cord with plug. For data logging purposes my visible LED shares the current with the LED from an opto-isolator. The opto-isolator output is brought out to a small connector. This gives me a logic level compatible fully isolated output (to 2.5 KV) that I can hook to a computer for logging usage data.

I agree with you that this is a superior method for sensing actual appliance usage without modifying the appliance.

Current transformers are neat and very useful. They are by far the primary means of measuring current for high current AC applications. One small caution is in order for those who are new to use of them. Current transformers with large winding ratios should never be allowed to operate with their secondary winding open-circuited. If operated with the secondary winding unloaded they can generate high voltages that can damage the winding insulation.

What the other Doug just said!

Commercial current have fixed loads very securely attached (for good reason). If the load goes open (like say you wiggle the wires....) that low resistance, which was reflected as impedance at the primary (current sense winding) modified by the turns ratio, goes to infinity. This means the reflected impedance on the sense winding also goes very very high, meaning that the voltage divide between the sense transducer and the load (say the washer) gets stood on it's head. All the mains voltage 'shows up' across that single turn in the transducer. 120 VAC gets multiplied up by several hundred by the turns ratio and shows up as *thousands* of volts. Equipment damage will be the smaller problem, we'll be lucky indeed not to damage the hapless experimenter.

That said, with proper caution, I too 'like it'. Thanks once again, MrAl!

Doug Owen

Dougs /ubbthreads/images/graemlins/smile.gif,

This interesting effect you pointed out is something I hadn't taken time to ponder. Thanks for the illumination. For a benchtop experiment a while back, I had a normal 120VAC to 12.6VAC @1A transformer with a lot of room left over on the core. I used this as a temporary current transformer by winding two turns over the secondary. The secondary was loaded with a sense resistor, but I left the primary (120VAC primary, that is) open. I'm guessing that since the 12.6V secondary was loaded and showing ~7VAC, the 120VAC windings would have measured ~70VAC had I thought to look at them.

If, however, I had left both the 12.6 and the 120VAC windings open - your posts suggest the 120VAC winding (I'm picking in this winding because of the high turns ratio compared to the 12.6V winding) may have output very high voltage? I can grasp the concept from the impedance standpoint - do either of you have any handy math for estimating open circuit voltage? Basically, in the open winding configuration, the arrangement is analogous to an automotive ignition coil, huh?

Hello again,

Thanks to the two Dougs (Doug e Doug) for bringing up these cautionary
points. Very important!

The golden rule is this:
Always make sure the secondary winding of a current
transformer has some resistance across it, and that
this resistance can never become disconnected even
for a short time period. The resistance must be
low enough to prevent high output pulses from
occurring (see below).

As for knowing the output voltage with an open circuit,
it may be a bit hard to calculate.

Lets say we have 120vac 60Hz source, 12 ohm load (appliance),
3 turns on the primary, and 150 turns on the secondary,
0.3125 square inches of core area, core is typical
grain oriented steel normally used for these things,
and start with a 10 ohm resistor on the secondary.
The turns ratio of the transformer is 150/3=50.

With the resistor properly connected, we have a fairly linear
circuit, with the input current being about 10 amps and
10/50=200ma. 200ma across 10 ohms gives 2 volts rms output at
60 Hz. For reference, 2 volts output means 2/50=40mv across
the input. The input resistance looks like a 0.004 ohm resistor.

Now what happens if the 10 ohm resistor becomes disconnected...

Lets say at first the resistor shoots up to 1Megohm,
that puts the input impedance at 400 ohms now. The
voltage divider effect with the 12 ohm appliance puts
97% of the source voltage (120vac) across the primary
3 turns ! This works out to 116 volts across 3 turns,
and the transformer starts to act as a voltage transformer,
so the output voltage reaches 116*50=5800 volts?
But wait, we have other things to consider...

Using Faradays law, we end up with a neat little formula
for these kinds of transformers:

[EQ1]
E=FAN/175
where
E is the rms voltage across the winding
F is the frequency in Hertz
A is the core area in square inches
N is the number of turns.

This formula tells us the maximum sustaining ac sine voltage
that can be supported by a given number of turns around the
core for a winding.

Lets plug in F, A, and N and see what the max E the primary can
support for this core at 60 Hz. The primary is 3 turns, and
F is 60, and A is 0.3125, so E comes out to equal
0.3214 volts rms, or a peak of about 0.45 volts.
After that, the core saturates badly and the input impedance falls
very low.

So what does this mean?
This means that the primary can not support 116 volts rms with only
3 turns because the primary core saturates and the voltage across
it will tend to zero.
But wait again, this is the voltage at 60 Hz, what about some
higher frequency?
At a higher frequency the primary CAN support 116 volts rms, but
because the 60Hz wave only goes through zero 120 times per second
we will only see roughly one half cycle of the higher frequency
occur once every half cycle of the 60Hz wave. This will look like
a pulse of relatively high amplitude that occurs near the zero
crossing of the 60Hz wave. Also, the primary wont see 116 volts
(164 volts peak) as figured because the driving wave is still
only a 60Hz wave, so the amplitude will be limited by the envelope
formed by the 60Hz, 164v peak sine wave.
To make this problem a little easier, lets assume the higher frequency
wave is a sine wave also (it's not in real life, but we'll approximate
it as such for now). Now we use the low freqency envelope as the
locus of possible solutions for our peak high freq sine wave we
want to find the peak voltage for as well as the frequency so we can
determine an estimate of the pulse.
The level of the sine input at 60Hz will be simple:
Ep=164*sin(377*tp/4)
where tp is the period of the HF wave to be determined,
so
[EQ2]
Ep=164*sin(377/4F)

Solving EQ1 with Ep=1.4142*FAN/175 for F, we get:
F=175*(Ep/1.4142)/AN
so
[EQ3]
F=175*Ep/(1.4142*A*N)

Now substituting EQ2 into EQ3 we get

F=175*[164*sin(377/4F)]/(1.4142*A*N)
or
[EQ4]
F=175*116*sin(377/4F)/(A*N)

Equating EQ4 to zero we get:
[EQ5]
175*116*sin(377/4F)/(A*N)-F=0

Solving for F (i used a nonlinear solver) we get:
F=1428Hz

To get the peak, we solve:
[EQ6]
Epk=164*sin(377*tp) {sin of radians}
where tp=1/4F
For 1428 Hz we get tp=1.75e-4
so
Epk=10.8 volts peak ac 1428Hz

or a 350us pulse every half cycle of the 60Hz input wave.
(4.2% duty cycle of a half 60Hz cycle)

Now for the long awaited output voltage...

With 10.8 volts peak on the input 3 turns we get an output of
540 volts peak for somewhat less then 350us once every half cycle
of the 60Hz input wave!
Since this represents about a 4 or 5 percent duty cycle it
sure is low but could still do damage.

Note we havent included core loading effects, but i'd still
assume a much higher then normal output pulse.


Conclusion:

When the output resistance goes very high although we dont get
the full effect of the full 120vac input and turns ratio, we
still can get an output voltage of very unexpectedly high value
which could easily ruin external equipment. Precautions are
certainly mandatory, but dont let that stop you from building
one because although there are precautions these things are
used all the time.

See again golden rule of current transformers above first though


Take care,
Al