wall@fortune.UUCP (Jim Wall) (03/22/85)
*** REPLACE THIS LINE WITH YOUR CIRCUIT ***
I have a tough question here. I need a circuit to indicate
the presence of a load on an AC line. Wait, it gets more difficult.
The load can vary from 100 watts to 1000 watts, and the output
signal must be a TTL level. Also, it cannot drop the 110 volt
AC line by very much. One more thing: the ground of whatever
circuit is used will be tied to the neutral line of the AC, so complete
isolation is not possible.
I have been thinking along the lines of an inline inductor,
you know: V = L(di/dt); as opposed to an inline resistor. The
resistor will drop too much voltage at 10 Amps or not produce enough
voltage at 1 amp.
Any ideas? This is trying to be low cost, so the simpler the
better. Good Luck Mr. Phelps. Your terminal will self destruct in
10 seconds.
-Jim (10 9 8 ...) Wall
...amd!fortune!walljs2j@mhuxt.UUCP (sonntag) (03/22/85)
> I have a tough question here. I need a circuit to indicate > the presence of a load on an AC line. Wait, it gets more difficult. > The load can vary from 100 watts to 1000 watts, and the output > signal must be a TTL level. Also, it cannot drop the 110 volt > AC line by very much. One more thing: the ground of whatever > circuit is used will be tied to the neutral line of the AC, so complete > isolation is not possible. How about wrapping a few turns around one of the AC power lines, and connecting the ends to the coil of a low-voltage AC relay? You could pick up your TTL levels off of the contacts of the relay. I'm not sure how many turns you'd need, but I doubt if you'd need more than 10-20 if your relay was reasonably wimpy. Well, you *did* ask for *simple* designs, didn't you? -- Jeff Sonntag ihnp4!mhuxt!js2j "War is peace."-the ministry of truth
jj@alice.UUCP (03/23/85)
It's not necessary to have any connection at all. You can wrap a few turns on a small core (ferrite) that's been sliced in half, and then glue it back together around ONE of the conductors in the wire. It varies as to what voltage you get, but it's usually well within what any grunge op-amp can amplify to a big enough signal to rectify and drive a schmidt trigger. You want to put a pair of diodes across the coil, and a fairly large resistor in series with in (and then take the signal across the diodes). I really don't have time to draw it out, but the problem is conventional and fairly easily solved... JJ -- TEDDY BEARS NEED SECURITY BLANKETS ONCE IN A WHILE! "... John? Who'd of thought it! ..." (allegra,harpo,ulysses)!alice!jj
dsi@unccvax.UUCP (Dataspan Inc) (03/24/85)
Use a current transformer * AROUND * the conductor in question, and
an op-amp with a diode connected in the feedback loop for the required
impedance transformation and rectification. If you want to get terribly
fancy, a comparator at the output of said integrated and rectified current
transformer will allow you to set the TTL "load trippage" level or whatever.
The design of such transformers at AUDIO frequencies should be available
from any decent textbook on AC machines, etc. These kinds of current sampling
devices are often used at radio frequencies for sampling the magnitude and
phase of antenna current; etc. Of course, be sure to put whatever transformer
you make around just ONE conductor.
dyazben@umd5.UUCP (03/25/85)
Does not sound that difficult. I suppose I am paranoid, but anything
connected to both line 125 and TTL gives me the willies, so I would use
a (gasp) relay. You might be able to make do with an optoisolator though,
if you distrust moving parts.
The key here is that you said it would draw from 100 to 1000 watts. If the
minimum power drawn is 100 watts, you will be pulling at least 3/4 of an
amp. I would sense the voltage across a SMALL shunt (like 1/100 ohm) with
a sensitive relay. If the shunt is that small, even at 1000 watts, about
10 amps, the droop will only be 1/10 of a volt.
Strangely enough, I am looking at building a similar circuit for a friend's
video system. I need to sense if the VCR is on, and if so supply power to
the TV set and stereo system. The object of the game is to make the whole
thing go on and off with the VCR remote control box. I will only have to
do this if the VCR he gets does not have a switched outlet.
They used to have these boxes commercially available, for turning your
stereo off when the turntable gets to the end of the record and shuts off.
I sensed the current drain of the turntable through a special outlet.
I don't think they had them big enough to handle 10 amps though - that
would be one heck of a stereo :-)
LEDs are current driven devices and don't need that much. How about
paralleling the LED in an optoisolator with a shunt that would put the
maximum of 50 mills through the LED when the load is drawing its rated
maximum of 10 amps, and use the output of the optoisolator as the TTL side?
I don't know if 5 mills (at 1 amp) is quite enough. Perhaps something
like a zener diode or constant-current diode could be used to fix this.
We are looking at something like:
-----------------/\/\/\/\/\/\--------------------
AC line | R1 | LOAD
| |
--\/\/\/----|<---------
R2 opto
isolator
where R1 is 1/100 ohm and R2 is about 2 ohms. Get the rateings of the
particular optoisolator you want to use and arrange the resistors such
that when load current is max, optoisolator current is at its rated max.
Then see if it works down to 100 watts (one amp).
By the way, the circuit as drawn will chatter at 60 cycles. I suggest
you remove this with a long time-constant filter at the TTL level...
Although you might be able to do something at line levels, why take the
risk...
--
Ben Cranston ...{seismo!umcp-cs,ihnp4!rlgvax}!cvl!umd5!zben zben@umd2.ARPAzben@umd5.UUCP (03/25/85)
Oops, I forgot that LEDs don't shine at all until you overcome the band-gap
energy (which is .7 volts for silicon). Relays don't have these kinds of
problems... :-)
Well, if you make the shunt big enough to drop .7 volts at 1 amp, it will
drop 7 volts at 10 amps, and that is probably too much. The only other
approach I can suggest is to find an old filament transformer (remember
those?) of 10 amps capacity (maybe 5 or 6 volts) and use it backwards as
a current sensing transformer. Something like:
-----------------------()()()()()()()()()()--------------------
Line ==================
()()()()()()()
| |
sense here
With the 5 volt side in series with the load and some kind of sensor on
the 110 volt side. Problem areas:
Secondary, being designed for 5 volts, might not have 110 volts worth of
insulation to the case of the transformer. This would be a bad failure
mode...
Impedance of the sensor circuit would be reflected back into the line
circuit by the turns ratio of the transformer. You should make this
impedance as small as possible.
A note on removing 60 cycle chatter - use TTL level to continuously
retrigger a 74121 multivibrator set up to maybe 1/10 to 1 second time
constant. Its a one chip solution.
--
Ben Cranston ...{seismo!umcp-cs,ihnp4!rlgvax}!cvl!umd5!zben zben@umd2.ARPAzben@umd5.UUCP (03/25/85)
-----+--------/\/\/\/\/\/\/\/\---------+-------- Line | R1 | Load | | +--\/\/\/\/\/-----()()()()()()----+ R2 ============ Isolation transformer ()()()()()() | | ------- R3 |\ | +--| ~ + |--+--/\/\/\/\--+----| >-- TTL out | | | | | |/ | | | --- + \---\ Schmitt trigger | | | --- C1 Z1 / \ Gate (74??) | | | | --- | | | | | +-------------| ~ - |--+------------+ ------- Full wave Bridge This circuit has the advantage that the impedance to the load can never be greater than the value of R1. R2 limits the fraction of the load current that goes thru the isolation transformer. The bridge rectifies and C1 filters the 60 cycle. R3 limits the current thru the zener diode Z1. The zener diode Z1 (~4v) ensures that the input to the gate can never get higher than the +5 supply rail. Gosh, I wonder if it would really work.... :-) -- Ben Cranston ...{seismo!umcp-cs,ihnp4!rlgvax}!cvl!umd5!zben zben@umd2.ARPA
jpexg@mit-hermes.ARPA (John Purbrick) (03/25/85)
The toroid is the way to go. But you don't need to cut it--just pass one lead (not both!) of the AC line through the hole, and wrap "some number" of turns of wire around the toroid then lead the ends off to the inputs of an op amp or better still an instrumentation amplifier. You will have created a crude transformer, where a magnetic field in the toroid will be induced by current in the ac line, then coupled into the "secondary" winding. Turning the output of the amplifier into a ttl level is left as an exercise for the reader. But I'm with the folks who get nervous as all hell about connecting in any way to the ac line! What if the damn thing ever gets plugged in backward? A more complicated version of this circuit would involve sawing a slot in the toroid and slipping a hall-effect (magnetic field) sensor into it, like the Sprague UGN3501T--that way you get a voltage output. I've done this and it works pretty well with both AC and DC.
ron@brl-tgr.ARPA (Ron Natalie <ron>) (03/26/85)
> I have a tough question here. I need a circuit to indicate > the presence of a load on an AC line. Wait, it gets more difficult. > The load can vary from 100 watts to 1000 watts, and the output > signal must be a TTL level. Also, it cannot drop the 110 volt > AC line by very much. One more thing: the ground of whatever > circuit is used will be tied to the neutral line of the AC, so complete > isolation is not possible. > How about building a current sensor by running (one side of) the cord through a coil. You could then sample the output with a comparator circuit. This is how my PDU in the computer room works. -Ron
dr@ski.UUCP (David Robins) (03/27/85)
> > I have a tough question here. I need a circuit to indicate > > the presence of a load on an AC line. Wait, it gets more difficult. > > The load can vary from 100 watts to 1000 watts, and the output > > signal must be a TTL level. Also, it cannot drop the 110 volt > > AC line by very much. One more thing: the ground of whatever > > circuit is used will be tied to the neutral line of the AC, so complete > > isolation is not possible. > > > How about building a current sensor by running (one side of) the cord > through a coil. You could then sample the output with a comparator > circuit. This is how my PDU in the computer room works. > > -Ron You can also do two other things. Popular Electronics, and Radio-Electronics have had circuits for stereo system power switches. One techniques is to run the load current through a pair of diodes (large enough to carry the current) in parallel , but reversed. What this does is to show the diode voltage drop (about 1 volt or so) on bot h halves of the AC cycle, whenever the load is on. This voltage is therefore 0 when off, and about 1 volt over most of the 360 degree sine-wave cycle. A differential-input sensor picked up this voltage and used it to close a relay, to power the rest of the stereo system when the turntable was turned on. A second technique was to use a transformer in series with the load to act as an isolated current transformer. I belive they used a 120 to 6.3 v. power transformer. Current from the load was passed through the transformer's SECONDARY (the 6.3 v. side). Since the load resistance dropped the voltage, the transformer saw little voltage. This current in the secondary created a voltage on the primary, which was then sensed. The secondary current rating of the transformer should be high enough to carry your load current. I like the first solution better, because the voltage to be sensed is not related to the load current, in the case of the varying loads you referred to. If you need references to the articles, I *THINK* I cut them out of the magazines, and can dig them up. Send me mail if you need them. -- David Robins, M.D.; Smith-Kettlewell Institute of Visual Sciences 2232 Webster St; San Francisco CA 94115 415/561-1705 {ucbvax,dual,sun}!twg!ski!dr dual!ptsfa!ski!dr
jans@mako.UUCP (Jan Steinman) (03/28/85)
Here's one I cooked up for turning all the bits and pieces of my computer on whenever I turn on the desk lamp. It's independent of power level (up to the current rating of the rectifiers), has a moderate, but constant drop (~3.0 volts), has no moving parts, and uses readily available parts. (Radio Shack variety) Bridge Rectifier +-------+ | BR1 | o---------------+~ ~+------------------o Line | + - | Load +-------+ +-------+ +----------+---|--------|+ +|------o TTL current sink | | | | | | D1 v | | IC1 | | - +-\/\/\--|- -|----+ | | | R1 +-------+ | | D2 v | v TTL system ground | - | | C1 | | +---)|-----+---+ The bridge rectifier, D1, and D2 should have a current rating greater than the the maximum load current. BR1, D1, and D2 should have PIV ratings of at least 250 for safety. C1 is not critical: 1uf @ 5VDC will do. R1 should be chosen to provide the needed current to the opto-isolator IC1, 180 ohms should do. Most LEDs (like the type used in opto-isos) have a threshold of about 1.3 volts, therefore two silicon drops (D1 and D2) are required in order to get the opto-iso to fire. I used a MOC3010 opto-iso (which has a triac output) to trigger an 8 amp triac and built the whole thing INSIDE a Radio Shack outlet box. Different opto-isos provide simple (or darlington) open-collector outputs. This circuit works reliably from miliwats up to the current limit of the rectifiers, 20 amp bridges and rectifiers are not uncommon or expensive. -- :::::: Jan Steinman Box 1000, MS 61-161 (w)503/685-2843 :::::: :::::: tektronix!tekecs!jans Wilsonville, OR 97070 (h)503/657-7703 ::::::
padpowell@wateng.UUCP (PAD Powell) (03/29/85)
In article <2338@mit-hermes.ARPA> jpexg@mit-hermes.ARPA (John Purbrick) writes: >The toroid is the way to go. But you don't need to cut it--just pass one lead >(not both!) of the AC line through the hole, and wrap "some number" of turns >of wire around the toroid then lead the ends off to the inputs of an op amp >or better still an instrumentation amplifier. You will have created a crude Ahem. Why not simply use a pot core? It is easier, comes in two pieces, and has a slot for your sensor. Also a nice little spool for winding a couple of dozen turns of wire on? Patrick Powell
fish@ihlpg.UUCP (Bob Fishell) (04/04/85)
> > > I have a tough question here. I need a circuit to indicate > > > the presence of a load on an AC line. Wait, it gets more difficult. > > > The load can vary from 100 watts to 1000 watts, and the output > > > signal must be a TTL level. > A second technique was to use a transformer in series with the load to > act as an isolated current transformer. I belive they used a 120 to > 6.3 v. power transformer. Current from the load was passed through > the transformer's SECONDARY (the 6.3 v. side). *** AC T YOUR AGE *** Following this discussion has gotten me interested in such circuits, and I have, indeed, put one together that does the trick, using an ordinary power transformer, as the excerpt above suggests. I had an old 3 amp power transformer laying around, so I thought I'd experiment with a circuit wherein the transformer's 6.3 v coil is in series with the load: ,-------- AC ---------:||( ^ line --: )||( | | )||( Vsec | )||( | | --'||( v | | `-------- `--load Vsec, open circuit, is fairly high with a 100-watt bulb serving as the load, but a 150-ohm resistor pulls it down to a level suitable for driving CMOS. Voltage drop accross the primary (the 6.3v coil) is negligible, and there's no heat buildup after several minutes of operation. The complete circuit full-wave rectifies Vsec and uses it to drive a power transistor which, in turn picks up a relay that controls another 110v circuit. A 500-ohm pot controls the base current so that the relay is picked up when the load current (and hence Vsec) reaches a desired level. I use it to control power to my TV set, which lacks remote control, by the power supplied to my VCR, which has a wireless remote. /_\_ Bob Fishell ihnp4!ihlpg!fish
gv@hou2e.UUCP (A.VANNUCCI) (04/05/85)
>> I have a tough question here. I need a circuit to indicate >> the presence of a load on an AC line. Wait, it gets more difficult. >> The load can vary from 100 watts to 1000 watts, and the output >> signal must be a TTL level. > A second technique was to use a transformer in series with the load to > act as an isolated current transformer. I belive they used a 120 to > 6.3 v. power transformer. Current from the load was passed through > the transformer's SECONDARY (the 6.3 v. side). What you really need is a transformer with a very-low-voltage, high-current secondary. For example, a transformer with a 110 V primary and a 2 V, 10 A secondary. Place the secondary in *series* with your AC load and connect the primary to a 600 ohm resistor. This is equivalent to placing a resistor in series with your AC load, whose value is 600/(55^2), where 55 is the transformer ratio. With this setup the voltage drop for your load will always be small, at most 2 V when your load is pulling 10 A; however your secondary will provide you with a sizable AC voltage even when your load is pulling a low current: from 5 V rms when you load is pulling 500 mA up to a maximum of 110 V when the load is pulling 10 A. The source impedance of this signal is 600 ohm, i.e. the resistance of the resistor applied to the transformer primary (which is actually acting as a secondary). Well, but you want a TTL level, and a 110 V signal when the load is pulling 10 A is a touch too high for TTL chips. So, put a limiter (a zener diode will do fine) in parallel to your 600 ohm resistor. Actually, now that you have the limiter there you don't even need the 600 ohm resistor any more. If there is any current flowing through your AC load, there will be a proportional current (reduced by the transformer ratio) flowing through the limiter. The voltage waveform across the limiter will be just about exactly a square wave between the two voltage levels where the limiter limits. If you use a 4 V zener (a good choice to get a TTL level) with the anode connected to ground (together with one end of the transformer primary) the other end will oscillate between ~ -.7v and ~ +4V. Since the voltage across the zener is a square wave the power dissipated in the zener can be calculated by multiplying the average (not rms average) current by the (unsigned) average voltage. The worst case occurs when the load is pulling 10 A rms. With a transformer ratio of 55 the zener must dissipate ((10*sqrt(2)*2/pi)/55)*(4+.7)/2 = 385 mW no big deal !! An extra bonus of using the zener is that now the voltage drop for your AC load is truly negligible. If the voltage across the transformer "primary" oscillates between -.7 V and 4 V, the voltage across the "secondary" (which is in series with the AC load) will oscillate between -.7/55 V and 4/55 V !!! Also, the transformer does hardly any work; if you use a 110 V transformer all that iron in the core is really wasted as the primary only sees a few volts. So, you might as well make your own transformer. Use a ferrite core with a couple of turns of wire big enough to carry 10 A and 100 or so turns of smaller wire. Giovanni Vannucci AT&T Bell Laboratories HOH R-207 Holmdel, NJ 07733 hou2e!gv