Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10 5/3/83 based; site hou2e.UUCP Path: utzoo!linus!philabs!cmcl2!seismo!harvard!talcott!panda!genrad!decvax!harpo!whuxlm!whuxl!houxm!hou2e!gv From: gv@hou2e.UUCP (A.VANNUCCI) Newsgroups: net.analog Subject: Re: AC current monitor Message-ID: <529@hou2e.UUCP> Date: Thu, 4-Apr-85 19:13:38 EST Article-I.D.: hou2e.529 Posted: Thu Apr 4 19:13:38 1985 Date-Received: Sun, 7-Apr-85 09:07:32 EST References: <5139@fortune.UUCP> <9482@brl-tgr.ARPA> <167@ski.UUCP>, <344@ihlpg.UUCP> Organization: AT&T Bell Labs, Holmdel NJ Lines: 64 >> 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