Schematic Interpretation  - Page 1
Forum section: electronics / projects

author: lion
author_rank: Contributor
timestamp: September 14, 2016, 04:42:20 pm
content: Can you tell me what are those diodes doing here in this current sensor project? And the capacitor?Source:

author: MosherIV
author_rank: Super Contributor
timestamp: September 14, 2016, 04:47:38 pm
content: D1,2, 7 and 6 form a rectifier for the AC.C1 is to smooth out the recitifed AC into DCD1, 2 and 3 look like there are there as a voltage clamp (stops the voltage going above 2.1V - 3 x 0.7V )Well at least that is my interpretation.

author: rstofer
author_rank: Super Contributor
timestamp: September 14, 2016, 05:13:39 pm
content: Pretty clever circuit!For the + half of the sine wave, D4, D1, D2, D3 and D6 pass it through to the output with 5 diode drops (probably not important).For the - half of the sine wave, D5, D3, D2, D1 and D7 pass it through to the output with 5 diode drops.In the meantime, D1, D2 and D3 have enough combined drop to charge the capacitor to perhaps 2.1V (give or take) which will be enough to turn on the LED.

author: tatus1969
author_rank: Super Contributor
timestamp: September 14, 2016, 06:27:14 pm
content: but this is not supposed to be thecurrent sensor, I hope. Don't see that function here...

author: retrolefty
author_rank: Super Contributor
timestamp: September 14, 2016, 06:35:10 pm
content: Quote from: tatus1969 on September 14, 2016, 06:27:14 pmWell not current measurement that is for sure, however as a signal that current is flowing or not flowing signal, possibly.

author: CatalinaWOW
author_rank: Super Contributor
timestamp: September 14, 2016, 06:50:55 pm
content: Quote from: MosherIV on September 14, 2016, 04:47:38 pmI think you meant D4, 5, 6 and 7 form the rectifier. I find your approach the easiest way to understand the circuit function. It provides an isolated 5 V logic signal telling whether AC power is on or off. I can think of several applications where this would be useful. I am a little surprised that there isn't a resistor in one of the input leads to limit the current through D1, 2 and 3. Must mean that the specific application had a narrow range of AC power to monitor.

author: rstofer
author_rank: Super Contributor
timestamp: September 14, 2016, 07:18:10 pm
content: This circuit is intended to supply some amount of current to a high voltage load (mains supplying a well pump, for example). The input is AC and, except where the signal gets squared off near zero crossing, the output is AC. The diodes ALL need to pass the full load current, whatever it may be.On the way by, there is a DC voltage created across the capacitor equivalent to 3 diode drops. This provides power for the LED at a sane voltage. Perhaps 2.1V or so.

author: rstofer
author_rank: Super Contributor
timestamp: September 14, 2016, 07:39:21 pm
content: Incidentally, I'm not convinced C1 is oriented correctly...

author: MosherIV
author_rank: Super Contributor
timestamp: September 14, 2016, 07:50:47 pm
content: QuoteLooks correct to me, do the trick where you follow the current round the rectifier for positive and negative volts.

author: tatus1969
author_rank: Super Contributor
timestamp: September 14, 2016, 08:09:47 pm
content: Quote from: rstofer on September 14, 2016, 07:39:21 pmthat - aside of it should be a +, probably lost by image resolution reduction.

author: tatus1969
author_rank: Super Contributor
timestamp: September 14, 2016, 08:14:46 pm
content: Quote from: retrolefty on September 14, 2016, 06:35:10 pmso my guess would be something like a master-slave power strip that activates the slave outlet when a connected device draws current from the master outlet. But the price of 5 diode drops is huge in my opinion, there should be smarter ways to do it.

author: rstofer
author_rank: Super Contributor
timestamp: September 14, 2016, 08:15:07 pm
content: Quote from: MosherIV on September 14, 2016, 07:50:47 pmLook at D4 - the current is coming out of the cathode and heading right to the capacitor. I would think that would need to be the + end of the cap.

author: rstofer
author_rank: Super Contributor
timestamp: September 14, 2016, 08:17:57 pm
content: Quote from: tatus1969 on September 14, 2016, 08:14:46 pmThe opto could run to a uC connected to a LAN and send an SMS message to your cell phone when the basement flood pump starts up. This is a very cool circuit!The LED only turns on when there is current flow. No current flow, no voltage drop across the 3 diodes, no LED.

author: tatus1969
author_rank: Super Contributor
timestamp: September 14, 2016, 08:20:30 pm
content: Quote from: rstofer on September 14, 2016, 08:15:07 pmIt is correct, that is a bitmap export glitch.

author: rstofer
author_rank: Super Contributor
timestamp: September 14, 2016, 08:32:53 pm
content: Quote from: tatus1969 on September 14, 2016, 08:14:46 pm5 diode drops is meaningless in terms of 'mains' voltages. Maybe 120 VAC comes out to be 115 VAC - probably not a big deal. 235 VAC vs 240 VAC?2 of the drops steer the AC around the circuit and 3 are used to generate enough voltage for the LED. Not only do the 3 diodes get a DC component for the LED (which might not be necessary if the emitter circuit is run to a low pass filter) but they also detect current flow.

author: tatus1969
author_rank: Super Contributor
timestamp: September 14, 2016, 08:53:53 pm
content: Quote from: rstofer on September 14, 2016, 08:32:53 pmI am thinking power wise. 5V * 16A = 80W in my power strip example.

author: rstofer
author_rank: Super Contributor
timestamp: September 14, 2016, 09:16:18 pm
content: Quote from: tatus1969 on September 14, 2016, 08:53:53 pmThat's true! It would make a nice little space heater if it was carrying much of a load. Not to mention the size of 20A diodes!

author: JS
author_rank: Frequent Contributor
timestamp: September 14, 2016, 10:57:32 pm
content: I'drather use an AC optocouple, H11AA1 is the one I've used to sense voltage on a load, rather than current on a line. Then you just add 2 diodes in parallel each direction (one wouldn't cut the 1.5V forward voltage on the coupler).Same circuit would work using a conventional optocoupler, 2 diodes in parallel, only one in the oposite direction (if the offset isn't a problem) The only problem is the smoothing cap, which I think for any case should be put in the other side, Time constant long enough, 3 cycles? 50Hz for the conventional optocoupler, 100Hz for the AC optocoupler. You can get away with a much smaller cap. I used 47µF IIRC.For higher currents a shunt could be used, we are stuck with 2V now so 20A would make 40W anyway. The shunt needs to handle all that power, and the sensor will end ignoring lower currents which would be detected if just using beefier diodes.In any case, how much does the components on this costs, and how much does a proper hall current sensor costs? The drop/power is not even considered and you can measure quite well how much it is. The most problematic part I guess is to handle the analog output but in the worst case a comparator does the trick, with easily adjustable sensibility.JS