How Does This Current Sensor Close its switch?

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OK, here it is as described above.
This is an untested design, but will probably work...

The bridge rec type is not critical, eg. a 1A integrated bridge or four 1N4000 series diodes should be fine.

 
Lets take rjenkinsgb's Schematic.
CTs will saturate if the voltage gets too high across the part. I would use 1A Schottky Diodes to reduce the voltage.
A CT acts much like a constant current source. If the load is 3A and the CT is 1000:1 then the secondary will make 3mA. The CT will push that 3mA somewhere.
I am looking at a SS relay G#VM-21. It will close with less than 3mA and 1.33V on the LED. (0.7mA typical 3mA worse case) The data sheet I linked to hours ago has more or different data. It shows the 0.5mA to 3mA turn on current but also shows the part staying on with only 0.1mA of current. This gives me the idea that C2 could be very small. (almost not needed)

If the SS relay or opto isolator is directly across the diodes then the voltage will be less than 2V and C2 could be rated at 6V or less.

The SS relay can handle 30mA max or 1A for a very short time. I think there should be a 3.3V or 4.7V Zenner (1 watt) diode across C2 to take the current if current gets too high. (short & trip the breakers)

R2, should be adjusted to get the right current. If there was no R2 the relay will trip at very low current.

RonS
 
Ive been building an order with Tayda for the past few days. They do not have 4N32M but they do stock TLP240GA (120mA 400V) would this be a good replacement? TLP240GA

 
ronsimpson said:
R2, should be adjusted to get the right current. If there was no R2 the relay will trip at very low current.
Click to expand...
The tools im using are chop saws, planers, routers, ect. Most run at 8-15 amps 120v. the relay could trip at 3 amps so that would be about 30ma.
 
I left out the 1K resistor across the optoisolator LED, that needs adding..


ronsimpson said:
CTs will saturate if the voltage gets too high across the part. I would use 1A Schottky Diodes to reduce the voltage.
Click to expand...
The load sense range was given earlier as 8A to 15A, so 8mA to 15mA out. Schottky diodes certainly would not hurt, but I don't think they are essential.

The capacitor and 100R are to damp switching transients, as well as hold the opto on rather than it pulsing.
Power tools may take several times their rated current for short time when switched on.

I'd say a 6.8V zener across the cap for protection, if anything/

CTs are (or were) often used in thyristor drives for input current sensing, using resistors as loads and a few volts out. The CT output voltage should be no more than 5V at maximum load, using a 1000:1 CT & the values I gave.

ThomsCircuit said:
the relay could trip at 3 amps so that would be about 30ma.
Click to expand...
The CT is supposed to be 1000:1, so one mA per amp. Substituting a 100:1 type would destroy all the components...


The TLP240GA has a lower rated LED than the 4N32; if you use that, do definitely fit a zener across the capacitor to restrict the voltage - 3.9V should be OK to keep it under 30mA.

Note that a bigger capacitor would not hurt, eg. a 220 or 470uF, whichever is convenient.
 
rjenkinsgb said:
The CT is supposed to be 1000:1, so one mA per amp. Substituting a 100:1 type would destroy all the components...
Click to expand...
My mistake. The CT i intend to use is 1000:1
I will create a schematic with all the revisions and post for discussion. I am very greatful for everyone's help.
 
CT DESIGN 01
Not sure where to put ZD1 (4.7volt 1W) & D5 (40volt 1AMP)
Ive decided to use 4N32
CT1 part number ( Vitec 57PR1831 )
 
ronsimpson said:
You know the 4N32 is not like a relay but like a transistor.
Click to expand...
That i did not know.
What i know is which apples hold up during baking and that my chocolate covered macaroons taste just like an almond joy.
Ill place ZD1 parallel to C1 and repost to confirm
 
all resistors are rated at 1Watt
the cap is 16v nothing special. if it helps i can get some BI-Polar 16v by ELNA
 
That current transformer has a maximum load resistance of 72 Ohms. That represents 1.8 V at the maximum input current.

The transformer is physically quite small. I think that means that the magnetic core of the transformer will saturate if the voltage exceeds 1.8 V at 50 Hz.

If you want to use that transformer, I would suggest using an AC input opto isolator, such as a TLP182 https://toshiba.semicon-storage.com/ap-en/semiconductor/product/optoelectronics/detail.TLP182.html. You could connect the input of that straight to the current transformer.

A resistor can be added in parallel to alter the current at which the TLP182 turns on.

The TLP182 output is DC, and you would need a DC power supply, which would be switched by the TLP182, to turn on the 4N32.
 
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