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How Does This Current Sensor Close its switch?

Current Measurement Range : 0.25 - 200 A
Current Sensor type : Current Switch, AC
Electrical Wiring Size : 24 - 14 AWG (0.2 - 2.1 mm²)
Measurement Type : AC Current

This switch (file hawkeye 800)runs about 15$.
Id like to see if I can build it.
Besides the cost this one is big. Id like to make it smaller. For my purpose 20amps is all that i would require.
Ive looked at IC types but they require DC voltage to function. I like how this uses the current from the device passed through the cores winding. I mean im unsure how it uses that current and closes a switch. I found this device opened (file current amplifier...) but i cannot tell how it functions.
My goal is to understand how this works and make one that does the following...
detects up tp 20amps of AC using the "hall effect"
while the current is 0 the switch is open
when it does detect current the switch closes
when current is 0 the switch opens again
It needs no external power to function
I will be connecting it to SW1 on pcb (file workshop)
This circuit (when SW1 is closed) will close relay K1 after a 0-6 second delay. when SW1 opens relay K1 will open after a 5-15 second delay.
thank you.
 

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ronsimpson

Well-Known Member
Most Helpful Member
What you show uses a Current Transformer, which can produce some power. You propose using a Hall Effect which will use power not produce.
 
The above part by VITEC produces or transforms volts depending on how much amperage is passing through the center. I thought that process was called a hall effect. Now part 57P1831 creates .072 volts per amp. So a question would be what transistor or mosfet would i need to bring the voltage up high enough to trip a switch.? What switch would i use?
 
Last edited:

MacIntoshCZ

Active Member
Current transformer for measuring is easy. If you have 1A on primary 1 turn, 30 turns on secondary - then on secondary will be 30times lower current. U = R*I.
So just pick resistor equal to turns on secondary and you have values equal -> Iprimary = Usecondary
But be cautious about losses. P = R*I*I , or P= U*U/R....
 

rjenkinsgb

Well-Known Member
Most Helpful Member
There are two different things here:

Traditional current transformers are a wound transformer, or a wound secondary and a hole for the primary wire to pass through.
They are AC only. They are most often used for current measurement, either feeding a lower current AC ammeter or a rectifier and load resistor to give a (rough) DC voltage proportional to AC current.


They are extremely dangerous if incorrectly used - they are normally supplied with the secondary shorted and must never be in a live power circuit without a secondary load, as the open circuit voltage can be ludicrously high, many thousands of volts.


The other style is an electronic current transformer.
They typically have a toroidal core with a gap in it, containing a hall effect sensor IC. Those need power and give a voltage or current proportional to the primary current.

Cheap ones just have a sensor, high accuracy ones also have a winding and an internal amplifier used to hold the hall output at zero, then the winding current is precisely proportional to the primary current.

(There is also a so-called "DC current transformer" system used in the 50s - 70s which is wound device that actually works as a magnetic amplifier, with the DC current affecting core saturation. It's nothing at all like the ones above).


You already have a powered circuit; you could use either type with the appropriate load circuit and a comparator IC plus preset pot to set the current that switches at, to feed the rest of your circuit.
 

Les Jones

Well-Known Member
Most Helpful Member
You question is not clear. Do you want the switch contacts to be open at anything less than 20 amps and closed at anything more than 20 amps ? How precise does the threshold have to be ? (For example +/- 1 amp, +/- 10 mA) How much current do the contacts have to carry ? Giving some information on what you are using this device for would be helpful.

Les.
 

ronsimpson

Well-Known Member
Most Helpful Member
You already have a powered circuit;
It needs no external power to function
ThomsCircuit wants to have no external power source.

It appears Tom wants us to design (reverse engineer) this product.

The Current Transformer is 2 inches across. It looks like it I tried to make some thing similar using smaller CTs but the power level was too small. So stay with a 2 inch CT! If you can find one. CT Coilcraft I think this CT has a turn ration of 1000:1. I don't think you can get/make the CT for $7.50 with out the circuit board.
I pulled all the data I can find on " Hawkeye 800 " and it appears they use a solid state relay. SS relay
 

eTech

Well-Known Member
The above part by VITEC produces or transforms volts depending on how much amperage is passing through the center. I thought that process was called a hall effect. Now part 57P1831 creates .072 volts per amp. So a question would be what transistor or mosfet would i need to bring the voltage up high enough to trip a switch.? What switch would i use?
hi

How about explaining what you want to accomplish? Then we can recommend some solutions.
 
There are two different things here:
The information you have shared is extremely helpful as i have read about both types and you have put them in perspective for me. thank you. The style i need is the current transformer as it will produce enough dc voltage to energize a small relay thus closing SW1 on (file workshop)#1
 

MacIntoshCZ

Active Member
The information you have shared is extremely helpful as i have read about both types and you have put them in perspective for me. thank you. The style i need is the current transformer as it will produce enough dc voltage to energize a small relay thus closing SW1 on (file workshop)#1
I dont think i understand what you want.
CT is not used for powering stuff. Unless you dont want them to emit light...
 
You question is not clear.
So sorry. Im such a novice. I will try again.
Let me keep this simple and say i want to create a product like the (hawkeye 800) from post #1
I have 2 of them they are pretty big. Ive used them in the (file workshop) post #1. they work well but like i said they are big and costly.
My goal is to A understand how it works the B make one.
 
It appears Tom wants us to design (reverse engineer) this product.
i do. i do. i like the CT you recommended and i appreciate your research in the hawkeye800.
now im new at this but ive (with the groups help) made a few wonderful circuits i have learned that a transistor can help trigger a relay when voltage is too low. So i feel im getting close here.
For my application im using this in a workshop with 8-14 amp power tools. So the minimum amps needed to trigger the relay would be no less than 8amps. Does that make sense? Meaning a minimum of 8 amps passing through the windings is necessary to trigger the relay.
 

rjenkinsgb

Well-Known Member
Most Helpful Member
member ronsimpson says they do produce some power based on the amount of amps passing through the windings.
Yes; but but "power " as an input to instrumentation; very low levels.

Rectified and limited, you may get enough to turn on a high gain opto isolator.

Remember that the secondary current (not voltage) is directly proportional to the primary current. If you want a 20:1 range if input, the opto drive limiting circuit must be able to cope with 20x higher current than needed for the opto device to turn on, using some form of shunt limiter.


And again, be extremely careful, they are not something to experiment with if you do not fully understand them.

A current transformer _must_ always have a low impedance path for the secondary, so the voltage cannot build up. If the secondary winding current path is interrupted, it will try and become something like a tesla coil & give lethal voltages, until it destroys itself.

Do not use push-in prototyping board or connections, everything must be screwed or soldered so there is no possibility of a disconnection - and everything must be rated for far more than the worst-case secondary current.
 
How about explaining what you want to accomplish? Then we can recommend some solutions.
Hello friend. Great to hear from you.
You assisted me with this workshop project a year ago. I made a few of these. You suggested that i look into hall effect sensors to trigger the switch. that led me the hawkeye800. i purchased it and applied it to the project and it worked well. I simply want to build something similar to the hawkeye800.
I currently use this in my shop now. A line from my chopsaw (9amp) is passed through the hawkeye and when i start the saw the hawkeye energises its solid state relay and closes its switch. two screw terminals atop the hawkeye are connected to SW1 of my workshop circuit which in turn close that circuit. Thats it.
 

rjenkinsgb

Well-Known Member
Most Helpful Member
OK, if you only need a 2:1 current range (rather than the vast range of that commercial switch), that makes it far simpler and more practical.

Get a 1000:1 CT and connect a 2K2, 1W resistor directly and permanently across the output terminals before you do anything else.
That will act as a safety load and prevent dangerous voltages appearing.

Also connect the AC side of bridge rectifier, feeding eg. a 100uF 25V capacitor from the DC side; and from that connect a 100 Ohm resistor and the LED of eg. a 4N32 (darlington) opto isolator in series across the cap. Add a 1K across the LED to discharge the cap with no input.

The transistor output of the opto should switch on with a suitable current through the primary. That should substitute a switch in any DC electronics control circuit upt 24V or so and 50 - 100mA maximum.
 
that makes it far simpler and more practical.
Nice. id really like to make this if for just the experience alone. i would like to apply what you have suggested but i would need a least a drawing to get started. i could get the parts and test it out. I do understand the 2K resistor across the CT but for the rest id need a sketch
 

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