Need to design a current comparator

[steve_khor]

I an new member to this forum and would appreciate your help on the following problem.

I need to design a circuit as a switch to compare two currents, the conditions are,
if c >= 2A then switch on
c < 2A then switch off


The voltage supply of my circuit application is 240 V while my current that I required was 63A.

 

[Reloadron]

OK, if a current exceeds 2 amps you want to switch something on and if the current is less than 2 amps you want to switch something off. Not sure where the 240 V @ 63 Amps figures in but you have a load a little over 15 KW in there. I guess for openers the first question is the 2 Amps AC or DC current?

Ron

 

[steve_khor]

is AC current... erm, is not necessary for 63 amps, as long as the circuit's power supply is 220V ~ 240V. Do you have any references or circuit examples so that I can modify？ Many thanks

 

[Hero999]

Here's an idea.

A current transformer with an amplifier/precision rectifier, followed by a filter capacitor and a comparator driving a relay via a transistor.

It's by no means a complete final solution, some of the component values will depend on your choice of current transformer.

 

[Reloadron]

Personally I would start with an AC current sensor like this one with a range of maybe 0 to 5 Amps. I would look at an output of 0 to 5 Volts so an input of 0 to 5 Amps input is equal to a 0 to 5 Volts output. I would run the 0 to 5 Volts out into a comparator circuit with a reference voltage set for 2 Volts which would be 2 Amps through the sensor.
When the output of the sensor was below 2 Volts the output of the comparator could be logic low and when the current exceeds 2 Amps (2 Volts Out) the comparator output goes high. Then use the comparator to drive whatever you want.

Regardless of the route you take you will likely need a power supply to provide low voltage DC for the measurement circuits. Regardless of how you choose to do it the starting point will be an AC current sensor that will convert an AC current into a signal you can work with.

What I am suggesting is one of likely several ways to go about this. Maybe others here have some thoughts.

The circuit Hero suggest using a CT (Current Transformer) is another viable option.

Ron

 

[steve_khor]

Here's an idea.

A current transformer with an amplifier/precision rectifier, followed by a filter capacitor and a comparator driving a relay via a transistor.

It's by no means a complete final solution, some of the component values will depend on your choice of current transformer.

The two main components I will use AC1030 30 Amp Current Transformer and the Power Relay HHC71E

 

[steve_khor]

Personally I would start with an AC current sensor like this one with a range of maybe 0 to 5 Amps. I would look at an output of 0 to 5 Volts so an input of 0 to 5 Amps input is equal to a 0 to 5 Volts output. I would run the 0 to 5 Volts out into a comparator circuit with a reference voltage set for 2 Volts which would be 2 Amps through the sensor.
When the output of the sensor was below 2 Volts the output of the comparator could be logic low and when the current exceeds 2 Amps (2 Volts Out) the comparator output goes high. Then use the comparator to drive whatever you want.

Regardless of the route you take you will likely need a power supply to provide low voltage DC for the measurement circuits. Regardless of how you choose to do it the starting point will be an AC current sensor that will convert an AC current into a signal you can work with.

What I am suggesting is one of likely several ways to go about this. Maybe others here have some thoughts.

The circuit Hero suggest using a CT (Current Transformer) is another viable option.

Ron

Thanks for your suggestion, but I try the Hero's idea 1st, because I concern on current method.. is that the software called Multisim suitable used to simulate?

 

[steve_khor]

Sorry to your patient, can I know more explanation on your circuit diagram , which part is design for controlling my condition ( c >= 2A, c < 2A)
The current amplifier I used was AC1030, 30 Amp Transformer and power relay HHC71E

 

[Hero999]

I Googled for information about the relay but they come in a range of coil voltages, what's the coil voltage rating of your relay?
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There's no current amplfier, that's part of what my circuit does, it's just a passive current transformer, here's a link to the datasheet for anyone else who's interested.
https://www.electro-tech-online.com/custompdfs/2010/04/AC1030.pdf

 

[steve_khor]

I Googled for information about the relay but they come in a range of coil voltages, what's the coil voltage rating of your relay?
Buy High Power Electromagnetic Relay,Clion Relay Maker,HHC71E

There's no current amplfier, that's part of what my circuit does, it's just a passive current transformer, here's a link to the datasheet for anyone else who's interested.
https://www.electro-tech-online.com/custompdfs/2010/04/AC1030-1.pdf

Oops, sorry for my mistake .. I wrote it wrong, it is current transformer AC 1030, they just mentioned about the Power Coil.. I'm not so quiet sure for coil voltage rating.. the Power Coil for DC 3.6W and AC 6.5 VA.. Well, can I find out the coil voltage rating from the Power coil value?

 

[Hero999]

Yes, it's very important you know the coil voltage.

 

[steve_khor]

Yes, it's very important you know the coil voltage.

From my information here,

the Rated Coil Power : 2.5W and 5.5 VA

From calculation: P = VI
If the current required on my circuit is 2A, then Rate coil voltage is
2.5 = V (2)
V = 1.25V
Can I figure out like this?

Sorry if that is wrong way, I was poor in electronic and electrical knowledge..

 

[Hero999]

Sorry, that's wrong, nice try though.

2A is the current through the transformer and is nothing to do with the relay.

The relay power is 3.6W if it has a DC coil and 6.5VA if it has an AC coil.

You can't calculate the voltage because we don't have the current required or the resistance/impedance of the relay coil.

Just for your information VA and W aren't the same thing, W is real power and VA is apparent power, the AC relay coil probably has the same real power consumption as the DC version, it's power factor is just 0.55 so the current is higher. Google power factor for more information.

 

[Reloadron]

The CT (Current Transformer) you have chosen is a pretty standard CT. A single current carrying conductor is passed through the center hole and acts as the primary. The transformer has a 1000:1 ratio Input to Output. Therefore with a full scale input of 30 Amps the output current will be 30 mA into a burden load of 100 Ohms. See the charts on the data sheet, they cover that. If you start increasing the burden resistance the CT like any CT becomes inaccurate.

Therefore with a 30 Amp primary current the secondary current is 30 mA through 100 Ohms so you get a 3 volt drop across the 100 Ohm resistor. Now your 0 to 30 Amps is equal to 0 to 3 Volts. The frequency will be whatever the primary frequency is. If a primary current os 2 Amps is passed the output will be 2 mA through the 100 Ohm burden resistor resulting in a 200 mV drop. You now have 200 mV RMS to work with.

Unfortunately that 200 mV now needs to be amplified before you can convert it to DC. Once amplified that voltage is converted to a DC level and sent on to a comparator circuit and compared to a reference voltage.

Something to consider is that your CT is based on primary amp turns. Passing the conductor through the hole is one amp turn. That same conductor can be looped through the hole several times to increase the output. Each loop doubles the output. So if 2 amps primary with a single turn yields 200 mV then 4 amp turns would yield 800 mV which becomes easier to amplify and helps eliminate noise.

Additionally the same line of CT you plan to use also comes in this 5 amp model so why bother with a 30 amp CT when you goal is to measure a 2 amp trip unless your main conductor will well exceed 5 amps. Matter of fact they make a 5, 10, 15, 20 and 25 amp model of the same CT.

Ron

 

[steve_khor]

Sorry, that's wrong, nice try though.

2A is the current through the transformer and is nothing to do with the relay.

The relay power is 3.6W if it has a DC coil and 6.5VA if it has an AC coil.

You can't calculate the voltage because we don't have the current required or the resistance/impedance of the relay coil.

Just for your information VA and W aren't the same thing, W is real power and VA is apparent power, the AC relay coil probably has the same real power consumption as the DC version, it's power factor is just 0.55 so the current is higher. Google power factor for more information.

erm, I read from your link, one of the information shown that:

Specifications of coils DC 6-220V (Customizable Voltage )
AC 6-380V (Customizable Voltage )

is that a rated voltage coil value?

 

[steve_khor]

The CT (Current Transformer) you have chosen is a pretty standard CT. A single current carrying conductor is passed through the center hole and acts as the primary. The transformer has a 1000:1 ratio Input to Output. Therefore with a full scale input of 30 Amps the output current will be 30 mA into a burden load of 100 Ohms. See the charts on the data sheet, they cover that. If you start increasing the burden resistance the CT like any CT becomes inaccurate.

Therefore with a 30 Amp primary current the secondary current is 30 mA through 100 Ohms so you get a 3 volt drop across the 100 Ohm resistor. Now your 0 to 30 Amps is equal to 0 to 3 Volts. The frequency will be whatever the primary frequency is. If a primary current os 2 Amps is passed the output will be 2 mA through the 100 Ohm burden resistor resulting in a 200 mV drop. You now have 200 mV RMS to work with.

Unfortunately that 200 mV now needs to be amplified before you can convert it to DC. Once amplified that voltage is converted to a DC level and sent on to a comparator circuit and compared to a reference voltage.

Something to consider is that your CT is based on primary amp turns. Passing the conductor through the hole is one amp turn. That same conductor can be looped through the hole several times to increase the output. Each loop doubles the output. So if 2 amps primary with a single turn yields 200 mV then 4 amp turns would yield 800 mV which becomes easier to amplify and helps eliminate noise.

Additionally the same line of CT you plan to use also comes in this 5 amp model so why bother with a 30 amp CT when you goal is to measure a 2 amp trip unless your main conductor will well exceed 5 amps. Matter of fact they make a 5, 10, 15, 20 and 25 amp model of the same CT.

Ron

Actually the goal I want to hit is 63 A that's why I chosen the AC1030 current transformer (30A normal, 75A maximum) . I just presume 2A as example, sorry for what I'm not saying clearly.

After that I need to design a circuit to switch on/off (used Power Relay HHC71E)

 

[Reloadron]

I found this PDF sheet on those relays. The selectable coil voltages means they have several flavors. I was unable to locate a US or Euro distributor for these creatures. Personally I would be looking for another relay or SSR but that is just me.

Ron

 

[Reloadron]

Actually the goal I want to hit is 63 A that's why I chosen the AC1030 current transformer (30A normal, 75A maximum) . I just presume 2A as example, sorry for what I'm not saying clearly.

After that I need to design a circuit to switch on/off (used Power Relay HHC71E)

If that is the case I would be using a 100 Amp CT and not a 30 Amp running at its maximum. Overall current transformers are inexpensive. Actually I would use a CT with a DC output as I originally posted then feed a comparator.

Now I much better understand the goal here, I was focused on that 2 amps.

As to your power relay you can't turn it on and off till you have a coil voltage determined. I can't find much beyond the last PDF data sheet I linked to. The suffix of the part number determines coil voltage as to DC coils from what I see.

Ron

 

[Hero999]

Have you already purchased the relay?

I assumed you already had. If not, it's good since I can select a suitable coil voltage for you.

 

[steve_khor]

I already purchased this power relay unfortunately.... From the data sheet information given, the Specification of Coil ~ DC 6 - 220 V and AC 6 - 380V .. is that all right?

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