Need to design a current comparator

[Reloadron]

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?

Buy High Power Electromagnetic Relay,Clion Relay Maker,HHC71E

When you purchased the relay you had to purchase it specifying a coil voltage. You either bought an AC or DC coil with a specified coil voltage. Look at the entire part number. The suffix (end of part number) should specify the coil voltage. Example:

HHC71E(JQX-59F)-1Z-12VDC

The example would be a 12 volt DC coil. That is what Hero needs to know to plan on the relay coil voltage. Everything between the dashes in the part number spells things out. Did you even read the last PDF I linked to for that relay? Somewhere on the relay is a full part number, what do you see?

Ron

 

[steve_khor]

the link you attached was Chinese version, from that data sheet information, it shown HHC71E(JQX-59F)-1Z-12VDC..

 

[Hero999]

Then it's a 12VDC relay, was that so hard?

Fortunately, that makes it easy, I'll get back to you later.

 

[Hero999]

Here's the final circuit.

The relay will actually trigger at a slightly lower current 1.8A which is what you need because the current could be slightly lower than 2A.

All resistors can be carbon film 5% tolerance, except for R1 and R2 which need to be 1% tolerance metal film.

 

[steve_khor]

Ok, so I would like to design a comparator devices 1st.

I try to design a comparator by using the Multisim software, yy idea was, put the Voltage Reference as negative value (pin 2), then waiting for the Input Signal (pin3). The comparator I used was LM311H. But something was not right after I simulated. The output (Pin 7) still got signal out even my input signal(pin3) is zero.

What's wrong with the Multisim software??

 

[Hero999]

Probably nothing.

Try simulating the circuit I posted previously and it will work.

The only thing you might have a problem with is the current transformer but you can replace it with an AC voltage with an RMS voltage of 100mV (141mV peak) for every Amp of current through the transformer, for example if you have a 2A load, it needs to be 200mV or 283mV peak.

 

[steve_khor]

Here's the final circuit.

The relay will actually trigger at a slightly lower current 1.8A which is what you need because the current could be slightly lower than 2A.

All resistors can be carbon film 5% tolerance, except for R1 and R2 which need to be 1% tolerance metal film.

Sorry, I just saw your reply after I do my own simulation. Well, I've one question to ask, if I change my specification.. AC needs 30 Amp, then my relay condition was (c <= 0 disconnect, c > 0 connect)

 

[Hero999]

All right, I'll have to explain how the circuit works.

The relay is activated when the voltage on Cf is greater than 2.5V.

UA1 forms a non-inverting precision rectifier with a gain determined by R1 and R2.
[latex]Av = 1+ \frac{R1}{R2}[/latex]

The output voltage from your current transformer is equal to the current multiplied by 0.1V.
V = 0.1×I

The peak voltage will be √2 times the RMS value.

The voltage across Cf is equal to the following formula:

V = I√2×0.1×Av

For 30A I'd probably increase the voltage reference to 5V which would mean using the LM336-5

At 30A the peak voltage from the transformer will be 0.1×30√2 = 4.2V.

I'd probably make the gain of U1A equal to 1.2 which would mean replacing R1 and R2 with 3k and 15k respectively making the gain equal to 1.2.

Now when the current is 30A the voltage across Cf will be 5.1V


The relay will then turn on at

 

[steve_khor]

All right, I'll have to explain how the circuit works.

The relay is activated when the voltage on Cf is greater than 2.5V.

UA1 forms a non-inverting precision rectifier with a gain determined by R1 and R2.
[latex]Av = 1+ \frac{R1}{R2}[/latex]

The output voltage from your current transformer is equal to the current multiplied by 0.1V.
V = 0.1×I

The peak voltage will be √2 times the RMS value.

The voltage across Cf is equal to the following formula:

V = I√2×0.1×Av

For 30A I'd probably increase the voltage reference to 5V which would mean using the LM336-5

At 30A the peak voltage from the transformer will be 0.1×30√2 = 4.2V.

I'd probably make the gain of U1A equal to 1.2 which would mean replacing R1 and R2 with 3k and 15k respectively making the gain equal to 1.2.

Now when the current is 30A the voltage across Cf will be 5.1V


The relay will then turn on at

Thanks for your explanation

I cant find any voltage references in Multisim, any type of components else that I can use?

 

[Hero999]

I'll give you a clue.

This is the problem with not fully understanding the circuit or what the simulator does.

Have you tried improvising?

What could you replace the voltage reference with?

Hint: think about what a voltage reference does.

 

[steve_khor]

This is the problem with not fully understanding the circuit or what the simulator does.

Have you tried improvising?

What could you replace the voltage reference with?

Hint: think about what a voltage reference does.

Well, I can use Vcc ~ 5V instead of the voltage reference your circuit had...

 

[Hero999]

Yes, just replace the voltage reference with a voltage source of the same voltage - it's common sense when you think about it.

 

[steve_khor]

I used 30A as my load ..

Due to my current transformer ratio condition 1000 : 1
So my output current was 30mA .
Then my output voltage would be 30m x 100 = 3V

The voltage transformers from the Multisim label was NLT_PQ_4_10, NLT_PQ_4_12, NLT_PQ_4_120 etc ..... is that the last value of the label is shown the voltage? so can I use this ~ Non Linear Transformer "NLT_PQ_4_10" ?


N the second problem I faced on the component was the power relay.. cannot get any power relay in Multisim software. What I'm find for the suitable relay was called Current_Controller as shown below

**broken link removed**

I've to adjust the current_controller value (1mA 0mA) as well, so it make me confuse..Threshold Current and Hysteresis Current

 

[Hero999]

You're right about the current transformer.

I'm not going to answer your question about Miltisim and the relay because you should be able to figure it out for yourself.

Hint: the simulator doesn't have to be exactly the same as your circuit, you don't even have to use a relay, if you don't want to.

 

[steve_khor]

I've faced some problem when the signal through the comparator ..

The output of the comparator shown the two results when the Voltage Threshold (V-) is greater than Voltage Input Signal (V+).. The results were Vrms = 0, Vdc = 5.0

It supposed 0 when there was no signal in Voltage Input Signal after compared.

Is that Vrms belong to V (AC) ?

 

[steve_khor]

What kind of voltage we are looking for when the both voltages are comprising in the IC comparator? (My answer is Vdc, bt thn about Vrms?)

After compared, which type of voltage we should take it from the Vout? Vdc o Vac o Vrms?

 

[Hero999]

Read through the calculations in my previous posts & you'll be able to figure it out.

As I said before the peak voltage is √2 times the RMS voltage, the voltage across Cf is the peak voltage from the transformer multiplied by the gain of the non-inverting op-amp stage.

There's one thing I missed in my explanation: the purpose of R4 and R5.

They provide the comparator with some hysteresis so the turn on voltage will be slightly higher than the off voltage which will ensure that the relay doesn't chatter.

Hysteresis = Vout×(R4/R5)

Vout is the suppply voltage to the comparator minus the losses in the output stage, for convenience let's call it 10V.

The hysteresis is the difference between the turn on and turn off voltages and should be higher than the ripple on Cf.

Suppose R5 = 1M and R4 = 15k as per the schematic.

Hysteresis = Vout×R4/R5 = 10×15/1M = 150mV

Suppose Vref is 5V.

Von = Vref+hysteresis/2
Voff = Vref-hysteresis/2

hysteresis/2 = 150mV/2 = 75mV

Von = 5 + 75mV = 5.075V
Voff = 5 - 75mV = 4.925V

The ripple accross Cf should be <150mV which it was the last time I simulated it, if it's higher, increase the value of R4.

 

[steve_khor]

I've checked the IC LM358, this IC feature is not design for comparator isn't it?

 

[Hero999]

It's an op-amp but it can also be used as a comparator.

I suggest you get someone to help you with this and doing a fair amount of research before doing this project, no offence but you don't have enough experience and there's a limit to what I can help you with via the forum, although I shall continue to do my best to help, just as long as to research every question you have before asking me.

 

[steve_khor]

I get my result finally thanks Hero... But I will do research on it..