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dreamer2012
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« on: June 02, 2011, 03:32:30 03:32 »

Hello,
i need help with the attached design.
I try to detect a burned 110 vac / 10 amps heater, using a hall effect sensor ACS712 by Allegro micro.
The sensor works on ac and give 2.5 volts for 0 amps , 0.5vdc  for -10 A  and 5 vdc for +10 A.
I decided to use an op amp as integrator and take only to up portion of the voltage, from 2.6 vdc to 5 vdc. This is a pulsating voltage if i ignore the lower part.
I built the attached simulator, but it does not works for me.
If i insert an AC ammeter to measure the amps , simulator does not start and reports errors.
I am not sure if the 110vac source is correct - i am brand new to isis so i just guessed.
The oscilloscope, some time display in a separate windows, but other times i cannot make it to show the wave forms.
Please help
Thank you
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TomJackson69
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« Reply #1 on: June 02, 2011, 06:18:13 06:18 »

igeorge,

Try this:

That is your circuit, I have not modify, only move the pot and fix the trigger on the scope so you can see the waveform.

Tom
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« Reply #2 on: June 02, 2011, 09:32:21 09:32 »

Place a resistor in series with an LED (the resistor limits the current through the LED based upon your expected supply voltage) and put those across your heater element. Now use a phototransistor to detect when the LED is on, indicating an open heater element. If the heater is good, it will short out the LED - it'll be dark.
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« Reply #3 on: June 02, 2011, 08:21:35 20:21 »

Hi Ion,

To detect a broken heater I use one or two diodes in series with the heater, when the heater draws enough current then the voltage drop on the diode(s) will be enough to lit a led of an optocoupler. An anti parallel diode provide other half cycle of the current to the heater. Attached the simulation.

-ichan
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« Reply #4 on: June 03, 2011, 12:05:46 00:05 »

Put the LED across the TRIAC and the resistor to the neutral return. The TRIAC won't fire if the heater is open and if that resistor is trickling 10mA "through" it. When it does fire, the LED will be shorted off.

Don't be shy to push the little button with the red star beside it on SONSIVRI to maybe get more people to jump in. You also may draw more people out of the woodwork to help if you post jpg instead of dsn...I don't have the software installed that you are using, as many may not.
« Last Edit: June 03, 2011, 12:09:53 00:09 by solutions » Logged
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« Reply #5 on: June 03, 2011, 06:55:53 06:55 »

Hello, Thanks for suggestion. In fact I did what you suggested - I am referring of using the LED -aka Optocoupler in my particular case, instead of using my original ACS712.
In what concern pouting the led across the triac it is not feasible in my case.
Let me explain why
The SSR is incorporated in the heat controller module - Sealed plastic box.
The only thing that i have access is the Hot wire of the main, Neutral, and heater wire coming out from the SSR.
.....
On the PCB , i will have my detector portion , all the Optocoupler will be inputs to a PIC micro, and a port expander, and the PIC will send messages through RS232 to PLC , if a heater or more are burned out, indicating which one is defective.
I attached the picture of my working solutions, but if you read my previous post i am not happy with hardware. Should be a simpler way to do it.
Sorry about the little red start , but i forgot about it.
Ion

LED goes from hot (main) to heater wire - that's directly across the TRIAC.  The hot wire and the heater wire are easily accessible on your board and don't care about that sealed plastic box.  Just an idea....
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« Reply #6 on: June 03, 2011, 01:26:19 13:26 »

Hello Ichan,
Please be so kind and explain to me what i am doing wrong with your design.
I changed the alternator frequency from 6 Htz to 60 Htz and now the lamp does not light anymore.
When i put it 6 is OK, but for 60 is not lighting.
Regards
Ion

I do not know the answer, maybe others know. At the first trial the circuit didn't work, by blind guessing i lowered the freq then it works... but don't worry it is works on the real life and no problem with triac, I've used it several times. 10 Amp indeed a problem on power dissipation, i never try it above 5 amp.

The ACS should work do not rely on isis simulation for serious work, but it will be too costly if used only for detecting the burned heater. I would try to use current transformer, just try to wind some wires on a small torroid core and do some experiment.

-ichan
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« Reply #7 on: June 04, 2011, 06:36:00 18:36 »

igeorge,

Like Ichan have said and I also think using current transformer is much simpler than other method. You can wind your own or buy already made from Jameco or Digikey for about $3 to $6 USD.

All you have to do is run one wire that connected to the heater element through the current transformer. The current run through the wire will induces a magnetic field in the current transformer core, and the second winding of the current transformer (the number of turns that you wind on the torroid core) will generate a voltage proportional to the turn ratio of the primary and secondary of the current transformer.

Here is a formula for winding the current transformer:

Np*Ip = Ns*Is

Where:
Np is number of turn on the primary (ONE turn in this case)
Ip current of the primary
Ns is number of turn on the secondary (you calculate to get the output voltage thatís high enough to light the LED or feed to A/D of uC)
Is is the current generate from the secondary.

Another equation may help: Ns = Ip/Is    (Ip = 10 in your case)

If you donít like wind it on your own then buy one to experiment. Make sure the output current is high enough so you can place a resistor across the secondary to get the measure voltage.

Happy experiment,

Tom


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« Reply #8 on: June 04, 2011, 11:45:41 23:45 »

igeorge,

Your selected CT give very low output voltage. 50mV is not even get into the diode to rectify it.

You will need something called an IDEAL RECTIFIER that involve am op-amp and a diode and voltage supply.

I will have to look into it and maybe I will draw one in Proteus for you to try out.

Tom
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« Reply #9 on: June 05, 2011, 12:14:54 00:14 »

Amplify first then rectify.

Opamp non-inverting amplifier (200 - 500x gain) + Diode 1n914 + Cap 10u + Resistor 10K across the cap.

-ichan

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« Reply #10 on: June 05, 2011, 01:00:09 01:00 »

Something like below.

-ichan
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« Reply #11 on: June 05, 2011, 01:51:15 01:51 »

igeorge,

Here is a circuit in Proteus for you to try out. It works in Proteus.
Try it on the real circuit, and post what you find.

Tom

Sorry Ichan, I am just behind you a bit.
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« Reply #12 on: June 05, 2011, 02:44:08 02:44 »

Hi igeorge,

I have modified the circuit to simulate with LM324 as your requested.

Also the input is 10mVAC to 100mVAC (or higher). The output is about 2.5VDC; you can feed this directly to the PIC but, if you are not comfortable with it, than do like I suggested in the SCH.

Post what you find in the real circuit,

Tom
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« Reply #13 on: June 05, 2011, 11:14:32 23:14 »

Tom,
You are a life saver.
Works great on simulator.
I will make it tonight and post the result tomorrow
 Thank you again for help
Ion


Posted on: June 05, 2011, 02:51:11 02:51 - Automerged

Good morning Tom,
Will you be so kind to look at this change I did.
I do not have the CT in my hand tonight, but i want to try the design
So, i did change the resistor in the op amp to 1.5 Mohm
In this case, i read for 2.5 volts input , 1.73 out, and for 4.5 input, i read 2.7
Is this correct ?
What i am looking to achieve here, using the ACS712 eval board from Allegro, that any voltage over 2.6 to give me a valid PIC input, and under 2.6 to give me anything which the Pic will take it as 0
From data sheet of the pic, at 5 v supply, the minimum High is 2 volts.
Please do not spend time, just share your opinion.
I also have lm393, 339 and 311 comparators which i believe will do a better job as not being necessary to amplify any voltage.
Just set threshold 2.6v, and anything above that will be ONE and bellow that will be ZERO.
 

igeorge,

The setup is Non-inverting amplifier

Vpin1 = Vin(1+(R1/R2))     ***(plus V diode)***

If you make R2 > R1 then Vo < Vin and is not good for this application.

In this application, R1 must >>> R2 (greater greater greater)

Also this setup is for applify the output of the current transformer, not for voltage 2.5 V or 4.5 V.

The change you made will not works.

If you want to use the original idea, that using a hall effect sensor ACS712 than different setup is needed.

Tom


Posted on: June 05, 2011, 11:08:14 23:08 - Automerged

Hello Tom,
I made this circuit with a generic comparator from library. I have LM339N by National, LM393N from ST Microelectronics, and LM311N from National, but i did not found the model file for them
Anyway, the circuit works fine, with only two issues:
When i open the switch, the Led goes off, and the voltmeter show me 0.00V but the trace GREEN one stay on for 2 more seconds before it drops.
I will see what happen in real life when i will test it.
The biggest concern is for 2.5 volts, when the output is supposed to be ZERO, it show me a pulsating square wave. I assume it come from the filtering capacitor. The led is off, but for a PIC, depending when it read it , can be 1 or zero.
Do you have any suggestion how to make it a steady zero volts line and not a square wave ?
Thank you for all the effort you put in helping me for this project. I posted the question in many forums, including the EDA board but no answers.
Thanks
Ion


igeorge,

For voltage comparator, this should works.

I have no idea where that square wave comes from. I like to see it on the real circuit. If it appears on the real circuit, than self osc is happened.

Try to build a real circuit, donít count on the Proteus.

Tom
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« Reply #14 on: June 06, 2011, 04:29:29 04:29 »

igeorge,

Ok to try on the hair dryer, but what is Io = 800??

Should it be like about 13 Amps?

Your dryer should have two power levels, use lower wattage to see what the read out.

Tom
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« Reply #15 on: June 06, 2011, 05:31:51 05:31 »

I did a change in my design based on your original file.
I put the generators to have a 2.5 volts offset in the positive side, so i can get the proper wave form.
The voltage has to get the + 2.5 volts as ground and going up and down according with the voltage curve . +2.5 volts is zero amps.
Regards
Ion

Try this: "S_311_offset_Ideal_Diode.DSN"

It works with 2.5V, and > 2.5V

LED off when in = 2.5V
LED ON when in > 2.5V
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« Reply #16 on: June 06, 2011, 06:48:55 06:48 »

Sorry Ichan, I am just behind you a bit.

Not a problem Tom.

I might need a good (workable lowest cost) solution for this kind of problem next time, now i am wondering if a sensitive hall effect switch like Allegro A3213 can be triggered by let say 5 Amps current flows through  a wire?

-ichan

Posted on: June 06, 2011, 06:40:32 06:40 - Automerged

what is funny is that for low, the voltage decrease, but when i put the blower in high i have 3.5 volt going to the pin , after diode and it is crazy, in high before the diode i have 2.5, the same like with the blower off.

How is the output of the ACS712?

-ichan
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« Reply #17 on: June 07, 2011, 06:02:23 18:02 »

igeorge,

Post your newest circuit that you are using (exact circuit) and explain of what happen when one light bulb, two light bulb ON/OFF. Can you be more detail? What kind of the light bulb?

We will deal with the PIC later.

Tom
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« Reply #18 on: June 08, 2011, 01:11:19 01:11 »

igeorge,

I think, the correct way is to measure the output of the ASC712 when you have the load ON and OFF. I just want to verify the output measurements to see if its matching the spec.

After you have the reading of the output than you can adjust the circuit later to make it works.

 Remember to measure with meter and scope (for AC and DC). What is the reading of the scope and volts meter?

Tom
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« Reply #19 on: June 08, 2011, 10:12:27 22:12 »

igeorge,

I see the video, the LED is ON/OFF when you turn the light ON/OFF. That is one light still on. Is that correct? What happen when both light are off?

If no current goes through the ACS than the output should get 2.5V, and, any current > 0, the output should goes > 2.5V. If what I understand correct, than the comparator should works.

I wish you can discribe it in more detail so I can understand what is the problem.

Tom
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« Reply #20 on: June 09, 2011, 12:03:23 00:03 »

What if you use a window comparator, so you only get an output say when signal is between 2.4 & 2.6 volts you have logic 1 (or voltage output).
This is used in fire alarms to monitor
1. no detector fitted or end of line resistor missing (voltage 16 to 24V)
2. healthy circuit (voltage 14 to 15V)
3. fire (10V to 5V)
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« Reply #21 on: June 09, 2011, 10:25:16 10:25 »

Hi igeorge;
If you can tolerate some delaying for your output, to eliminate the pulsating
simply increase the time constant of your rectifier circuit (R5 to say 100k).
This is just an idea, try it!
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« Reply #22 on: June 09, 2011, 07:36:50 19:36 »

Pickit2 and Tom,
I made the windows comparator which works good between 2,6v to 5 volts. My problem is still here when the source it is between 0 to 2.4 volts - which is half of the time at 60 hz
Any suggestion how to make that pulsating or sine dc to a straight line ?
I i solve that issue , everything will be fine.
Regards
Ion

"I i solve that issue , everything will be fine."
So, you did solve the problem?
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« Reply #23 on: June 09, 2011, 09:03:36 21:03 »

igeorge,

Try this modified circuit. It works.

Read the notes on the circuit, the notes is about the main goal of this project. And the main goal is to detect if the heat element is working or broken. If the element is working, the current is in the ten or so. if the heat element is broke, the current will be zero. If you are trying to detect some unknow current or very low current then LOT MORE work have to be done. I think, the best way is to stick with the original goal. Later if the circuit work, you can modify for lower current.

Try to simulate this circuit, it works good.
Than try with current about 0.5Amp to see the output osc.

That is why R14 is there for.

Tom
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« Reply #24 on: June 09, 2011, 11:18:53 23:18 »

george,

If I know the heater are 250W and 450W than I would concerntrate at lower current. Anyways, if you have to use the light bulb to test the circuit; use four 60W for 240W and five 100W for 500W to see if it work. That is about 2Amps and 4Amps.

You can modify the LOAD RESISTOR to simulate 2Amps and 4Amps on Proteus. Use R14 to adjust the ref. point.

Should works

Tom
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