Need help with connecting a CT to a PIC - Electronic Circuits Projects Diagrams Free

tyru · 7th May 2008, 01:16 PM

Need to develop a circuit that will allow a PIC microcontroller to monitor the current to a auger motor for JAM detection.

The motor is 120 VAC and rated 0.5 FLA with a peak lock rotor current of 2.5A and a jam trip point of 1 A. I need a circuit to convert the AC amps to DC Volt (0- 5 VDC) for the pic input pin and A/d converter.

Found this application circuit at the CR micro site. http://www.crmagnetics.com/pdf/ancrct-4.pdf

CR Magnetics sent me thier burden test for the http://www.crmagnetics.com/8400.pdf. Indicates that the secondary voltage needs to be maintained less than 2V to prevent saturation.

I need to determine the values of the resistors along with teh final reistor and capacitor to get a 0-5V DC signal as the current in the primary to the CT goes from 0 to 2.5 A.

crutschow · 7th May 2008, 05:15 PM

You might look at the thread at http://www.electro-tech-online.com/e...ent-shunt.html. It covers some of the same issues.

tyru · 8th May 2008, 12:33 AM

Quote:

Originally Posted by crutschow

You might look at the thread at http://www.electro-tech-online.com/e...ent-shunt.html. It covers some of the same issues.

I've been following that thread but I didn't want to thread steal.

crutschow · 8th May 2008, 08:56 PM

The LM324 shown in the circuit will only go to about +3.8V output with a +5V supply. To get 5V output you will either need a supply voltage higher than +5V or use an op amp with a rail-to-rail output. What supplies do you have?

tyru · 8th May 2008, 09:18 PM

12 VDC and 5 VDC

crutschow · 10th May 2008, 05:58 AM

Attached is a schematic and simulation of the CR Magnetics circuit. I selected values to give +5V DC output with 2.5A AC and a 1000:1 turns ratio current coil (which gives 2.5mA of current at the circuit input). The input resistor gives 1.5V peak signal at the coil output. I also added a resistor at the -input to the first op amp since it was loading the signal on the negative peak (due to operating the amp with only a positive supply). The op amp operates off +12VDC.

I did modify the output filter so that it responds to the average value of the waveform and not the peak since that is less affected by noise and waveform distortion. The output ripple is less than 2mVpp with the filter values shown, which should have no effect on your circuit operation.

Of course a simulation is no guarantee that the circuit will work but it should be fairly close to the actual circuit. You may have to tweak the resistor values R4 or R5 (on the CR Magnetics schematic) to get the desired gain if the output voltage is not quite right.

Any questions, let me know.
Cur Sens Cir.jpg

tyru · 12th May 2008, 01:18 PM

These are some comments that I recieved from CR Magnetics related to my application.

"I would kick up the voltage to 100 mV and use a 20 ohm on the burden resistor. I would kick up R2 and R3 to 10K to minimize input voltage differential effects. (50 mV/100 ohms = 0.5 mA error on a 5 mA signal).

I would decrease R5 to 500K for a gain of 50.

Please also note that using the 1000 turns is the ideal transfer ratio, however, no CT is perfect, and corrective factors are given in the spec sheet. Even with this adjustment, you will get variance part to part, and for the most complete and accurate system, an adjustment pot is usually added to calibrate. Without calibration, you will probably get a sensor that is +/- 5% part to part.

Filtering depends on acceptable ripple. Full wave rectifying on 60 Hz results in 120 Hz ripple on a DC offset. Assuming single pole filtering 20db per octave, you will get 50 mV of ripple using a filter with a corner frequency of 12 Hz. (V/100). 1 divided by 12 Hz = time constant = .083 = 1/(2 * pi * f * R * C). Assuming C = 10 uF, R would be around 2K. 4.7 uF and 10K would be a pretty good filter."

Crutschow: What simulation program are you using?

crutschow · 12th May 2008, 03:20 PM

Quote:

Originally Posted by tyru

These are some comments that I recieved from CR Magnetics related to my application.

"I would kick up the voltage to 100 mV and use a 20 ohm on the burden resistor. I would kick up R2 and R3 to 10K to minimize input voltage differential effects. (50 mV/100 ohms = 0.5 mA error on a 5 mA signal).

I don't understand their comment. I used a burden resistor to give 1.5V (based upon their info that said keep it below 2V) and they want to "kick up the voltage to 100mv". Seems like kick it down is more appropriate.

I also don't know what "input voltage differential effects" means.

Crutschow: What simulation program are you using?

I use Electronic Workbench. It's now available from National Instruments. It's a very easy program to use.

tyru · 14th May 2008, 04:37 PM

Great program. Used it to simulate the circuit using the suggested values from CR Magnetics but I couldn't get it to work using the LM324. However, when I replaced it with a LM124, circuit seems to behave as expected.

The schematic by CR Magnetics shows two Zener diodes across the CT. Any idea what they do? They don't seem to effect anything with the resulting voltages.

Leftyretro · 14th May 2008, 04:57 PM

Quote:

Originally Posted by tyru

Great program. Used it to simulate the circuit using the suggested values from CR Magnetics but I couldn't get it to work using the LM324. However, when I replaced it with a LM124, circuit seems to behave as expected.

The schematic by CR Magnetics shows two Zener diodes across the CT. Any idea what they do? They don't seem to effect anything with the resulting voltages.

Most likely to protect the CT secondary windings in case the secondary load is opened but with still current passing through the CT primary hole. With such large turns ratio an unterminated CT can generate hugh voltages that can damage the CT winding.

Lefty

tyru · 14th May 2008, 05:45 PM

Thanks Lefty.

I'm assuming that something is wrong in the simulation program with the LM324 or I had a bad connect somewhere. They appear to be about the same chip.

Can I assume that I can use a LM324 (which I have) instead of the LM124 in the circuit?

crutschow · 14th May 2008, 09:05 PM

Quote:

Originally Posted by tyru

I'm assuming that something is wrong in the simulation program with the LM324 or I had a bad connect somewhere. They appear to be about the same chip.

Some of the LM124/324 models do not simulate properly with the negative power lead grounded. (Simulations are only as good as the models and some models are incorrect. Unfortunately you only find that out when the simulations don't seem to work as they should.) I believe the National Semiconductor models do work properly.

Can I assume that I can use a LM324 (which I have) instead of the LM124 in the circuit?

Yes. They are the same expect for maximum operating temperature and slight differences in some of the parameters, which should not be significant in this circuit.

tyru · 14th May 2008, 09:51 PM

After experimenting a bit i find that any of the LM124/LM324AN or N series work fine in electronic work bench but if I replace it with an AJ part, the output either goes to 23V or is very erratic. Very Weird.

mneary · 14th May 2008, 10:19 PM

First thing to suspect is an error in the LM124AJ spice library for your tool set. But if you don't plan to use the ceramic dual inline package, I wouldn't worry about it.

crutschow · 15th May 2008, 12:38 AM

As I previously noted, the circuit as designed by CR Magnetics has an output proportional to the peak of the current waveform. I suggest you use my modification on the output filter so it responds to the average value, which is less sensitive to noise and distortion.

7th May 2008, 01:16 PM	(permalink)
tyru New Member	Need help with connecting a CT to a PIC Need to develop a circuit that will allow a PIC microcontroller to monitor the current to a auger motor for JAM detection. The motor is 120 VAC and rated 0.5 FLA with a peak lock rotor current of 2.5A and a jam trip point of 1 A. I need a circuit to convert the AC amps to DC Volt (0- 5 VDC) for the pic input pin and A/d converter. Found this application circuit at the CR micro site. http://www.crmagnetics.com/pdf/ancrct-4.pdf CR Magnetics sent me thier burden test for the http://www.crmagnetics.com/8400.pdf. Indicates that the secondary voltage needs to be maintained less than 2V to prevent saturation. I need to determine the values of the resistors along with teh final reistor and capacitor to get a 0-5V DC signal as the current in the primary to the CT goes from 0 to 2.5 A. Last edited by tyru; 11th May 2008 at 12:46 AM.

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7th May 2008, 05:15 PM	(permalink)
crutschow Experienced Member	You might look at the thread at http://www.electro-tech-online.com/e...ent-shunt.html. It covers some of the same issues.

8th May 2008, 08:56 PM	(permalink)
crutschow Experienced Member	The LM324 shown in the circuit will only go to about +3.8V output with a +5V supply. To get 5V output you will either need a supply voltage higher than +5V or use an op amp with a rail-to-rail output. What supplies do you have?

8th May 2008, 09:18 PM	(permalink)
tyru New Member	12 VDC and 5 VDC

10th May 2008, 05:58 AM	(permalink)
crutschow Experienced Member	Attached is a schematic and simulation of the CR Magnetics circuit. I selected values to give +5V DC output with 2.5A AC and a 1000:1 turns ratio current coil (which gives 2.5mA of current at the circuit input). The input resistor gives 1.5V peak signal at the coil output. I also added a resistor at the -input to the first op amp since it was loading the signal on the negative peak (due to operating the amp with only a positive supply). The op amp operates off +12VDC. I did modify the output filter so that it responds to the average value of the waveform and not the peak since that is less affected by noise and waveform distortion. The output ripple is less than 2mVpp with the filter values shown, which should have no effect on your circuit operation. Of course a simulation is no guarantee that the circuit will work but it should be fairly close to the actual circuit. You may have to tweak the resistor values R4 or R5 (on the CR Magnetics schematic) to get the desired gain if the output voltage is not quite right. Any questions, let me know. Cur Sens Cir.jpg

12th May 2008, 01:18 PM	(permalink)
tyru New Member	These are some comments that I recieved from CR Magnetics related to my application. "I would kick up the voltage to 100 mV and use a 20 ohm on the burden resistor. I would kick up R2 and R3 to 10K to minimize input voltage differential effects. (50 mV/100 ohms = 0.5 mA error on a 5 mA signal). I would decrease R5 to 500K for a gain of 50. Please also note that using the 1000 turns is the ideal transfer ratio, however, no CT is perfect, and corrective factors are given in the spec sheet. Even with this adjustment, you will get variance part to part, and for the most complete and accurate system, an adjustment pot is usually added to calibrate. Without calibration, you will probably get a sensor that is +/- 5% part to part. Filtering depends on acceptable ripple. Full wave rectifying on 60 Hz results in 120 Hz ripple on a DC offset. Assuming single pole filtering 20db per octave, you will get 50 mV of ripple using a filter with a corner frequency of 12 Hz. (V/100). 1 divided by 12 Hz = time constant = .083 = 1/(2 * pi * f * R * C). Assuming C = 10 uF, R would be around 2K. 4.7 uF and 10K would be a pretty good filter." Crutschow: What simulation program are you using?

14th May 2008, 04:37 PM	(permalink)
tyru New Member	Great program. Used it to simulate the circuit using the suggested values from CR Magnetics but I couldn't get it to work using the LM324. However, when I replaced it with a LM124, circuit seems to behave as expected. The schematic by CR Magnetics shows two Zener diodes across the CT. Any idea what they do? They don't seem to effect anything with the resulting voltages.

14th May 2008, 05:45 PM	(permalink)
tyru New Member	Thanks Lefty. I'm assuming that something is wrong in the simulation program with the LM324 or I had a bad connect somewhere. They appear to be about the same chip. Can I assume that I can use a LM324 (which I have) instead of the LM124 in the circuit?

14th May 2008, 10:19 PM	(permalink)
mneary Experienced Member	First thing to suspect is an error in the LM124AJ spice library for your tool set. But if you don't plan to use the ceramic dual inline package, I wouldn't worry about it.

15th May 2008, 12:38 AM	(permalink)
crutschow Experienced Member	As I previously noted, the circuit as designed by CR Magnetics has an output proportional to the peak of the current waveform. I suggest you use my modification on the output filter so it responds to the average value, which is less sensitive to noise and distortion.