# Making Current Sensing Circuit More Reliable

Discussion in 'General Electronics Chat' started by cadamhill, Jan 23, 2016.

Oct 10, 2015
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I am working to create a simple Current Transducer Circuit that will work for an industrial application. I included a current transducer (LTS 15-NP) and I am looking to keep the output signal between -5 and +5 volts AC peak to peak (approximately +/-3.6 V RMS). As you can see, I have incorporated (not knowing if there is a better solution) a couple of 1N4148's to clamp any stray source voltages that may otherwise burn up the pcb. Keeping in mind that I don't want to do anything that will change the voltage signal from Rs (sensing feed), what can I do to make the circuit more resilient and robust? What should be done to make sure it is safer and protected, prior to initial testing? Thanks for any help or suggestions. PS ...I understand there is no gain on the Op Amp.

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2. ### #12 Expert

Nov 30, 2010
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First, use a different op-amp. The 741 is 50 years old and needs 200 na input bias current which limits Rs to 250 ohms for a 1% accuracy budget. Upon changing to a higher impedance input, raise Rs to a much higher value and use a pair of back-to-back zener diodes in the 5.1V to 6.8V range to clamp the input voltage relative to ground, not the supplies.

It is normal practice to use low current diodes to clamp voltages in a 5 volt system but you have +/- 12 volts. If you clamp to that, random noise inputs can be way outside the input range of the 741 chip or the one I'm going to suggest.

How about a TL071? Rs can be anything up to 250,000 ohms and the input voltage range will be +/- 8 volts.
So, raise Rs to maybe 10,000 ohms then add equal to or less than 2.6 nf from the positive input to ground to cut the high frequency hash that might be in the environment. That will keep you within a 1% error budget at 60 Hz.

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3. ### tsan Member

Sep 6, 2014
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Is your transducer LEM LTS-15 NP? LEM LTS-15 NP is a hall effect current transducer and the output voltage of the sensor is max 5 V because it's supply voltage is 5 V. If I read the datasheet correctly output voltage is 2,5 V with zero current. With +15A output voltage is 2,5V + 0,625V=3,125 V.

4. ### AnalogKid Distinguished Member

Aug 1, 2013
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200 nA times 250 ohms is 50 uV of error. This is 0.001% of 5 V, the OP's peak signal voltage. Seems good enough to me.

As a current limiter, 70 ohms is too small a value. A 1 K resistor will keep transient currents to much lower values while introducing a possible max error of only 200 uV. The tradeoff is that the resistor forms a lowpass filter with the junction capacitance of the transient suppression diodes. If changing to zeners, make sure this corner freq is at least 10 times your highest frequency of interest.

ak

5. ### #12 Expert

Nov 30, 2010
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Is this kind of like, "leave out the nanofarads, the zeners come equipped with nanofarads"?
But seriously...Vishay seems reticent about posting their picofarads. Can you provide a reference?

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6. ### Lestraveled Well-Known Member

May 19, 2014
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As @tsan pointed out, the LEM TSL-15-NP output is referenced to +2.5 volts. This makes your simulation wrong. In order to get a bi-polar output you need to sum a 2.5 V reference voltage into your op-amp. (The LEM LTSR-15-NP outputs that reference voltage. You might consider changing sensors.) (I have used the LEM LTSR-6-NP.)

Edit: I finally found the LTS app note.

• ###### LEM LTSR app note.pdf
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7. ### Lestraveled Well-Known Member

May 19, 2014
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Also, Your sim says you are measuring 50 amps. So,
50 Amps (RMS) X 1.414 = 70.7 amps peak,, 70.7 amps X .0416 = 2.94 V +2.5V = 5.44 volts This is outside of the sensors output range.

The spec sheet says 48 amps max. (This means peak. So...)

48 amps(peak) X .707 = 33.9 amps (RMS) If you need to measure over 40 amps you need to use a different sensor. (I suggest the LEM HAL 50-S).

Oct 10, 2015
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I'm sorry. Its actually going to be an LEM LA 55-P . I cannot find a transducer that is similar to this, in multisim so I used this one, thinking it was similar.

Oct 10, 2015
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Im sorry for the mixup. It is an LA 55P transducer being used.

10. ### Lestraveled Well-Known Member

May 19, 2014
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OK, huge difference.
The LA-55P is a current output sensor, not a voltage output. So, the 70 ohm resistor (Rs) must go to ground, not the input of the op-amp. The sensor has a 1000:1 current scaling, which means, 1amp In, generates .001 amp Out. The resistor Rs scales the current into a voltage, so it must go to ground.

So, to make this a little easier, add a 70 ohm resistor from the sensors output pin to ground, and change the current 70 ohm resistor to a 1K. Simple enough?

Question: Why the 70 ohm resistor? It causes a kind of odd scaling factor; 10 amps in equals .7 volts out. Why not a 100 ohm resistor? This would give you a 10:1 current to voltage scaling; 10 amps in equals 1 volt out.

Oct 10, 2015
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Simplified it certainly! I was just going by the example listed in a tech document I found. I agree with you on that. One thing I don't understand however... I thought the LA 55-P current transducer converted a current to a voltage signal that represents a linear relationship of the output current? What I'm trying to achieve at the end of the day is to take that output current and convert it to a voltage signal from -5 V to +5 V so I can run it through a bipolar ADC. The ADC would allow me to recreate the line side current signal (from perhaps an electric motor) in digital form so I can send it to a database. ? What am I missing if you don't mind me asking? Thanks for your help by the way.

12. ### Lestraveled Well-Known Member

May 19, 2014
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OK
Look at the data sheet:
- The third term down is called, "Measuring Resistance". This describes the range of resistances that meet the specs for voltage, temperature and accuracy. This is the resistor that is placed between the output pin and ground.
- The forth term is,"Secondary Nominal Current RMS". This is the output current where you get the best performance (accuracy, response,...).
- The fifth term is, "Conversion Ratio". This tells you relationship between the current through the aperture and the current out of the output pin. It is just a ratio of current in verses current out.

If you have no resistor to ground the voltage at the output pin would be pegged at ether max positive or negative voltage rails.

Your first current sensor (LTS-15-NP) was voltage out sensor. In it's data sheet it did not specify ratio, it specified "Sensitivity" in mV/Amp.

Some hall effect current sensors are voltage output devices and some are current output devices.

A current output device is better if the sensor is located a long distance away from the electronics. It also allows for greater scaling control. This is helpful when matching a sensor output range to an ADC input. Use that feature to your advantage.

13. ### #12 Expert

Nov 30, 2010
16,704
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I got caught answering a person who asked for what he wanted instead of what he needed.
Darn! Missed again.

14. ### AnalogKid Distinguished Member

Aug 1, 2013
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I've used a few thousand of these. We were not happy with the total error budget over -25 to +85, so I wrapped a gain and offset trim circuit around it. The output flexibility is terrific.

ak

Oct 10, 2015
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If I set the transducer at .001 transresistance, I get a very low output current compared to the cut sheet.

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Oct 10, 2015
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did you just use something like a 10K gain trim on the output?

17. ### Lestraveled Well-Known Member

May 19, 2014
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Oct 10, 2015
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Ok, I see what I'm doing wrong. I'm trying to model my circuit using a generic transducer found in multisim. Its a different type. Im going to have trouble modeling this circuit it looks like. Not sure how to set up the transducer and LEM does not offer matlab to add to multisim. Ill just have to build it I guess. You've been very helpful.

May 19, 2014
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1,218