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My aim is to convert current obtained from CT to precise voltage that can be applied as input to ADC. enter image description here

As shown in the diagram , a conductor carries current of up to 63 A passes through the Current Transformer(CT) .The CT CT 1273-A1-RC has turns ratio as 1:2500 and current range of 0 to 100A . I have made one simple circuit in which there is shunt resistor in parallel and a diode in series with CT. The current passing through CT is from power distribution unit i.e. used in server rooms. So it is ac current . $$I_s=(I_p)/2500$$

I_s=(63/2500)=0.0252. My requirement is minimum-maximum voltage at the input of ADC should be in the range from 0 to 3.3 V . So for shunt\burden resistor R_shunt= 3.3/0.0252=50 ohm,

Problem I am facing is how to obtain precise voltage (DC value) and also is there any need of additional circuitry in the circuit should be made so that it can be applied as input to the internal A to D converter of STM32L011x3 and display on seven segment display .So can anybody help me regarding it?

NOTE: I have done few calculation please see 1,2,3 for the selection of components. Can anyone crosscheck the values and tell me whether am I going right or wrong?

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    \$\begingroup\$ Just a comment: NEVER put a series diode on a current transformer secondary like that. You should ALWAYS have a burden resistor directly across the secondary. Back to back zener clamps and other protection can be added as well, but AFTER the burden resistor. You don't want to be around open circuited high power CT's. \$\endgroup\$ – R Drast Dec 23 '16 at 10:58
  • \$\begingroup\$ @R Drast 1.is rectification of ac voltage/current is necessary before applying it to input of ADC? 2. can't we just put Shunt resistor of suitable value (here say 130ohm as Is=0.0252A,V(in_max_to_ADC)=3.3V across CT and apply this input to ADC directly? \$\endgroup\$ – user3559780 Dec 29 '16 at 11:31
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    \$\begingroup\$ Yes, unless your ADC can handle digitizing the AC waveform. The burden resistor location is a safety issue. \$\endgroup\$ – R Drast Dec 29 '16 at 12:10
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Do not use an inline diode to rectify a CT output, you will never be able to calibrate it, and it is non-linear.

Use an op-amp rectifier.

There are endless examples in application notes and on the web. Never put anything other than the burden resistor and clipping zeners on the output of the CT.

enter image description here

For those interested in in more detail, this may be helpful. The burden on a Current transformer is critical to it providing accurate output. The burden is calculated to provide a defined VA load, and it is the current and not the output voltage that is the most accurate reflection of the input current value. Raising the voltage produced on the output of a CT reduces it's accuracy. Adding diodes in series provides a VA offset that reduces the overall accuracy and linearity and would never be done in instrument grade measurements. Typically even low grade installations look for better than 1% CT accuracy, and many need 0.3% class measurements.

Adding another reference for operation of a precision rectifier (though it's a dual supply op amp) from TI that talks through how the diode Vf is compensated. These type of op amp rectifiers work down to just a few mV of AC input signal.

Update: added the feedback document for the OP as a design stepthrough.

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  • \$\begingroup\$ The LM324's in the picture should work but treat them as an example. LM324's are ok but need + and - 5V to work well and are rubbish near the supply rails. As you mention 3.3V I assume that's supply for your system which wont work for the LM324's \$\endgroup\$ – Spoon Dec 23 '16 at 8:05
  • \$\begingroup\$ @Jack Creasey in the given PDF it is said that " R2 should be chosen at least 10 times greater than R1 for proper accuracy." But what is the reason behind it.? \$\endgroup\$ – user3559780 Dec 23 '16 at 17:37
  • \$\begingroup\$ @Spoon if so is there any alternative for LM 324, that can work properly at 3.3 V? \$\endgroup\$ – user3559780 Dec 23 '16 at 17:39
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    \$\begingroup\$ @user3559780. When you include the diode within the feedback loop of an op amp you actually don't 'lose' any voltage due to the diode. In the schematic shown, as the output of the op amp goes positive, while the diode does not conduct there is no (it's open circuit high gain) feedback resistor. It's only when the op amp actually forward biases the diode that the feedback works. So the op amp slews past the diode Vf until feedback current flows to bring the gain back to 1. I added another reference to the answer that talks through the way it works. \$\endgroup\$ – Jack Creasey Dec 26 '16 at 6:40
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    \$\begingroup\$ @user3559780. In the schematic I provided, the diode is protection only...nothing to do with rectification. If you need to put Zeners on the CT output, then you'd select a voltage that clips above your output voltage. The reason for the 2 * LM358 is because the first is the precision rectifier, the second the peak detector (which you wanted, with a high value tau). I would not by choice ever use a peak detector since it hides data from me (distorted current waveform, power factor). I would scale the CT output directly into the A/D. \$\endgroup\$ – Jack Creasey Dec 29 '16 at 17:14
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In theory this could work (though you would need another diode to load the transformer on the negative half cycles) since the CT is approximately a constant current source when appropriately loaded, however it would have to be a different (much larger core) CT.

As you can see below, with the give CT there will be very little signal available at the output.

schematic

simulate this circuit – Schematic created using CircuitLab

The recommended maximum burden for this transformer to meet specs is 7.5\$\Omega\$, which means that it can only output 300mV RMS at 100A in. The diode drop is included in that maximum output voltage, which leaves you with very little signal even with a Schottky diode.


Your best bet probably is to load the part with 7.5\$\Omega\$ or so, giving you 300mV@100A, and amplify the AC signal to a couple volts (of the order of 10:1) with an op-amp then apply an AC-coupled precision rectifier (full wave or half wave). This can be done with a single supply if you AC-couple the amplifier or bias it to the middle of your 3.3V supply.


Below (from a TI document) is another way to do with a generated -230mV supply (so it can operate without an external negative supply). You can modify the feedback of the amplifier U2A (by increasing the value of R2- to 10K or 15K) to get the required output voltage, and of course you need the burden resistor R4 to be the correct value- less than 140mV RMS, so about 3.5 ohms for 100A full scale. A 5V supply is shown but it will operate from 3.3V.

enter image description here


Edit: To respond to a query in the comments- below is an LTSpice simulation to aid in understanding my first circuit above. Current is steered by the diodes either through the 1R resistor via D1 for positive current or through D2. The 55mV peak voltage can be low-pass filtered and amplified.

enter image description here

The simulation has a 1:2500 turns ratio ideal CT measuring 100A RMS from a 2.2 ohm resistor on a 220VAC RMS supply. The presence of the diodes has a negligible effect on the response.

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    \$\begingroup\$ Pethany. You can't load a CT with any form of avalanche, zener or diode device without a burden directly connected as it will produce a simple square wave (precisely because it will act like a constant current source). They have to be resistively loaded to work as designed. For the very same reason you end up with dangerous voltages on the output terminals if there is no burden. \$\endgroup\$ – Jack Creasey Dec 23 '16 at 18:45
  • \$\begingroup\$ @JackCreasey We don't really care about the waveform at the transformer output just so long as it doesn't get too high voltage relative to maximum burden (which could result in saturation). See edit above for LTSpice simulation. The inverse diode is there to conduct the negative half-cycles because the CT should always be loaded adequately. \$\endgroup\$ – Spehro Pefhany Dec 23 '16 at 20:50
  • \$\begingroup\$ @Sphero Pethany. Of course we care about the waveform!! The voltage of the waveform DEFINES the current flowing in the primary. A square wave tells you absolutely nothing about the primary current. \$\endgroup\$ – Jack Creasey Dec 23 '16 at 22:48
  • \$\begingroup\$ @Sphero Pethany. Perhaps this might help you: netaworld.org/sites/default/files/public/neta-journals/… or this electronics-tutorials.ws/transformer/current-transformer.html \$\endgroup\$ – Jack Creasey Dec 23 '16 at 22:52
  • \$\begingroup\$ @Sphero Pefhany. I don't deny that your circuit with diodes in series will produce an output voltage, but it will not be accurate (particularly at low currents). Once you put a diode in series the CT you distort the voltage output. \$\endgroup\$ – Jack Creasey Dec 23 '16 at 23:08
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enter image description hereWe can use " Current Sensor Module" such as ACS712TELC-05B . The output of current transformer direct connect to input of module.

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    \$\begingroup\$ Maybe, but the OP would learn nothing from that. \$\endgroup\$ – Transistor Nov 22 '18 at 22:32
  • \$\begingroup\$ Apologize me. OP? \$\endgroup\$ – Amir Yousefi Nov 22 '18 at 23:15
  • \$\begingroup\$ Sorry, Original Poster - the author of the question. \$\endgroup\$ – Transistor Nov 22 '18 at 23:18

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