TECHNICAL FAQS

  • How can magnetization of the current transducer be avoided?

    To avoid magnetization of a current transducer, the following measures can be taken:

    1. Prevent Overload Currents:

    • Avoid High Overload Currents: Magnetization occurs when the transducer’s core is exposed to excessive currents that can saturate the magnetic circuit. To minimize this, ensure that the current transducer is not subjected to currents exceeding its rated capacity. Proper protective devices, such as fuses or circuit breakers, should be in place to limit high overload currents.

    2. Ensure Continuous Power Supply:

    • Maintain Power Supply: Interruption or a lack of power supply to the transducer can result in an undesirable magnetic offset. To prevent this, ensure a reliable power supply and avoid unnecessary interruptions.

    3. Use a Transducer with Lower Magnetic Remanence:

    • Select Materials with Lower Residual Flux: Some magnetic materials used in transducers have higher residual flux (magnetic remanence), leading to higher magnetic offsets. Choosing transducers made from materials with low remanence can reduce the likelihood of magnetization.

    4. Demagnetization (Degaussing) Process:

    • Perform Periodic Demagnetization: If magnetization occurs, a degauss cycle can be used to restore the transducer’s core to its initial state. This involves applying a low-frequency AC current through the core, gradually decreasing the amplitude to remove residual magnetization. This process should be performed after any overload or interruption event.
    • Reversal of Primary Current: Another method of demagnetization is to apply a reverse current to the primary winding, effectively neutralizing the magnetic offset.

    5. Regular Monitoring:

    • Monitor for Magnetic Saturation: Regularly checking the transducer’s performance can help in detecting any offsets early. If a magnetic offset is detected, performing demagnetization can restore accuracy and prevent long-term performance degradation.

    By taking these precautions, magnetization of the current transducer can be minimized or avoided, ensuring accurate and stable performance over time.

  • Can a transducer be demagnetized?

    Process of Demagnetization:

    To demagnetize a transducer’s magnetic core, the following steps are typically used:

    1.AC Excitation: A low-frequency AC source drives the magnetic core through its entire B-H hysteresis loop (the magnetic field vs. magnetization curve), applying alternating current to cycle the core’s magnetization from positive to negative saturation.

    2.Gradual Decrease of Excitation: After cycling the core through the B-H loop for at least 5 full cycles at maximum amplitude, the amplitude of the AC excitation is reduced gradually. The rate of reduction should be no faster than 4% per cycle to ensure a smooth return of the core’s magnetization to zero (the origin of the B-H curve).

    3.Time Requirements: Typically, a demagnetization cycle will require around 30 cycles at a low frequency, such as 500 milliseconds at 60 Hz.

     

    Special Considerations for Closed-Loop Devices:

    For closed-loop transducers (devices that have a compensation coil), additional care is needed because the compensation coil can interfere with the demagnetization process by negating its effects. This interference can prevent the core from reaching its neutral state. Therefore, extra steps may be required to ensure proper demagnetization.

    Partial Demagnetization:

    Alternatively, partial demagnetization can be achieved by applying a specific signal of opposite polarity to the core’s magnetization. However, this method is more complex because it requires precisely determining the correct amplitude and duration of the signal to effectively reduce the magnetization without overshooting or undercompensating.

    In practice, if a specific application is well understood, it is possible to determine the necessary correction signal empirically and apply this whenever demagnetization is needed.

    Demagnetization is essential for maintaining the accuracy and reliability of a transducer’s measurements, especially after strong magnetic influences or long-term use.

  • Regarding the effects of different Vref connections on the output.

    Let’s take a product with a supply voltage of +5V and an output of 2.5V ± 0.625V with a precision of 1% as an example.
    Our Vref pin serves two functions: one is to act as a reference point for output, commonly used in differential input acquisition devices; the other function is for single-ended input acquisition devices, where it is used to control the sensor output, making the static output (2.5V) more accurate. It only affects the static output (2.5V when there is auxiliary power and no input signal) and does not affect the amplification (0.625V).
    Due to the discreteness of product parameters, sensors cannot be perfectly identical and ideal, and there will always be some degree of error. For most of our products, the static reference point error is within ±15mV. This means the static output of 2.5V (with 0V or GND as the reference point) will be between 2.485V and 2.515V, which is considered within the acceptable range.
    For example, for a product with a power supply of +5V and an output of 2.5V ± 0.625V with a precision of 1%:

    If the output is referenced to 0V (GND), we calibrate it as follows:
    First, check if the Vref voltage meets the requirements (standard is 2.5V ± 1%, generally no more than 0.5%).
    The second step is to adjust the static output to between 2.485V and 2.515V (i.e., 2.5V ± 15mV).
    The third step is to apply the input signal to the rated input and adjust the output to within 2.5V + (0.625V ± 1%).
    Finally, check if the output at 10% of the rated input is accurate. If there is significant error, adjust the zero-point component for correction. Then, check if the zero-point and full-scale values meet the requirements. If everything is correct, the product moves on to the next process step.

    If the output is referenced to Vref, we adjust it as follows:

    First, check if the Vref voltage meets the requirements (standard is 2.5V ± 1%, generally no more than 0.5%).
    The second step is to adjust the static output to between 2.485V and 2.515V (i.e., 2.5V ± 15mV).
    The third step is to connect the black probe of the multimeter to the Vref point, apply the input signal to the rated input, and adjust the output to within 0.625V ± 1%.
    Finally, check if the output at 10% of the rated input is accurate. If there is significant error, adjust the zero-point component for correction. Then, check if the zero-point and full-scale values meet the requirements. If everything is correct, the product moves on to the next process step.

    If a product adjusted using the first method is used in a differential input acquisition device, the acquired signal will be inaccurate and may even produce negative output errors.
    If a product adjusted using the second method is used in a single-ended input acquisition device, the acquired signal will also be incorrect and may result in overflow errors.

    By default, our products are calibrated using the first method (referenced to 0V/GND), as most customers request this.

    Example 1: A product calibrated using the first method, with Vref = 2.510V and Vout = 2.505V in the static state. This product is qualified, and the precision is acceptable. However, when used in a differential input acquisition device, since Vout is lower than Vref, the static output will be negative, which is incorrect.
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    Example 2: A product calibrated using the second method, with Vref = 2.510V and Vout = 2.520V in the static state. This product is also qualified, as the zero-point error does not exceed 15mV. However, when used in a single-ended input acquisition device, the output of 2.520V exceeds the calibrated 15mV error, and the full-scale value will also be higher, which could reach up to 2% in extreme cases, affecting the customer’s use.
    Now that some customers have reported this type of error, we are asking them to provide calibration requirements to correct the issue. Sales representatives are also being asked to label products based on different calibration methods to ensure correct production for customers.
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  • How to calculate transformer short circuit current?


    To calculate the input current of a transformer, you can use the following formula based on the transformer’s power rating, input voltage, and efficiency. The input current depends on the power demand on the secondary side (load), and this can be calculated for both single-phase and three-phase transformers.

    1. Single-Phase Transformer Input Current Calculation

    For a single-phase transformer, the input current IentradaI_{\text{input}} can be calculated using the formula:

    Ientrada=PoutVprimary×ηI_{\text{input}} = \frac{P_{\text{out}}}{V_{\text{primary}} \times \eta}

    Where:

    • IentradaI_{\text{input}} = Input Current (in Amperes)
    • PoutP_{\text{out}} = Output Power (in Watts or VA)
    • VprimaryV_{\text{primary}} = Primary Voltage (in Volts)
    • η\eta = Efficiency of the transformer (expressed as a decimal)

    If the efficiency is not given, it can be assumed based on typical transformer efficiency (for instance, 0.98 for a high-efficiency transformer).

    2. Three-Phase Transformer Input Current Calculation

    For a three-phase transformer, the input current IentradaI_{\text{input}} is calculated with a similar formula but with an additional factor for three-phase power:

    Ientrada=Pout3×Vprimary×ηI_{\text{input}} = \frac{P_{\text{out}}}{\sqrt{3} \times V_{\text{primary}} \times \eta}

    Where:

    • IentradaI_{\text{input}} = Input Current (in Amperes)
    • PoutP_{\text{out}} = Output Power (in Watts or VA)
    • VprimaryV_{\text{primary}} = Primary Line-to-Line Voltage (in Volts)
    • η\eta
    • 3\sqrt{3} = 1.732, a factor used for three-phase power

    3. Example Calculation (Three-Phase Transformer)

    Suppose you have a 500 kVA transformer, with a primary voltage of 11 kV (11,000 V) and an efficiency of 98%.

    Step 1: Determine Output Power:

    For a 500 kVA transformer, the power PoutP_{\text{out}} is 500,000 VA (since 1 kVA = 1,000 VA).

    Step 2: Apply the Formula:

    Ientrada=500,0003×11,000×0.98I_{\text{input}} = \frac{500,000}{\sqrt{3} \times 11,000 \times 0.98}

    Ientrada=500,0001.732×11,000×0.98I_{\text{input}} = \frac{500,000}{1.732 \times 11,000 \times 0.98}

    Ientrada=500,00018,577.17626.92AI_{\text{input}} = \frac{500,000}{18,577.176} \approx 26.92 \, \text{A}

    So, the input current for the primary side of the transformer is approximately 26.92 A.

    Important Notes:

    • The output power PoutP_{\text{out}} is usually given in kVA or VA. To convert from kW to VA, use the apparent power formula Pout=kWpower factorP_{\text{out}} = \frac{\text{kW}}{\text{power factor}} if necessary.
    • Efficiency η\eta
    • For three-phase systems, make sure you use the line-to-line primary voltage.

    This calculation helps determine how much current is drawn on the primary side of the transformer based on the output load.

  • What is current transformer Burden?

    In a current transformer (CT), the burden refers to the total impedance (resistance and reactance) presented by the connected measuring devices, such as ammeters, relays, or meters, to the secondary winding of the CT. It is usually expressed in volt-amperes (VA) at a specific secondary current (typically 1A or 5A) and can include the impedance of the connecting wires and instruments.

    Key Points:

    1. Burden Components: The burden consists of the impedance of the connected devices (like meters, protection relays, and wires) which draw power from the CT’s secondary winding.
    2. Burden Rating:
      • CTs are rated for a maximum burden in VA. For example, a CT rated for 10 VA means that the connected load should not exceed 10 VA.
      • If the burden exceeds the CT’s rated value, it can cause errors in current measurement or protection malfunctions because the CT may not be able to maintain its accuracy or properly step down the current.
    3. Importance of Proper Burden:
      • The burden affects the accuracy of the CT. If the burden is too high (i.e., the impedance of the connected devices is too large), the CT will not provide an accurate current transformation, leading to incorrect measurements or protection relay malfunctions.
      • On the other hand, if the burden is too low, the CT may operate closer to saturation, which also leads to inaccuracies.
    4. Practical Consideration:
      • When installing a CT, it is crucial to calculate the burden of all connected devices and ensure it falls within the CT’s rated burden for proper performance.

    In summary, the burden is the load imposed on the secondary winding of the current transformer, and ensuring it is within the CT’s rated capacity is essential for accurate current measurement and safe operation.

  • How to connect current transformer?

    1. Single-Phase Circuit:

    • Connection Mode: A single current transformer is connected in series with the live wire or neutral wire of the circuit.
    • Application: This is used for measuring single-phase current, power, and providing protection against single-phase faults, such as ground faults or short circuits.
    • Diagram:

    2. Incomplete Star (V Connection) for Two Current Transformers (Three-Phase Circuit):

    • Connection Mode: Two current transformers are connected in a “V” configuration (incomplete star), where the primary windings are connected to two of the three-phase conductors. The secondary windings are then connected together to form a measuring point.
    • Application: This connection is used to measure three-phase current, power, and to protect the system from phase-to-phase short-circuit faults. Although only two transformers are used, it provides sufficient information about the three-phase system for protection and measurement in a balanced system.
    • Diagram:

    3. Two-Phase Current Difference Between Two Current Transformers (Phase Difference Protection):

    • Connection Mode: Two current transformers are placed in two different phases (e.g., Phase A and Phase B), and their secondary windings are connected to each other to detect any current difference between these phases.
    • Application: This configuration is used for measuring one-phase current and for protecting against phase-to-phase short-circuit faults. It is based on detecting differences in the phase currents, which indicates a fault between phases.
    • Diagram:

    4. Star (Wye) Connection Mode for Three Current Transformers (Three-Phase Four-Wire System):

    • Connection Mode: Three current transformers are connected in a star (wye) configuration. Each CT is placed on one phase of a three-phase system, and the secondary windings are connected to a common point (neutral).
    • Application: This setup is suitable for measuring three-phase current, power, and protecting the system in a three-phase four-wire system. It can detect any form of fault, including ground faults, phase-to-phase faults, and phase-to-neutral faults.
    • Diagram:
  • What is Residual Current Sensor?

    A Residual Current Sensor (RCS) is an electrical safety device designed to detect residual current (also known as leakage current) in a circuit. Residual current occurs when there is an imbalance between the current flowing into a system (through the live wire) and the current returning from the system (through the neutral wire). This imbalance indicates that some current is leaking to the ground, often due to insulation failure, faulty wiring, or a person coming into contact with the electrical system.

    How Does a Residual Current Sensor Work?

    1. Current Imbalance Detection: The RCS measures the difference between the current in the live wire and the neutral wire. Under normal operating conditions, the current flowing through these two wires should be equal. If there is a discrepancy, it means current is leaking somewhere.
    2. Detection Threshold: If the difference between the live and neutral current exceeds a certain threshold (usually set to a few milliamps, such as 10 mA or 30 mA, depending on the application), the sensor detects it as residual current.
    3. Activation of Safety Mechanism: When residual current is detected, the sensor typically signals a protective device, such as a Residual Current Device (RCD) or Ground Fault Circuit Interrupter (GFCI). These devices will quickly disconnect the power to prevent hazards like electric shock or fire.

    Key Features of a Residual Current Sensor:

    • High Sensitivity: RCS devices can detect very small current differences, often in the range of milliamps (mA).
    • Non-Intrusive: The sensor can monitor current flow without interrupting the circuit.
    • Rapid Response: It triggers safety mechanisms immediately upon detecting leakage, minimizing risks.
    • Precisión: Modern RCS units provide precise detection and monitoring of residual current to avoid nuisance tripping.

    Applications:

    1. Household Electrical Safety: RCS is widely used in household RCDs and GFCIs to protect users from electric shocks, particularly in wet areas like kitchens and bathrooms.
    2. Industrial Systems: In industrial environments, RCS devices monitor equipment to detect insulation failures or potential fault conditions that could lead to electrical hazards.
    3. Renewable Energy: RCS is also employed in renewable energy systems, such as solar power installations, to ensure safe operation and prevent electrical leakage to ground.

    Benefits:

    • Prevention of Electric Shock: RCS can detect when current flows through a person or faulty wiring to the ground, preventing serious injury.
    • Fire Prevention: By detecting leakage currents caused by insulation failure, the RCS can help prevent electrical fires.
    • Compliance with Safety Standards: Residual current sensors are essential for meeting electrical safety regulations in homes, offices, and industrial settings.

    In summary, a Residual Current Sensor is a critical safety device that continuously monitors circuits for leakage current, helping to protect people and property from electrical hazards by detecting imbalances in current flow and triggering protective mechanisms.

  • What is a Rogowski coil and how does it work?

    Rogowski coil also called differential current sensor, it is a “hollow” circular coil, arranged around the conductor, so that an alternating magnetic field generated by the current induces a voltage in the coil. The coil is actually a ransformer coupled to the conductor under test, and the voltage directly output from the coil is proportional to the rate of change of current, e.g. @50Hz/1kA Vout=85mV, @60Hz/1kA Vout=85*60/50=102mV. If you want to obtain the current waveform or current value that doesn’t matter to frequency also need to add the integral circuit to achieve 90° phase shift compensation and frequency equalization.

    RF series is a flexible current sensor based on Rogowski coil principle, which is in small size, light weight and easy to install and offers a choice of different sizes, which can also be customized design of the customer requires a special order. It has no magnetic saturation and a shielding layer to resist the influence of external magnetic field, so stable measurements can be achieved in the range of low current to hundreds of kA. The flexible rogowski coil is an extremely comfortable solution for current measurement,
    particularly suited to current monitoring and electrical retrofitting, can be used in many cases where a traditional current transformer (CT) is not available, or can replace it.Systems that use an ADC chips(ADS131M04) or a power metering ICs(ADE7753) that support the Rogowski coil principle are even more advantageous.

  • How to Choose a Current Transformer?
    1. Determine the Type of Current to be Measured:
      • Identify whether the current to be measured is AC (Alternating Current) or DC (Direct Current). Most current transformers are designed for AC, so if you need to measure DC current, you will need to select a specific DC current transformer.
    2. Determine the Current Rating:
      • Assess the maximum current that needs to be measured. Current transformers are designed to measure a specific range of currents, so selecting the correct range is essential for accurate measurements.
    3. Determine the Accuracy Requirement:
      • Different applications require different levels of accuracy. For example, power generation and transmission typically require higher accuracy than some industrial applications. Determine the level of accuracy required for your application and select a current transformer that meets that requirement.
    4. Determine the Burden Requirement:
      • The burden is the maximum load that the current transformer can drive, and it is usually expressed in VA (Volt-Amperes). Choose a transformer with a burden that is sufficient for your application.
    5. Consider the Type of Output Required:
      • Determine the type of output required, whether it is a voltage or current output, and select a transformer with the appropriate output.
    6. Consider Physical Size and Mounting Requirements:
      • Choose a transformer that is suitable for the physical size and mounting requirements of your application.
    7. Verify Compatibility:
      • Make sure the current transformer is compatible with your measurement equipment and that it meets any relevant safety standards.
  • What is the difference between current transformer and voltage transformer?
    • Current Transformers (CTs): Measure and step down actual for use in measurement, monitoring, and protection systems. They are connected in series with the circuit, providing a lower current proportional to the high current in the system for safe, accurate measurement.
    • Voltage Transformers (VTs): Measure and step down tensión for similar purposes. They are connected in parallel to the circuit, delivering a reduced voltage proportional to the high voltage for safe analysis and control.

    In essence, CTs are used for current measurement and VTs for voltage measurement, both essential for protection, control, and monitoring in electrical power systems.

  • Why current transformer secondary is always short circuited?

    The secondary of a current transformer (CT) is always short-circuited (or connected to a low-impedance load like a meter or relay) for several important reasons related to safety and proper functioning:

    1. Voltage Rise on an Open Secondary:

    • Dangerous Voltage: If the secondary winding of a CT is left open while current flows in the primary winding, it behaves like a step-up transformer. Since the primary winding carries high current and the secondary winding has many turns, an open circuit can cause the voltage in the secondary to increase to dangerously high levels.
    • Insulation Breakdown and Safety Hazard: The high induced voltage can exceed the insulation rating of the transformer, leading to insulation breakdown and a potential shock hazard to personnel or equipment damage.

    2. Core Saturation:

    • Magnetic Core Saturation: In normal operation, the secondary current creates a magnetic field that opposes and cancels out most of the magnetic flux created by the primary current. If the secondary is open, no current flows in the secondary, and the magnetic flux in the core increases to very high levels.
    • Core Overheating: The excessive magnetic flux can saturate the CT core, leading to overheating and potential damage to the transformer. This can result in permanent damage to the CT, reducing its accuracy and reliability.

    3. Accurate Measurement:

    • Current Proportionality: CTs are designed to measure current, and they rely on the secondary current being proportional to the primary current (following the turns ratio). If the secondary is open, this relationship is broken, and no accurate current measurement can be made.
    • Protection System Failure: CTs are often used in protective relays to monitor overcurrent or fault conditions. An open secondary may prevent the proper operation of protective devices, potentially leading to system failures or unprotected faults.

    4. Reduced Burden on the Secondary:

    • Minimizing Burden: A CT is designed to operate with a very low burden (load) on its secondary side to maintain accuracy. Short-circuiting the secondary or connecting it to a low-impedance load ensures that the CT operates within its designed parameters, providing accurate current measurements without overloading the core or secondary windings.
  • How to read current transformer (CT) with arduino?

    To connect a CT sensor to an Arduino, the output signal from the CT sensor needs to be conditioned so it meets the input requirements of the Arduino analog inputs, i.e. a positive voltage between 0V and the ADC reference voltage.

    This give the example of an Arduino board working at 5 V and of the EmonTx working at 3.3 V. Make sure you use the right supply voltage and bias voltage in your calculations that correspond to your setup.

    The circuit consists of two main parts, their functions are to change the c.t’s current into a voltage of the correct amplitude, and position this voltage in the centre of the ADC’s input range.

    The voltages and currents shown are for a 5 V Arduino, with a 0 – 5 V range for the analogue input, about 1.6 V rms for a sine wave. For the emonTx V2 & V3 and the emonPi, the analogue input range is 0 – 3.3 V, so the midpoint voltage is 1.65 V and the analogue input voltage swings between 0 and 3.3 V (approximately 1 V rms for a sine wave). For the emonTx4 and emonPi2, the analogue input range is 0 – 1 V and is intended for use with 0.333 V rms output current transformers, which do not need a burden. so this resistor is omitted.

    ct-sensors-circuit1.png

    Detailed Guide

     

     

     

  • Why current transformer secondary should not be open?

    The secondary of a current transformer (CT) should never be left open when current is flowing through the primary because doing so can create dangerous high voltages. Here’s why:

    1. Induced High Voltage:

    • A current transformer operates on the principle of electromagnetic induction. The current in the primary winding induces a current in the secondary winding. If the secondary is left open, there is no current flowing through it, which prevents the magnetic field from opposing the primary current.
    • As a result, the magnetic flux in the core increases dramatically, which leads to a very high voltage being induced in the secondary winding. This high voltage can damage the CT insulation or create a serious safety hazard for anyone nearby.

    2. Saturation and Core Damage:

    • With no load (open secondary), the core of the CT may become saturated due to the high magnetic flux. Prolonged exposure to this condition can permanently damage the CT core by altering its magnetic properties, reducing its accuracy and performance over time.

    3. Safety Hazards:

    • The high voltage produced across the open secondary can be lethal and may damage connected equipment. In some cases, the voltage can even cause arcing, which presents a significant fire or electrical shock risk.

    4. Inaccuracy in Measurements:

    • If the CT secondary is open, it cannot perform its primary function, which is to accurately measure current. With the secondary circuit open, there is no reliable way to measure the current in the primary circuit, which can lead to incorrect readings and relay malfunctions in protection systems.

     

  • How to read current sensor with arduino?

    Take the HSTS016L model as an example, it is capable of measuring current values ​​from 10A to 200A. Using the split core current sensor type, we can turn the CT on and off without changing the existing system, and just use a clip to put the measuring cable inside the CT. The output voltage of this sensor is 2.5V +/- 0.625V, which is a good accuracy. Although the voltage output range of this sensor is between 1.875V and 3.125V, it is also suitable for 3.3V analog sensors, especially Arduino Nano and NodeMCU microcontrollers.

     

    The sensor has 4 output pins: RED (5V input), BLACK (0V Gnd), YELLOW (Analog Output), and WHITE (Analog for Calibration). Sometimes the supply voltage may not be exact 5V thus we will need additional Analog Pin to measure the exact middle point from the White Pin. For Arduino UNO, there are 6 analog input pins (A0-A5) where we can use one of the pins to measure AC current and 1 more to use for calibration purpose. The analog input pins will map input voltages between 0 and 5V into integer values ​​between 0 and 1023 with resolution of 4.9mV per unit (5.00V / 1023 units).

     

    We connect RED to 5V input, BLACK to ground, YELLOW to analog pin A1, and WHITE to analog pin A2. It is highly recommended to use a 12V power adapter to power the Arduino Uno and the sensor. Using a 5V power supply over USB will have an initial offset value, you may need to manually add the offset value when uploading the code.

     


    // Which pin to measure Current Value (A0 is reserved for LCD Display Shield Button function)
    int currentAnalogInputPin = A1;  
    // Which pin to calibrate offset middle value
    int calibrationPin = A2; 
    // Key in value to manually offset the initial value
    float manualOffset = 0.00;
    // If using “Hall-Effect” Current Transformer, key in value using this formula: mVperAmp = maximum voltage range (in milli volt) / current rating of CT
    float mVperAmpValue = 12.5;        
    // Analog input pin maximum supply voltage, Arduino Uno or Mega is 5000mV while Arduino Nano or Node MCU is 3300mV
    float supplyVoltage = 5000;
    /* to read the value of a sample for offset purpose later */
    float offsetSampleRead = 0;        
     /* to read the value of a sample including currentOffset1 value*/
    float currentSampleRead  = 0;     
      /* to count time for each sample. Technically 1 milli second 1 sample is taken */
    float currentLastSample  = 0;     
      /* accumulation of sample readings */
    float currentSampleSum   = 0;     
      /* to count number of sample. */
    float currentSampleCount = 0;     
        /* to calculate the average value from all samples, in analog values*/
    float currentMean ;            
      /* square roof of currentMean, in analog values */
    float RMSCurrentMean ;             
       /* the final RMS current reading*/
    float FinalRMSCurrent ;           
    void setup()                                 /*codes to run once */
    {
            Serial.begin(9600);                 /* to display readings in Serial Monitor at 9600 baud rates */
    }
    void loop()
    {
            /* 1- AC & DC Current Measurement */
            if(micros() >= currentLastSample + 200)                     /* every 0.2 milli second taking 1 reading */
              {
               /* read the sample value including offset value*/
               currentSampleRead = analogRead(currentAnalogInputPin)-analogRead(calibrationPin); 
              /* accumulate total analog values for each sample readings*/
               currentSampleSum = currentSampleSum + sq(currentSampleRead) ;
               /* to count and move on to the next following count */
               currentSampleCount = currentSampleCount + 1; 
               /* to reset the time again so that next cycle can start again*/
               currentLastSample = micros(); 
              }
            /* after 4000 count or 800 milli seconds (0.8 second), do this following codes*/
            if(currentSampleCount == 4000) 
              {
                /* average accumulated analog values*/
                currentMean = currentSampleSum/currentSampleCount;
                 /* square root of the average value*/
                RMSCurrentMean = sqrt(currentMean);
                 /* calculate the final RMS current*/
                FinalRMSCurrent = (((RMSCurrentMean /1023) *supplyVoltage) /mVperAmpValue)- manualOffset;
                  /* if the current detected is less than or up to 1%, set current value to 0A*/
                if(FinalRMSCurrent <= (625/mVperAmpValue/100))
                { FinalRMSCurrent =0; }
                Serial.print(” The Current RMS value is: “);
                Serial.print(FinalRMSCurrent,decimalPrecision);
                Serial.println(” A “);
                currentSampleSum =0;                                 /* to reset accumulate sample values for the next cycle */
                currentSampleCount=0;                                 /* to reset number of sample for the next cycle */
              }
    }
  • Why does a transformer not work with direct current?

    A transformer does not work with direct current (DC) because its operation relies on changing magnetic fields, which DC does not provide. Here’s a detailed explanation of why transformers are ineffective with DC:

    1. Principle of Operation:

    • Transformers work based on Faraday’s Law of Electromagnetic Induction, which states that a changing magnetic field in the core induces an electromotive force (EMF) or voltage in the secondary winding.
    • This changing magnetic field is produced by an alternating current (AC), which constantly fluctuates and reverses direction, causing the magnetic flux in the transformer core to vary.

    2. No Changing Magnetic Field in DC:

    • DC provides a constant current that does not change direction or magnitude over time.
    • Since there is no fluctuation in the current, the magnetic flux in the transformer core remains constant. Faraday’s Law requires a time-varying flux to induce a voltage in the secondary winding, so no voltage is generated when DC is applied.

    3. Core Saturation:

    • When DC is applied to the primary winding, it creates a constant magnetic flux in the transformer core, which can quickly lead to core saturation.
    • In core saturation, the core’s magnetic material is fully magnetized, and it cannot carry any additional flux. This results in a significant increase in primary current, which can cause overheating, insulation failure, and potentially damage the transformer.

    4. High Heat Generation:

    • Because the resistance of the transformer windings is relatively low, a constant DC current can cause a large amount of power dissipation in the form of heat. This can quickly overheat and damage the windings and insulation of the transformer.

    5. Absence of Induced Voltage:

    • In AC, the continuous change in current induces a voltage in the secondary winding through the process of mutual induction. Since DC does not create this changing current, no voltage is induced in the secondary winding, meaning the transformer cannot step up or step down the voltage with DC input.

    Conclusion:

    • Transformers are designed to work with alternating current (AC) because the changing nature of AC creates the necessary varying magnetic field for voltage induction. When DC is applied, the lack of a varying magnetic field prevents the transformer from functioning, and prolonged use with DC can lead to overheating and failure.
  • Why use current transformer?

    A current transformer (CT) is a vital electrical device used in high-power systems for the accurate measurement of alternating current (AC). It operates on the principle of electromagnetic induction, enabling the safe reduction of high current levels to a manageable value for measurement or protection. Below is a detailed explanation of its construction and working principles:

    1. Construction:

    A current transformer consists of two main windings: the primary winding and the secondary winding.

    • The primary winding is typically connected in series with the high-current-carrying conductor, whose current needs to be measured.
    • The secondary winding is connected to measuring instruments like ammeters, protective relays, or metering devices. The primary winding often consists of just one turn, while the secondary winding has a significantly higher number of turns.

    2. Turns Ratio:

    The turns ratio between the primary and secondary windings is a crucial parameter and is usually very high, often in the range of 100:1 to 5000:1. This means the secondary winding has many more turns compared to the primary winding. This ratio determines the magnitude of the current transformation and helps scale down the high current to a level suitable for standard measuring devices.

    3. Electromagnetic Induction:

    When an alternating current flows through the primary winding, it generates a corresponding alternating magnetic field around the conductor. According to Faraday’s Law of Electromagnetic Induction, this changing magnetic flux induces a proportional voltage in the secondary winding. The magnitude of the induced voltage and the resulting current in the secondary is dependent on the turns ratio.

    4. Step-Down Effect:

    The CT effectively steps down the high current from the primary side to a much lower current on the secondary side. While the turns ratio governs the reduction, the CT ensures that the secondary current is a scaled, proportional version of the primary current. For instance, a current transformer with a turns ratio of 1000:1 will produce 1 ampere in the secondary circuit when 1000 amperes flow in the primary conductor. This step-down effect enables standard measuring instruments, which are designed for low current ranges, to monitor the current accurately.

    5. Burden:

    The burden refers to the load connected to the CT’s secondary winding, which can be an ammeter, protective relay, or another measuring device. The impedance of this burden significantly affects the CT’s accuracy and performance. For accurate measurement, the burden impedance must match the CT’s specified range. If the burden is too high, the accuracy may be compromised, and in extreme cases, it may even damage the CT.

    6. Current Ratio:

    The current ratio of a CT is defined as the ratio of the primary current to the secondary current and is determined by the turns ratio. It is a fixed value for any given CT. For example, in a CT with a 1000:1 turns ratio, 1000 amperes in the primary circuit will produce 1 ampere in the secondary circuit. This consistent current ratio allows for the easy conversion of high primary currents into lower secondary currents for safe monitoring and protection.

    7. Accuracy Class:

    Current transformers are categorized into various accuracy classes based on their precision under specified conditions. The accuracy class defines the maximum permissible error in the measurement. The common accuracy classes are 0.1, 0.2, 0.5, 1, and 3, with class 0.1 offering the highest accuracy. CTs with higher accuracy are typically used in metering applications, whereas those with lower accuracy may be sufficient for protection purposes.

    The current transformer operates by stepping down high AC currents in a power system to a safer and manageable level, using the principle of electromagnetic induction and a high turns ratio. The CT’s secondary current is proportional to the primary current, which enables accurate measurement and protection of electrical systems. Proper matching of the CT’s burden and adherence to its accuracy class ensures reliable performance in various applications.

  • How to measure DC current using hall effect sensor?

    Components:

    1. Hall Effect Sensor: Senses the magnetic field generated by the current in a conductor and outputs a voltage proportional to the magnetic field strength.
    2. Power Supply: Provides power to the sensor and a reference voltage for calculations.
    3. Load: Completes the circuit, allowing current to flow through the conductor.
    4. Multimeter: Measures the output voltage from the Hall effect sensor.

    Steps to Measure DC Current:

    1. Connect Components: Connect the Hall effect sensor, power supply, and load according to the manufacturer’s instructions.
    2. Place Sensor Near Conductor: Position the sensor so that the magnetic field from the conductor is perpendicular to the sensor’s detection surface.
    3. Power On: Turn on the power supply to energize the circuit and the Hall effect sensor.
    4. Measure Output Voltage: Set the multimeter to measure voltage and connect it to the Hall effect sensor’s output.
    5. Record the Output Voltage: The measured voltage is directly related to the strength of the magnetic field produced by the current in the conductor.
    6. Calculate the Current: Using the sensor’s sensitivity (usually given in mV/Gauss or V/Tesla) and the reference voltage from the power supply, calculate the current using the proportional relationship between the magnetic field and current.

    Important Considerations:

    • Calibration: Hall effect sensors may need calibration to account for temperature variations and other factors to ensure accurate current measurement.
    • Linear Response: Ensure that the Hall effect sensor operates within its linear range for best results.

    By following these steps, you’ll be able to accurately measure DC current with the help of a Hall effect sensor.

  • How to measure AC current using hall effect sensor?

    Components:

    1. Hall Effect Sensor: Detects the magnetic field generated by the AC current flowing through the conductor and outputs a voltage proportional to the magnetic field.
    2. Load: A load (e.g., a resistor) completes the circuit, allowing current to flow.
    3. Oscilloscope: Measures the output voltage from the Hall effect sensor, allowing you to visualize the AC waveform.

    Steps to Measure AC Current:

    1. Connect Components:
      • Connect the Hall effect sensor and load according to the manufacturer’s instructions. The sensor requires correct wiring to the power supply and proper placement near the current-carrying conductor.
    2. Position the Sensor:
      • Place the Hall effect sensor near the AC current-carrying conductor, ensuring that the sensor is positioned perpendicular to the magnetic field created by the current in the conductor.
    3. Connect the Oscilloscope:
      • Attach the oscilloscope probes to the output of the Hall effect sensor. Set the oscilloscope to measure AC voltage. This allows you to observe the voltage waveform generated by the changing magnetic field.
    4. Turn On the Power Supply:
      • Activate the AC power supply, allowing current to flow through the conductor and generate a magnetic field around it. The Hall effect sensor detects this field.
    5. Observe the Output:
      • On the oscilloscope, you will see an AC voltage waveform. The amplitude of this waveform corresponds to the strength of the magnetic field, which is proportional to the AC current in the conductor.
    6. Calculate AC Current:
      • Use the sensibilidad of the Hall effect sensor (often specified in units such as mV/Gauss or V/Tesla) to convert the measured voltage into a corresponding current value. The formula will depend on the sensor’s sensitivity rating and the amplitude of the AC voltage waveform.

    Important Considerations:

    • Calibration: Hall effect sensors may exhibit variations due to temperature and magnetic field strength, so calibration is important to ensure accurate current measurement.
    • Waveform Analysis: Since you are measuring AC current, the waveform on the oscilloscope will show both the magnitude and frequency of the AC current. Ensure your oscilloscope settings (time and voltage divisions) are appropriate for the expected waveform.
    • Safety: Working with AC current can be hazardous, so always follow proper safety precautions to avoid electrical shocks or equipment damage.

    By following these steps and ensuring proper calibration, you will be able to accurately measure AC current using a Hall effect sensor.

  • How to connect current sensor with nodemcu?

    Connecting a current sensor with a NodeMCU involves a few steps. Here’s a general overview:

    1.  Choose a current sensor: There are many types of current sensors available, such as Hall effect sensors or shunt resistors. Select a sensor based on your requirements and available resources.
    2. Identify the sensor pins: Look for the pinout diagram or datasheet of your selected current sensor and identify the pins for power supply, ground, and signal output.
    3. Connect power supply and ground: Connect the power supply pin of the sensor to the 3.3V pin of the NodeMCU, and connect the ground pin of the sensor to the GND pin of the NodeMCU.
    4. Connect signal output: Connect the signal output pin of the sensor to any available analog input pin of the NodeMCU, such as A0.
    5. Code the NodeMCU: Write code to read the analog input from the current sensor and process the data. For example, you can use the analogRead() function to read the voltage at the analog input pin and convert it to current using the sensor’s sensitivity. Then, you can use the WiFi capabilities of the NodeMCU to send the data to a cloud service or to a local server.

    Here’s an example code snippet that reads the voltage at pin A0 and converts it to current using a Hall effect sensor with a sensitivity of 100 mV/A:

    Arduino Copy code

    const int sensorPin = A0;
    const float sensitivity = 0.1// 100 mV/A
    void configuración() {
    Serial.begin(9600);
    }
    void bucle() {
    int sensorValue = analogRead(sensorPin);
    float voltage = sensorValue * (3.3 / 1023); // convert ADC value to voltage
    float current = voltage / sensitivity;
    Serial.imprimir(“Current (A): “);
    Serial.println(current);
    delay(1000);
    }

    Note: This is just an example code snippet and may need to be modified based on your specific requirements and sensor characteristics.

  • What’s the current transformer working principle?

    A current transformer (CT) operates on the principle of electromagnetic induction, similar to a standard transformer. Its primary purpose is to step down high currents from a power system to a lower, manageable value for metering, protection, and control devices. Here’s a breakdown of the working principle of a current transformer:

    1. Basic Construction:

    • A CT typically consists of:
      • Primary winding: The conductor carrying the high current, which can be a single turn or just the system conductor passing through the CT core.
      • Core: Made of ferromagnetic material, it concentrates the magnetic field created by the current in the primary winding.
      • Secondary winding: Wound around the core, with many more turns than the primary. This winding connects to meters, relays, or other measurement/protection devices.

    2. Working Principle: Electromagnetic Induction:

    The CT works based on Faraday’s Law of Electromagnetic Induction, which states that a changing magnetic field within a conductor induces a voltage in another conductor near it.

    • Primary Current:
      • When high current flows through the primary winding (or conductor), it generates a magnetic field around the primary conductor.
      • This magnetic field is concentrated by the CT core and links to the secondary winding.
    • Induced Current in the Secondary:
      • The varying magnetic field generated by the primary current induces an electromotive force (EMF) in the secondary winding, which causes a current to flow through it.
      • The secondary current is proportional to the primary current but reduced by the turns ratio between the primary and secondary windings.
    • Turns Ratio:
      • The ratio of the number of turns in the secondary winding to the number of turns in the primary winding determines the reduction in current.
      • For example, if the primary consists of one turn (a conductor passing through the core), and the secondary has 1000 turns, a primary current of 1000 A would result in a secondary current of 1 A (1000:1 ratio).

    3. Operation in a Closed Secondary Circuit:

    • The secondary winding must always be closed (either connected to a load or short-circuited) when the primary current is flowing. This allows the magnetic field created by the primary current to induce a proportional current in the secondary.
    • The secondary winding is typically connected to low-impedance measuring devices, such as ammeters, relays, or protection systems.

    4. Purpose and Applications:

    • Current Measurement: CTs reduce the high current in the primary circuit to a smaller, standardized value (e.g., 1 A or 5 A), which can be safely measured by instruments.
    • Protection: CTs provide current input to protective relays that trigger circuit breakers in case of overcurrent or fault conditions.
    • Isolation: CTs electrically isolate the high-voltage primary circuit from the low-voltage measuring or protective devices for safety.

    5. Accuracy Classes:

    CTs are designed to operate with high accuracy, so the secondary current is directly proportional to the primary current within a specified accuracy class (e.g., 1% or 0.5% error). Accuracy is critical in both metering and protection applications.

    Summary of Working:

    • The primary current creates a magnetic field in the CT core.
    • This varying magnetic field induces a current in the secondary winding.
    • The secondary current is a scaled-down version of the primary current, based on the turns ratio.
    • The secondary current can then be measured or used for control/protection purposes.

    This allows the safe and accurate measurement of high currents in power systems, protecting both personnel and equipment.

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