Monitoring 220V AC Load Current using Current Transformer [Circuit Diagram]

In this post I have explained a circuit which can be used for monitoring or controlling AC 220V load current by sensing the load current through a contactless current transformer.

Figure 1 below depicts a circuit diagram designed for a non contact load current sensing through a current transformer, which seemed to work well. It is intended to run on a basic +12 volt regulated or unregulated power source, such as a "wall wart" adapter.

Basic Working Theory

One of the wires of the load is passed through the current transformer. The current through the load wire is magnetically induced into the secondary winding of the current transformer.

This induced current is stepped up by the current transformer and is sent to op amp amplifier stages where it is amplified to finally operate a relay.

The relay switches OFF the load when the current exceeds the cut off threshold, as set by the op amp.

What is a Current Transformer

T1 denotes the current transformer, whose toroid core was salvaged from an old computer power supply.

A current transformer or a CT basically consists of secondary winding having many number of turns whose ends are configured with an external amplifier.

The primary side is created simply by allowing one of the wires of the load power supply to pass through the sensing transformer core.

The current passing through the load supply wire is induced into the secondary winding of the sensing transformer, which is proportionately stepped up by the secondary winding and supplied to the external amplifier for further amplification and detection.

The amplifier circuit amplifies the signal to operate a relay or a switching device for executing the necessary corrections.

Just like any other transformer, a current transformer primary and secondary voltages can be calculated using the following basic formula:

Es/Ep = Ns/Np

where,

• Es = Secondary Voltage,
• Ep = Primary Voltage,
• Ns = Number of secondary turns,
• Np = Number of Primary turns

How turns ratio can be calculated for a current transformer? It can be calculated using the following formula.
I_p / I_s = N_s / N_p
where, I_p is the current in the primary, I_s is the current in the secondary, N_p is the number of primary turns, N_s is the number of secondary turns.

In our current sensing transformer, the secondary is made up of about 100 turns of 22 gauge super enameled magnet wire.

The primary is formed by passing the hot side of the load line through the core. This implies that instead of a core suitable to radio frequencies, you should use a core which can tolerate lower frequencies.

The ferrite core selected for this project wasn't ideally suitable; it was probably built for a 20 kHz switching supply, but it still functioned perfectly.

Because the core soon saturates, the output waveform is a very pointed-looking, instead of a smooth sine wave.

How the Circuit Works

To safeguard the opamp, D1 and D2 clip the output waveform, which is capacitively associated to the op-amp input via C1.

R1 and R2 are positioned to bias the non-inverting input to half the supply voltage, while R3 and R5 work similarly for the inverting input.

A comparator is formed using IC 1 a. Its output is rectified through the diode D3 and smoothed using the capacitor C2 and R6, which generate a time constant to ensure that the circuit is neither activated or impacted in any way due to voltage spikes or an unusual waveform from a controlling load.

This alternating current voltage activates second comparator, which powers the output transistor, Q1.

D4 safeguards Q1 from reverse voltage spikes emanating from the relay coil, while the transistor Q1 works like a switch to toggle on the relay.

R4 was introduced so that the circuit could be fine-tuned to meet modest low current loads.

If you want to utilize the circuit with heavy loads (cutting tools, for example), you may omit R4 and just hook the junction of R3 and R5 to pin 2 of IC1a.

R4 is set such that the output of IC1 a is close to zero volts when the control load is switched off and the relay switches in reliably when the load is turned on.

When an oscilloscope is hooked up to IC1a pin 1, and R4 set to the center of the range, it generated a waveform, which ensured that when the load was switched off, pin 1 potential returned to zero.

R9 and R10 serve in lowering the low-level saturated output voltage of U1b, preventing Q1 from turning on at that moment.

You may come across some opamps that simply won't drop low enough to complete the job on their own.

The relay used in this circuit must be rated at 12 V. For switching heavier loads, you may want this relay to be a bigger relay (often referred to as a contactor).

Enclose the circuit, particularly the 120 VAC mains components, in a safe, grounded metal box.

Fixing the box to the metal frame of your table saw could most likely fulfill the purpose.

For operating a heavy load like the table saw, a line cable may be more suitable with an outlet in the box, similar to a single-outlet extension cord, with the black hot wire passing through the torroid prior to getting firmly attached to the hot side (brass color) of the receptacle.

Everything can be protected and contained in this manner. In this case, the simplest method to handle the power supply is to create one from the beginning so that it could be contained inside the enclosure.

A wall-wart power supply option might not be a good idea, since this might end up tumbling out of its socket due to saw vibration or causing a trip hazard if connected into an external socket.

Another Current Monitor Circuit

A 220 V AC current-monitoring circuit with an audible and visual warning output is depicted in the figure above.

The difference in the output voltage of the transformer may be balanced out by adjusting the gain of the op-amp using R6.

The circuit can monitor AC current levels between less than 1 amp and more than 5 amps.

In this variable-gain voltage amplifier circuit, one op-amp from an LM324N quad opamp package is connected.

The gain can range from zero to roughly one hundred. The output of the amplifier is coupled with a DC rectifier circuit composed of D1, D2, C2, and C3.

One of the six buffers in a 4049UB IC is linked to the input of the positive DC output signal.

A constant AC output voltage is provided at the secondary of the current transformer by current flowing in the primary, and a positive voltage is supplied at the input of the 4049UB buffer.

Because the IC buffer is an inverter, the output is low when the input is positive. Therefore, neither the LED nor the piezo buzzer are turned on.

The input of the inverter IC becomes low when the load circuit fails to maintain current flow, allowing the positive voltage across C3 to discharge via R5 to almost ground level.

The LED is lit and the alarm is activated as the 4049UB's output increases from a low to a high level.

When the load current is at its lowest feasible level, the gain of the op-amp must be adjusted to provide a minimum of 7 volts DC at the input (pin #3) of the 4049UB.

By doing this, the alarm won't sound when the load is using its minimum amount of current.

How to Build the Current Transformer

The current transformer for the above circuit is a simple design as shown in the following schematic:

We wind the two winding over a plastic bobbin.

We then insert and place the bobbin in the middle leg of the two E-cores.

We use 6 turns of 18 SWG super enameled copper wire for the current sensing winding which is wired in series with the 220V or 120 V AC load. This forms the primary winding of the current transformer.

We use upto 100 turns of 30 SWG enameled copper wire for the secondary winding which is connected with the op amp circuit.

The number of secondary turns required is determined by the primary side current and the output voltage needed by the monitoring alarm circuit.

About 300 secondary turns are needed for low current levels lower than 1 amp to about 3 amps, while fewer turns are needed for higher current levels.