Electronic ### Briefly describe the difference between current transformer and voltage transformer

Design a current transformer. The use of current transformers can reduce the loss when measuring the primary current of the converter, such as high-power switching power supplies, because the current is too large, it is necessary to use a current transformer to monitor the current to reduce the loss.

Design a current transformer. The use of current transformers can reduce the loss when measuring the primary current of the converter, such as high-power switching power supplies, because the current is too large, it is necessary to use a current transformer to monitor the current to reduce the loss.

What is the difference between a current transformer and a general voltage transformer? This question is difficult for even experienced magnetic component designers to answer. The basic difference is that the transformer tries to transform the voltage from the primary side to the secondary side, while the current transformer tries to transform the current from the primary side to the secondary side. The voltage of the current transformer is determined by the load.

We can better understand the working principle of current transformer through a practical design example.

Assuming that a current transformer is used to measure the primary side current of the converter, the primary side 10A current corresponds to 1V voltage. Of course, we can use a resistance of 1V/10A=100mΩ to measure, but the loss caused by the resistance is 1V×10A=10W. Such a large loss is unacceptable for almost all designs. Therefore, a current transformer should be selected, as shown in Figure 1. Of course, in order to reduce the winding resistance, we take the number of turns of the primary side as 1 turn, and in order to reduce the current to a relatively low level, the number of turns on the secondary side should be more. If the number of turns on the secondary side is N, from Ohm’s law, (10/N)R=1V, and the power consumed in the resistor is P=(1V)^2/R. We assume that the power consumed is 50mW (that is, we can use a 100mW resistor), which requires R not to be less than 20Ω. If a 20Ω resistor is used, the number of secondary turns N=200 can be obtained by Ohm’s law.

Now let’s look at the magnetic core, assuming that the diode is an ordinary diode, the on-state voltage is about 1V, and the current is 10A/200=50mA. The transformer output voltage is 1V, plus the diode’s on-state voltage of 1V, the total voltage is about 2V. When working at a frequency of 250kHz, the magnetic induction intensity on the magnetic core will not exceed the period of 4us, which is definitely less than one cycle. Since the time of current flowing through the primary side cannot exceed the switching period (otherwise, the magnetic core cannot be reset). Therefore, Ae can be very small, and B will not be very large. In this example, the size of the magnetic core cannot be determined by the loss requirement or the flux saturation requirement. It is more likely to be determined by the isolation voltage between the primary and secondary sides. If the isolation voltage is not required, the size of the magnetic core is generally determined by the volume occupied by the 200-turn winding. You can use a No. 40 wire to flow a peak current of 500mA, but this wire is too thin and ordinary transformer manufacturers will not wind it for you.

Practical Tips Unless you must use it, do not use wires with a size smaller than 36 gauge wire under normal circumstances. Now let’s analyze why we can’t use a voltage transformer to replace the current transformer? We already know that the secondary voltage is only 2V, so the primary voltage is 2V/200=100mV. If the input DC voltage is 48V, then the 10mV voltage on the primary side of the current transformer is negligible for the 48V voltage-then you can get 50mA on the secondary side, and there is almost no effect on the primary side. Assuming another situation (unrealistic), the input DC voltage of the primary side is only 5mV, then the primary side of the transformer cannot have a voltage of 10mV. At the same time, because the primary side impedance (such as the reflected secondary side impedance) is also relatively large, it is decided Therefore, it is impossible for the secondary side to generate a current of 50mA. Even if the entire 5mV voltage is applied to the primary side, the secondary side can only generate a voltage of 200×5mV=1V: it cannot generate enough voltage on the conversion resistor. Therefore, the voltage transformer can only be used as a transformer and cannot be used to detect current.

From another point of view: Although the voltage of the input power supply is 48V, the magnitude of the current flowing through the current transformer is not determined by the 48V voltage on the primary side, but by other factors.

Current transformers are voltage transformers with impedance limitations.

Finally, let’s take a look at the error situation of the current transformer? The answer lies in the basic definition of the current transformer: what is induced is current.

Practical hints that the resistance of the diode and the secondary winding in the current mutual inductance will not affect the current measurement, because (as long as the impedance is not infinite) the current in the series circuit is equal everywhere and has nothing to do with the series components.

In actual work, it does not matter whether a Schottky diode is used as a rectifier diode: the low on-state voltage of the diode only affects the transformer, not the current transformer.

If the inductance of the secondary side of the transformer is too small, the measurement error will increase. That is, the excitation inductance is too small. Assuming that we require the maximum error of the measured current to be 1%, and the primary current is 10A, then the secondary current is 50mA, which means that the required excitation current (secondary) should be less than 50mA×1%= 500μA. The excitation current does not flow through the conversion resistance, and we cannot detect this current, so the error increases. We can calculate the minimum value of the secondary inductance.