The difference between current transformer and general voltage transformer

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The difference between current transformer and general voltage transformer

Posted Date: 2024-02-03

The use of current transformers can reduce the loss when detecting the primary current of the converter, such as high-power switching power supply. Because the current is too large, it is necessary to use a current transformer coil to monitor the current to reduce losses.

What is the difference between a current transformer and a general voltage transformer? This is a question that even the most experienced magnetic component planners find difficult to answer. The basic difference is that a transformer attempts to convert voltage from the primary to the secondary, while a current transformer attempts to convert current from the primary to the secondary. The voltage size of the current transformer is selected by the load.

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

Assume that a current transformer is used to detect the primary current of the converter. The primary current of 10A corresponds to a voltage of 1V. Of course, we can use a 1V/10A=100mΩ resistor to detect, but the loss caused by the resistor 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.

Figure 1 Using current sensing transformers to reduce losses

Of course, in order to reduce the winding resistance, we set the number of turns on the primary side to 1 turn. At the same time, in order to reduce the current to a relatively low level, the number of turns on the secondary side should be more. Assuming that the number of turns on the secondary side is N, we can get 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 to be not less than 20Ω. If a 20Ω resistor is used, the number of secondary turns N=200 can be obtained from 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 output voltage of the transformer is 1V, plus the on-state voltage of the diode is 1V, the total voltage is about 2V. When operating at a frequency of 250kHz, the magnetic induction intensity on the magnetic core will not exceed

Among them, 4us is the time of one cycle, which is definitely less than one cycle in reality.

Because the time when the current flows through the primary side cannot span the switching cycle (otherwise, the core cannot be reset). So Ae can be very small, and B will not be very large. In this example, the specifications of the magnetic core cannot be confirmed by the loss requirements or the magnetic flux sufficiency requirements. It is more likely to be confirmed by the isolation voltage between the primary and secondary sides. Assuming that there is no requirement for isolation voltage, the size of the magnetic core is generally determined by the volume occupied by the 200-turn winding. You can use 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.

Helpful tip: Unless necessary, do not use wires smaller than 36 gauge wire under normal circumstances.

Now let's analyze why a voltage transformer cannot be used instead of a current transformer? It is known that the secondary voltage is only 2V, so the primary voltage is 2V/200=100mV. Assuming the input DC voltage is 48V, then the 10mV voltage on the primary side of the current transformer is negligible to the 48V voltage - then you can get 50mA current on the secondary side with almost no impact 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 impossible to generate a current of 50mA if the secondary side is selected. Even if the entire 5mV voltage is applied to the primary side, the secondary side can only generate a voltage of 200×5mV=1V: a satisfactory voltage cannot be generated on the conversion resistor. Therefore, the voltage transformer can only be used as a transformer and cannot be used to detect current.

From another perspective: Although the voltage of the input power supply is 48V, the size 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 impedance-bound voltage transformers.

Finally, let's take a look at the error status of the current transformer? The answer lies in the basic definition of a current transformer: it induces current.

It is a useful reminder that the resistance of the diode and secondary winding in the current mutual inductance will not affect the detection of the current, because (as long as the impedance is not large) the current in the series circuit is flat everywhere, regardless of the components in series.

In practice, 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 detection error will increase. That is to say, the excitation inductance is too small. Suppose we require that the maximum error of the detected current is 1%, and the primary current is 10A, then the secondary current is 50mA, which means that the required excitation current (secondary side) should be less than 50mA×1 %=500μA. The excitation current does not flow through the conversion resistor, and we cannot detect this current, so the error increases. The minimum value of the secondary inductance can be calculated:

The current number of turns is 200. We need a magnetic ring with AL=16mH/200=400nH. Just use an ordinary small ferrite magnetic ring. This kind of ferrite magnetic ring is easy to find.

Review Editor: Huang Fei

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