Let me show you how to calculate a current transformer!
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We will design a current transformer. Using a current transformer can reduce the loss when measuring the primary current of the converter. For example, in a high-power switching power supply, the current is too large, so a current transformer coil is needed 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 to answer even for experienced magnetic component designers. The basic difference is that the transformer tries to convert the voltage from the primary to the secondary, while the current transformer tries to convert the current from the primary to the secondary. The voltage of the current transformer is determined by the load.
We can better understand the working principle of the current transformer through a practical design example.
Suppose a current transformer is used to measure the primary current of the converter. The primary current of 10A corresponds to a voltage of 1V. Of course, we can use a resistor of 1V/10A=100mΩ to measure it, but the loss caused by the resistor is 1V×10A=10W, which is unacceptable for almost all designs. Therefore, a current transformer should be selected, as shown in Figure 1.
Figure 1 Using current detection transformer to reduce losses
Of course, in order to reduce the winding resistance, we take the number of turns on 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 relatively large. If the number of turns on the secondary side is N, Ohm's law shows that (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 no less than 20Ω. If a 20Ω resistor is used, Ohm's law shows that the number of turns on the secondary side is N=200.
Now let's look at the magnetic core. Assuming that the diode is an ordinary general 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 working at a frequency of 250kHz, the magnetic induction intensity on the magnetic core will not exceed
4us is the time of one cycle, which is definitely less than one cycle in reality.
Since the time that the current flows through the primary side cannot exceed the switching cycle (otherwise, the core cannot be reset). Therefore, Ae can be very small, and B will not be very large. In this example, the size of the core cannot be determined by loss requirements or flux saturation requirements, but is more likely to be determined by the isolation voltage between the primary and secondary sides. If there is no requirement for isolation voltage, the size of the core is generally determined by the volume occupied by 200 turns of winding.
You can use a No. 40 wire to flow a peak current of 500mA, but this wire is too thin and general transformer manufacturers will not wind it for you. Practical Tips: Unless you have to use it, generally do not use wires with a specification smaller than No. 36 wire.
Now let's analyze why a voltage transformer cannot be used to replace a current transformer? It is already known 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 - that way you can get 50mA current on the secondary side with almost no effect on the primary side.
Assume 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, and because the impedance of the primary side (such as the reflected secondary side impedance) is also relatively large, it is determined that the secondary side cannot generate a current of 50mA.
Even if the entire 5mV voltage is added 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, not 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 of the primary side, but by other factors.
The current transformer is a voltage transformer with impedance limitation.
Finally, let's take a look at the error of the current transformer. The answer lies in the basic definition of the current transformer: it senses current.
Practical tips The diode in the current transformer and the resistance of the secondary winding will not affect the current measurement, because (as long as the impedance is not infinite) the current in the series circuit is equal everywhere, regardless of the series components.
In actual work, it does not matter whether or not 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%, the primary current is 10A, then the secondary current is 50mA, which means that the 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. We can calculate the minimum value of the secondary inductance
Now the number of turns is 200, and we need a magnetic ring of AL=16mH/200=400nH. Ordinary small ferrite magnetic rings can be used. This kind of ferrite magnetic ring is easy to find.
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