Use capacitance to do transformer how to calculate
According to this feature, if we connect a resistive element in series with a 1uF capacitor, the voltage obtained at both ends of the resistive element and the power dissipation it generates depends entirely on the characteristics of the resistive element.
Although the current flowing through the capacitor has 70mA, but in the capacitor does not produce power, because if the capacitor is an ideal capacitor, the current flowing through the capacitor for the virtual part of the current, the work done by the reactive power.
The principle of capacitor buck is not complicated. His working principle is to use the capacitor at a certain AC signal frequency to produce capacitive reactance to limit the maximum operating current. For example, in the 50Hz frequency conditions, a 1uF capacitance generated by the capacitive reactance of about 3180 ohms. When 220V AC voltage is applied to both ends of the capacitor, the maximum current flowing through the capacitor is about 70mA.
Similarly, we can also 5W / 65V bulb and 1uF capacitor series connected to the 220V / 50Hz AC, the bulb will be lit, and will not be burned. Because the 5W / 65V bulb operating current is also about 70mA. Therefore, the capacitor buck is actually using capacitive current limiting. While the capacitor actually functions as a limiting current and dynamically distributes the voltage across the capacitor and the load.
Capacitance buck method: For example, we will be a 110V / 8W bulb and a 1uF capacitor in series, received 220V / 50Hz AC voltage, the light bulb is lit, the normal brightness will not be burned. Since the 110V / 8W bulb requires a current of 8W / 110V = 72mA, it is consistent with the current limiting characteristics of the 1uF capacitor.
