Capacitors are indispensable components in circuit design. Because it is an energy storage element, it is used more in power filtering, and the capacitance value is relatively large. The capacitor is open for DC, and the impedance decreases with increasing frequency. r=1/(s*c)，s=jw。In the DC steady state, how does the voltage applied to the two capacitors change when they are connected in series?Let's discuss it on a case-by-case basis.
The first type has only two capacitors connected in series
Figure 1: The voltage applied to the two capacitors is 24V.
In this case, at the moment when the circuit is energized, it is AC for the capacitor. The voltage applied to C1 and C2 is inversely proportional to its capacitance. Because rc1=1 (s*c1), rc2=1 (s*c2);(rc1, rc2 represent the resistance of c1, c2);The voltage of C1 = RC1 (RC1+RC2), and the voltage of C2 = RC2 (RC1+RC2). In the diagram above, since c1=c2 are both 1ufTherefore, the voltages at both ends of C1 and C2 are equal to 12V. In the actual circuit, there will be some small differences in the voltages of C1 and C2, because their capacitance, equivalent series inductance, and equivalent series resistance will be different, and they are not completely equal in the ideal state.
SectionII. II. IITwo capacitors are connected in series, C2 has a 10m resistor in parallel
As shown in Figure 2: DC voltage is unchanged or 24V. The resistor R2 is disconnected when the power is turned on, and when the power supply is stable, R2 is merged to C2. Or use a multimeter instead (the input impedance of the multimeter is generally 10m). The positive pen of the multimeter is connected to the VO, and the negative pen is grounded. At this point, you will notice that the multimeter reading starts to drop slowly, that is, the voltage at both ends of the C2 is dropping. What is the voltage of C1 doing at this time?The voltage at both ends of C1 is rising. When C2 drops a few volts, C1 rises a few volts. Until C2 drops to zero and C1 rises to 24V. Because the voltage at both ends of the C1 and C2 series has not changed, it is still 24V. Why does the voltage at both ends of C2 become zero, because in the DC state, the capacitance is open and the impedance is infinite;But C2 has a 10m resistor in parallel. According to the resistor divider, the final voltage of C2 is zero. If R2 is removed at this time, the voltage of C2 is still zero volt, and C1 is still 24V. Because the impedances of C1 and C2 are infinite. The rc constant is also infinity. The voltages of C1 and C2 remain the same. In fact, due to the presence of the air impedance, the impedance of the capacitor's insulating medium, C2 cannot be consistent with zero, but it rises more slowly.
SectionThreeTwo capacitors are connected in series, C1 has a 10m resistor in parallel
This state is the same as in Figure 2, so I will not repeat it here.
SectionFourTwo capacitors are connected in series, C1 and C1 both have a 10m resistor in parallel
As shown in Figure 3, each capacitor in this structure has a resistor, which is the resistor equalization. The voltage across the capacitor is proportional to the resistance value of the capacitor in parallel. This is talking about the voltage at steady state. At the moment of power-up, the voltage across the capacitor is related to the total impedance of each segment. As shown in Figure 3: The value of VO is the total impedance of VO to ground, and the total impedance of VO to 24VO dividuation. This kind of circuit generally takes the value c1=c2,r1=r2;It is mostly used in voltage doubling circuits.