How to calculate the withstand voltage value of two capacitors with different withstand voltage valu

The total capacitance of the two capacitors in series is generally known, i.e., c=c1*c2 (c1+c2).

The total withstand voltage (DC voltage) after series connection cannot be simply calculated by a formula:

First of all, it is necessary to ensure that the voltage divider value on a single capacitor is less than the allowable value of the capacitor, otherwise once the capacitor breaks down, the total voltage must be added to another capacitor, if the capacitor withstand voltage value is lower than the total voltage, the second capacitor is also facing the danger of breakdown.

The actual voltage on a series capacitor is inversely proportional to its capacitance, i.e., the voltage distributed with a large capacitance is low, and the voltage distributed with a small capacitance is high. If the capacity is the same, then the voltage is the same.

The formula for calculating the voltage division of two capacitors in series is as follows:

First of all, let the total voltage be U, and the voltages on C1 and C2 are U1 and U2 respectively, then.

u1=c2*u/(c1+c2)

u2=c1*u/(c1+c2)

If the total voltage is 550V, the voltage of C1 is U1 = C2 * U (C1 + C2) = 366V, which is significantly higher than the allowable withstand voltage value of C1. Once C1 breaks down, C2 can withstand 300V, which is also dangerous.

Therefore, after the capacitors are connected in series, the maximum withstand voltage value after series connection can be obtained by calculating and comparing the capacitance of the component and the allowable withstand voltage value.

FYI.

The capacity of the two capacitors in series is equal to the capacity of the two capacitors multiplied and divided by the capacity of the two capacitors, i.e., C1XC2 (C1+C2). Well, thanks.

The same capacitor is connected in series or parallel, so how to calculate the withstand voltage value and capacitance?

will increase. After the power supply is fixed, the same two capacitors are connected in series, and the voltage at both ends of the capacitor after series connection is 1 2 of the voltage value of a single capacitor, and the withstand voltage is doubled, but the capacitance is only half of the original.

After the capacitor is connected in series, the withstand voltage will increase, if two identical capacitors are connected in series, the withstand voltage is twice that of a single capacitor; If the capacities are not the same, after power-on, the voltage drop with small capacity is large, and the voltage drop with large capacity is small; In most applications, the same capacitor is always used in series in circuits with relatively high operating voltages; If it is true that capacitors of different capacitances are required in series to obtain the required capacitance, the withstand voltage will be considered as a small capacitance.

Why does the withstand voltage value increase when the capacitor is connected in series? Will there be less capacity?

When the same two capacitors are connected in series, the voltage that can be withstood by the two terminals after series connection is twice that of a single capacitor. In a series circuit, the total voltage in the circuit is the sum of the shunt voltage, so the double voltage value is evenly distributed to each capacitor, which is the original withstand voltage value of the capacitor. However, in practical application, due to the different leakage resistance of the capacitor, it will cause the difference in voltage distribution, resulting in a high voltage at both ends of a capacitor, and a voltage equalizing resistor can be connected at both ends of each capacitor, which can generally be selected according to the voltage level of tens of thousands of ohms to hundreds of kiloohms.

Let's start with two voltage levels.

The same type of capacitor, for example, two electrolytic capacitors, rated voltage.

100V, if these two capacitors are connected in parallel, after parallel connection, the total withstand voltage value is still 100V;

If these two capacitors are connected in series, it is necessary to increase the voltage equalization resistance and voltage equalization to ensure that the voltage on the two series capacitors is equal, and after series connection, the total withstand voltage value is 100V+100V=200V;

If the pressure is not uniform, the total withstand voltage cannot reach 200V.

If two capacitors with different voltage levels are connected in parallel, the withstand voltage value shall be subject to the capacitor with low voltage;

If two capacitors with different voltage levels are connected in series, the situation is complicated; You can think for yourself.

The situation is even more complicated if two capacitors of different types are connected in series, and in fact, there is no such application;

It is possible to connect two capacitors of different types in parallel, and the withstand voltage value is subject to the capacitor with low voltage.

There are four 4UF capacitors, the withstand voltage value is 500V, you can first connect the two in parallel, and then connect them in series to 1000V, but in series, you must add a voltage divider resistor, and the voltage is equalized.

For example, a 48V electric vehicle is connected by four 12V batteries, each battery originally rated at 12A, and the current of 12A remains unchanged after series connection. 48v*12a=576 w .

In order to increase the power, people often connect one battery in series, becoming five with a total voltage of 60V, and the current is still 12A. 60 V*12A=720W, the power has become larger, the car has more power, and it runs faster.

After the capacitor is connected in series, the capacity is halved, and the withstand voltage is increased by half, which requires the same materials to be connected in series and parallel.

The withstand voltage value of the capacitor in parallel remains unchanged, and the capacitance is added.

First of all, calculate the partial voltage value of each capacitor according to the capacitance, and then see if each capacitor can withstand it.

If a capacitor of the same capacity is connected in series, the withstand voltage value of a single capacitor is multiplied by the number of series.

The same capacitor is connected in series or parallel, so how to calculate the withstand voltage value and capacitance?

The same capacitor is connected in series or parallel, so how to calculate the withstand voltage value and capacitance?

The electronic engineer taught you how to rectify the filter capacitor and how to calculate the maximum withstand voltage value.

Hello: After the capacitor is connected in series: u=u1+u2....un

Therefore, if the capacitor has the same withstand voltage value, it will be n times the withstand voltage of a single capacitor after being connected in series.