Why is the large capacitor connected in parallel at the output of the power supply not short-circuited?
The role of capacitors at the output of the power supply, the parallel connection of resistors at the output of the power supply, the effect of adding a resistor to the output of the power supply, the effect of connecting an inductor to the output of the power supply, the effect of connecting a diode at the output of the power supply, the effect of adding an inductor at the output of the power supply, the effect of connecting a capacitor at the output of the power supply in parallel, and the effect of connecting an inductor to the output of the power supply. How to filter the power frequency ripple at the output end, the electrolytic capacitor at the output end of the power supply, and the diode model connected in parallel at the output end of the power supply
A large capacitor is connected in parallel at the output of the power supply. At the moment when the large capacitor is powered on, for example, the large capacitor is connected to the load, and at the moment when the power supply supplies power to the load.
At the moment of power on, the power supply is short-circuited,
The short circuit is equal to the supply voltage divided by the wire resistance plus the equivalent series resistance of the capacitor. These two resistors are very small, so the current at the moment of power-on is very large.
For a load with a large capacitor connected in parallel to the input, we call it a capacitive load. When the power supply supplies power to the capacitive load, the instantaneous short circuit may be as high as dozens of times the normal operating current.
When supplying power to capacitive loads, we need to consider the multiple of overcurrent, the instantaneous overcurrent capability of the power supply, and even the overcurrent capability of the circuit breaker.
For loads controlled by relays, it is also necessary to consider selecting relays that are suitable for capacitive loads, so as to avoid the short-circuit circuit at the moment of power-on that will fuse the contacts of the relay together, making it impossible to disconnect normally.
If the capacitance is too large, there may be power output protection, or even circuit breaker overcurrent tripping.
After the power is turned on, the output voltage of the power supply is basically constant. According to the relationship between the current flowing through the capacitor and the two ends of the capacitor is Cdu/dt, only when the voltage changes, there will be current flowing through the capacitor, so the current flowing from the power supply Only the operating current of the load, there is no longer a short circuit situation.
Why, as long as the choice is proper, the power supply can still work normally even if it is short-circuited?
At the moment of power-on, according to the unit step response in circuit theory, from the one-variable ordinary differential equation, the voltage across the capacitor can be solved as u=us*(1-exp(-t/(R*C)).
And the current flowing through the capacitor is i=us/R*exp(-t/(R*C)).
Among them, R is the equivalent series resistance of the wire resistance plus the capacitor, and C is the capacitance of the capacitor.
From these two equations, it can be seen that the current flowing through the capacitor decays rapidly exponentially.
For example, R is generally tens of milliohms, and C is usually several thousand uF, which can decay to a very small current in about a few milliseconds.
So the short-circuit time is very short, maybe a few microseconds to a few milliseconds.
All power supplies have the ability of instantaneous over-current, and generally carry out short-circuit protection according to the relationship of inverse time limit. When it does not exceed n times of its rated current, it will not be protected immediately, but will be delayed for a period of time that is inversely proportional to the over-current multiple. for protection.
