The L and C components are called “inertial components”, that is, the current in the Inductor and the voltage across the capacitor have certain “electric inertia” and cannot change suddenly. The charging and discharging time is not only related to the capacity of L and C, but also related to the resistance R in the charging/discharging circuit. “How long is the charging and discharging time of a 1UF capacitor?” If you don’t talk about resistance, you can’t answer.

The L and C components are called “inertial components”, that is, the current in the inductor and the voltage across the capacitor have certain “electric inertia” and cannot change suddenly. The charging and discharging time is not only related to the capacity of L and C, but also related to the resistance R in the charging/discharging circuit. “How long is the charging and discharging time of a 1UF capacitor?” If you don’t talk about resistance, you can’t answer.

The time constant of RC circuit: τ=RC
When charging, uc=U×[1-e(-t/τ)] U is the supply voltage
When discharging, uc=Uo×e(-t/τ) Uo is the voltage on the capacitor before discharging
Time constant of RL circuit: τ=L/R
LC circuit is connected to DC, i=Io[1-e(-t/τ)] Io is the final stable current
The short circuit of the LC circuit, i=Io×e(-t/τ)]Io is the current in L before the short circuit

How to calculate the charging and discharging time of the capacitor?This article makes it clear

Let V0 be the initial voltage value on the capacitor;
V1 is the voltage value that the capacitor can finally be charged or discharged;
Vt is the voltage value on the capacitor at time t. but:
Vt=V0 +(V1-V0)×[1-e(-t/RC)]
t = RC × Ln[(V1 – V0)/(V1 – Vt)]
For example, a battery with a voltage of E is charged to a capacitor C with an initial value of 0 through R, V0=0, V1=E, so the voltage charged on the capacitor at time t is:
Vt=E ×[1-e(-t/RC)]
For another example, capacitor C with an initial voltage of E is discharged through R, V0=E, V1=0, so the voltage on the capacitor at time t is:
Vt=E × e(-t/RC)
For another example, the capacitor C with an initial value of 1/3Vcc is charged through R, and the final value of the charge is Vcc. How long does it take to charge to 2/3Vcc?
V0=Vcc/3, V1=Vcc, Vt=2*Vcc/3, so t=RC × Ln[(1-1/3)/(1-2/3)]=RC × Ln2 =0.693RC
Note: Ln() is a logarithmic function with e as the base

How to calculate the charging and discharging time of the capacitor?This article makes it clear

Provide a common formula for constant current charge and discharge: SVc=I*St/C. Provide a common formula for capacitor charging: Vc=E(1-e(-t/R*C)). RC circuit charging formula Vc=E(1-e(-t/R*C)).

Regarding what kind of capacitor is better for the capacitor used for the delay, it cannot be generalized, and the specific situation is analyzed in detail. The actual capacitance is added with parallel insulation resistance, series lead inductance and lead resistance. There are more complex modes-causing adsorption effects and so on. for reference.

E is the amplitude of a voltage source, through the closing of a switch, a step signal is formed and the capacitor C is charged through the resistor R. E can also be the high-level amplitude of a continuous pulse signal whose amplitude changes from a low level of 0V to a high-level amplitude.

The voltage Vc at both ends of the capacitor changes with time as the charging formula Vc=E(1-e(-t/R*C)).

In the formula, t is the time variable, and the small e is the natural exponential term. For example: when t=0, the 0th power of e is 1, and Vc is calculated to be equal to 0V. It conforms to the law that the voltage at both ends of the capacitor cannot change suddenly.

The commonly used formula for constant current charging and discharging: SVc=I*St/C, which comes from the formula: Vc=Q/C=I*t/C.

For example: Set C=1000uF, I is a constant current source with a current amplitude of 1A (that is, its output amplitude does not change with the output voltage) to charge or discharge the capacitor. According to the formula, it can be seen that the capacitor voltage increases or decreases linearly with time. Many triangle waves or sawtooth waves are generated in this way.

According to the set value and formula, it can be calculated that the rate of change of the capacitor voltage is 1V/mS. This means that a 5V change in capacitor voltage can be obtained in 5mS; in other words, it is known that Vc has changed by 2V, and it can be calculated that it has experienced a time history of 2mS. Of course, both C and I in this relationship can also be variables or reference quantities. For details, please refer to related textbooks. for reference.

How to calculate the charging and discharging time of the capacitor?This article makes it clear

First, suppose the amount of charge of the capacitor plates at time t is q, and the voltage between the plates is u. According to the loop voltage equation, we can get:

Uu=IR (I means current), and because u=q/C, I=dq/dt (where d means differential), we get:
Uq/C=R*dq/dt, that is, Rdq/(Uq/C)=dt, then calculate the indefinite integral on both sides, and use the initial conditions t=0, q=0 to get q=CU[1-et/(RC)】

This is the function of the change of the charge on the capacitor plate with time t. By the way, RC is often referred to as the time constant in electrical engineering.

How to calculate the charging and discharging time of the capacitor?This article makes it clear

Correspondingly, using u=q/C, the function of the plate voltage change with time is immediately obtained, u=U[1-e -t/(RC)].

From the obtained formula, only when the time t tends to infinity, the charge and voltage on the electrode plate reach stability, and the charging is considered to be over.

But in practical problems, since 1-et/(RC) quickly tends to 1, after a short period of time, the change in the charge and voltage between the capacitor plates has been minimal, even if we use highly sensitive electrical instruments. It is impossible to notice that q and u are changing slightly, so it can be considered that the balance has been reached at this time, and the charging ends.

To give a practical example, suppose U=10 volts, C=1 picofarad, and R=100 ohms. Using the formula we derived, we can calculate that after t=4.6*10(-10) seconds, the plate voltage has reached 9.9 volts. It can be said to be a flashy moment.

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