Calculation and Selection of Capacitor Charge and Discharge Time in Power Supply

First, the capacitor charge and discharge time calculation

1. The L and C components are called "inertial components". That is, the current in the inductor and the voltage across the capacitor have a certain "electrical inertia" and cannot suddenly change. The charge and discharge time is not only related to the capacity of L and C, but also related to the resistance R in the charge/discharge circuit. “How long does the 1UF capacitor charge and discharge it?” and cannot answer without resistance.

Time constant of RC circuit: τ=RC When charging, uc=U×[1-e^(-t/τ)]

U is the Power Supply voltage; when discharging, uc=Uo×e^(-t/τ)

Uo is the voltage on the capacitor before discharge

Time constant of RL circuit: τ=L/R LC circuit connected to DC, i=Io[1-e^(-t/τ)]

Io is the final steady current; the short circuit of the LC circuit,

Io is the current in L before the short circuit

2. Let V0 be the initial voltage value on the capacitor; V1 is the voltage value that the capacitor can be charged or put into; Vt is the voltage value on the capacitor at time t. Then: Vt = V0 + (V1-V0) × [1-exp (-t/RC)] or t = RC × Ln [(V1 - V0) / (V1 - Vt)] For example, the battery with voltage E passes R charges the capacitor C with an initial value of 0, V0 = 0, V1 = E, so the voltage on the capacitor when charged to t is: Vt = E × [1-exp(-t/RC)] Again, the initial voltage The capacitor C for E is discharged through R, V0 = E, V1 = 0, so the voltage on the capacitor when placed at t is: Vt = E × exp (-t/RC) Another example is that the initial value is 1/3Vcc capacitor C charges through R, the final charging value is Vcc. What is the time required to charge 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: The above exp() denotes the exponential function with e as the base; Ln() is the logarithm function with e as the base

3. Provide a common formula for constant current charge and discharge: ?Vc=I*?t/C. Then provide a common formula for capacitor charging: Vc=E(1-e-(t/R*C)). In the RC circuit charge formula Vc=E(1-e-(t/R*C)): -(t/R*C) is the negative exponent term of e. The capacitor used for the delay capacitor is relatively good and cannot be generalized. The actual capacitance is added with parallel insulation resistance, series lead inductance and lead resistance. There are more complicated modes - causing adsorption effects and so on. for reference.

E is the magnitude of a voltage source. By closing a switch, a step signal is formed and the capacitor C is charged through the resistor R. E can also be a high-level amplitude of a continuous pulse signal whose amplitude changes from a low level of 0V to a high level. The variation of the voltage Vc across the capacitor with time is the charging formula Vc=E(1-e-(t/R*C)). Among them: -(t/R*C) is the negative exponent term of e, which cannot be shown here and requires special attention. Where t is the time variable and e is the natural exponent term. For example: When t=0, e is 0 to 0, and Vc is calculated as 0V. Meet the law that the voltage across the capacitor cannot be abruptly changed. The common formula for constant current charge and discharge is: Vc=I*?t/C, which is derived from the formula: Vc=Q/C=I*t/C. For example: Let C = 1000uF, I is a constant current source of 1A current amplitude (ie: its output amplitude does not change with the output voltage) to charge or discharge the capacitor, according to the formula can be seen, the capacitor voltage increases or decreases linearly with time, Many triangular waves or sawtooth waves are produced in this way. According to the set value and formula can be calculated, the rate of change of the capacitor voltage is 1V/mS. This means that the 5V capacitor voltage change can be obtained in 5mS; in other words, it is known that Vc changes by 2V, and it can be calculated that it has gone through a 2mS time history. Of course, both C and I in this relation can also be variables or reference quantities. Details can refer to the relevant materials to see. for reference

4. First set the capacitor plate charge at t is q, the voltage between the plates is u. According to the loop voltage equation can be: Uu = IR (I represents the current), and because u = q / C, I =dq/dt (where d is the derivative), after substitution, we get: Uq/C=R*dq/dt, that is, Rdq/(Uq/C)=dt, and then we get the indefinite integral on both sides, and use the initial condition: t =0, q=0 gives q=CU [1-e^-t/(RC)] This is the function of the charge on the capacitor plate over time t. By the way, electrical engineering often calls RC a time constant. Correspondingly, using u = q/C, the function of the plate voltage over time is immediately obtained, u = U [1-e^-t/(RC)]. Judging from the formulas obtained, only when the time t approaches infinity, the charge and voltage on the pole plate reach stability, and the charging is finished. However, in practical problems, since 1-e ^-t/(RC) tends to move toward 1 soon, after a short period of time, the change in charge and voltage between the capacitor plates is negligible, even if we use very high sensitivity. The electrical instrument is also unaware that the q and u changes slightly, so at this point it can be assumed that the balance has been reached and the charge is over. To give a practical example, suppose that U = 10 volts, C = 1 picofarad, and R = 100 ohms. Using our formula, we can calculate that after t = 4.6 * 10 ^ (-10) seconds, the voltage of the plate has reached 9.9 volts. It can be said that it is a moment of speed.

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