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# Selection and Calculation of Capacitance in Power Supply(I)

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Calculation of charge and discharge time of capacitors.

1. The L and C elements are called “inertial elements”, that is, the current in inductance and the voltage at both ends of capacitors which has certain “electric inertia”. It cannot be changed suddenly. The charge-discharge time is related not only to the capacity of L and C, but also to the resistance R in the charge / discharge circuit. “how long is the charge and discharge time of the 1UF capacitor?”No resistance, no answer.

Time constant of RC Circuits:τ=L/R.

i=Io[1-e^(-t/τ)] is the supply voltage, and Uo × e ^ -t / τ) Uo is the upper voltage of the capacitor before discharge.

Time constants of RL Circuits: τ / L / R.

LC circuit is connected to DC / i=Io[1-e^(-t/τ)]  Io is the short-circuit of the final stable current / LC circuit, Io is the middle current of L before short-circuit.

2. Let V0 be the initial voltage value on the capacitor, V1 be the voltage value that the capacitor can eventually charge or put, and Vt be the voltage value of the capacitance at t time. Then:

Vt=V0 +（V1-V0）× [1-exp(-t/RC)] or  t = RC × Ln[(V1 - V0)/(V1 - Vt)] For example, a battery with a voltage of E is charged via R to a capacitance C of 0 to V0 / 0 / V1E, so the voltage charged to the capacitance at t time is:Vt=E × [1-exp(-t/RC)].

For example, the capacitance C with the initial voltage E is discharged through R, and the  V0=E，V1=0, so the voltage placed on the capacitance at t time is:  Vt=E × exp(-t/RC)
For example, the initial value of the 1/3Vcc capacitor C through the R charge, the final value of the charge VCC, asked what is the time required to charge to the 2/3Vcc?  .

V0=Vcc/3，V1=Vcc,Vt=2*Vcc/3，故 t=RC × Ln[(1-1/3)/(1-2/3)]=RC × Ln2 =0.693RC

Note: expa（） above means that the exponential function based on e is a logarithmic function based on e