Calculation of capacitor charge and discharge time

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Calculation of capacitor charge and discharge time [Copy link]

1. L and C components are called "inertia components", that is, the current in the inductor and the voltage across the capacitor have a certain "electrical inertia" and cannot change suddenly. The charging and discharging time is not only related to the capacity of L and C, but also to the resistance R in the charging/discharging circuit. "How long is the charging and discharging time of a 1UF capacitor?", it cannot be answered without talking about resistance. The 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 The time constant of RL circuit: τ=L/R When LC circuit is connected to DC, i=Io[1-e(-t/τ)] Io is the final stable current When the LC circuit is short-circuited, i=Io×e(-t/τ)] Io is the current in L before the short circuit

2. Let V0 be the initial voltage value on the capacitor; V1 Vt is the voltage value on the capacitor at time t. Then: Vt = V0 + (V1-V0) × [1-e(-t/RC)] or t = RC × Ln[(V1 - V0)/(V1 - Vt)] For example, a battery with a voltage of E charges a capacitor C with an initial value of 0 through R, V0 = 0, V1 = E, so the voltage on the capacitor at time t is: Vt = E × [1-e(-t/RC)] For another example, a capacitor C with an initial voltage of E discharges through R, V0 = E, V1 = 0, so the voltage on the capacitor at time t is: Vt = E × e(-t/RC) For 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 base e 3. Provide a common formula for constant current charging and discharging: ⊿Vc=I*⊿t/C. Provide another common formula for capacitor charging: Vc=E(1-e(-t/R*C)). RC circuit charging formula Vc=E(1-e(-t/R*C)). As for what kind of capacitor is better for delay capacitors, it cannot be generalized and needs specific analysis. The actual capacitor is attached with parallel insulation resistance, series lead inductance and lead resistance. There are more complex modes - causing adsorption effect, etc. For reference. E is the amplitude 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 the high level amplitude of a continuous pulse signal whose amplitude changes from a low level of 0V to a high level amplitude. The law of change of the voltage Vc across the capacitor with time is 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 0V. It conforms to the law that the voltage across the capacitor cannot change suddenly. The commonly used formula for constant current charging and discharging: ⊿Vc=I*⊿t/C, which comes from the formula: Vc=Q/C=I*t/C. For example: Let C=1000uF, I be a constant current source with a current amplitude of 1A (i.e., 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, and many triangular waves or sawtooth waves are generated in this way. According to the set values and formula, it can be calculated that the rate of change of the capacitor voltage is 1V/mS. This means that a 5V capacitor voltage change can be obtained in 5mS; in other words, it is known that Vc has changed by 2V, and it can be inferred that it has experienced a time history of 2mS. Of course, C and I in this relationship can also be variables or reference quantities. For details, please refer to relevant textbooks. For reference. 4. First, let the charge on the capacitor plates at time t be q, and the voltage between the plates be u. According to the loop voltage equation, we can get: Uu=IR (I represents current), and because u=q/C, I=dq/dt (d here represents differential), we can get: Uq/C=R*dq/dt, that is, Rdq/(Uq/C)=dt, and then find 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 charge on the capacitor plates with time t. By the way, RC is often called the time constant in electrical engineering. Accordingly, using u=q/C, we can immediately get the function of the plate voltage changing with time, u=U【1-e -t/(RC)】. From the formula we get, only when time t tends to infinity, the charge and voltage on the plate will reach stability and charging will be considered complete. However, in practical problems, since 1-et/(RC) tends to 1 very quickly, after a very short period of time, the change in charge and voltage between the capacitor plates is already very small. Even if we use highly sensitive electrical instruments, we cannot detect the slight change in q and u. Therefore, it can be considered that balance has been reached and charging is complete. For a practical example, assuming U=10 volts, C=1 pF, 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 was truly a lightning-fast moment.