Capacitance charge and discharge time calculation: The 1L and C components are called “inertial componentsâ€, that is, the current in the inductor and the voltage across the capacitor have a certain “electrical inertia†and cannot be changed suddenly. 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 is the charge and discharge time of a 1UF capacitor?", you can't answer without talking about resistance.
Time constant of RC circuit: Ï„ = 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: Ï„ = L / R
The LC circuit is connected to DC, i=Io[1-e(-t/Ï„)] Io is the final stable current
Short circuit of LC circuit, i=Io×e(-t/τ)] Io is the current in L before short circuit
2 Let V0 be the initial voltage value on the capacitor;
V1 is the voltage value at which the capacitor can eventually be charged or placed;
Vt is the voltage value at the time of 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 R to a capacitor C with an initial value of 0, and V0=0, V1=E, so the voltage on the capacitor at time t is:
Vt=E × [1-e(-t/RC)]
For another example, the capacitor C whose initial voltage is 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 by R, and the final charging value is Vcc. What is the time required 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 the logarithmic function of e as the base
3 Provide a common formula for constant current charge and discharge: ⊿Vc=I*⊿t/C. Provide a common formula for capacitor charging: Vc=E(1-e(-t/R*C)). The RC circuit charging formula Vc=E(1-e(-t/R*C)). Regarding the capacitor used for the delay, what kind of capacitor is better, can not be generalized, the specific situation is analyzed. The actual capacitor is supplemented 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 that is closed by a switch to form a step signal and charge capacitor C through 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)). Where t is a time variable and small e is a natural exponent 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 across the capacitor cannot be abruptly changed. A common formula for constant current charge and discharge: ⊿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 with a current amplitude of 1A (ie, its output amplitude does not change with the output voltage) to charge or discharge the capacitor. According to the formula, the capacitor voltage increases or decreases linearly with time. A lot of triangle waves or sawtooth waves are produced in this way. According to the set value and formula, the rate of change of the capacitor voltage is 1V/mS. This means that a 5V capacitor voltage change can be obtained with a time of 5 mS; in other words, it is known that Vc has changed by 2 V, which can be inferred and has experienced a time history of 2 mS. Of course, both C and I in this relationship can also be variables or reference quantities. For details, please refer to the related materials. for reference.
4 Firstly, the amount of charge of the capacitor plate at time t is q, and the voltage between the plates is u. According to the loop voltage equation:
Uu=IR (I represents current),
And because u=q/C, I=dq/dt (where d is the differential),
After the substitution, you get:
Uq/C=R*dq/dt,
That is, Rdq/(Uq/C)=dt, then the two sides are indefinitely integrated, and the initial condition is used: t=0, q=0 to get q=CU[1-et/(RC)] This is the capacitor plate. The function of the charge as a function of time t. Incidentally, electrical engineering often refers to RC as the time constant.
Correspondingly, using u=q/C, the function of the plate voltage as a function of time is obtained immediately.
u=U[1-e -t/(RC)]. From the formula obtained, only when the time t tends to infinity, the charge and voltage on the plate are stabilized, and the charging is over.
However, in practical problems, since 1-et/(RC) tends to 1 very quickly, after a short period of time, the change in charge and voltage between the plates of the capacitor has been negligible, even if we use highly sensitive electrical instruments. I can't detect that q and u are changing slightly, so I can think that the balance has been reached and the charging is over.
As a practical example, suppose U=10 volts, C=1 picofarad, R=100 ohms, which can be calculated using our deduced formula. After t=4.6*10(-10) seconds, the plate voltage has reached 9.9 volts. It can be said that it is a moment of rushing.
Ningbo Autrends International Trade Co.,Ltd. , https://www.supermosvape.com