First-order RC Circuits:

2. First-order RC Circuits: In this section, we consider transients in ckts that contain Independent dc sources, resistances, and a single capacitance.

3. Discharge of a capacitance through a resistance Prior to t = 0, the capacitor is charged to an initial voltage Vi. Then, at t = 0, the switch closes and current flows through the resistor, discharging the capacitor.

4. Writing a current equation at the top node of the ckt after the switch is closed yields Multiplying by the resistance gives This is a first-order linear differential equation. The solution for vc(t) must be a function that has the same form as its first derivative. The function with this property is an exponential.

5. We anticipate that the solution is of the form in which K and s are constants to be determined. Substitute this solution in to the previous equation, we get Solving for s, we obtain which leads to the solution

6. The voltage across the capacitor cannot change instantaneously when The switch closes. This is because the current through the capacitance is ic(t) = Cdvc/dt. In order for the voltage to change instantaneously, the current in the resistance would have to be infinite. Since the voltage is finite, the current in the resistance must be finite, and we conclude that the voltage across the capacitor must be continuous.

7. Therefore, we write vc(0+) = Vi in which vc(0+) represents the voltage immediately after the switch closes. Substitute this in to the equation: Hence, we conclude that the constant K equals the initial voltage across the capacitor. Finally, the solution for the voltage is

8. The time interval is called the time constant of the ckt. In one time constant, the voltage decays by the factor e-1 = 0.368. After about 5 time constants the voltage remaining on the capacitor is negligible compared with the initial value.

9. Charging a capacitance from a DC source through a resistance A dc source is connected to the RC ckt by a switch that closes at t = 0. we assume that the initial voltage across the capacitor just before the switch closes is vc(0-) = 0. Let us solve for the voltage across the capacitor as a function of time.

10. We start by writing a current equation at the node that joins the resistor And the capacitor. This yields Again, we have obtained a linear first-order differential equation with constant coefficients.

11. The solution is found to be Again, the product of the resistance and capacitance has units of seconds and is called the time constant t = RC. The plot is shown below. Notice that vc(t) starts at 0 and approaches the final value Vs asymptotically as t becomes large. After 1 time constant, vc(t) has reached 63.2 percent of its final value. For practical purposes, vc(t) is equal to its final value Vs after about five time constants. Then we say that the ckt has reached steady state.

12. Lessons We have seen in this section that several time constants are needed to charge or discharge a capacitance. This is the main limitation on the speed at which digital computers can process data. It is impossible to build ckts that do not have some capacitance that is charged or discharged when voltages change in value. Furthermore, the ckts always have nonzero resistances that limit the currents available for charging or discharging the capacitances. Therefore, a nonzero time constant is associated with each ckt in the computer, limiting its speed.

13. DC steady state Consider the equation for current through a capacitance If the voltage vc(t) is constant, the current is zero. In other words, the capacitance behaves as an open ckt. Thus, we conclude that for steady-state conditions with dc sources, capacitances behave as open ckts.