First-order RC Circuits:

In this section, we consider transients in ckts that contain Independent dc sources, resistances, and a single capacitance.

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 theresistor, discharging the capacitor.

Writing a current equation at the top node of the ckt after theswitch is closed yields

0)()(

R

tv

dt

tdvC CC

Multiplying by the resistance gives

0)()(

tvdt

tdvRC C

C

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. Thefunction with this property is an exponential.

We anticipate that the solution is of the formst

C Ketv )(in which K and s are constants to be determined.

Substitute this solution in to the previous equation, we get

0 stst KeRCKseSolving for s, we obtain

RCs

1

which leads to the solutionRCt

C Ketv /)(

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

Therefore, we write vc(0+) = Vi in which vc(0+) represents the voltageimmediately after the switch closes. Substitute this in to the equation:

KKeVv iC 0)0(

Hence, we conclude that the constant K equals the initial voltage acrossthe capacitor. Finally, the solution for the voltage is

RCtiC eVtv /)(

The time interval RC 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 isnegligible compared with the initial value.

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 theswitch closes is vc(0-) = 0. Let us solve for the voltage across the capacitor as a function of time.

We start by writing a current equation at the node that joins the resistorAnd the capacitor. This yields

0)()(

R

Vtv

dt

tdvC sCC

sCC Vtvdt

tdvRC )(

)(

Again, we have obtained a linear first-order differential equation withconstant coefficients.

The solution is found to be /)( tssC eVVtv

Again, the product of the resistance and capacitance has units of seconds and is called the time constant = 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) isequal to its final value Vs after about five time constants. Then we saythat the ckt has reached steady state.

Lessons

We have seen in this section that several time constants are needed tocharge or discharge a capacitance. This is the main limitation on the speed at which digital computers can process data. It is impossible tobuild 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 forcharging or discharging the capacitances. Therefore, a nonzero timeconstant is associated with each ckt in the computer, limiting its speed.

DC steady state

Consider the equation for current through a capacitance

dt

tdvCti C

C

)()(

If the voltage vc(t) is constant, the current is zero. In other words, thecapacitance behaves as an open ckt. Thus, we conclude that forsteady-state conditions with dc sources, capacitances behave as open ckts.