Adjust the input voltage slide so Vstep is 10.0 volts. Leave the R and C values set at 2.24. Press the Numeric button and observe the output voltage and circuit current as functions of time.
You should note that a constant voltage, suddenly applied at t = 0, does not produce an instantaneous output. The circuit current, however, immediately jumps to a value equal to Vstep/R because the voltage across the capacitor is initially zero.
Note also that the output voltage builds rapidly at first and then slows as the output voltage builds. The initial rate of increase of the output voltage is 2 volts per seconds. Since Vo is initially zero, you can see from equation (4.3) that the initial rate is 2 volts per second.
At this rate, the output voltage would reach the step input value in 5 seconds. However, this does not happen because the rate slows as the output voltage builds. You will recall from the block diagram discussion of this circuit given in Section 3.3 that this slowing of the rate is due to the output voltage feeding back and reducing the current flowing into the capacitor. As this current is reduced, the charge on the capacitor asymptotically builds to the value of the input voltage.
It is important to note also that in 5 seconds (one time constant) the output voltage is about 6.5 volts, or roughly 65% of its final value. In 10 seconds (2 time constants), the output voltage is about 8.8 volts or 88%, and in 15 seconds (3 time constants) the output reaches about 9.6 volts, or 96% of its final value. This form of response of this circuit is called the “step response” of a first-order ordinary differential equation.
Adjust the Vstep slider to 5 volts and leave the values of R and C set to 2.24. Press the Numeric button and observe the response. Calculate the out voltage as a percentage of the input voltage at 5, 10, and 15 seconds. Compare these percentage results with those when Vstep was set to 10 Volts. Are they similar?
You should find the output voltage percentages for a Vstep of 5 volts to be very close as the percentage for Vstep of 10 volts. No matter what the Vstep value, a first-order ODE will have exactly the same characteristic shape.
Put the value of the step input voltage back to 10 volts. Set the value of R to 3.0 ohms and the value of C to about 1.68 farads. Press the Numeric button and note the response. Did the output voltage response change? Did the initial value of the circuit current change?
You should recall that the product RC has units of time. It is known as the time constant and given the symbol τ. In the simulator this value is shown to the right of the R and C sliders labeled tau. Even though you varied R and C, you varied them in such a way that the product of the two was still 5 seconds. Consequently, the output voltage response remained the same. On the other hand the initial value of the circuit current dropped because we increased R from 2.4 to 3.0 ohms. So the initial current Vstep/R decreased. The value of C has no effect on the magnitude of the initial circuit current.
Keep the value of the step input voltage at 10 volts. Set the value of R and C to values that result in value of tau equal to 2 seconds. Before pressing the Numeric button, write down how long you think it will take the output voltage to reach about 96% of the input voltage. After you write your answer, push the Numeric button, note the response, and check your answer.
By decreasing tau to 2 seconds we increased initial rate of change of the output voltage and made the overall response of the system faster. In about 3 time constants the output of a first order system will reach about 96% of the input value. So in this assignment, we should expect the output voltage to reach 96% in six seconds. If you did not get the answer, go back to the simulator and adjust R and C so tau is 0.5, 1, 2, 3, 4, 5 and 6 seconds then observe the time it takes for the output voltage to reach about 95% of the input voltage. That time should be approximately 3 times the value of the time constant.
Table 4.2 give the exact percent value of the output for a first-order linear ODE versus time (expressed in terms of time constants). Many times manufactures product data is list response time of their components as a step response to an input signal. Given such data you can estimate the time constant of a first order linear ODE that you could use in a dynamic model of that component.