Block B - Natural Response

Objectives

Determine a capacitance C by measuring voltage decay in an RC circuit.
Measure the current transient in an overdamped and an underdamped RLC circuit.

Background

Natural response is the behavior of an electrical or mechanical system due to internal energy storage. Forced response is the behavior of a system due to an external energy source. Natural responses always die out with time due to the finite energy storage while forced responses can be maintained indefinitely by the continuing input of energy.

The simplest systems have one energy storage element, either a capacitor C or inductor L, or their equivalent in a mechanical system. They also have a resistance R to dissipate the energy. These systems can be described by a first-order differential equation, hence are called first-order systems. In the RC circuit, we are interested in the voltage decay across the capacitor, while the current decay in the inductor is of more interest in the RL circuit. A capacitor charged to a voltage V_o and then connected to a resistance has a voltage decay

v_R = V_o e^(-t/RC) .........(1)

Similarly, the current i in an RL circuit with an initial current of I_o decays according to the equation

i = I_o e^(-Rt/L) .........(2)

With two storage elements, the circuit behavior becomes more complicated. Instead of a simple monotonic decay, it is possible to have a decaying sinusoid. The differential equation is second-order so the solution must have two terms in it. For a series RLC circuit, the solution for current is

.........(3)

where A₁ and A₂ are determined by the initial conditions and s₁ and s₂ are given by:

...............(4)

The plus sign goes with s₁ and the minus sign goes with s₂.

The values of R, L, and C are all real and positive, but the values of s₁ and s₂ may be real, complex, or purely imaginary, depending on the quantity under the radical sign (called the discriminant). If the discriminant is positive, the values of s_1,2 will be real, negative, and distinct. This is called the overdamped case. If the discriminant is zero, the two values will be real, negative, and identical (the critically damped case). If the discriminant is negative, the values will be complex numbers, or, in the special case where R = 0, purely imaginary (the underdamped case).

If the roots are complex, it is convenient to write the solution in a different form.

.........(5)

where

.........(6)

.........(7)

The solution contains sinusoidal functions of time with an exponentially decaying amplitude, a `damped sinusoid'. The factor is the natural frequency of the oscillation in radians per second. The factor is the damping coefficient (per second). The solution is bounded by a function which includes e^-t, which is called the envelope of the response.

The constants A₁ and A₂ or B₁ and B₂ are determined by using the rules that current cannot change instantaneously through an inductor, and the voltage across a capacitor cannot change instantaneously. Consider the case of the series RLC circuit shown in Fig. 1c where the switch is closed at time t= 0. The initial inductor current is zero and the initial capacitor voltage is V_o. From (3), for t= 0,

0 = A₁ + A₂ ........(8)

since e^s₁^t = 1 at t= 0. Therefore

A₁ = -A₂ .........(9)

We need one other boundary condition to complete the solution. By Kirchhoff's law, v_L = v_C - v_R. But v_R = iR = 0 at t = 0+ since the current has not had time to change. Also, v_L = L di/dt. We take the derivative of i in (3), multiply by L, and set the right hand side equal to the capacitor voltage at t = 0+. This gives

L[s₁ A₁ + s₂ A₂] = V_o .........(10)

Substituting A₂ = -A₁, we can solve for A₁ as

.........(11)

Using a similar process for the case where s_1,2 are complex, we find that

.........(12)

The solution process always starts with the calculation of s_1,2. If they are real and negative, we use (3) and (11). Otherwise, we use (5) and (12). We could use mechanical switches for the circuits shown in Fig. 1 to measure the experimental waveforms on the scope, but such switches tend to be noisy during the switching operation. If the switch has significant resistance during the switching transient, the effective circuit is changed and likewise the solution. It is also difficult to get a clean, sharp pulse for triggering the scope (telling the scope when to start recording the waveform). For this reason we will use a solid state switch called the power MOSFET (Metal Oxide Semiconductor Field Effect Transistor). Smaller MOSFETs are used for many amplifier and digital applications, but power MOSFETs are mostly used as switches.

The basic symbol for the power MOSFET is shown in Fig. 2. It has three terminals, a drain, a source, and a gate. When the gate is several volts positive with respect to the source, a channel forms between the drain and source, allowing current to flow. There is no channel when the gate is at zero volts so the impedance between drain and source is very high. Thus the device acts as a voltage controlled switch. Compared with SCRs (Silicon Controlled Rectifiers), which will be seen later in the course, power MOSFETs are faster but have somewhat more limitations regarding maximum open circuit voltage and maximum short circuit current.

Smaller MOSFETs do not have the built-in diode that power MOSFETs have. This diode is the result of fabrication requirements for the larger devices. It means that the switch is only an open circuit to one polarity of applied voltage. A single power MOSFET could not be used as a switch in an AC circuit because the diode would conduct during the negative half cycle. We are switching DC current so this is not a factor.

A complete RC circuit with power MOSFET switching is shown in Fig. 3. The battery (or laboratory power supply) is connected to the capacitor through switch S₁ for charging. S₁ is opened and S₂ is closed to connect the capacitor to the remainder of the circuit. This is conveniently done by a single-pole double-throw switch. We will use such a switch made with black plastic in the experiment. The capacitor is connected to the resistance R by pushing a small push button switch S₃ to supply gate voltage to the MOSFET. The MOSFET becomes an effective switch S_PM. The voltage across the resistance R is observed on the scope by the connections CH1 and GND. The current in the circuit can be calculated from i = v/R.

Discussion and Calculations

Assume C = 10 µF, R = 10 k ,and V_B = 16 volts in Fig. 3. The resistance of the power MOSFET is less than one Ohm, hence can be ignored. If C is charged to 16 volts, S₁ is opened, S₂ is closed, and S₃ is closed to complete the circuit, compute v_R for t = 0+, 0.02, 0.06, and 0.1 seconds. Sketch v_R versus t.
Solve (1) for C. What quantities need to be measured in order that C can be determined experimentally using this equation? Suppose R = 10 k and V_o = 16 V and you suspect C to be near 10 µF . What oscilloscope settings should be used for the vertical amplifier and for the time base, in volts/div and time/div?
Solve the circuit of Fig. 4 for i and v_R for t 0 using the values shown if the 10 µF capacitor is charged to 20 V by closing and then opening S₁, and is then discharged through the RL circuit by closing S₂ and S_PM. Evaluate v_R for several values of time, say 0.2, 0.6, 1.0,and 2.0 ms if the initial capacitor voltage is 20 volts. Sketch the curve of v_R versus time.

Instructional Activity in Class

Locate the small fixture with a (nominal) 10 µF tantalum capacitor. Take it to the Digibridge in the rear of the room and measure the capacitance. Record in your notebook.

Use your digital multimeter to adjust one of the 20 k resistors mounted on the bench to 10 k.

Use the variable DC voltage supply that should be on the bench. Connect the negative (-) terminal to the scope's ground (located below the scope screen and indicated by the ground symbol) and the positive (+) terminal to CH1. Set CH1 to 2 V/div and the time base to 20 ms/div. Adjust the position controls for CH1 until the trace starts at the left edge of the screen and the ground symbol lies at the bottom of the screen. Adjust the power supply until the sweep is exactly at the top of the screen. Since there are 8 vertical divisions at 2 V/div, we should have the power supply set at 16 V.

Connect the circuit of Fig. 3. Use the black knife switch for S₁ and S₂ and the power MOSFET in a black plastic box for S_PM. Set the scope to trigger on the up-slope. Set the trigger mode to single sweep. Charge C by closing the knife switch one way and then connect C to the rest of the circuit by throwing the switch the other way. Set up the scope to AUTO STORE the signal. Press the red button on the power MOSFET switch. If the oscilloscope does not trigger (trace does not appear) check that CH1 is "ON" and that RUN or STORE appears on the upper right-hand corner of the scope's screen. Remember that you must recharge the capacitor after every attempt to get the discharge curve. If it still does not trigger, call the instructor.

The curve should start in the upper left corner of the scope's screen. If it does not, try capturing the curve another time or two, and then recheck your zero reference and 16 V level. If you cannot get within say a tenth of a major division, note where the actual starting point is on your sketch.

Read the voltage values at 0, 0.02, 0.06, and 0.1 seconds using the CURSOR button and knob on the scope. Prepare a table with the left column for time, the middle column for experimentally observed voltages at each time, and the right column for predicted voltage values, using the measured capacitance value and the same procedure as in Discussion & Calculations part 1. Experiment values should be within about 10 percent of the theoretical ones.

Calculate values for the capacitance using the measured decay rate and the technique of Discussion & Calculations part 2, for 0.02, 0.06, and 0.1 sec. How close are these values to each other and to the value measured on the Digibridge?

Assemble the circuit in Fig. 5. Use one of the (approximately) 200 Ohm variable resistors on the bench. Set the resistor to about half value, at the center of its range. Measure the resistance of the 1 Ohm resistor. Set CH1 to 50 mV/div and adjust the position of the zero reference to two divisions up from the bottom of the screen. Set the time base to 1 ms/div or 2 ms/div. Make sure the trace starts on one of the vertical lines on the left side of the screen. Charge C to about 20 volts from a variable DC power supply by closing and then opening S₁. Close S₂ and then S_PM and observe v_R on the storage oscilloscope. Some adjustment of the trigger controls may be necessary. Get a family of curves on the storage oscilloscope for R approximately equal to 0, 50, 100, 150, and 200 . (Just set the variable resistor at appropriate positions on its dial.) Sketch curves in your notebook for the damped sinusoid and monotonic decay cases, noting the corresponding value of R. How close does the trace correspond to the calculated curve of Discussion & Calculations part 3 for R = 200 Make a short table showing corresponding values of predicted and measured v_R for values of t= 0.2, 0.6, 1.0 and 2.0 ms. Ask the instructor to come over so you can display and explain the family of curves to him or her.

Conclusion

Compare experimental results with theoretical ones for Lab Activity 1. What might be done to improve the results?
Compare experimental results from Lab Activity 2 with theoretical ones from Discussion & Calculations part 3. Why did we use a 240 Ohm resistor for calculations in Fig. 4 in place of the 200 Ohm resistor in the circuit of Fig. 5?
What do you predict would happen if we did not include the 1 Ohm resistor in Fig. 5 and set the variable resistor to zero? What difference would it make if the inductor was IDEAL (not lossy)? Can you suggest some safety precautions for handling capacitors based on these observations?
Discuss the desired damping characteristics of some other systems in which damping has a significant influence on performance. Some possible systems: automobile shock absorber, child's swing, bungee jumping, others?