EE 11L Lab 3

Experiment 3

Transient Response of First Order Circuits

Max Woo
12/8/14
Lab 1D, Professor Babaie

Objectives

The objectives of this lab are to investigate the behavior of first order capacitative and inductive circuits, and their responses to voltage changes. In addition, we will try to construct a first-order circuit to match certain characteristics. Lastly, we will predict and observe the initial conditions of a circuit containing both inductors and capacitors.

Theory

One of the major concepts that we will be using throughout the entire lab is the fact that the voltage through a capacitor cannot change instantaneously, and the current through a inductor cannot change instaneously. This means for any system of resistors, capacitors, and inductors, one can find the instantaneous voltages and currents across each component by replacing capacitors with voltage sources and inductors with current sources and solving the resulting system. This also applies to initial conditions, where one can solve for a circuit’s initial conditions right after a change (e.g. switch closed, voltage step-up) by replacing capacitors with voltage sources with the value equal to the voltage in the capacitor right before the change, and likewise for the inductor.

The first part of our experiment involves a first-order capacitative circuit, more specifically an RC circuit (resistor and capacitor in series with voltage source). We will be looking at the voltage across the capacitor when the voltage changes from 0 to $V$. We know that a capacitor follows the equation:

$i(t) = C \frac{dv(t)}{dt} (Eq. 1)

Thus, KVL (Kirchoff’s Voltage Law) analysis gives us a differential equation of the form:

$V = V_R + V_C = iR + V_C = C \frac{dV_C (t)}{dt} + V_C$ (Eq. 2)

We know that the voltage of a capacitor cannot change instantaneously. Thus, when the voltage steps up from 0 to $V$, the capacitor will first start at 0 and then slowly approach $V$ as charge flows from one capacitor plate to the other. Solving Eq. 1 for the initial conditions $V_C (0) = 0$ and $V_C (\infty) = V$, we get

$V_C (t) = V (1 - e^{\frac{-t}{RC}})$ (Eq. 3)

Note that the $RC$ term is called the “time-constant” of the RC circuit, and is the amount of time for the voltage to increase to $V (1 - \frac{1}{e}) \approx 0.632 * V$.

The second part of the experiment involves an RL circuit (a resistor and inductor in series with a voltage source), with the voltage again changing from 0 to $V$. We know that an inductor follows the equation:

$v(t) = L \frac{di (t)}{dt}$ (Eq. 4)

Using KVL analysis for the RL circuit gives us the differential

$v(t) = iR + L \frac{di (t)}{dt}$ (Eq. 5)

We know that the current in an inductor cannot change instantaneously. Thus, when the voltage switches from 0 to $V$, initially the inductor will block all current, but as the time goes on the impedance of the inductor approaches 0. As the inductor’s impedance reaches 0, it acts like a short circuit, making the entire circuit act like a resistor and voltage source in series, for which we know the current from Ohm’s law $I = \frac{V}{R}$. From these conclusions, we know to solve the differential with the initial conditions $i(0) = 0$ and $i(\infty) = \frac{V}{R}$, which gives us:

$i(t) = \frac{V}{R} (1 - e^{\frac{R}{L} -t})$ (Eq. 6)

Note that the time constant in this case is $\frac{R}{L}$, and represents the amount of time for the current of the circuit to increase to $1 - \frac{1}{e} \approx 0.632$ of it’s final current.

Part 1: RC Circuit Analysis

Procedure

First, construct the circuit shown in Fig. 1. Use a $47 \mu F$ capacitor and $680 \Omega$ resistor.

Fig. 1: First RC Circuit for part 1

Use the myDAQ to supply $V_{IN}$, using the function generator to output a 1V square wave with a 0.5 voltage offset. Use a period of over $5\tau$ (where $\tau$ is the calculated time constant) to give the circuit enough time to reach steady state after each voltage change. Use the myDAQ to monitor the voltage across the capacitor as the voltage steps up from 0V to 1V (it may help to use one channel to directly monitor $V_{IN}$ and use an edge trigger on that channel to start recording once the voltage steps up). Experimentally measure the time constant by measuring the amount of time it takes for the voltage to reach approximately 63.2% of the final voltage value (starting from when the voltage steps up).

Reduce the period of the square wave to $\tau / 4$ and observe what happens.

Then construct the circuit in Fig 2 using the same resistor and capacitor, and measure the voltage response across the resistor. Measure the time constant again, but this time using the voltage of the resistor.

Fig. 2: Second RC circuit for part 1

Data

Using what we know from the Theory section, we can find the predicted time constant $\tau = RC = 31.96 ms$. Thus we know to use a frequency of around $1 / 0.15 \approx 6.66 Hz$ when doing measurements.

After the voltage step-up, the capacitor charged from 0V to 1V, so we know that we are looking for the amount of time it takes to get from 0V to 0.632V. We measured this to be $27.76 ms$.

When we reduced the square-wave’s period to $1/4 \tau$, the graph between each step looked like a truncated version of when the square-wave’s period was $5 \tau$, with the voltage never able to reach it’s steady state values of 0V and 1V. A screenshot of the oscilloscope is shown below:

Fig. 3: screenshot of the oscilloscope when the period is around $1/4 \tau$. Each voltage division is 500mV, and each time division is 5ms. Channel 0 (green) is the output of the function generator. and channel 1 (blue) is the voltage across the capacitor

The voltage response of the resistor was as expected, the difference between the square wave input and the capacitor output (due to KVL). It jumped to 1 volt when the square wave steps up, and then exponentially decayed to 0V. Thus, the time constant would be 63.2% of the progress from 1V to 0V, or the time from the voltage step-up to when the voltage is 0.368V. This was measured to be $26.88 ms$

Data Analysis:

The theoretical and measured time constant were both already found during the Data section, shown below together with the error:

Results 1: Measured and theoretical time constant for RC capacitor

To find the predicted resistor time constant, we know the voltage across a capacitor in an RC circuit (Eq. 3), so we can calculate the voltage across the resistor using KVL:

$V_R = Ve^{\frac{-t}{RC}}$ (Eq. 7)

We also calculated in the Data section that the time constant is the time between $t = 0$ and $V_R = 0.368$. Solving for when $V_R = 0.368$ gives us $t = 31.96 ms$, giving us $\tau = 31.96 ms$, the same time constant as the capacitor. The theoretical and measured results are compared below:

Results 2: Measured and theoretical time constant for RC resistor

Discussion

We first noticed that the theoretical time constant for the resistor and capacitor were the same, which makes sense because the voltage response of the resistor was just the difference between the square wave input and the capacitor voltage response. While the voltage response of the capacitor approached the source voltage quickly at first and then more slowly the closer it got, the voltage across the resistor started at the source voltage and then dropped quickly at first before slowing down as it approached 0.

For the time constants of both the resistor and capacitor, the measured values fell within the error bounds of the theoretical values, verifying our results. However, one source of error could have come from the sampling rate of the oscilloscope, restricting our time measurements to discretely spaced values.

When we increased the frequency of the input until the period was smaller than the time constant, we noticed that the voltage of the capacitor never reached either steady-state of 0V and 1V. Instead, it fluctuated between the 0V and 1V, with the voltage response between each voltage change looking like a truncated version of the voltage response when the period was larger than the time constant.

Part 2: RL Circuit Analysis

Procedure

Construct the circuit shown in Fig. 5, using a $150 mH$ inductor and $1 k\Omega$ resistor.

Fig. 5: RL circuit for part 2

Measure the resistance of the inductor. Then calculate the time constant using the information in the Theory section. Use the myDAQ to input the same 1V square wave as in part 1, and adjust the period of the square wave to be $\approx 5\tau$. While we can’t directly measure the current of the circuit, we know from Ohm’s law that the voltage across the resistor is directly proportional to the current of the circuit. Thus, we know the time constant of the current is the same as the time constant of the voltage across the resistor. To find this, we measure the initial and steady-state voltages across the resistor after the voltage step-up, and measure the time it takes for the voltage to reach 63.2% of the final voltage.

Also measure and observe the voltage response of the inductor.

Data

We measured the resistor of the inductor to be $260 \Omega$. We then calculated the theoretical time constant to be $\tau = L/R = 11.90 ms$.

We then measured the initial and steady-state voltages of the resistor after the voltage step-up, and got $V_R(0) = 0 mV$, $V_R (\infty) = 793.44 mV$. Thus, we needed the time between the voltage step-up and when the voltage across the resistor reached $793.44*0.632 = 501.45 mV$, which we measured to be $\tau = .110 ms$. We know that this is the same as the time constant for the current across the inductor.

We also looked at the voltage response across the inductor. A screenshot of the oscilloscope is shown below:

Fig. 6: Voltage response of the inductor to the square wave

Data Analysis

The measured and theoretical values of the time constant were both found in the Data section, and are compared below:

Results 3: The measured and theoretical time constants for the current across the inductor.

Discussion

When we calculated the theoretical time constant of the circuit, we made sure to factor in the resistance of the inductor. This is because the inductor was not ideal, and thus the internal resistance factored into the total resistance of the RL circuit. Our measured time constant fell within the error bounds of the theoretical time constant, verifying our results.

We noticed that the while the current response of the inductor looked similar to the voltage response of the capacitor in the RC circuit (starting at 0 and sharply approaching the steady-state value and then slowing down the closer it got), the voltage response behaved like the voltage response of the resistor in the RC circuit (starting high and sharply approaching 0, slowing down the closer it got).

Part 3: DC Switching Analysis

Construct the circuit shown in Fig. 6 with the values:
$R1 = 1k$, $R2 = 3.3k$, $R3 = 2.2k$, $L1 = 150mH$, $L2 = 150mH$, $C1 = 22 \mu F$, $C2 = 1 \mu f$.
Remember to measure the resistance of both inductors.

“Fig. 7: Circuit for part 3

Set up the function generator of the myDAQ to input a 1V square wave offset by 2.5V, so that the voltage switches between 2 and 3 volts. Use a slow frequency to ensure the system reaches steady state before each voltage change, something around 1 Hz. Denoting the time where the voltage switches from 3 to 2 volts as $t = 0$, and the time right before the voltage switches back to 3 volts as $t = \infty$. Measure the experimental values of the following:
$\ $
$V_{C1} (0-), V_{C1} (0+), V_{C1} (\infty)$
$V_{C2} (0-), V_{C2} (0+), V_{C2} (\infty)$
$V_{R1} (0-), V_{R1} (0+), V_{R1} (\infty)$
$V_{R2} (0-), V_{R2} (0+), V_{R2} (\infty)$
$V_{R3} (0-), V_{R3} (0+), V_{R3} (\infty)$
$V_{L1} (0-), V_{L1} (0+), V_{L1} (\infty)$
$V_{L2} (0-), V_{L2} (0+), V_{L2} (\infty)$

Calculate the predicted values for each of them measured values, and compare.

Data

We measured the resistor values of each inductor to be $R_{L1} = 250\Omega$ and $R_{L2} = 260\Omega$. We then measured the voltages specified in the procedure, shown below (note that the voltages across the inductors had to be measured indirectly, using the voltages across the components parallel to the inductors, and factoring in the internal resistance of each inductor)

Table 1: measured values for Part 3. All values given in Volts

Data Analysis

To calculate the theoretical values across each component, we start with the steady state voltage before the input drops from 3V to 2V. At steady state, capacitors act like open circuits and inductors act like short circuits, so the circuit can be modeled with the following diagram:

Fig. 8: Steady state circuit before voltage drop

The total resistance can be easily calculated using equivalent resistance laws (from Lab 1), which gives us

$R_{total} = R_{L1} + R2 + \frac{R3 R_{L2}}{R3 + R_{L2}} = 3782.52 \Omega$

Thus, the total current (from Ohm’s law) is $I = 3/R_{total} = 0.7931 mA$. Using this, we can use Ohm’s law to calculate the voltages across $R_{L1}$ and $R_2$, which also give us the voltages of $C_!$ and $C_2$ respectively. We can get the voltage across $R3$ using the relationship $R3 = 2 - V_{C1} - V_R2$. Lastly, we know the voltages across both inductors are 0 at steady state, and because $C2$ acts like an open circuit at steady state, then the voltage across $R1$ is also 0. The full theoretical results for $V(0-)$ are shown below:

Results 4: Theoretical results for V(0-), compared against the measured values. All values given in Volts

To calculate the theoretical values for V(0+), we first make the component replacements described in the Theory section, replacing capacitors with voltage sources at the capacitor’s V(0-) voltage, and replacing inductors with current sources at the inductors I(0-) current. The resulting circuit is shown below:

Fig. 9: Circuit model after voltage drop

The values of $I_{L1} (0-)$ and $I_{L2} (0-)$ are equal to the currents across $R_{L1}$ and $R_{L2}$ respectively, which can be calculated from the values calculated for the circuit before the voltage drop, giving us $I_{L1} (0-) = 8.237 mA$ and $I_{L2} (0-) = 6.943$. We can then use KVL analysis using the current loops shown in Fig 6 to get a system of equations (note that the $i_3 = I_{L2}$, so that was directly substituted in)

$V_{C1} + V_{C2} + R1 (i_1 - i_2) + R3 (i_1 - I_{L2} ) = 2$ (Eq. 8)
$R2 i_2 + R1 (i_2 - i_1) = V_{C2}$ (Eq. 9)

Substituting known values and solving Eq. 8-9 gives us $i_1 = 0.4551 mA$ and $i_2, = 0.7144 mA$. Now we have all the information needed to calculate the V(0+) values. We know that the V(0+) values are equal to the V(0-) values for the capacitors. We also know to use Ohm’s law to calculate the resistor voltages: $V_{R1} = R1 (i_1 - i_2)$, $V_{R2} = R2 i_2$, $V_{R3} = R3 (i_1 - I_{L2} (0-) )$. Lastly, for the inductor voltages, we can use Ohm’s law to calculate the voltages across $R_{L1}$ and $R_L2$, and then calculate them using $V_{L1} = V_{C1} - R_{L1} I_{L1}$, $V_{L2} = V_{R3} - R_{L2} I_{L2}$. The results are shown below:

Results 5: Theoretical results for V(0+), compared against measured results. All values given in Volts

For $V(\infty)$, we simply use the steady-state method used for $V_(0-)$, but replace the 3V voltage source with a 2V voltage source. The results are shown below:

Results 6: Theoretical results for $V(\infty)$, compared against measured results. All values given in Volts

Discussion

For all of our results, the measured values fell within the error bounds of the theoretical values, confirming our calculations and measurements. However, many of the results still contained error, showing that the experiment strayed from ideal results.

Part 4: First Order Circuit Diagram

Procedure

We need to construct a first-order circuit that matches the following voltage response:

Fig. 10: Desired voltage response for part 4

To do this, we first can notice that the voltage response looks like the voltage response of the capacitor in an RC circuit, which we observed in Part 1. Thus, we just need to calculate the time constant that matches the time constant in the desired voltage response, $\tau = 9.68 ms$ (shown in Fig. 10 as dT, the time for the voltage to increase from 0V to 0.632V, which is 63.2% of the steady state voltage of 1V).

We know $\tau = RC$, so while we can choose any capacitor, for convenience use the one used in Part 1, the $47 \mu F$ capacitor. We can then calculate that we need a resistor of value $R = 9.68 / 0.047 \approx 200 \Omega$. Construct an RC circuit using these two components, and compare the measured voltage response to the desired voltage response.

Data

Using the circuit constructed in the procedure, we measured an experimental time constant of 11.44 ms.

Data Analysis

The desired and measured time constants are compared below:

Results 7: Desired and measured time constants for part 4

Discussion

The measured time constant fell within the error bounds of the theoretical time constant. This verifies our calculations and measurements, and validating our analysis of RC circuits.

Conclusion

To conclude, in every part of our experiment, our measured values fell within the error bounds of our theoretical values, confirming our equations and calculations for first order circuits (given in the Theory section). We were able to verify the formulas for the time constants of both RC and LC circuits, and also confirm our predictions behind the initial condition behaviors of capacitors and inductors.