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| university:courses:engineering_discovery:lab_2 [01 Mar 2016 17:33] – Jonathan Pearson | university:courses:engineering_discovery:lab_2 [03 Jan 2018 19:35] (current) – [Observations and Conclusions] Doug Mercer |
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| ===== Summer Camp Lab 2 ===== | ===== Introduction to RC Circuits ===== |
| ===== Introduction to Electronic Components and Equipment ===== | {{ analogTV>4800540010001}} |
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| ==== Introduction ==== | ==== Introduction ==== |
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| Resistors, capacitors, and inductors are the basic passive building blocks of many circuits. These elements can be used to control timing and frequency response. In this experiment we will investigate how circuits comprised of resistors and capacitors behave when their voltages and currents are switched. We will watch the voltage increase on a capacitor as it charges over a relatively long period of time, then observe shorter-term effects with a resistor-capacitor (RC) circuit. We will also look at what happens when we use a diode to block current flow, causing a capacitor to hold a voltage instead of discharge, noting that the capacitor cannot hold the charge forever since neither it nor the diode are ideal. | Resistors, capacitors, and inductors are the basic passive building blocks of many circuits. These elements can be used to control timing and frequency response. In this experiment we will investigate how circuits comprised of resistors and capacitors behave when their voltages and currents are switched. We will watch the voltage increase on a capacitor as it charges over a relatively long period of time, then observe shorter-term effects with a resistor-capacitor (RC) circuit. |
| ==== Objective ==== | ==== Objective ==== |
| To study how RC circuits behave with switched input voltages. To see how a diode can be used to prevent a capacitor from discharging by blocking current flow in one direction. To see that there are no ideal circuit elements, and diodes do not perfectly block current and capacitors do not hold charge forever. | To study how RC circuits behave with switched input voltages. Following completion of this lab you should be able to explain how a capacitor charges when a step voltage is applied to it through a resistor, describe the shape of the voltage across a capacitor when a square voltage wave is applied to it through a resistor, calculate the time constant of a RC circuit and know to what level the capacitor charges in one time constant, determine one of the more commonly-accepted time intervals that it takes to fully charge a capacitor when a step voltage is applied to it through a resistor, observe polarity requirements of electrolytic capacitors, and give a basic explanation of dielectric absorption. |
| ==== Materials and Apparatus ==== | ==== Materials and Apparatus ==== |
| * Resistor and capacitor code handouts | * Resistor and capacitor code handouts |
| * (1) 47 μF capacitor from the ADALP2000 Analog Parts Kit | * (1) 47 μF capacitor from the ADALP2000 Analog Parts Kit |
| * (1) 1 μF capacitor from the ADALP2000 Analog Parts Kit | * (1) 1 μF capacitor from the ADALP2000 Analog Parts Kit |
| * (1) 1N914 diode from the ADALP2000 Analog Parts Kit | |
| ==== Procedure ==== | ==== Procedure ==== |
| - Construct the following circuit on the solderless breadboard; note the polarity of the 47 μF capacitor{{ university:courses:engineering_discovery:lab_2_image_1.png?300 }} | - Construct the following circuit on the solderless breadboard; note the polarity of the 47 μF capacitor{{ university:courses:engineering_discovery:lab_2_image_1.png?300 }} |
| - Observe the voltage on Channel B increase slowly | - Observe the voltage on Channel B increase slowly |
| - Change the output voltage from Channel A to 0 V and measure how long it takes the voltage on Channel B to reach 0 V (theoretically the capacitor voltage never gets to 0 V, but we can get a good estimation within the measurement limitations of the M1K) | - Change the output voltage from Channel A to 0 V and measure how long it takes the voltage on Channel B to reach 0 V (theoretically the capacitor voltage never gets to 0 V, but we can get a good estimation within the measurement limitations of the M1K) |
| - Modify the circuit as shown below; keep the anode of the diode unconnected for now{{ university:courses:engineering_discovery:lab_2_image_2.png?300 }} | |
| - Set the constant output voltage from Channel A to 5 V and connect the wire from Channel A to the anode of the diode while observing the voltage on Channel B | |
| - Does the voltage on Channel B make it all the way to 5 V? Why or why not? | |
| - Change the voltage from Channel A to 0 V and observe the voltage on Channel B. Does it decrease to 0 V at the same rate as it did without the diode? Why or why not? | |
| - Construct the following circuit on the solderless breadboard{{ university:courses:engineering_discovery:lab_2_image_3.png?300 }} | - Construct the following circuit on the solderless breadboard{{ university:courses:engineering_discovery:lab_2_image_3.png?300 }} |
| - Set up Channel A to source a 100 Hz square wave that swings between 0 V and 5 V. | - Set up Channel A to source a 100 Hz square wave that swings between 0 V and 5 V. |
| - Increase the frequency of the square wave. What happens to the voltage waveform across the capacitor? | - Increase the frequency of the square wave. What happens to the voltage waveform across the capacitor? |
| ==== Theory ==== | ==== Theory ==== |
| A capacitor is comprised of two conductors, typically called “plates,” separated by an insulating material called a “dielectric.” In electric circuits the capacitors are used in a way in which equal and opposite charges are created on each conductor. The charges are created by accumulation or depletion of free electrons in the conductors. The charges produce an electric field in the dielectric, thereby developing a voltage across the plates. For a fixed charge, Q, on the plates, the voltage, V, across the plates is determined by the capacitance value, C, according to the following relationship. | A capacitor is comprised of two conductors, typically called “plates,” separated by an insulating material called a “dielectric.” In electric circuits the capacitors are used in a way in which equal and opposite charges are created on each plate. The charges are created by accumulation or depletion of free electrons in the conductors. The charges produce an electric field in the dielectric, thereby developing a voltage across the plates. For a fixed charge, Q, on the plates, the voltage, V, across the plates is determined by the capacitance value, C, according to the following relationship. |
| {{ university:courses:engineering_discovery:lab_2_equation_1.png?70 }} | |
| If we want to determine how the voltage and current are related, it is useful to rearrange the equation as follows.{{ university:courses:engineering_discovery:lab_2_equation_2.png?70 }} | <m>V = Q/C</m> |
| We can get current from Q by taking the first time derivative of each side, noting that C is constant. | |
| {{ university:courses:engineering_discovery:lab_2_equation_3.png?180 }} | If we want to determine how the voltage and current are related, it is useful to rearrange the equation as follows. |
| We can also write this explicitly for the rate of change in voltage. | |
| {{ university:courses:engineering_discovery:lab_2_equation_4.png?105 }} | <m>Q = CV</m> |
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| | Since current is the time derivative of Q, taking the first time derivative of each side - noting that C is constant - produces an expression that relates current to the rate of change of the voltage in a capacitor. |
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| | <m>{dQ(t)}/dt = i(t) = C{dV(t)}/dt</m> |
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| | This can be written explicitly for the rate of change in voltage. |
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| | <m>{dV(t)}/dt = {i(t)}/C</m> |
| We can use this equation to study the charging of the capacitor. If we had a constant current source, I, the capacitor would charge at a constant rate of I/C volte per second. In the RC circuit, we have a voltage source in series with a resistor instead of a current source. When the 5 V is initially applied to the circuit, it all appears across the resistor because the initially zero voltage across the capacitor cannot change instantaneously (unless, of course, we had an infinite current available). In the first circuit the 5V across the resistor produces a current of 5V/100K amps by Ohm’s law, which begins to charge the capacitor. As the capacitor charges, however, the voltage across the resistor drops by Kirchhoff’s Voltage Law, thereby decreasing the current and the rate of change in capacitor voltage. This process produces an exponential slowing in the rate of increase of the capacitor voltage. | We can use this equation to study the charging of the capacitor. If we had a constant current source, I, the capacitor would charge at a constant rate of I/C volte per second. In the RC circuit, we have a voltage source in series with a resistor instead of a current source. When the 5 V is initially applied to the circuit, it all appears across the resistor because the initially zero voltage across the capacitor cannot change instantaneously (unless, of course, we had an infinite current available). In the first circuit the 5V across the resistor produces a current of 5V/100K amps by Ohm’s law, which begins to charge the capacitor. As the capacitor charges, however, the voltage across the resistor drops by Kirchhoff’s Voltage Law, thereby decreasing the current and the rate of change in capacitor voltage. This process produces an exponential slowing in the rate of increase of the capacitor voltage. |
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| The charging rate in a RC circuit depends on the RC product, which is typically called the “time constant,” represented by the Greek letter tau. In one time constant, a charging capacitor will move 63% of the way from its current voltage to the voltage applied through the resistor. The capacitor voltage, V<sub>C</sub>, in a RC circuit with an applied step voltage of V<sub>T</sub> is governed by the following equation. | The charging rate in a RC circuit depends on the RC product, which is typically called the “time constant,” represented by the Greek letter tau. In one time constant, a charging capacitor will move 63% of the way from its current voltage to the voltage applied through the resistor. The capacitor voltage, V<sub>C</sub>, in a RC circuit with an applied step voltage of V<sub>T</sub> is governed by the following equation. |
| {{ university:courses:engineering_discovery:lab_2_equation_5.png?175 }} | |
| In one time constant the capacitor reaches approximately 63% of the applied voltage, and in five time constants it reaches approximately 99.3% of the applied voltage. The time it takes to fully charge a capacitor, from a practical perspective, is often taken to be five time constants. In the first circuit, the time constant was (100 KΩ)*(47 μF) = 4.7 seconds, making the time to fully charge the capacitor equal to about 23.5 seconds. Time constants are the same for charging and discharging, so it would take about 23.5 seconds ti fully discharge the capacitor when the voltage out of Channel A was reduced to zero volts. | |
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| When the diode was placed in series with the resistor, it allowed current to flow in the direction of the arrow contained in the diode symbol, allowing the capacitor to charge. The capacitor, however, could not charge to 5V because there is an approximately 0.7V drop across the diode when it is carrying typical currents. This means that the capacitor can only charge to about 4.3 V. | <m>V_C(t) = V_T(1 - e^{-t/tau})</m> |
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| When the voltage sourced from Channel A was dropped to 0 V, the diode became “reverse biased,” meaning that it would prevent current from flowing. It cannot completely block the current, but it reduces the current to a very small level. Because of this, the capacitor “holds” its charge when the source voltage drops to zero instead of discharging according to the time constant rule. It cannot hold the charge forever, though, because the diode does not perfectly block reverse current flow and the capacitor dielectric is not a perfect insulator. | In one time constant the capacitor reaches approximately 63% of the applied voltage, and in five time constants it reaches approximately 99.3% of the applied voltage. The time it takes to fully charge a capacitor, from a practical perspective, is often taken to be five time constants. In the first circuit, the time constant was (100 KΩ)*(47 μF) = 4.7 seconds, making the time to fully charge the capacitor equal to about 23.5 seconds. Time constants are the same for charging and discharging, so it would take about 23.5 seconds ti fully discharge the capacitor when the voltage out of Channel A was reduced to zero volts. |
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| ==== Observations and Conclusions ==== | ==== Observations and Conclusions ==== |
| * A capacitor in a simple RC circuit moves to 63% of the difference between its current value and a step voltage applied to the circuit in one time constant | * A capacitor in a simple RC circuit moves to 63% of the difference between its current value and a step voltage applied to the circuit in one time constant |
| * A commonly accepted time for a capacitor to fully charge is equal to five time constants | * A commonly accepted time for a capacitor to fully charge is equal to five time constants |
| * A diode can be used a simple electronic switch | |
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| | **Return to [[university:courses:engineering_discovery|Engineering Discovery Index]]** |
