ENFUSE M DC-AC Inversion Experimental Procedue



Module: Energy/Environment: DC-AC Power InversionExperimental ProcedureName: Score: ____________ / 100Partner: Overall Objectives of this ModuleCircuits Concepts:Reinforce understanding of power and energyUnderstand DC-AC inversion, how electronic circuits perform inversion, and the energy efficiency of inversionDevelop an understanding of energy efficiencyUnderstand the PV solar panel as a power sourceUnderstand why the grid uses ACContinue to learn to use basic lab instrumentation, including a function generator, oscilloscope, and multimeterSystems Concepts:1. Learn how a system is implemented as an electronic circuit to perform a specific function2. See how a system can decomposed into subsystems to aid both understanding and debuggingGradingPart 1 / 15Part 2 / 15Part 3 / 20Part 4 / 20Part 5 / 15Part 6 / 15Important ConceptsYou will construct and explore an inverter that converts DC power to AC power. The resulting AC power can be applied to something called a load, which is EE-speak for something that does something useful with the power. So we can focus on the inverter system, in this experiment our load is a simple resistor.Alternating current (AC) circuits are circuits in which voltages and currents are not constant, but change with time.The pattern in which voltages and currents change with time is called a waveform or a signal.Ideally, AC waveforms are sinusoidal. As we will see, we can develop inverters that generate an approximation of a sinusoidal waveform.Power is the rate of change (or derivative) of energy. Conversely, energy is the time integral of power.BackgroundIn this experiment, you will implement a system known as an inverter to perform a function: the conversion of DC power to AC power. Probably the most interesting thing about inverters is that they can take a constant-voltage input and produce a voltage on the load that not only varies, but rapidly swings between positive and negative voltages.To implement this system, you will integrate three subsystems, as shown in Figure 1. The first is a clock source that generates a clock waveform (or signal) that determines the frequency of the AC waveform; we will use a function generator as a clock source. The clock signal drives the second subsystem, which is a control subsystem that determines when the current direction is switched. The control subsystem will be implemented using a digital logic chip with two flip-flops. The third subsystem is the H-bridge. We will implement this with six MOSFET transistors. An H-bridge consists of four legs that control the flow of current, with a pair of legs connected to each side of the load. In an ideal circuit, only one MOSFET would be needed in each leg. But because of details (beyond the scope of this course) of how the selected MOSFETs work, each leg of the final circuit needs one more MOSFET to control one of the MOSFETs that actually carries the current to the load.ClockSourceControlH-BridgeDC InputACOutputClockWaveformControlSignalsCVEnergySourceCVLoadInverter SystemFigure SEQ Figure \* ARABIC 1. The system-level diagram of the inverter system shows its three subsystems as blocks and the waveforms/signals connecting them as arrows (power signals are bolded). The inverter transforms DC power from the energy source into AC power that can be delivered to the load. The control subsystem generates two signals that control the H-bridge.A complete system requires one more subsystem to provide power to the inverter. In the last part of this experiment, you will supply input power to the system with a renewable energy source—a small photovoltaic solar panel.The inverter you will build is small, and can only handle a few hundred milliwatts of power. However, the same basic principles are also used in inverters that connect photovoltaic solar farms to the power grid. These can handle megawatts of power.Note: You may use some circuit components, such as flip-flops and MOSFETs, for the first time. Don’t worry about the details of how these electronic components work. If you are studying Electrical Engineering, you will learn about them in much more detail in subsequent courses. If you are a non-EE major, our focus on how circuits implement important systems will help you see how they are an important part of modern engineered systems.Resources Videos for this learning module (watch these before working through this laboratory experience)A basic function generatorA dual-channel oscilloscopeA basic multimeter that can measure voltage and currentA power supply that can deliver 5 V DC (1 A is enough current capacity)You will also need the electronic components in Table 1 (substitute part numbers that will also work are shown in parentheses):SubsystemPart NumberDescriptionQuantityControl subsystem7474TTL Dual D-Type Flip-Flop1H-bridgeBS170N-Channel Enhancement Mode Field Effect Transistor61N4007 (1N4001, 1N4002, 1N4003, 1N4004, 1N4005, 1N4006)Standard Recovery Rectifier2Various valuesResistors8Various valuesCapacitors5Renewable Energy Source (Optional)1W Solar Panel 80x1001 W Solar (PV) Panel (Vendor: )1Table 1. Parts list for this experiment.Part 1 – Control of the H-bridge transistorsTo generate the AC current, the circuit should automatically switch the flow of current through the load. Roughly, it should flow in one direction for one-half of an AC cycle, and the other direction for the other half, and repeat this forever.In this part, we construct and integrate two of the subsystems in Figure 1: the clock source that produces the clock waveform, and the control subsystem.We use the function generator to produce the clock signal. What should its frequency be? Since the AC waveform is 60 Hz, it repeats every 1/60 Hz = 16.67 ms (note that the inverse of time (s) is frequency (Hz), and vice versa). But the trick is that we need to switch the flow at four times that frequency, or 240 Hz.The clock waveform is not rich enough in content to drive the H-bridge: to change the direction of current through the load, we need to turn one dog-leg of the H-bridge off when the other is on, and vice versa. See Figure 2; the top waveform is the clock waveform, and the two other waveforms activate the H-bridge dog-legs.ClockQ1Q2Figure 2. Waveforms for the input (Clock) and outputs (Q1 and Q2) of the control subsystem.Here is what happens when we use these signals to control the H-bridge. When Q1 is high and Q2 is low, the load current will flow in one direction, and when Q1 is low and Q2 is high, the load current will flow in the opposite direction. However, we have a slight complication: what happens when both are high (or low)? In both cases, the voltage across the load is zero, so no current will flow. Thus the two states (Q1=1 & Q2=0) and (Q1=0 & Q2=1) are zero-current transition periods between the states (Q1=1 & Q2=1) and (Q1=0 & Q2=1) when the load current flows.166243066675Figure 3. Implementation of the inverter control subsystem using two D-type flip-flops integrated into a single chip. The numbers outside the chip boundary (labeling the wires) are the physical pin numbers; the labels inside the chip boundary describe the logical function; for example, Q1 is the output of flip-flop 1, and in our circuit, it is connected to the input D2 of flip-flop 2. (We will see later that Q1 is also the control signal for one dog-leg of the H-Bridge, while Q2 is the control signal for the other dog-leg.) Notice that /Q2 (the inverted output of the second flip-flop) is connected to the input D1 of the first flip-flop.Now that we have a model for the signals that must be generated, we must develop a circuit to generate them. It is an implementation of our control subsystem for the AC-DC inverter. It uses two flip-flops; see the schematic in Figure 3. It will be familiar to you if you’ve studied digital logic. If not, you can just build it and see how it works.Procedure1. Set up the function generator to produce a square wave at a frequency of 120 Hz. The low voltage should be zero volts (0 V), and the high voltage should be 5 V. Verify the waveform on the oscilloscope.178117510877552. On your breadboard construct the control circuit according to the schematic in Figure 3. See Figure 4 for a diagram of the 7474 flip-flop.3. Connect the function generator output (CLK_IN in the schematic of Figure 3) to the clock inputs of the flip-flop. Make sure the negative output of the function generator is connected to the circuit ground.Figure 4. Packaging of the 7474 dual D-type flip-flop. Note that pin 1 is directly below the dimple at the left-hand side of the physical package. Note: while this is a very typical package for the 7474, you may have a different package; check the datasheet for your device.4. Using the oscilloscope, verify that the signals look like those in Figure 2. (First compare each Q output with the clock waveform, and then compare the two Q1 outputs.)Have your lab instructor or aide review your results and sign below:Signature ________________________________ Date ___________________Part 2 – H-BridgeHere we construct the circuit that rapidly changes the direction of the current through the load. The switches in each leg are MOSFETs, denoted by Q1 through Q6 in Figure 5. The load is connected between OUT1 and OUT2. When current flows from OUT1 through a load to OUT2 (Figure 5), Q1 and Q6 are on. When the current flows in the opposite direction, Q5 and Q3 are on. Think of MOSFETs Q2 and Q4 as inverters, so, e.g., when Q2 is on, Q1 is off.Note: Take time to understand how the schematic in Figure 5 implements an H-bridge, even though it does not look like an H.Warning: Do not confuse different Q’s in the schematics for the inverter system. The label Q is used for (i) flip-flop outputs in the control subsystem and (ii) MOSFETs in the H-bridge subsystem. In more detail: According to convention, Q1 and Q2 are outputs of the flip-flops in the control subsystem. Alarmingly, Q is the conventional prefix for transistors in a circuit, so Q1 and Q2 are also two of the MOSFETs in the H-bridge. Use context to determine whether Qx refers to a flip-flop output or a MOSFET. Engineering is sometimes messy, but we make it work.Figure 5. Schematic for the H-bridge subsystem.The MOSFETs are in TO-92 packages, as shown in Figure 6 (left). Use Figure 6 (right) to map the D (drain), G (gate) and S (source) pins on the physical device to the schematic representation in Figure 5 (don’t worry about the diode connecting the source and drain in Figure 6 (right).Figure 6. (Left) Physical package of the MOSFET. (Right) Schematic of the MOSFET. Procedure1. Build the circuit in Figure 5 on your breadboard. To assure a clean layout, keep in mind that in the next part of this experiment the Q1 and Q2 signals will be connected to the Q1 and Q2 outputs (respectively) of the flip-flip using resistors.2. Note that you will connect the 5 V supply to all four DC_INPUT points on the schematic. (In the next part, you will use the 5 V DC power supply to power both the control subsystem and the H-Bridge. In other words, the 5 V DC power supply is both “5 V” in Figure 3 and “DC_INPUT” in the schematic of Figure 5.)Part 3 – Integration of the SubsystemsHere you will connect the Control and H-Bridge subsystems to complete the design.Figure 7. Schematic of complete inverter, including control and H-bridge subsystems.Procedure1. Connect the control subsystem to the H-bridge. First, connect the flip-flop output Q1 to the gate of MOSFET Q2 using a 1K resistor. Similarly, connect flip-flop output Q2 to MOSFET Q4 using another 1K resistor. 2. Connect the H-bridge to the load. Connect a 1K resistor between AC_OUTPUT_1 and AC_OUTPUT_2.3. View the load voltage waveform on the oscilloscope. Sketch the waveform in the box below.4. How does the output waveform differ from a sinusoid? Also, suggest some reasons. Have your lab instructor or aide review your results and sign below:Signature ________________________________ Date ___________________Part 4 – Efficiency of InversionWhenever energy is transformed, some is lost; in an inverter, energy is dissipated as heat by resistors as well as the unavoidable resistance of the other electronic devices.How efficient is our inverter? Efficiency can be measured as the ratio ofthe output power to the input power. For example, if ? of the power is lost in the inversion process, then only 3/4 is generated as AC output power; so in this case the efficiency of the inverter would be 75%.Procedure1. Measure the input power (supplied by DC_INPUT): because the input is DC, use DC measurements of the current and voltage to obtain the power.2. Measure the output power dissipated by the load. Here, use the RMS measurement mode of the multimeter to account for the fact that the voltage and current are time-varying.3. Compute the energy efficiency. Because power is the rate of energy, we can compute energy efficiency as the ratio of the output power to the input power.Have your lab instructor or aide review your results and sign below:Signature ________________________________ Date ___________________Part 5 – Smoothing the AC WaveformThe output of the inverter does not look sinusoidal, but rather like stair steps. Sophisticated inverters use clever tricks to generate a smoother signal, but all inverters rely on some kind of filtering to do this. A good way to think about filtering is as a way to quickly store and release energy; when an input voltage drops rapidly, the filter boosts it, and when the voltage rises, the filter can absorb it. Here we explore filtering using a capacitor, which stores energy in an electric field.Note that from a design perspective, the capacitor is part of the inverter system, even though on our breadboard it looks like we are making the load reactive by putting capacitance in parallel with the resistance.ProcedureYou will put a capacitor in parallel with the load. Here is where you can experiment. Try 10 uF, 100 uF, and 500 uF capacitors, and interpret your results, noting not only what happens, but why.Have your lab instructor or aide review your results and sign below:Signature ________________________________ Date ___________________Part 6 – ReliabilityOne method to improve the reliability of a system is to build it with better (that is, more reliable) components. But this approach does not alleviate the problem of single-point failures, where the failure of one component causes the entire system to fail. One way to remove the possibility of single-point failures is to build redundant systems. In our case, we might use two inverters, each connected to the DC energy source and the load, with each running at half of its capacity.Analyze the advantages and disadvantages of redundant design.Have your lab instructor or aide review your results and sign below:Signature ________________________________ Date ___________________ Part 7 (Optional) – Inverters in Renewable Energy SystemsIf you have access to a 1 W solar PV panel, try connecting it to the DC_INPUT of your inverter and placing the panel in the sun. Monitor the DC_INPUT voltage with a multimeter. You will notice that the output is maximized when the panel is perpendicular to the sun’s rays. Using a protractor, measure the power output of the panel when it is at different angles to the sun.Question (multiple choice question): the panel power output is proportional to the (a) sine(b) cosine(c) arctangent(d) none of the aboveof the angle of the panel to the sun’s rays.Try to fit your data to the above three models. Describe and explain your answer below.Have your lab instructor or aide review your results and sign below:Signature ________________________________ Date ___________________ ................
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