EE 40 Final Lab: An Audio AmplifierPart 1: Power SupplyLab GuideIn this lab we will build and analyze a simple AC to DC converter. You will receive a breadboardand some parts that you may keep and take home. Take care to build this circuit carefully so that it willlast a while. Treasure it forever.1 Analyzing The TransformerThe transformer we will be using in lab is shown in Figure 1. The primary winding is connected directlyto the wall outlet. Place an oscilloscope probe across the secondary winding, plug the transformer into thewall, and sketch Vout. DO NOT SHORT THE SECONDARY WINDING.What is the maximum voltage you see at Vout? What is the minimum? How does the waveform differfrom your expectations, and why is it this way?2 Adding In The Bridge RectifierNow change your circuit to look like Figure 2. We’ve taken the output of the transformer and fed it througha bridge rectifier circuit. Sketch Vout.What is the maximum voltage you see at Vout? What is the minimum?What is the frequency of Vout?Why?10:1120 Vrms, 60 Hz(US Wall Outlet)VoutFigure 1: Transformer circuit.1EE 43/100 FINAL PROJECT:AN AUDIO AMPLIFIEREE 40 Final Lab: An Audio Amplifier 210:1120 Vrms, 60 Hz(US Wall Outlet)VoutFigure 2: Adding in the bridge rectifier...2200 uF10:1120 Vrms, 60 Hz(US Wall Outlet)Vout2.2 kFigure 3: Loading the bridge rectifier...3 Analyzing The Bridge RectifierUsually we place a capacitor on the output of the bridge rectifier. We then use this signal to power anothercircuit or electric device. The other circuit we power is known as the load and can usually be approximatedby placing a resistor on the output of the bridge rectifier.When we make these two changes to the circuit, we end up with something like Figure 3. Together, theR and C form a low pass filter. Sketch Vout.What is the average voltage seen at Vout?4 Bridge Rectifier RippleWe like to power our circuits or electronic devices with an constant DC voltage. However, it’s extremelydifficult to get a truly constant DC voltage. We can get an approximation to it however, and often we liketo compare DC power supplies based on how closely they can approximate this constant DC voltage.A useful metric to measure the quality of an AC to DC converter is the output voltage ripple, Vripple.Often the output of a DC power supply will be approximately the DC value we want, but it may rise aboveEE 43/100 - Final Project: Audio Amplifier, Part IEE 40 Final Lab: An Audio Amplifier 3VinVoutLinear Voltage RegulatorFigure 4: A simple linear voltage regulator.Figure 5: LM340T12 linear voltage regulator.or fall below the DC value from time to time. We say that the output voltage “ripples.” We define Vrippleas the peak to pick voltage of the ripple.Use your oscilloscopes to measure Vripplefor this very simple AC to DC converter. What is the frequencyof this ripple voltage?5 Linear Voltage RegulatorOne way to combat ripple voltage is to use a device called a linear voltage regulator (v-reg). This device isillustrated in Figure 4. In the most simplistic view, a v-reg can be thought of as two variable resistors. Voutis then obtained from Vinthrough a simple voltage divider circuit. The v-reg then adjusts the values of thetwo variable resistors until the output voltage is desirable. If the input voltage changes, it will dynamicallyadjust the two resistors so that the change is not seen on the output. Using a v-reg is a great way of cleaningup a power supply to generate a constant DC voltage.The voltage regulator we will use in this lab is an LM340T12. The 12 suffix means that this v-reg isdesigned to output 12 volts. The pinout of this voltage regulator is shown in Figure 5.Modify your circuit to look like Figure 6. Notice that the load resistance has been moved to the outputof the v-reg. This is because we will power our circuit or electronic device from the v-reg output. Recallthat the resistor approximates the load, so it needs to connect to the v-reg output just like another circuitor electronic device would.Sketch Vout. What is the average value (DC component) of Vout?WhatisVripple?EE 43/100 - Final Project: Audio Amplifier, Part ILM340T5LM340T555EE 40 Final Lab: An Audio Amplifier 42200 uF10:1120 Vrms, 60 Hz(US Wall Outlet)Vout2.2 kLM340T12gnd220 uFinoutFigure 6: Adding a linear voltage regulator...6 Response To A Changing LoadSuppose the v-reg is powering another circuit (the load). Then there will be some current flowing fromthe v-reg output to the load. Suppose the load suddenly changes. This will cause the current to change,and because the v-reg is just generating the voltage with a voltage divider (Figure 4), the output voltagewill change as well. A good v-reg will notice this change and quickly adjust the two variable resistors tocompensate. However, this compensation takes some time. Another way to measure the quality of an ACto DC converter is to measure how quickly it can respond to a changing load.Modify your circuit so that it looks like Figure 7. Note that the pinout for the MOSFET is shown inFigure 8.Sketch Voutand ILon the same axes. To get IL, measure the voltage across R and use Ohm’s law.Approximately how long does the v-reg take to stabilize the output voltage?7 EfficiencyWe define the efficiency of a power supply to be the ratio of the output power to the input power. It’sinteresting to find the efficiency of this simple AC to DC converter. Finding this efficiency does require afew circuit modifications. The modified circuit is shown in Figure 9.The output power is relatively easy to find. Simply measure the current going through the 50Ω loadresistor and use P = V2/R. Note: the load resistance changed here to provide a slightly higher outputpower.The input power is slightly harder to find. Insert a 1Ω shunt resistor as shown in Figure 9. Now, if wemeasure the voltage across this resistor we will actually be measuring the current flowing into the circuit.Over one period, make several meaurements of the input current ISand VS. Use the fact that P = IV toget the instantaneous power for each measurement. Now average all your values to get the average powerflowing into the circuit.After doing all this, you should be able to get an efficiency for the circuit to the right of the transformer.If we know that the transformer is 97% efficient, what is the total efficiency of this AC to DC converter?Why is this efficiency so low? Where did all the excess power go? How can we build a better AC to DCconverter?EE
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