470μF. resistances, then you simply chose resistor values to match this ratio. To find

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1 Ryan Hoover EE 310 Lab 3 Formal Report Introduction: In this lab my partner and I were designing and constructing a 5VDC power supply using the 120 VAC from any regular electricity outlet. To do this we had to first step down the voltage from the 120 VAC to 10 VAC, we did this using a transformer with a turns ratio of 12:1. Next we had to add the diode full rectifier circuit with a filter capacitor to give us a signal that now had a mean value of 12 VDC. Finally we had to add a zener diode voltage limiter circuit to limit the voltage in the output down to 5 VDC. Once all of these component circuits were completed testing the circuit is required to ensure that the power being supplied is truly 5 VDC. Circuit Design and Reasoning: Attached to this formal report is a schematic of the unregulated power supply, this circuit consists of the transformer and the diode rectifier circuit with a filter capacitor. The rectifier used in the power supply was chosen due to its simplicity, and due to its ability to flip the negative part of the input signal into a positive signal. This is the function of a rectifier circuit. Attached to this report is also a schematic of the voltage regulator zener diode circuit. The zener diode voltage regulator literally meant to limit the voltage at the vout terminal. This is done by hooking the zener diode in a fashion such that if the voltage across it is too large current will flow in the reverse bias direction. When there is reverse bias current flow the output voltage is limited to voltage specifications of the zener, in the case of our power supply circuit, 5 VDC. Supporting Analysis: Attached is a model of the transformer, calculations for PIV for the rectifier diodes, the chosen diodes, calculations for the resistors and filter capacitor, along with a multisim simulation of the diode current. To calculate the PIV for the rectifier you have to know the maximum current that could be flowing through that diode and to know the maximum voltage across that diode in its reverse bias state. Once you know all of these values you choose a diode that has twice the specification, this is to prolong the life of the components used in the power supply circuit. The maximum current was found to be 93.3mA from ohm's law, the resistor was chosen to dissipate 1W, so using P = V 2 /R we found R to be 144Ω. In order to solve for the capacitor you use the formula C = V m, using this formula you solve for C, which we found to be around 470μF. 2fRL V r In the multisim simulation of the circuit you can see that the maximum current in the diode isn t too far off the calculated maximum current. To find the parameters for and you set up a voltage division equation and get a ratio between the two R i R L resistances, then you simply chose resistor values to match this ratio. To find you assume that the I Zmax load resistor is an open circuit, then you ohm s law to determine the max current, and to find assume that the load resistor is a short circuit and solve using ohm s law again. Data: you I Zmin

2 During the experiment we measured the value of the transformer open circuit voltage and the Rw voltage. Our measured values for the open voltages are as follows; V ab = 10.0V, V BC = 9.9V, V ac = 20.4V. We then added a 168Ω resistor across V ac and measured the voltage, we found it to be 19.8V. Rw, which is the resistance of the transformer was found by using voltage division and the measured voltages within the transformer. Once solved, we found Rw to have a value of 2.63Ω, a small resistance but still there. Figures: 1) (Diode vs AC) Above is the waveform of just the rectifier circuit attached to the transformer. The input would just be the sine wave from the output of the transformer. With the rectifier attached the negative of the input is inverted.

3 2) (transformer vs Vc) Above is the figure of the DC signal vs the AC signal. This is showing the input from the transformer vs the voltage across the filter capacitor. As you can see the voltage across the capacitor is almost horizontal, therefore it is a DC signal. 3) (diode rectifier) Above is the figure of the diode rectifier waveform. You can see how the rectifier circuit takes the negative portion of the input waveform and inverts it. To make it into a viable power supply you just have to add a filter capacitor.

4 4) (IC regulator) Above is the waveform of the IC regulator. The value is constant because the output of the filter capacitor is almost a constant DC value. The average of this waveform is 5 VDC, exactly as we expect for the 5 VDC power supply. The difference of this waveform with and without a load is very small, less than half a volt, so it is not even visible on this figure. 5) (IC ripple) In the this waveform we used a 1 ohm resistor to find the ripple in the IC, this was done by looking at the current through the 1 ohm resistor.

5 6) (Multisim) Above is an image of the power supply circuit recreated in multisim, and near the bottom of the image you can see the diode current listed inside the tan box. Discussion: In this lab we chose components in order to keep the power consumption as low as possible, this was done to ensure the safety of the components used in the construction of the power supply circuit. Once we had the final power supply build we tested the regulation of the voltage provided, this was V o V o measured no load measured w/ load done by formula % reg = V o, when we did this we found our voltage regulation measured no load was 0.774%. We then removed the zener voltage regulator and added a IC voltage regulator. Using this IC we found the %reg to be 0.798%. Our zener voltage regulator was actually better at regulating the voltage then the IC, this is due to more closely selecting the correct components for a very specific task. The specifications of the power supply were an output voltage of 5.0V ± 0.5V this was well within our final measured values for Vout, with and without a load resistor attached. The required value for Vc was 12V, this was also well within our design, our Vc voltage was found to be 12.54V, this value was a littler closer to the required value, however it won t matter, when it is regulated it ll be limited to 5V. Summary and Conclusions: The point of this lab was to see what it takes to construct a 5 VDC power supply from all the circuits that make it up. The power supply requires a transformer to step down the voltage of the input 120 rmsv to 10 rmsv, then to invert the negative portion of the input signal you use a diode rectifier circuit with a filter capacitor, in our case we used a full rectifier circuit to get both parts of the input signal. Finally you need a voltage regulator circuit to limit the output voltage of the circuit, this is done with a zener diode, which has special characteristics, when a large enough reverse bias voltage is applied, there will be reverse bias current, thus limiting the output voltage. I learned quite a lot from

6 this lab, I was quite surprised how fairly uncomplicated it was to build a circuit that goes from 120 VAC to 5 VDC in a fairly straight forward manor. Attached are my lab notebook pages and remaining waveform graphs.