ECEN 1400 - Introduction to Digital and Analog Electronics

Peter Mathys, Spring 2014


Lab 11: Building a Power Supply for Your Clock

Quick Links


Goals of this Lab


This lab is an individual effort. Each student has to build their own power supply. Feel free to discuss your plans with others and help others or get help from them.


Introduction

Why not Use Batteries? A single LED needs about 10 to 20 mA of current to light up with reasonable intensity. Thus, a single 7-segment display needs about 100 mA when all segments are lit up. Four such displays will need close to 400 mA. The capacity of a 1.5 V AA battery is approximately 2000 mAh. Using four such batteries in series could thus power digital logic circuitry and four seven segment displays for only about 5 hours. This is feasible for a digital multimeter, but for a digital clock it would be very inconvenient (and expensive!) to have to change batteries every 5 hours. Thus, a better solution in the long run is to use a power supply that converts the 120 Vac line voltage from a regular power outlet to the 5 Vdc voltage needed for most digital logic ICs. The goal of this lab is to build and measure such a power supply and understand how it works.


2. Lab Experiments

The goal of this lab is to build a 5 Vdc power supply for your project. The schematic of the power supply looks as follows.

Schematic of 5V power supply

The transformer T1 converts the 120 Vac line voltage to approximately 9 Vac and at the same time electrically isolates the output of the power supply from the 120 Vac power line. The rectifier D1 is a bridge rectifier that consists of four diodes. The function of the rectifier is to take the absolute value of the sinusoidal 9 Vac waveform that comes from the secondary side of the transformer. Capacitor C1 then acts as an energy "reservoir" that stores energy when the instantaneous voltage from the bridge rectifier is high and releases energy when that voltage is low. When no load is connected at the output of the power supply the voltage across C1 is a pure dc voltage. But as the output of the power supply gets more loaded, the voltage across C1 consists of a dc component and an additional ac component. The latter is referred to as ripple voltage and it comes from periodic charge and discharge of the "reservoir" capacitor. The purpose of the regulator circuit IC1 is to remove the ripple voltage and produce a regulated and stable dc voltage at the output of the power supply. Note that this only works well if the voltage difference between the input and the output terminals of the regulator, referred to as the dropout voltage, is large enough. The capacitors at the output of the voltage regulator improve the transient response to rapid changes in the load current. Finally, diode D2 is included to prevent damage to regulator when high capacitance loads are present and the power supply is turned off. In this case the energy from the load capacitor is safely transferred through D2 to C1, rather than (in reverse direction) from the regulator output to the regulator input.

Here is the parts list for the power supply:

Kits that include all parts as shown in the picture below can be purchased for approximately $40 (incl. tax) from the E-store.

Parts in power supply kit

Important: When you build and test a power supply you have to work with line voltages of 120 Vac which are potentially dangerous if handled improperly. Don't take any risks! Unplug the power supply before you make any changes or touch any wires on the primary side. Insulate all connections on the primary side, e.g., using shrink tubing. Make sure that no blank wires that carry line voltages can touch the metal enclosure.

E1. Wire the Primary Side. The primary side of the power supply consists of the power cord, the fuse holder, and the primary winding of the transformer. Pull the power cord through the hole in the power supply enclosure before soldering it to the fuse holder and the transformer. Also cut some shrink tubing and pull the wire through it (without shrinking the tube yet) before soldering. The picture below shows the connections on the primary side after soldering but before shrinking the insulation tubes.

Wiring of the primary side of the power supply before shrinking tubes

The red and the white wire are the 120 Vac input wires. The blue wires are the 9 Vac output wires. The black wire on the primary side would be used if the line voltage were 230 Vac. This wire will produce 230 Vac when you connect the other two wires to 120 Vac (the transformer acts as a step-up transformer for this wire) and therefore the black wire on the primary side needs to be insulated well. The black wire on the secondary side also needs to be insulated, but it is less crucial since it carries only 4.5 Vac. The picture below shows the wiring of the primary side after using shrink tubing for insulation.

Wiring of the primary side of the power supply after shrinking tubes

Mount the transformer and the fuse holder with screws in the metal enclosure of the power supply as shown in the picture below. Also mount the strain relief on the power cord and push it into the feedthrough hole for the power cord.

Wiring and mounting of the primary side of the power supply completed

When the soldering, insulating and mounting on the primary side is complete, insert a fuse in the fuse holder and connect to power cord to a line outlet. Measure the voltage on the secondary side (blue wires) with the multimeter in the AC position. Then use the oscilloscope to display the voltage waveform on the secondary side. Sketch the voltage waveform and label it with the measured period T of the waveform, the measured average voltage and the measured peak-to-peak voltage. How is the voltage measured with the multimeter related to the peak-to-peak voltage measured with the oscilloscope?

E2. Wire the Secondary Side. On the secondary side a bridge rectifier is used to rectify (i.e., take the absolute value of) the 9Vac sinusoidal output from the transformer. Capacitor C1 is used as a RC lowpass filter (together with the resistance of the diodes and the transformer wire) to smooth out the rectified sinusoid. The resulting voltage waveform is mostly dc with an ac ripple on top of it. To make this into a constant 5Vdc voltage the L7805CV regulator IC is used. The function of this IC is to act like a variable series resistance that dynamically reduces the input voltage to a constant 5Vdc output voltage. Of course, the instantaneous input voltage must be larger (by a minimum dropout voltage of about 2...2.5V) than the desired output voltage at all times. The difference between the input and output voltage times the current flowing through the regulator is the power that must be dissipated by the voltage regulator. At the full 1A current this is about 3-4 W and the regulator must be mounted on a heatsink to dissipate this power safely. In this power supply this is achieved by mounting the regulator IC directly to the metal case. Note that the metal tab of the regulator IC is electrically connected to the ground terminal and if it is screwed to the enclosure, the whole case will be at reference ground potential (the coax power connector will also put the metal enclosure to reference ground potential). At the output of the voltage regulator IC a ceramic 0.1 uF capacitor and a 100 uF electolytic capacitor are connected to ground. The 0.1 uF capacitor has good high frequency properties and prevents the regulator from oscillating. The electrolytic capacitor acts as an energy "reservoir" during large changes in the output current. The arrangement of the electronic components on the secondary side of the power supply is shown in the picture below. Note that the position of the voltage regulator IC on the general purpose board is determined by the location of the mounting hole in the case.

Placement and wiring of the components on the secondary side of the
      power supply, component side view

Check the datasheets for the L7805CV regulator IC and the W04G bridge rectifier for the correct pin connections. The wiring on the backside of the general purpose board is shown below.

Placement and wiring of the components on the secondary side of the
      power supply, wiring side view

The rectifier is on the left side with the ac input terminals labeled with ~ and the dc output terminals labeled with their polarity (- and +). Make sure to observe the polarity of the electrolytic capacitors. The red and the black wire that will be connected to the DC output plug are connected to the wires labeled 5Vdc and the - on the board.

To complete the wiring of the secondary side, solder the blue wires from the transformer to the ac inputs (labeled with ~) of the rectifier and solder the red wire so that it connects to the middle pin of the DC output plug. The black wire then gets soldered to the tab of the DC output plug that connects to the outer part of the coaxial plug. The picture below shows he completed secondary side board assembly before mounting it in the metal enclosure.

Wiring of the components on the secondary side of the power supply completed

Next, screw the metal tab of the voltage regulator IC to the bottom of the metal enclosure as shown below. This completes the electrical assembly of the power supply.

Completed 5V Power Supply

For the following measurements the power supply needs to be connected to 120 Vac. Do not touch the fuse or the fuse holder while the power supply is connected to 120 Vac. Also, do not connect the oscilloscope or the oscilloscope probe to any part of the prinary circuit (which includes the fuse and the fuse holder).

While the power supply enclosure is still open, complete the following measurements. With no output load connected, use the multimeter to measure the ac voltage on the secondary side of the transformer, the dc voltage across the 1000 uF capacitor, and the dc voltage at the DC output plug. Use the oscilloscope to display the waveform of the voltage across the 1000 uF capacitor and the voltage at the DC output plug. Repeat these measurements with an output load connected to the DC output plug. This load can be your clock or simply load resistors (e.g., four 100 ohm resistors in parallel). Why is the dc voltage across the 1000 uF capacitor more than 9 Vdc if the transformer output voltage is 9 Vac? How much power is dissipated in the voltage regulator IC when a load is connected to the DC output plug? How is this power related to the load current?

As a final touch to your power supply you may want to stick on rubber feet as shown in the picture below so that the power supply doesn't slide around and doesn't scratch your furniture.

Underside of Completed 5V Power Supply with Rubber Feet