Lab 1 – Power Supplies

By

Krystle Haley

ECT246 Electronic Systems III with LabAnup Majumder

DeVry University OnlineMarch 5, 2013

Part A-Power Supplies

TCO #1:Given an application requiring an AC to DC regulated power supply discuss the block diagram of the supply, and how its internal subsystems are related, using the appropriate data sheets, determine the device’s operating parameters, calculate, simulate and measure the power supply’s electrical parameters.

This week you will build a regulated power supply. Each subsystem - transformer, bridge rectifier, filter and voltage regulator is examined and the voltages and waveforms are analyzed. You will troubleshoot the system for failures and write a short report. CAUTION: You are working near 115V ac voltage. Be very careful and if you have any questions, ask the professor before proceeding.

A. Draw and discuss the subsystems of an ac to dc regulated power supply.a. Draw and label each of the subsystems of a regulated power supply.

Transformer:

T1

Diode bridge/Bridge rectifier:

Capacitive Filter (Capacitor):

C1

100µF

Voltage regulator:

U1LM7805CTLINE VREG

COMMON

VOLTAGE

b. Describe briefly the purpose of each subsystem.

The transformer is used to step-up or step-down the AC voltage before it is used and isolates main supply from the rectifier (as it mentions in our textbook that “it couples the AC line input to the power supply…and then the output from the transformer is then applied to the rectifier.”

The bridge rectifier converts AC voltage in the secondary of the transformer to pulsating DC.

The capacitor reduces ripple voltage…It converts the DC ripple voltage into smooth DC.

The voltage regulator provides constant voltage across the power supply terminals even if there is change in the circuit current flow.

B. Use a bridge rectifier’s data sheet to determine its operating parameters.

a. Using the data sheet for a W04M or similar device, determine and record the pin configuration and maximum operating characteristics.

Package Drawing and Pin Configuration:

Maximum Ratings and Electrical Characteristics:

Peak Repetitive Reverse Voltage ( ): 400 V

RMS Reverse Voltage ( ): 280 V

Average Rectified Output Current ( ): 1.5 Amps

Forward Voltage ( ): 1.0 V

C. Given a full-wave bridge rectifier schematic, calculate the RMS and peak-to-peak output voltage of the transformer’s secondary. Draw and label the time domain output waveform of the transformer’s secondary side. Use the schematic to simulate the circuit. Record the transformer’s primary, secondary voltages and the dc load voltage. Record the input and output waveforms of the bridge. Compare the simulated results to the calculated values.

Schematic:

a. Use the schematic to determine or calculate the following values:

( )P pkV : 120√2=169.70 volts

( )S pkV : 12/0.707=16.97 volts

( )L pkV : 16.97-1.4 = 15.57 volts

aveV : 2*15.57/pi = 9.91 volts

b. Draw and label the input and output waveforms of the bridge.Typical input and output voltage wave forms are given below.

c. In Doc Sharing, use “Select View” and go to Week 1. Download the Multisim PS1.ms Multisim file or create the circuit. Run the simulation and record the following:

: 169.706 V

: 16.640 V; Being that the transformer secondary AC voltage measures 11.999V, if you divide 11.999 / 0.707 you get 16.97 V for Vs(pk)

: 15.240 V

: 9.227 V

d. Use the oscilloscope to view the input and output waveforms of the bridge. Copy and paste the waveforms below.

D. Explain ripple voltage and how a passive capacitor circuit is used to reduce the effect. Given a full-wave bridge rectifier schematic, calculate the ripple voltage. Simulate the circuit and record the unfiltered and filtered response of the bridge. Compare the results to the calculated values.

a. Explain the cause of ripple voltage and how a capacitor can reduce its effect.

Ripple voltage is due to insufficient suppression of the alternating waveforms within the power supply. A large ripple means less effective filtering and a smaller ripple means more effective filtering. When a capacitive filter is used, it converts the full-wave rippled output of the rectifier into a smooth DC output voltage.

Per our textbook, “the filtering action is based on the charge and discharge action of the capacitor”. The RC charge time of the filter capacitor must be short and the RC discharge time must be long to eliminate ripple voltage. In other words, the capacitor must charge up fast, preferably with no discharge at all.

b. Given the following circuit, assuming that Vdc approximately equals Vpk, calculate the expected ripple voltage. Vrms = 120/10 = 12vVs(pk)= 12v/0.707 = 16.97vVL(pk) = Vs(pk) – 1.4 = 15.57vIL = Vdc/RL = 15.57/120ohm = 129.75mAVr = ILt /C = 129.75*8.33 / 470uF = 2.30 mVpp

: 2.30 mVpp

c. From Doc Sharing, week 1 download the Multisim PS2.ms Multisim file or create the circuit. Simulate the circuit without the capacitor connected and record the ripple voltage. Connect the capacitor and repeat the simulation.

: Vm/2 = 15.57/2 = 7.785 V (without capacitor)

: 2.30mVpp as calculated above (with capacitor)

d. Describe the effect of the capacitor on the ripple voltage.

The calculations speak for itself….But once simulated, the ripple voltage decreases when the capacitor is connected. The capacitor stores energy during the conduction period and delivers this energy to the load during the non-conduction period. Therefore, the time in which current flows through the load is considerably prolonged and ripple is considerably reduced, as it showed in the waveform. The ripple voltage is the deviation of the load voltage from its average or DC value. The higher the capacitance value, the lower will be the ripple, but we cannot have high value of capacitance as it puts high demand of current on the supply. The current has to be supplied in very short time.

E. Discuss the operation of a voltage regulator IC. Identify the operating parameters from the data sheet. a. Locate the data sheet for a LM7805 voltage regulator and determine

its pin configuration, current and voltage rating.

Package Drawing and pin configuration:

Maximum Ratings and Electrical Characteristics:Input Voltage ( ): 35 V

Output Voltage ( ): 5 V (max = 5.2 V)

Quiescent Current ( ): 5 mA (max = 8 mA)

F. Simulate a full-wave bridge regulated power supply. Measure, and record the output voltages of the transformer, bridge rectifiers and regulator.

a. From Doc Sharing, week 1 download the Multisim PS3.ms file or create the circuit.

b. Simulate the circuit and record the following values.

: 62.198 V

: 9.854 V w/ oscilliscopeBeing that the transformer secondary AC voltage measures 6.997V, if you divide 6.997 / 0.707 you get 9.89 V for Vs(pk)

: 8.454 V

GND

: 5.001 V

: ?

G. Explain the operation of a full-wave bridge rectifier. In short form: The full-wave bridge rectifier consists of four diodes, D1, D2, D3 and D4, which are connected to form a bridge. The AC voltage to be rectified is applied between the diagonally opposite ends of the bridge through the transformer. Between the other two diagonally opposite ends of the bridge, the load resistance is connected. During the positive half-cycle of the secondary voltage, diodes D1 and D3 conduct being forward biased. The load current flows through the load in a downward direction, developing the output voltage across RL. During the negative half-cycle of the secondary voltage, diodes D2 and D4 conduct and again the current load flows through in a downward direction, resulting in a full-wave rectification output, which is a pulsating DC. I hope my understanding is correct!

According to my web source, for a better detail and using the picture above, assuming the transformer is working properly and there is a positive potential at point A and a negative potential at point B. The positive potential at point A will forward bias D3 and reverse bias D4. The negative potential at point B will forward bias D1 and reverse bias D2. At this time D3 and D1 are forward biased and will allow current flow to pass through them; D4 and D2 are reverse biased and will block current flow. The path for current flow is from point B through D1, up through RL, through D3, through the secondary of the transformer back to point B. This path is indicated by the solid arrows. Waveforms (1) and (2) can be observed across D1 and D3. Source: http://www.tpub.com/neets/book7/27c.htm

One-half cycle later the polarity across the secondary of the transformer reverses, forward biasing D2 and D4 and reverse biasing D1 and D3. Current flow will now be from point A through D4, up through RL, through D2, through the secondary of T1, and back to point A. This path is indicated by the broken arrows. Waveforms (3) and (4) can be observed across D2 and D4. You should have noted that the current flow through RL is always in the same direction. In flowing through RL this current develops a voltage corresponding to that shown in waveform (5). Since current flows through the load (RL) during both half cycles of the applied voltage, this bridge rectifier is a full-wave rectifier. Source: http://www.tpub.com/neets/book7/27c.htm

H. Use the Internet to find a different bridge rectifier package not discussed in class. Paste a picture of the package below. Specify

the following about the rectifier.

Package type: DF08S

Picture:

http://www.datasheetcatalog.org/datasheet/fairchild/DF08S.pdf

Maximum Peak Reverse Voltage: 800 VMaximum Average Forward Output Current: 1.5 A

I. The following circuit has the outputs indicated. How would you reduce the ripple voltage to approximately 500 mV? Show the calculations and explain your answer.

Answer:

Here, Vr = 500mV = 0.5V, f = 60 Hz, C =?,Vm = 16.26√2-1.4 =21.595, say 21.6 volts, R= 2.2kohms.

Therefore,

So we should connect a capacitor of 161.742µF to get ripple voltage as 0.5V

Part B-Soldering

Parts needed:1 - Power Supply parts kit

Tools needed:1 – Soldering iron1 – Safety glasses1 – Protective board

Instructions:In this activity, you will build a regulated power supply. The parts kit contains all of the necessary components to populate and solder a printed circuit board designed as a full-wave regulated 5-V power supply. Before beginning, read the following information on soldering.

Reading:http://www.aaroncake.net/electronics/solder.htm

Questions:

1. Soldering is defined as the joining of two metals by a fusion process.

2. The heat source is the soldering iron.

3. Soldering irons in the rage of 15 watts to 30 watts are good for most electronics circuits.

4. The best solder for electronic PCB is thin rosin core solder.

5. You should never use acid core solder on an electronic PCB.

6. When soldering, you should work in a well-ventilated area because the flux in the solder will release fumes as it is heated and these fumes are harmful to your eyes and lungs.

7. Since hot solder will splash, eye protection (e.g. safety glasses) is advised when soldering.

8. A clean surface is needed for best results.

9. Before soldering component leads, you should place the components onto the board.

10.To hold parts in place while you solder them, you should bend the leads on the bottom of the board at a 45 degree angle.

11.To help conduct heat to the components and PCB, you should apply a small amount of solder to the tip of the iron.

12. When soldering a component, the tip of the iron should rest against both the component lead and the board.

13. Normally, it should take 1 to 2 seconds to heat the component lead enough to for the solder to run smoothly.

14. Once the component lead and PCB are hot, you should touch the tip of the solder to the component lead and solder pad, and not to the tip of the iron.

15. If you move the connection before it cools, a cold joint could develop and you will need to re-solder the connection.

CAUTION:

You are working with hot surfaces. Be sure to wear proper eye-protections and utilize a soldering board. If you have any questions, ask the professor before you begin.

Step 1:Find an open area that provides enough room for you to work. Place the protective board on the table. Remove the parts from the kit and identify each.

_____ Power supply PCB Rev. 1.00

_____ AC Wall Adapter – 5.6VAC, 200mA_____ Bridge rectifier, W04, 1.5A_____ LM7805 -- +5 volt regulator_____ Resistor 120 Ohm, 1W

_____ 20 pos single row header (for P2 and J1-J4)

_____ Test Points

_____ Power connector

_____ Power Jack

_____ Elec. Cap 470uF

_____ Elec. Cap 10uF

_____ Soldering iron

_____ Solder

_____ Safety glasses

_____ Damp paper towels

Step 2:The drawing below shows the Power Supply PCB. Review each of the markings on the PCB. Preview the PCB and decide where each component is installed and locate it in the kit. Also, note the location of each Jumper and Test Point. Determine the component side of the PCB from the solder side.

Compare the PCB layout to the schematic below.

Step 3:

Place the protective eye wear on. Plug in the soldering iron and place it in the stand. Allow a few minutes for the iron to get hot. You can verify that the iron is at the correct temperature by touching the end of solder to the tip of the iron. If the solder melts quickly, the iron is ready. Cover the whole tip of the soldering iron completely on both sides and then wipe the soldering iron tip with the wet sponge. This is called “tinning the iron’s tip”. With a wet paper towel, remove any dust or dirt from the solder side of the PCB.

Step 4:Begin on the R2 end of the PCB. Bend the leads on the resistor to match the holes on the PCB. Insert the resistor into the mounting holes. Bend the leads of the resistor on the solder side to 45 degree angles to help hold the resistor while you solder it. Tin the iron’s tip. On the solder side of the PCB,

touch the tip of the iron to the junction of the PCB resistor pad and the resistor’s lead. After one or two seconds, touch the tip of the solder to the pad and allow the solder to run until the pad is covered. Do not touch the solder to the tip of the iron. If the solder does not flow smoothly, clean the tip with a damp paper towel and re-tin the tip. Remove the iron and the solder. Repeat the process for the other lead. Using side cutters, remove the excess resistor leads.

Step 5:Using proper safety and soldering techniques, continue to solder the remaining components to the PCB.

Note the following:

1. C1 and C2 are electrolytic capacitors and are polarized. Be sure to insert them into the PCB in the proper orientation.

2. The plus (+) and minus (-) on D1 must match the (+) and (–) in the PCB.

3. The test points and jumper connectors will get very hot when soldered. So, use needle nose plies to hold them while soldering.

4. Mount the P2 connector last. P2 mounts on the solder side of the PCB not on the component side.

Step 6:Review the soldering of each lead on the solder side of the PCB. If you see any small circles around any joint or dull colored joints, they may be cold solder joints. Use the iron only to reheat the joint. Look for solder splashes or bridges between connections and remove them with the tip of the soldering iron.

Step 7: Unplug the soldering iron and allow it to cool.

The completed regulated power supply PCB will look similar to the following picture.

Step 8: Double check all the parts are placed in the correct place, correct orientation and soldered properly. Measure, and record the output voltages and waveforms of the transformer, bridge rectifiers and regulator.

a. Prototype the circuit and record the following values.

: _________VAC [TP1 to TP2 with JP1, JP2, JP3 and JP4 open]

: _________ Vpp{With Oscilloscope} [TP5 to Ground- with all jumpers open]

: _________ VDC [TP5 to Ground- with JP1, JP3 & JP4 open and JP2 closed]

: _________ VDC [TP6 to Ground- with JP2, JP3 & JP4 closed, and JP1 open]