Experiment E7 DC Power Supply

Worst-Case Design for Half-Wave Rectifier Circuit

James J. Whalen

Fall 2000

Important Design Rules

1. Avoid destroying the Zener diode.

2. The design must work for any Zener Diode with
VZ = 12 V  10%.

3. The design must work with the load resistor connected or disconnected.

4. The design must work for an isolation transformer rms ac output voltage = 18 VAC  10%.

5. The design must work for resistors with a  10% tolerance.

6. The best design would have the best combination of the following factors:

A.  The highest allowable value of dc load current at 100% ac voltage.

B.  Low peak-to-peak ripple voltage under all conditions of load variation and ac voltage variation.

C.  Small change in dc load voltage under all conditions of load variation and ac voltage variation.

Procedure

1. Design the dc power supply and enter your design in your lab notebook. Be as complete as possible. Include the circuit schematic, design equations, and design values.

2. In your lab notebook create a table of design values that includes values for the following components: RL, RF, RS, CI, & CF.

3. Assemble the entire dc power supply and test it as an assembly.

4. Test the entire dc power supply at 100% ac voltage (18 VAC) with the load resistor connected. Measure the dc voltages and ac peak-to-peak ripple voltages at the nodes 1, 2 & 3 in Figure 1. Demonstrate to Staff your results. Staff must witness all six values of voltages and sign off on each in a space below the entry. Staff must also check and initial calculations for IZ + IL, IZ, & + IL.

5. Repeat Step 4 with load resistor removed. No demonstration required.

6. Repeat Step 4 at 90% ac voltage (16.2 VAC) with the load resistor connected. Staff must witness dc voltage and ac peak-to-peak ripple voltage at the node 3 and sign off on each in a space below the entry. Staff must also check and initial calculations for IZ + IL, IZ, & + IL. Is the Zener Diode cutoff?

7. Repeat Step 4 at 90% ac voltage with the load resistor removed. No demonstration required.

8. Repeat Step 4 at 110% ac voltage (19.8 VAC) with the load resistor connected. No demonstration required.

9. Repeat Step 4 at 110% ac voltage with the load resistor removed. Staff must witness dc voltage and ac peak-to-peak ripple voltage at the node 3 and sign off on each in a space below the entry. Staff must also check and initial calculations for IZ + IL, IZ, & + IL. Is the Zener Diode maximum current rating exceeded?

10.  Enter your result from Steps 4-9 in a TABLE: DC & AC RIPPLE VOLTAGES

11.  Also enter in the table the percent dc load voltage regulation. The percent dc load voltage regulation is calculated using the dc load voltage obtained in Step 4 as the standard and using the following equation:

Load Regulation = {VL (Step N)) - VL (Step 4)}  VL (Step 4) X 100% where N = 5 to 9.

12.  Summarize in 250 words or less the important results.

13.  Submit the copies of your lab notebook pages and any PSPICE or ELECTRONIC WORKBENCH simulations done in support of the design done prior to the lab.

Zener Diode Information

1N4742

Maximum Ratings

DC Power Dissipation 1 Watt

Derating Factor 6.67 mW per degree C

Junction Temperature 200 degree C

Electrical Characteristics at 25 C

Nominal Zener Voltage VZ @ IZT 12 V

Test Current IZT 21 mA

Max Zener Impedance ZZT @ IZT 9 ohms

Test Current IZK 0.25 mA

Max Zener Impedance ZZT @ IZK 700 ohms

Max Reverse Leakage Current IR 5 uA @ VR = 9.1 V

Max DC Zener Current IZM 76 mA

TABLE 1 VALUES OF DC VOLTAGES AT

NODES 1-3 AS A FUNCTION OF VAC

% / VAC
(rms) / Peak / Rectifier
Diode
Voltage / V1(dc)
0.9 X (Peak
- 0.7) / V3(dc)
 10% / V1(dc)- V3(dc)
V / V / V / V / V / V
100 / 18 / 25.5 / 0.7 / 22.3 / 12.01.2 / 10.31.2
90 / 16.2 / 22.9 / 0.7 / 20.0 / 12.01.2 / 8.01.2
110 / 19.8 / 28.0 / 0.7 / 24.6 / 12.01.2 / 12.61.2

Important Design Considerations

1. If the load resistor is removed, all the dc current must flow through the Zener Diode. That current must not exceed IZM = 76 mA. The worst case is at 110% ac voltage. The dc voltage drop across the resistors RF + RS is then given by V1(dc) - V3(dc) = (RF + RS) X IZ. The worst case occurs at 110% ac voltage where V1 (dc) - V3 (dc) has a maximum value given by max {V1 (dc)- V3 (dc)} = 12.6 + 1.2 = 13.8 V. To limit the dc current that flows through the Zener Diode to a value less than IZM = 76 mA, the value for (RF + RS) must exceed a minimum value for RF + RS given by

Min {RF + RS} = max {V1 (dc)- V3 (dc)}  IZM

= 13.8 V/76mA = 0.1816 kohm

Min {RF + RS} = 13.8 V/76mA = 0.1816 kohm

If we allow for a 10% tolerance on the resistors, then

Min {RF + RS} = 0.1816 kohm + 10% X 0.1816 kohm

= 0.1997 kohm

Min {RF + RS} = 0.1997 kohm

The nominal values available include 100 ohms. Thus an available choice is

RF = RS = 0.100 kohm = 100 ohm

2. Another important consideration is that the dc current through the Zener Diode must not fall below some minimum value IZmin. The worst case occurs at 90% ac voltage with the load resistor connected. The voltage V1(dc)- V3(dc) has a minimum value min{V1(dc)- V3(dc)} = 8.0 - 1.2 = 6.8 V. The current through RF + RS is a minimum and is given by

Min {IZ + ILOAD} = min {V1 (dc)- V3 (dc)}  (RF + RS)

Min {IZ + ILOAD} = 6.8V  0.200k = 34 mA

3. A value for IZ = 0.25 mA where the Max Zener Impedance ZZT = 700 ohms is obviously too low. The dc voltage regulation will be poor, and the ac peak-to-peak ripple voltage will be high. As an initial design decision a value for IZmin = 0.5 X IZT = 0.5 X 21 mA = 10.5 mA is selected.

IZmin = 10.5 mA

It should be noted that no design decision is final. The value selected for IZmin can be changed in a subsequent design.

4. The value for the load current ILOAD can be determined from

ILOAD = 34 mA - 10.5 mA = 23.5 mA

5. The value for the load resistor RL can be determined from

RL = 12V  23.5mA = 0.5106 kohm

RL = 511 ohm

The variable resistor available for RL should be set at 511 ohms. Use the DMM.

6. The resistors available have nominal values as given the Appendix.

7. The capacitors available have values 200 F, 100 F, and 50 F.

8. Initially the 200 F capacitor & 50 F capacitor are connected in parallel for the capacitor CI in Figure 1,

i. e. the capacitor CI = 250 F.

CI = 250 F = 0.250 mF

9. The maximum droop V across the capacitor across CI in Figure 1 can be estimated. See figure ? in Lecture Slides. Using the expression Q = CV where Q is the charge on a capacitor C and V the across the capacitor, the droop V can be calculated from

V = Q/C = It/C

where I = IZ + ILOAD is the discharge current for the capacitor. The worst case droop would be obtained where I = IZ + ILOAD has a maximum value max{IZ + ILOAD} . Since the value for the sum of the resistors (RF + RS) was chosen to limit the value of the current through them to a value IZ + ILOAD  IZM = 76 mA, the maximum value for I is IZM = 76 mA. The maximum droop max{V} is given by

Max {V} = IZM/60C

Max {V} = 76mA /(60 X 0.250mF) = 5.1 V

where t has been set equal to 1/f where f = 60 Hz. Actually, as shown in the Lecture slides, the value for t will be less than 1/f. For example, for V/V = 0.8 (20% droop), the value for t is given by t = 0.9/f. Thus the maximum droop calculated with the equations above would be over-estimated by 10%.

The filter capacitor also contributes charge to the current that flows through the Zener diode and the RL.

If the value of the filter capacitor CF = 100 F = 0.100 mF is added to the value of CF = 250 F = 0.250 mF, the value obtained is

CF + CI = 250 F + 100 F = 350 F = 0.35 mF

If the value CF + CI = 0.35 mF is used to calculate the droop, the droop value is given by

Max {V} = 76mA /(60 X 0.350mF) = 3.6 V

Which value is the better choice: 5.1 V or 3.6 V? That is a question best answered using PSPICE.

Since the capacitors have 20% tolerance, the droop calculated or predicted using PSPICE could differ from that measured by 20%.

10. The final design decision that must be made is how to divide the sum of the resistors (RF + RS). The filter capacitor CF and the filter resistor RF form a low pass RC filter and cause a reduction in the ac voltage between the input at node 1 and the output at node 2. The reduction in the ratio V2/V1 for a sinusoidal signal at frequency f is given by

V2/V1 = 1/{1 + (RF 2fCF)2}0.5

One of the value used for f can be f = 60 Hz. Harmonics at 120 Hz and higher are also present. They harmonics will be attenuated more. Previously, an initial assignment RF = 100 ohms was made. Using RF = 100 ohms, CF = 100 F, & f = 60 Hz, the value for V2/V1 is given by

V2/V1 = 1/{1 + (100X2X60X100)2}0.5 = 0.256

Since the harmonics are attenuated even more, the variation in voltage across CF will be even less then 0.256 times the variation in voltage across CI. Again this is a question best answered using PSPICE.

11. The Zener diode also causes a reduction in the ac voltage between the input at node 2 and the output at node 3. . The reduction in the ratio V3/V2 is given by

V3/V2 = {RLrd}  {RS + RLrd}

where rd is the dynamic resistance of the Zener Diode. In E6 Dynamic Impedance a value rd = 5.6 ohms at IZ = 10 mA was measured. Using RS = 100 ohms and RLrd = 5115.6 = 5.5, the value for V3/V2 is given by

V3/V2 = 5.5  {100 + 5.5} = 0.052

Thus the ac voltage at V3 should be more than an order of magnitude less than that at V2. Usually, voltage at V3 is a few mV. Averaging and manual adjustments of the cursors are the best experimental techniques to use to measure voltage at V3.

12. The ac voltage at V3 can be now be estimated. Multiple the results obtained for the droop in V1 and the voltage ratios V3/V2 & V2/V1. The results are

V3pp = 5.1V X 0.256 X 0.052 = 0.067 V = 67 mV

V3pp = 3.6V X 0.256 X 0.052 = 0.048 V = 48 mV

These results are much larger than that observed. . Again this is a question best answered using PSPICE.

13. The division of the sum of the resistors (RF + RS) should have as its goal making the overall reduction in the ac voltage large by making the product of the two voltage expressions small. That is make V3/V1 small where V3/V1 is given by

V3/V1 = 1/{1 + (RF 2fCF)2}0.5 X {RLrd}  {RS + RLrd}

The values for RF & RS are limited to what is available on the Heathkit Resistance Substitution Boxes. There are not too many combinations to test. It is recommended that neither RF nor RS be set equal to zero. Remember that RF + RS must exceed the minimum value calculated previously to prevent zapping the Zener Diode.

The first set of values recommended for E7 may be summarized as follows:

RF = RS = 0.100 kohm = 100 ohm

RL = 511 ohm

CI = 250 F

CF = 100 F

TABLE 2 DATA FOR DC POWER SUPPLY WITH FULL-WAVE RECTIFIER

The Fluke 8000A DMM was used to measure VT. The HP54600B CRO was used to measure V1(avg), V1(pp), V2(avg), V2(pp), and V3(pp). The Fluke 8010A DMM was used to measure the dc load voltage V3(avg). The load current IL was calculated using IL = V3(avg)/RL. The current IL + IZ was calculated using IL + IZ = {V3(avg) - V1(avg)} ¸ {RF + RS}. The Zener Diode current was calculated using IZ = IL + IZ – IL.

TABLE 2 DATA FOR DC POWER SUPPLY WITH FULL-WAVE RECTIFIER

C = RL CONNECTED & NC = RL NOT CONNECTED X DENOTES INITIALS REQUIRED

% / VT / RL / V1 / V1 / V2 / V2 / V3 / V3 / IL / IL+
IZ / IZ
VAC / rms / avg / pp / avg / pp / avg / pp
V / W / V / V / V / V / V / V / mA / mA / mA
100 / 18 / C
Staff / Must / Initial / X / X / X / X / X / X / X / X / X
100 / 18 / NC
90 / 16.2 / C
X / X / X / X / X
90 / 16.2
110 / 19.8 / C
110 / 19.8 / NC
X / X / X / X / X

PSPICE (or ELECTRONIC WORKBENCH) SIMULATION: Week 2

During the second week a PSPICE EXAM on the dc power supply circuit will be conducted. PSPICE will be used to simulate the half-wave rectifier dc power supply circuit and to determine values for all the voltages and currents measured.

The 50 F capacitor was connected in parallel with the 200 F. However, it may be that a better dc power supply results by connecting the 50 F capacitor in parallel with the 100 F capacitor and using the parallel combination as the filter capacitor CF. This is one of the alternative designs that can be checked using PSPICE. Other alternative designs compatible with the worst-case design procedure such as using 150 ohms & 47 ohms for RF & RS can also be checked.

References:

1. Sedra/Smith, “Microelectronics Circuits,” 4th ed. New York: Oxford University Press, 1998, Section 3.7 Rectifier Circuits, pp. 179-191.

2. EE 352 Fall 1999 Lecture Slides on Experiment No. 8 DC Power Supply

3. EE 312 Fall 2000 Lecture Slides on Experiment No. 7 DC Power Supply

Appendix: Resistance Values Available

There are 29 Heathkit boxes that are available. The Heathkit resistance substitution box resister choices are:

Low value:

15, 22, 33, 47, 68, 100, 150, 220, 330, 470, 680, 1k, 1.5k,

2.2k, 3.3k, 4.7k 6.8k 10k

A different resistance box is used to augment the 29 Heathkit boxes (20 boxes are available).

Low Values:

10, 47, 100, 220, 470, 1k, 2.2, 3.3k, 4.7k 6.8k 10k