Design of Power Supply & Selection of Component Value

Carolyn R. Jones

November 22, 2000

Class #: TuAM10

Partner: Nick Soda

EE 312

Introduction

In this experiment that we performed in the lab, we designed, constructed and tested a DC power supply. The power supply includes transformers, diodes and a filter. All were investigated in previous experiments. DC Power Supply (Half-Wave Rectifier Circuit) provides an opportunity to perform a worst-case design. Since design is the distinguishing characteristic of the engineering profession, ABET recommends the teaching of design at every opportunity. Unfortunately, there is no easy way to teach design. Design, by its very nature, requires iteration. Usually, several designs are necessary.

Design of Power Supply & Selection of Component Value

Shown in Figure 1 of the EE 312 & EE 352 Lab Manual on p. 62 is the power supply circuit. The circuit consists of an input section that includes an isolation transformer, a half-wave rectifier and a capacitor CI. Not shown is a ground connection to one terminal of the output side of the isolation transformer. The rms ac voltage at the output of the isolation transformer is VT. The voltage across the capacitor CI is measured at Node 1. The next section is the RC low-pass filter consisting of RF & CF. . The voltage across the capacitor CF is measured at Node 2.The third section is the Zener regulator section consisting of a current limiting resistor RS and a Zener Diode. The load resistor is labeled RL. The voltage across the load resistor RL is measured at Node 3.

Shown in Figure 2 of the EE 312 & EE 352 Lab Manual on p. 62 is the voltage waveform across the capacitor CI at Node 1. The voltage waveform consists of two parts: (1) sections of the positive part of sinusoids with a peak value 2 X VT; (2) sections showing drop that is approximated as a straight line. (Note: subtracting the dc voltage drop of ~ 0.7 V across the rectifier diode would lead to a more exact design.) The equation for the drop is

where R is the appropriate resistor that controls the discharge of capacitor CI. What value should be used for R will be discussed later. It is important to make the time constant RCI sufficiently large so that the percentage drop is not too large. A 20% to 30% droop is usually acceptable for a half-wave rectifier. Let t be the time that the capacitor CI discharges before being re-charged on the next positive part of the sinusoid. The value for t shown in Figure 2 is t = 1/f where f = 60 Hz. Actually the value for t is less than 1/f. For a specific droop the value for t can be calculated. For a 20% droop the value calculated for t was 0.9 X 1/f. The percentage droop is defined as 100 X {V1/ (2 X VT)} where V1 is the maximum drop in voltage v1(t). For a 20% to 30% drop, Equation (3-1) can be approximated as

The droop V1 is given by substituting t for t in Eq. (3-2) and subtracting the result from the peak voltage 2 X VT. The result is given by

The dc voltage at Node 1 is denoted by V1DC and is equal to the average value of the voltage across the capacitor CI and that is given approximately by

If the design value for the droop is 20%, then the dc voltage at Node 1 is given by

Equation (3-5) neglects the dc voltage drop across the rectifier diode. To account for that voltage 0.7 V should be subtracted from the peak value 2 X VT and that result multiplied by 0.9.

For a 20% drop Equation (3-3) indicates that

The values available for the capacitor CI are 200 F, 100 F, and 50 F. It was decided to use the 100 F capacitor for the RC filter section and to use the other two capacitors in parallel for CI. For CI = 200 F + 50 F = 250 F a value for the total resistance R can be calculated using Eq. (3-6) with t = 1/f where f = 60 Hz as follows

The resistor R is not quite equal to sum of the resistors RF + RS + RL on account of the Zener Diode in parallel with RL. If the dc resistance RZ of the Zener diode is defined to be RZ = VZ/IZ where VZ & IZ are the Zener diode dc voltage and current, then the resistor R is given by

The dc load current was specified to be 50 mA. For a 12 V Zener diode the appropriate value for RL was determined to be

The Zener diode dc current was selected to be IZ = 21 mA. For a 12 V Zener diode the appropriate value for RZ was determined to be

Using the values calculated with Equations (3-9) & (3-10), the value calculated for the RZRL was RZRL = 571240 = 170 . Inserting the value for RZRL = 170  into Eq. (3-8) yields a value for RF + RS.

Note: In the lecture slides the value obtained was RF + RS = 138 .

The last step is to decide upon the division of RF + RS in to separate parts. The ac voltage ratio V2ac/V1ac for the low-pass filter at a frequency f is given by

The RC low-pass filter was assigned a voltage reduction factor of 1/10 at a frequency of 60 Hz. Equation (3-12) was set equal to 1/10 and solved for RF. The result is given by

The value for RS is given by

The values available included 100 , 67 , and 47 . A decision was made to use RF = 100  and RS = 47. A value RS = 67  might have been a better choice. The design choices are listed in Table 3-1.

TABLE 3-1 DESIGN & ACTUAL VALUES FOR CIRCUIT COMPONENTS

Figure 3-1 Power Supply Circuit.

Figure 3-2 Voltage Waveform in Input Section.

These are the Waveform that we tested at these three points. The values that were taken for these three are the average voltage and the voltage peak to peak. The values for the output waveforms for the values 1, 2, and 3 are found in Table 1.

The values that the components had are shown in the figure below.

From these values we calculated and found the voltage peak to peak and the average voltage. From the voltages we were able to calculate the values of IL, IZ + IL, and IZ.

TABLE 1

This is the measured value for the DC power supply with full wave rectifier.

C = RL CONNECTED & NC = RL NOT CONNECTED.

Table 2

PSPICE was also used to simulate the half-wave rectifier dc power supply circuit and to determine values for all the voltages and currents measured.

The worst case that was found was

110% AC VZ1BD = 10.8VRLNCIZ = < max IZ

90% AC VZ1BD = 13.2VRLCIZ > 0

SUMMARY

In this experiment that we performed in the lab, we designed, constructed and tested a DC power supply. The power supply includes transformers, diodes and a filter. All were investigated in previous experiments. DC Power Supply (Half-Wave Rectifier Circuit) provides an opportunity to perform a worst-case design. Since design is the distinguishing characteristic of the engineering profession, ABET recommends the teaching of design at every opportunity. Unfortunately, there is no easy way to teach design. Design, by its very nature, requires iteration. Usually, several designs are necessary.

The initial power supply design might not perform well for a +10% & -10% variation in ac voltage and a 100% variation in dc load current. The Zener Diode could burn up at 110% ac voltage when the load resistor is removed. The Zener Diode could cut off at 90% ac voltage when the load resistor connected. . A second design was done in fall 1999 to deal with these problems. The design procedure developed is called a worst-case design.

We created a table of design values that includes values for the following components: RL, RF, RS, CI, & CF. We were required to show your table to the Staff before we were allowed to assemble the dc power supply. We assembled the entire dc power supply and test it as an assembly at 100% ac voltage (18 VAC) with the load resistor connected. We measured the dc voltages and ac peak-to-peak ripple voltages. After that we had to demonstrate to the staff our results. The staff witnessed all six values of voltages and sign off on each. They also check and initial calculations for IZ + IL, IZ, & + IL.

We repeated the experiment with the load resistor removed. We also repeated it with a 90% ac voltage (16.2 VAC) with the load resistor connected. One of the staff witnessed the DC voltage and AC peak-to-peak ripple voltage at the node 3 and sign off on each in a space allotted. The also checked and initialed the calculations for IZ + IL, IZ, & + IL.

We repeated the experiment again this time at a 90% ac voltage with the load resistor removed. Also at 110% ac voltage (19.8 VAC) with the load resistor connected. And again at 110% ac voltage with the load resistor removed. The staff witnessed the DC voltage and AC peak-to-peak ripple voltage. They also checked and initialed calculations for IZ + IL, IZ, & + IL.

We repeat 110% ac voltage with the load resistor removed. Staff witnessed again the DC voltage and AC peak-to-peak ripple voltage. They also checked and initialed calculations for IZ + IL, IZ, & + IL.

We entered our result from in the table for the values that we got from the DC and AC ripple voltages. We also entered in the table the percent DC load voltage regulation. The percent DC load voltage regulation is calculated using the dc load voltage obtained in above 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.

References

1. K. Etemadi, Laboratory Manual for EE 312 Basic Electronic Instruments Lab. & EE 352 Introductory Electronic Circuits Lab. Buffalo (NY): 1999, pp. 61-70.

2. J. Whalen, Lecture 8: Slides on DC Power Supplies. Buffalo (NY): 1999. (Slides available at htttp://www-ee.eng.buffalo.edu/~whalen/ee352

3. S. Wolf & R. F. M. Smith, Student Reference Manual for Electronic Instrumentation Laboratories. Englewood Cliffs (NJ): Prentice-Hall, 1990, 284-291 & 344-351.