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John Sopczynski EE 310 Section 4

DC Power Supply Design

Introduction The goal of this experiment was to design a DC power supply. Our team would be receiving 120 Vrms oscillating at 60 Hz as our input signal. This input is supplied out of a standard wall outlet. We were tasked with transforming this sinusoidal input signal into a -15V DC output signal with a margin error of plus/minus 0.5 Volts. Figure 1:

As shown above in Figure 1, the design of the Power Supply is broken down into 5 parallel branches of components: power transformer, rectifier, filter capacitor, voltage regulator, and the load. One of our first limitations was that we were to use a 12:1:1 transformer, which would supply us with 10 Vrms at each output of the transformer. For the rectifier section, only full wave rectifiers utilizing the 1N4004, 1N4934, or 1N4935 diodes were permitted. When designing out filter capacitor, we only had a small selection of electrolytic capacitors available to us. Available capacitors were 47, 100, 220, and 470 microFarads rated at 50 V as well as 1000 microFarad rated at 16 V. The nominal voltage (Vc) entering the voltage regulator section of the design must be -24 V. Additionally, the ripple voltage of our signal at this point of the design must be 15% of -24 V or -3.6 V. For our voltage regulator, we are limited to the following Zener Dodes: 1N4728, 1N4732, 1N4735, 1N4736, 1N4738, 1N4739, and 1N4742. Alternatively, we could use the following IC regulators: LM7805, LM7815, LM7905, or LM7815 to regulate our output voltage. Additionally, our load resistor must be chosen to dissipate approximately 1 Watt. In this report, I will walk you through how our team tackled these specifications and our design process that allowed us to produce a -15 V DC power supply.

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Power Transformer: The input into the Power Transformer is 120 Vrms 60 Hz sinusoidal signal. Our power transformer is a 12:1:1 transformer which allowed us to reduce our input to 10 Vrms depending on the output of our transformer. In order to understand the outputs of our transformer, which had outputs A, B, and C; we measured the open circuit voltages Vab, Vac, and Vbc. We found that:

Vab = 10.1 Vrms Vac = 20.8 Vrms Vbc = 10.1 Vrms

In order to effectively use the transformer, we must also know what the winding resistance of our transformer is. In order to do this, we connected a 2-Watt test resistor between the terminals A and C of our transformer and measured the voltage across this resistor. We would calculate the value of the test resistor by assuming that we would dissipate one Watt from terminals A to C. We also know that our open circuit voltage is 20.8 Vrms or 29.4 V. We plugged these two values into the power formula in order to find our value for Rtest.

𝑃𝑜𝑤𝑒𝑟 = 𝑉𝑎𝑐^2

𝑅𝑡𝑒𝑠𝑡 1 𝑊𝑎𝑡𝑡 =

(29.4𝑉)^2

𝑅𝑡𝑒𝑠𝑡 𝑅𝑡𝑒𝑠𝑡 = 865 Ω

This value was calculated to be 865 ohms. Additionally, we measured the voltage of Rtest, and this was found to be 20.6 Vrms. In order to find out winding resistance (Rw), we used voltage division.

𝑉𝑅𝑡𝑒𝑠𝑡 = (𝑉𝑜𝑐)(𝑅𝑡𝑒𝑠𝑡

2𝑅𝑤+𝑅𝑡𝑒𝑠𝑡) 20.6 𝑉𝑟𝑚𝑠 = (20.8 𝑉𝑟𝑚𝑠)

856 𝑜ℎ𝑚𝑠

2𝑅𝑤+865 𝑜ℎ𝑚𝑠

Solving this equation for Rw gives us 4.37 ohms. The open circuit voltages and winding resistance values allowed us to construct a more realistic of our power transformer. Now that we have a more accurate representation of our transformer, we can now construct the rectifier section. Rectifier: Our design specifications ask for our output voltage of the filter capacitor to be -24 Volts. This means we must set our transformer in such a way that allows us to reach this value. In order to do this, we must use Vac (20.8 Vrms) as our input voltage and a full-wave bridge rectifier configuration since we are limited to two input terminals. Even with the cut-in voltages of our diodes, we will still be able to obtain the peak voltage of -24 Volts. We must be careful to ensure that the diodes are placed with correct polarity in order to give us a negative waveform. The next decision we needed to make was to decide which diode to use. We decided to use a 1N4004 diode because it has a rated Peak Inverse Voltage of 400 V. Our circuit calls for a PIV of around 30 Volts, and you can clearly see that the 1N4004’s PIV is more than twice that. For future reference, the cut-in voltage of the diode is 1.0 Volts. This design uses a load resistor R’L that we choose to dissipate 1 W. In order to find the value of this resistor, we used the Power formula again.

𝑃𝑜𝑤𝑒𝑟 =(𝑉𝑐)2

𝑅′𝐿 𝑅′𝐿 =

(−24𝑉)2

1 𝑊𝑎𝑡𝑡 R’L = 576 Ohms

*Vc equals -24 volts because that is the peak voltage we wish to obtain We do not have a 576 Ohm Power resistor in our stock room. In order to combat this problem, we wired a 125 Ohm and 430 Ohm Power resistor in series with each other in order to create an equivalent 555 Ohm Power Resistor.

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Upon observing the output voltage across R’L, we found the peak output voltage (Vm) was equal to-27.6 Volts. Figure 2:

The Figure 2 shown above shows the voltage as measured across the R’L resistor. Figure 3:

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Figure 3 shown on the previous page shows the output voltage as measured across R’L when compared with the output voltage of the transformer, which is our rectifier input signal. The transformer voltage signal is shown in blue while the R’L voltage is shown in black. You can clearly see the effect of the full wave rectifier transforming the input sinusoidal signal into a completely negative waveform. The difference in the peak voltage of the transformer voltage and the R’L voltage is due to the cut-in voltages of the diodes. We measured the difference and found it to be around 2 Volts, which is consistent with the cut-in voltage of a 1N4004 diode. It is two volts since we have two diodes in series when using a full wave bridge configuration. Filter Capacitor: The next component of our design is the filter capacitor. This capacitor is to be placed in parallel with the output resistor of the rectifier circuit. This capacitor will allow us to filter our waveform into a signal that more closely resembles a DC signal. We want the voltage across this capacitor (Vc) to have a maximum value of -24 volts. With any filter capacitor, there is a ripple voltage due to the nature of a capacitor. The ripple voltage is allowed to be a maximum of 15% of our Vc (-24 Volts) or -3.6 Volts. We had to calculate the capacitance needed in order to limit our ripple voltage to -

3.6 Volts. The formula we used is: 𝐶 = 𝑉𝑚

2𝑓(𝑅′𝐿)(𝑉𝑟) where Vm = peak voltage = -27.6 Volts, f = frequency

= 60 Hz, R’L = load resistor of rectifier = 576 Ohms, and Vr = ripple voltage = -3.6 Volts. Plugging in the appropriate values stated above, we found our minimum Capacitor value to be 110.9 microFarads. The capacitor we chose to use was a 220 microFarad capacitor rated at 25 Volts. This capacitor will ensure that the ripple voltage of our output will not exceed 3.6 Volts. The voltage across the capacitor is the output voltage of this circuit (Vc). Vc is also the unregulated voltage of our design since we have yet to send this voltage signal through a voltage regulator circuit. The unregulated power supply schematic is as is shown below in Figure 4: Figure 4:

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Upon completion of our unregulated power supply, we made several key measurements that will help us understand the operation and functionality of the circuit. The first one we made was that we measured Vout = Vc with a DMM and we found the average or DC value to be -26.1 Volts. Additionally, we plotted the total output voltage using the oscilloscope. You can see the graph below in Figure 5. Figure 5:

As you can see from the graph. Our Vm = -27.2 Volts, our VL = -25.5 Volts and our Vr = 1.70 Volts. All of these values satisfy the specifications of our design. We also plotted the AC transformer secondary voltage waveform with the output voltage waveform still on the plot in order to compare the two waveforms which is shown in Figure 6. Figure 6:

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The transformer ac waveform is in blue while the capacitor waveform is in black. As you can see, we have greatly manipulated our signal from its original sinusoidal waveform to something that very closely resembles a DC signal. Additionally, we want to analyze the current across the diodes (Id). We first calculated the predicted value and then we measured an actual value by utilizing a 1 ohm resistor in series with one of the Diodes. By taking the voltage across the 1 ohm resistor, we were able to plot the current waveform on the oscilloscope. This is due to the fact that current = voltage/resistance, and since resistance is 1, our current now equals our voltage. When we measured the current of the of the diode, we found it to be -260 mA (260mV/1Ohm). You can see our measured waveform below in Figure 7 Figure 7:

In order to calculate a predicted value for the diode current, we used the following equation:

𝐼𝑑 =𝑉𝑚

𝑅𝐿(1 + 2𝜋√

𝑉𝑚

(𝑉𝑟)2)

Using Vm = -27.2, RL = 576 ohm and Vr = (-1.7)V, we found our Idmax to equal -886 mA. Our measured current is -260 mA, which is a large difference. Shown below is a MultiSim simulation of our circuit and a oscilloscope capture using a 1 Ohm resistor to create a voltage equal to current. Additionally, to verify that we were obtaining a correct value, we simulated the design on MultiSim in a similar fashion to how we tested for the Id waveform in our circuit. You can see our unregulated voltage supply design modified to test for Id in Figure 8 on the next page. In Figure 9, which is also on the next page, you can see the Id waveform as measured by the MultiSim oscilloscope.

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Figure 8:

Figure 9:

Our Voltage scale for the oscilloscope capture was 50 mV/Division. Taking this into account, you can clearly see that the maximum current is around -275 mA (-275V/1 Ohm). This very closely resembles our actual, measured value of Id. One of the main reasons for the difference between our calculated and observed value is due to internal resistance. The diodes and other devices do have an internal resistance that we are not taking into account. These internal resistances can greatly affect the amount of current flowing through a diode. Additionally, we are adding a 1 Ohm resistor to the circuit to create a waveform on the oscilloscope. This

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resistor will create a small voltage drop as well, which will also contribute to the reduction of the max current. In order to ensure that our power supply operates in a safe manner, we would like to calculate the power dissipation of each device in our power supply and compare that with the rating of each device. This will allow us to establish the safety margin of operation for each component. Figure 10 shows a table with various values of each component in our power supply in order to analyze their safety margins. Component Voltage Current Power Dissipation Rating Safety Margin

430 Ohm Resistor -20.2 V 47 mA 949 mW 2 Watts 1.05 Watts

125 Ohm Resistor -5.8 V 46 mA 269 mW 2 Watts 1.731 Watts

220 uF Capacitor -26.2 V 0 A 0 W 25 Volts -1.2 Volts

Diode -1 V -260 mA 260 mW 400 Volts 371 Volts

As you can see, most of our components have a large safety margin. However, the capacitor does not have a large safety margin. This is due to the fact that we were supplied with a capacitor rated at 25 Volts instead of the stated 50 Volts. The 50 Volts capacitor would have been ideal in ensuring a larger safety margin for our design. Voltage Regulator: Now that we have completed the unregulated power supply, we must find a way to regulate the voltage level in order to eliminate the ripple voltage even more. We wish to regulate our voltage down to -15 Volts. In order to do this, we must use to diodes in series, the 1N4736 and the 1N4738 diode. We must use them in series because the 1N4736 is rated for 6.8 Volts and the 1N4738 is rated for 8.2 Volts. By putting them in series, we are able to combine their voltages and obtain a final regulation voltage of -15 Volts. One of the most important things we must do for the regulator is identify our maximum Zener Diode current. We do this by utilizing the power formula:

𝐼𝑧𝑚𝑎𝑥 = 𝑃𝑧

𝑉𝑧 𝐼𝑧𝑚𝑎𝑥 =

0.5 𝑊𝑎𝑡𝑡

15 𝑉𝑜𝑙𝑡𝑠 Izmax = 33.3 mA

Whenever we choose our resistor Ri which is in connection between the capacitor and the regulator circuit, we need to use the Izmax value in order to ensure a safety margin while operating our circuit.

𝑅𝑖 = |𝑉𝑐−𝑉𝑧|

𝐼𝑧𝑚𝑎𝑥 𝑅𝑖 =

⌈25.9−15⌉

33.33 𝑚𝐴 Ri = 327 Ohms

We also need need to pick a load resistor value that will give us a minimum current of 30% times Izmax. This miniumum current is equal to 10 mA. In order to pick the correct resistor value, we used the

following formula: 𝑅𝐿 = 𝑉𝑧

(0.7)𝐼𝑧𝑚𝑎𝑥=

15 𝑉

23.33 𝑚𝐴= 643 𝑂ℎ𝑚𝑠

This resistor value will be called the full load. In order to calculate the regulation percentage, you would use the following formula:

𝑅𝑒𝑔𝑢𝑙𝑎𝑡𝑖𝑜𝑛 % =𝑉(𝑛𝑜 𝑙𝑜𝑎𝑑) − 𝑉(𝑓𝑢𝑙𝑙 𝑙𝑜𝑎𝑑)

𝑉(𝑛𝑜 𝑙𝑜𝑎𝑑)=

(−14.95) − (−14.68)

−14.95= 1.81%

This means that there is not much of a change in our voltage level when you vary the load resistance.

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With our full load still attached to the regulator, we observed the total output of our circuit, which can be seen below in Figure 11. Figure 11:

The very small peak to peak value of 280 mV show that there is almost no ripple and we have created a strong DC signal. In order to ensure that we have eliminated the ripple seen in the capacitor voltage, we compared the above graph with the AC component of Vc, which is shown below in Figure 12: Figure 12:

The AC component of VC is seen in the darker blue and the AC component of our output signal is seen in the light blue. Since our scaling is the same for both graphs, we can clearly see a significant decrease in the ripple from Vc to our Vout. By comparing the peak to peak values of these two voltage signals, we were able to see that the ripple of Vc is 25.5 times as big as Vout. This is a large number and shows just how significantly we decreased the ripple in our signal. The final Zener Diode regulator circuit design was construct in MultiSim and can be seen in Figure 13 on the next page.

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Figure 13:

Not only did we implement the regulator with Zener Diodes, but we implemented it with IC Regulators. An IC regulator is one of the easiest ways to provide a regulated DC signal from an unregulated DC supply. Since my group was trying to achieve a DC signal of -15 V, we chose the LM7915 IC regulator. This will allow us to get the needed -15V. Due to the fact that we were switching our regulation method, we needed a new load resistor. We wanted the load current to equal our maximum Zener Diode current from the previous part. This means that IL = 33.33 mA and VL = 15 V. Using Ohm’s law, we can conclude that the needed load resistor is 450 Ohms. In the same fashion as the previous part, we want to observe the % Regulation. In order to do this, we need to measure the No Load Voltage and the Full Load Voltage. These values are -15.12 volts for both which gives us a % Regulation of 0%. What this means is that the IC Regulator is very adept at handling varying loads. We want to look at the total output voltage with our oscilloscope in order to ensure that we are getting a -15 Volt DC output signal. The graph shown in Figure 14 shows that we have obtained the needed signal. Figure 14:

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In order to ensure that we have effectively eliminated the ripple, we compared the AC component of the output with the AC component of the voltage across the capacitor, which can be seen below in Figure 15. Figure 15:

The blue trace represents the AC component of our output and the light blue trace represents the AC component of our capacitor voltage. As you can clearly tell, we have effectively reduced our ripple and obtained a strong DC signal. By comparing our peak to peak values, we are able to see that we reduced the ripple by a factor of 31, which is a significant decrease. As easy the IC Regulator is to implement, it is not perfect. If you increase the load current enough, a ripple will begin to form in the output signal. In order to identify the point where this began to occur, we incrementally decreased RL from our original value of 450 Ohms down to a value of 360 Ohms, which is where we noticed a small ripple beginning to form in our output signal. You can see the ripple in the graph shown below in Figure 16. Figure 16:

d

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The blue trace still represents out output signal. You can see a ripple forming by all of the noise surrounding the trace. It is not a clearly defined like before. We maintained the same scale and retained Vc in order to provide some reference. What happens as IL is increasing is that the current through the regulator is decreasing, which in turn, decreases the voltage that the IC regulator is actually regulating. This is what allows the ripple to leak through and appear in our output. Now that we have ensured proper operation of our IC regulator, we construct a model on MultiSim, which can be seen below in Figure 17. Figure 17:

For our final Vout, dependent on the design, we got a Vout of -15.2 Volts (IC regulator) and Vout = -14.9 Volts (Zener Diode). Both of these output voltages satisfy the specification of -15V +/- 0.5 Volts. Utilizing the specifications given to use throughout our design is what allowed us to ensure that our output voltages would meet the required specifications. CONCLUSION: In this design, we learned how to design a power supply that would supply a specified voltage level across a load. We wanted to attain a -15V DC signal as our output, however, the skills learned in this design would allow us to attain 5V, 15V, -5V, as well as our -15V. In order to obtain an output signal that satisfies our specifications, it was vital that we ensure our calculations were accurate throughout the entire design process. We did receive some assistance in our design due to the specifications telling us Voltages, Currents, and Resistor values to have at certain stages of our design dependent on which design we were trying to accomplish. It is vital that you use power resistors since we have such a large amount of power being dissipated by these resistors, that you will blow a capacitor if you do not ensure proper power dissipation. The skills we learned in this design will be greatly beneficial in the coming years. Many Op Amps utilizing +/- 15 Vdc power rails. Additionally, many digital logic devices use +/- 5Vdc power supplies in order to operate. The fact that we know how to construct our own power supplies will greatly benefit us if we do not have access to an already constructed power supply. In conclusion, I now have learned a very important skill that will be greatly beneficial in the coming years. Additionally, I also realized the importance of accurate calculations and documentation throughout a design in order to ensure that we met all of our specifications.