Ankitha Miryala
Section 7
PSU ID: 950603219
TA: Alexander Cocking
Formal Design Report
EE 310 Electronic Circuit Design
Fall 2013
Experiment 3
Power Supply Design Project
Introduction:
This lab is designed to learn the making of the power supply devices.
During week one we designed the input transformer, the rectifier
and the filter circuits of a simple linear regulated power supply.
During the second week of lab we designed a zener diode voltage
regulator and compared it with 3-terminal IC voltage regulator. The
purpose of this lab is to build a complete AC to DC power supply
that is capable of supplying filtered and regulated DC voltage to a
load.
Design specifications and inputs
Our team was assigned Design C with power supply specifications as
follows:
DC Output voltage Vo = Vl =+15± 0.5 V
DC Voltage at the regulator input, VC =24V
Vr = 15%Vc
RL to dissipate 1 watt
AC input voltage, frequency: 120Vrms ±5%, 60Hz
Block diagram of a Power Supply:
CIRCUIT DESIGN AND REASONING
Schematic: Power Transformer
Schematic: Rectifier Circuit
Schematic: Voltage Regulator with zener diode
Circuit reasoning for the type of rectifier
Here, we choose the “Full-wave Bridge Rectifier” for Design C
Because of the design constraints we used the full-wave rectifier
model. And we choose bridge rectifier over the center-tapped
model because we had to obtain VC = 24V as the voltage across the
capacitor which could be obtained by a full-wave bridge rectifier
model only. Through the center-tapped model the maximum
voltage of only 12V can be obtained.
VM = 𝑉𝑝𝑒𝑎𝑘 −2𝑉𝛾 is the maximum voltage obtained for a bridge model.
Schematics of the unregulated power supply: Multisim
Initial Power Supply Circuit
Rw
V1
T1
4.016Ω
120 Vrms
60 Hz
0°
Rtest
416Ω
12:2
Rw_
4.016Ω
Rectifier Schematic
Rw
4.016Ω
V1
D1
D2
1N4004G
1N4004G
T1
120 Vrms
60 Hz
0°
Rtest
416Ω
12:2
Rw_
4.016Ω
D3
D4
1N4004G
1N4004G
C1
220µF
Voltage Regulator Schematic
Rw
4.016Ω
V1
D1
D2
1N4004G
1N4004G
T1
120 Vrms
60 Hz
0°
Rtest
416Ω
12:2
Rw_
D3
D4
1N4004G
1N4004G
4.016Ω
C1
Probe1
I(p-p): 390 mA
I(rms): 113 mA
I(dc): 41.2 mA
Freq.: 120 Hz
220µF
Ri
312.2Ω
D7
D6
D5
1N4733A 1N4733A 1N4733A
R_L
643.5Ω
Schematic of an IC Voltage Regulator
MC7815 IC voltage regulator chip
Explanation of the regulator circuitry
For the voltage regulator circuit we used the voltage across the
capacitor, VC as the input voltage for the regulator circuit. . We
choose 1N4744 diodes, which are of 5.1V. We used three of them in
series. The three diodes and a resistor, Ri is connected in series with
each other and parallel to the capacitor. Finally a load resistor, RL is
placed in parallel to the series arrangement of the three diodes. This
voltage regulator circuit is used to eliminate the ripple in the output
of the filter capacitor and produces an output with almost no ripple.
For the IC regulator we replace the diodes with the MC7815 IC
voltage regulator chip. Here, we observe that there is almost no
ripple as we see a straight line. Also, as we change the current (or
the resistance) we see that the ripple does not change and remains
constant.
SUPPORTING ANALYSIS
Model Transformer
Rw
4.016Ω
V1
T1
120 Vrms
60 Hz
0°
12:2
Rw_
4.016Ω
Calculate PIV for rectifier & choose diode: Rectifier
We choose the 1N4004 diode because it has a PIV of 400V,
which is more than two times the value of the PIV required for the
experiment.
c. For design C, the Bridge-wave Rectifier looks like below:
d. Oscilloscope capture attached.
The output waveform is displayed and we find the peak-peak
voltage,
VM = 27.2V
Now, Vr = 15% VC =
15
100
∗ 24 = 3.6𝑉
Therefore, the ripple voltage should be maintained at 3.6V.
Choose resistor and calculate filter capacitor
Here, RL’ =
(𝑉𝑐)2
1𝑊
=
242
1
= 576 Ω is the Actual value of the resistance.
We choose RL = 500 Ω from the stock room.
𝐶=
𝑉𝑀
2𝑓𝑅𝐿′ 𝑉𝑟
=
27.2
2
∗ 60 ∗ 576 ∗ 3.6 = 109.31µ𝐹
We used a capacitor of 220 µ𝐹
Average capacitor DC voltage = 𝑉𝐶 = 𝑉𝑀 −
Overall Measurements
𝑉𝑟
2
= 27.2 −
3.6
2
= 25.4𝑉
a. Average DC output using DMM = VDC = 25.3V
b. Oscilloscope capture saved, wherein:
VM = 26.6V
VL = 24.2V
Vr = VM - VL = 2.4V
The secondary voltage VS and the capacitor voltage VC and
simultaneously compared on the same graph in the oscilloscope
capture attached:
Predict maximum diode current, iD
We used a 1.1Ω resistor of 2 watt in series with one of the diodes in
the bridge-rectifier circuit and measured the current iD in the loop.
From the oscilloscope capture, ID, MAX =
=
𝑉𝑀
𝑅𝐿
𝑉
[1 + 2𝜋√2𝑉𝑀 ]
27.2
𝑟
27.2
[1 + 2𝜋√2∗3.6]
576
= 0.623 mA
ID, MAX actual < ID, MAX theoretical
176 mA < 623 mA
Here, the actual value is much less than the theoretical value
because the formula for the theoretical value has many
approximations based on the ideal conditions of the device
component. But in reality the ideal factors are not satisfied and also
because there will be experimental and measurement errors to a
certain extent.
Multisim simulation of diode current
Regulator parameters: IZmin, IZmax, Ri, RL
Verified the circuit is working properly
V0 = +15± 0.5 V (under no load conditions RL = ∞)
a. We choose 1N4744 diodes, which are of 5.1V. We used 3 of them.
b. IZ, MAX =
c. Ri =
𝑃𝑍
𝑉𝑍
𝑉𝐶 −𝑉𝑍
𝐼𝑍,𝑀𝐴𝑋
=
=
0.5𝑊
𝑉𝑍
=
25.4−15𝑉
33.3𝑚𝐴
0.5𝑊
15𝑉
= 33.33𝑚𝐴
= 312.3Ω
d. 𝐼𝑍,𝑀𝑖𝑛 = 30% ∗ 𝐼𝑍,𝑀𝐴𝑋
= 0.3*𝐼𝑍,𝑀𝐴𝑋
= 0.3*33.3 mA = 9.99 mA
𝑅𝐿 =
𝑉𝑍
15𝑉
=
= 643.5Ω
𝐼𝑍,𝑀𝐴𝑋 − 𝐼𝑍,𝑀𝑖𝑛 33.3 − 9.99𝑚𝐴
DATA:
Transformer open circuit voltages and Rw:
1. Power Transformer
a. Measurements of the terminal voltage of the transformer using
the DMM:
VAC = 20.4 Vrms
VAB = 9.9 Vrms
VBC = 9.9 Vrms
b. Now, VAC = 20.4 Vrms = 28.846V
Assume Rtest in the circuit
Current across Rtest = I =
Power = IV = 28.85𝑉 ∗
28.85𝑉
𝑅𝑡𝑒𝑠𝑡
28.85𝑉
𝑅𝑡𝑒𝑠𝑡
= 1𝑤𝑎𝑡𝑡
Rtest = 832.2 Ω
We used Rtest = 810 Ω from the stock room.
Now, we measure VAC with and without the Rtest
Without Rtest
VAC = 20.4 Vrms
With Rtest
VAC = 20.2 Vrms
Current in the circuit =
20.2𝑉𝑟𝑚𝑠
810Ω
= 24.9mA
Voltage across 2Rw =0.2 Vrms
So, 2𝑅𝑤 =
Rw =
8.032
2
0.2𝑉𝑟𝑚𝑠
24.9𝑚𝐴
= 8.032Ω
= 4.016Ω
Calculate & compare regulations with IC voltage reg.
Schematic:
We have constructed the circuit accordingly.
The capacitor voltage will serve as the secondary voltage for the
zener diode now.
The schematic for the full circuit on the breadboard is:
The Ri and RL for this circuit are according to the above-calculated
values
For
Calculated
Measured
Ri
312.3 Ω
310 Ω
RL
643.3 Ω
620 Ω
With Load
14.8V
Without Load
15.2V
e.
Zener
Output Voltage
Oscilloscope captures attached
%Regulation =
𝑉𝐿 (𝑛𝑜 𝑙𝑜𝑎𝑑)−𝑉𝐿 (𝑓𝑢𝑙𝑙 𝑙𝑜𝑎𝑑)
𝑉𝐿 (𝑛𝑜 𝑙𝑜𝑎𝑑)
∗ 100 =
15.2−14.8
15.2
∗ 100% = 2.63%
Task 2: IC Regulator
a. We pick RL such that
𝑉𝐿
𝑅𝐿
𝑅𝐿 =
= 𝐼𝑍,𝑀𝑎𝑥
𝑉𝐿
𝐼𝑍,𝑀𝑎𝑥
=
15𝑉
= 450.5Ω
3.3𝑚𝐴
b.
IC Regulator
Output Voltage, VL
%Regulation =
With Load
14.8V
𝑉𝐿 (𝑛𝑜 𝑙𝑜𝑎𝑑)−𝑉𝐿 (𝑓𝑢𝑙𝑙 𝑙𝑜𝑎𝑑)
𝑉𝐿 (𝑛𝑜 𝑙𝑜𝑎𝑑)
Without Load
14.9V
∗ 100 =
14.9−14.8
14.9
∗ 100% = 0.67%
c. Oscilloscope capture attached
d. Comparison capture attached
e. As we increase IL (i.e.) by reducing RL down to even 20% in the IC
Regulator we observe that the ripple does not change must. Thus
satisfying the purpose of the IC Regulator (i.e.) the ripple voltage
does not change and remains almost constant.
The %Regulation of the IC Regulator is very small compared to the
Shunt regulator. Also the difference in the ripple with and without
load for the IC regulator is therefore very hard to observe.
Plot of rectifier output
Plots of transformer outputs and Vc
Diode current waveform
Fully rectified AC output and ripple voltage of filter
capacitor
Zener/shunt regulator output waveforms VC and Vripple
Shunt regulator output waveform w/ ripple
IC regulator voltage and VC
IC regulator voltage
Power dissipations & margins
𝑉2
𝑃=
𝑅
Across RL , Power =
Across Ri , Power =
14.82
643.5
24.32
312.5
= 0.3404𝑊
= 1.89𝑊
These are the power dissipations for the circuit we constructed. The
elements we used in our circuit had a power rating which was much
higher than these values in order to prevent breakdown in the
circuit.
DISCUSSION
Power considerations
For all the elements in the circuit (i.e.) the resistors, capacitors and
diodes we have made sure that we used the ones which have a
power rating higher than their calculated or specified power
dissipation values. For we designed our circuit for 0.5W diodes but
have used 1W power rated element taking extreme conditions into
consideration.
Error calculations and reasons
Power supply output error:
𝑒𝑟𝑟𝑜𝑟 =
25.3𝑉 − 24𝑉
∗ 100% = 5.41%
25.3𝑉
Reason: Due to inaccuracy in the measurements by the
oscilloscope and the internal faults in the amplification
value of the transformer we used.
Zener Regulator Voltage:
𝑒𝑟𝑟𝑜𝑟 =
15𝑉−14.8𝑉
15𝑉
∗ 100% = 1.33%
Reason: This is because the formal gives the value of the
ideal diode characteristics but in reality the voltages in the
saturation region are not straight lines instead they have a
slight slant upwards.
IC Regulator Voltage:
𝑒𝑟𝑟𝑜𝑟 =
15𝑉 − 14.9𝑉
∗ 100% = 0.66%
15𝑉
Reason: The error is very small as it is supposed to be
because there is hardly any ripple. And the voltage with
load and without load was very hard to differentiate even
during the lab. This small error is probably due to the internal
characteristics of the IC we used.
Comparison of outputs to design specifications
According to our design C, we were supposed to obtain an output
of +15± 0.5 V and we were successful in obtaining values, which fall
very well within the range.
For the Zener Voltage regulator we obtained an output of 14.8V with
a small error in the percentage regulation as we expected.
For the IC Voltage regulator we obtained an output of 14.9V with a
hardly any error in the percentage regulation as it is supposed to be.
Therefore we can say that an IC regulator will be the better choice
for a voltage regulator that gives us an output with almost no ripple.
SUMMARY AND CONCLUSIONS
Finally we were successful in designing the required voltage
regulator that gives the filtered and regulated DC voltage of 15V
from the given AC voltage of 120Vrms and a frequency of 60Hz.
We have first passed it through a power transformer to obtain a 24V
output. Then we rectified the signal using a full-wave bridge rectifier
to obtain only a positive-side sine wave. Then we pass it through the
filter capacitor to obtain a particular voltage maximum boundary.
Then we use a voltage regulator to produce an even DC voltage
that is a constant output. Conclusively this can be done best by an
IC voltage regulator that gives a constant DC output voltage with
almost no ripple.
Pre-labs and lab notebook pages attached