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The following is the procedure used to design the 220VAC to 28VDC power supply that outputs a maximum of 5A using a transformer and rectifier.
The design was broken down to the following parts:
The transformer. The bridge rectifier. The smoothing capacitor. The voltage regulator.
The following formula was used to calculate the secondary voltage rating of the transformer: Vrms * sqrt(2) = Vout + Vr(p-p) + 2*Vdiode; where, Vrms is the rms voltage output of the transformer. Vout is the output voltage of the power supply(28VDC). Vr(p-p) is the peak to peak ripple voltage of the capacitor output. The desired value used is 2V. Vdiode is the forward biased voltage drop across the rectifier diode. At 5A, the voltage drop across the KBU1510 bridge recifier(refer to the bridge rectifier section) diode is 1.0V(as per the datasheet). Inserting the given values into the formula,the transformer to be used should hav a minimum rms voltage output of 23Vrms. A 220:24Vrms transformer was chosen for this design.
The bridge rectifier to be used had to meet the following requirements:
- It should have a minimum peak inverse voltage (PIV) rating of 68V(24Vrms * sqrt(2) * safety factor of 2).
- It should have a minimum maximum current rating of 10A(5A * safety factor of 2).
- It should have a minimum maximum rms forward voltage of 48V(24V * safety factor of 2). Using these specifications, the KBU1510 bridge rectifier was chosen. It has a PIV rating of 1000V, a maximum current rating of 15A and a maximum rms forward voltage rating of 700V.
To determine the maximum heat dissipated by the bridge rectifier, operation of the bridge rectifier at full load i.e. maximum current, was considered. The maximum power dissipated by each diode when forward biased is given by the voltage drop across the diode(1.0V) multiplied by the load current(5A), giving a value of 5W. Given that two diodes will be forward biased at any instance of a half sine wave cycle of the transformer output voltage, this value is multiplied by two to give 10W. To determine the maximum temperature raise above the ambient temperature when in full load of the bridge rectifier, the thermal resistance per element(2.7C/W) was multiplied with the maximum power dissipated by the bridge rectifier per half sine wave cycle(10W), getting the value 27C. Taking the ambient temperature as 25C, the overall temperature of the bridge rectifier in theory will increase up to 52C at full load. It was therefore recommended to attach a heat sink to the rectifier.
The value of capacitance for the required capacitor was calculated as follows: C = Iload / (2 * f * Vr(p-p)) where Iload is the maximum load current(5A), f is the frequency of the transformer signal(50Hz) and C the capacitance value of the capacitor. The recommended capacitance value was obtained as 25mF. Alternatively, using an arbitrary capacitance value of 10mF would result in a peak to peak voltage rating of 5V. Analyzing the PSU simulation designs with different capacitance values shows how the ripple voltage on the output voltage behaves. The voltage rating of the capacitor had to be greater than 32V(24Vrms * sqrt(2) - 2*Vdiode(1.0V)) without considering the load regulation. Taking a load regulation of 10%, this value rises up to 36V(32 * 110%). The current rating considered the ripple current of the capacitor which was taken as twice the maximum load current(10A). Therefore, a capacitor with a capacitance value between 10-25mF, a voltage rating of 50V or larger and a current rating larger than 10a can be used for this power supply.
The voltage regulator should be able to accept an input voltage within the range of 28-32V and output a steady 28VDC voltage. Considering the drop out voltage of 4V(Vin(max)(32V) - Vout(28V)) and a current of 5A, the power to be dissipated if a linear regulator were to be used would be 20W. When this value was multiplied with the Theta-JA rating of most linear regulators, it was found that the temperature of most linear regulators would rise to above 100C, making them unreliable. Given the large dropout voltage, these linear regulators would also have a low efficiency. Therefore, it was recommended that a buck converter be used as a voltage regulator. It should have a power rating of at least 140W(28VDC * 5A), be able to take in the input values of the rectifier system(28-32VDC) and output a steady 28VDC. It is highly recommended that it be attached to a heat sink. Several such converters are available on Aliexpress.
Using the following designs specifications, eagle, proteus and PSU design files were generated. On approval, the design is to be converted to a fabricated power supply jig workstation.