The objective of this project is to get a stabilised output voltage of 12v and a current of 100mA.
As Mauritius have a 230v AC voltage distribution with a frequency of 50Hz.
The VN30.15/10525 transformer fit with a frequency range of 50hz to 60hz and a 230v input.With a 15v Output.The 2,3 volt higher output are necessary for the other components like the power diode and the resistors.
We then start making a bridge “full wave rectifier” which turns the AC to a DC voltage.
We use a MBR140SFT1G with a PIV of 40V.To prevent the diode to burn with a charge from the transformer and at the same time from the capacitor=15v*2 we need a PIV of 30v or higher.
The bridge composed of 4 diodes ,only 2 work at the same time depending if the ac current coming is positive or negative.As each power diode use 0.36v and only 2 are active at the same time so 0.36*2=0.72v.The transformer give an output of 15 v so 15-0.72=14.28v.Which means 14.28v remains.
We then need a capacitor:
(Load current*time of one cycle)/ripple voltage
(120ma*10ms)/1v=1200 microfarad
We assume we want a ripple voltage of 1 volt.
So we use a 39D128G015FL4.
As told earlier we have 14.28v remaining.
12 volt is needed in the Zener diode so the remaining 2.28v need to go on the resistance.
As 20ma is used by the Zener diode and 100ma by the output.We have an input of 120ma and we need a resistance which can take 2.28v.0.27w power dissipation also is required because 0.12a*2.28v=0.2736w
R=V/I
2.28v/0.12a=19ohm
So we use a A-2642 18ohm resistor with its 5% tolerance value it reach 18.9ohm and it has also a 0.5w power dissipation
Can someone just check for me if it is good.
I also didn't understand well the PIV for the 4 power diode in the full wave rectifier you need always the double.The transformer only have an output of 15v why is a piv of 30v necessary?
I also saw that the capacitor have a voltage value is it important?
As Mauritius have a 230v AC voltage distribution with a frequency of 50Hz.
The VN30.15/10525 transformer fit with a frequency range of 50hz to 60hz and a 230v input.With a 15v Output.The 2,3 volt higher output are necessary for the other components like the power diode and the resistors.
We then start making a bridge “full wave rectifier” which turns the AC to a DC voltage.
We use a MBR140SFT1G with a PIV of 40V.To prevent the diode to burn with a charge from the transformer and at the same time from the capacitor=15v*2 we need a PIV of 30v or higher.
The bridge composed of 4 diodes ,only 2 work at the same time depending if the ac current coming is positive or negative.As each power diode use 0.36v and only 2 are active at the same time so 0.36*2=0.72v.The transformer give an output of 15 v so 15-0.72=14.28v.Which means 14.28v remains.
We then need a capacitor:
(Load current*time of one cycle)/ripple voltage
(120ma*10ms)/1v=1200 microfarad
We assume we want a ripple voltage of 1 volt.
So we use a 39D128G015FL4.
As told earlier we have 14.28v remaining.
12 volt is needed in the Zener diode so the remaining 2.28v need to go on the resistance.
As 20ma is used by the Zener diode and 100ma by the output.We have an input of 120ma and we need a resistance which can take 2.28v.0.27w power dissipation also is required because 0.12a*2.28v=0.2736w
R=V/I
2.28v/0.12a=19ohm
So we use a A-2642 18ohm resistor with its 5% tolerance value it reach 18.9ohm and it has also a 0.5w power dissipation
Can someone just check for me if it is good.
I also didn't understand well the PIV for the 4 power diode in the full wave rectifier you need always the double.The transformer only have an output of 15v why is a piv of 30v necessary?
I also saw that the capacitor have a voltage value is it important?
