I am in the process of making a regulated power supply to convert AC(220V, 60Hz) to DC(9V, 500ma). I have worked out all the calculations and would appreciate if someone could double check just to make sure they are alright.

My setup is as follows

AC(220,60Hz) -> Transformer(12V,500ma) -> Bridge Rectifier -> Voltage Regulator(7809)

  1. the 78xx series requires at least 2V to work => i need at least 11 as rectifier output

2.the ac input might vary by 10% ..

peak transformer voltage = 12 \$\cdot\$ \$\sqrt{2}\$ = 16.97V

min transformer voltage = 16.97 \$\cdot\$ 0.9 = 15.27V

vout-bridge = vmin-transformer - (2 \$\cdot\$ 0.7) - Vripple
vripple = 15.27 - 1.4 - 11 = 2.87V
dt * load current / C = 2.87V
C = 8ms * 500mA / 2.87V
C = 1393 μf

can someone confirm this is correct ??

thank you for the help !

  • \$\begingroup\$ @stevench ... Makes sense but then I am planning to connect a micro controller To t.. The current drawn by which would be max 200ma but could be lower as well. So that would mean the ripple voltage would vary depending on the current being drawn at the moment. So how do I ensure that this variation is small enough to be taken care of by the voltage regulator? \$\endgroup\$
    – Ankit
    Mar 14, 2012 at 14:49

1 Answer 1


Your calculations are correct, though it's safer to calculate with 1V drop for the diodes, as the following graph for the 1N4001 diode shows.

enter image description here

0.7V is only OK for currents less than 10mA.

If your load is 500mA you need more than that from the transformer. The 500mA is drawn from the rectified voltage, i.e. the 17V. This means power drawn is 17V \$\cdot\$ 500mA = 8.5W, which means 0.7A at 12V RMS. A safe value for the transformer's current rating is twice the DC current, so I would pick a 12V AC, 1A transformer.

  • \$\begingroup\$ @stevench Thank you for confirming and pointing out the 1v drop. another question i have is the current part in transformer rating. when it says 12V, 500Ma i presume it means at 12 max 500Ma can be drawn by the load. is that correct ? what happens if my load is expected to draw around 200mA .. do the above calculations change ? \$\endgroup\$
    – ankit
    Mar 14, 2012 at 9:36
  • \$\begingroup\$ @ankit - I added to my answer. If you draw only 200mA only a 560\$\mu\$F is required, as per your calculations. \$\endgroup\$
    – stevenvh
    Mar 14, 2012 at 10:01
  • \$\begingroup\$ @Ankit - (reply to what you posted as an answer) If you draw less current ripple will be smaller, meaning that the lowest voltage will be higher. Hence the 2V margin for the 7809 will certainly be met. If 200mA is your worst case, calculate for that and you'll be safe for lower currents as well. \$\endgroup\$
    – stevenvh
    Mar 14, 2012 at 15:35
  • \$\begingroup\$ @stevench sorry abt that .... damn ipad ! thank you for the answer \$\endgroup\$
    – ankit
    Mar 14, 2012 at 16:22

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