2
\$\begingroup\$

I am using a bridge rectifier power supply with a 7805 regulator IC for a microcontroller application. A capacitor (2200uF) is used for removing additional ripple. But when the microcontroller uses more power the DC output shows some ripple, the entire circuit draws about 150mA. How can I remove it completely?

\$\endgroup\$
  • \$\begingroup\$ There are many things to consider. You may need the rectifier to supply more power. It could perhaps also help with a larger capacitance at the regulator input. And perhaps a smaller one at the microcontroller. \$\endgroup\$ – HKOB Feb 13 '15 at 6:44
  • 1
    \$\begingroup\$ If you mean the 120 hz (not high-frequency) ripple, you will need a bigger cap, although 2200 uF is pretty big. Are you seeing the ripple at the output of the regulator? What is the current drawn by the microcontroller? Is the supply voltage 5V? What is the AC voltage level at the input of the rectifier? \$\endgroup\$ – biggvsdiccvs Feb 13 '15 at 6:44
  • \$\begingroup\$ Yes I mean 50 Hz ripple in my 5v power supply output. Microcontroller drawn about 150mA. Iam using a 9v(1Amp) transformer. \$\endgroup\$ – apswyd Feb 13 '15 at 9:30
  • \$\begingroup\$ I mean all circuit drawn about 150mA \$\endgroup\$ – apswyd Feb 13 '15 at 9:37
  • 1
    \$\begingroup\$ How big is the ripple? You can never completely remove any, just scale it down until it can't be noticed anymore. \$\endgroup\$ – PlasmaHH Feb 13 '15 at 9:47
3
\$\begingroup\$

A fullwave rectified power supply below will have ripple due to charging and discharging of the bulk capacitors.

(source)

enter image description here

(source) enter image description here

where the ripple voltage is $$V_{pp}=\dfrac{I}{2fC}$$ and I is the load current, C is the capacitance and f is the mains frequency.

So as the load current increases so does the ripple. To reduce this ripple you need to increase the bulk capacitance.

This is followed by a 7805 which should reduce that ripple by its RR (ripple rejection), which is about 60dB for the 7805. So Vpp in your first calculation should be reduced by about 1000.

If this is not happening for you check the dropout of the 7805 is respected, i.e. you need the input to be at least 2.0V more than the output i.e 7V min and this needs to be the case even under full load because the transformer output voltage will reduce under load.

\$\endgroup\$
  • \$\begingroup\$ I think that was my mistake, Actually my project was not properly connected to ground. Thanks for all your reply. \$\endgroup\$ – apswyd Feb 18 '15 at 4:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.