0
\$\begingroup\$

Design giving the necessary ratings and characteristics the necessary for the transformer, bridge rectifier capacitor and regulator , a 12v , 2A regulated power supply .

the available mains supply is 230v+10% @50hz

so far I have :

Regulator - Standard voltage regulator will require 2v to be maintained across it to prevent drop out , therefore input voltage should not drop below 14v

Transformer - Va atleast 50% & volt drop = 24v

Capacitor - the minimum peak voltage to the regulator input

(24-10%) x 1.414 - 2 = 28.8v

maximum peak voltage

(24+10%+10%)x1.414-1.4 = 38.9v

The minimum peak full hand voltage defines the minimum of ripple voltage (28.8-14) = 14.8

suitable reservoir capacitor ?

I am uncertain if these figures are even correct , I think I am close but could do with some guidance

thank you

\$\endgroup\$
  • \$\begingroup\$ You left out the all-important value of the capacitance on the output of the full wave rectifier. You pick the tradeoff between the capacitance and voltage dip, then work from there to find the other specs. Of course if any of them turn out too onerous, you go back and adjust your initial values. Note also that this type of regulator will make a lot of heat, which you have to consider how to deal with. \$\endgroup\$ – Olin Lathrop May 21 '15 at 16:38
  • 1
    \$\begingroup\$ I would use a transformer with a much lower secondary voltage than you are. That is a lot of extra power to dump using a linear regulator. \$\endgroup\$ – R Drast May 21 '15 at 16:42
  • \$\begingroup\$ C = 2A / 50Hz * 1V whence C = 2/50 Farads = 10,000 uF is that what I am missing ? \$\endgroup\$ – Chris May 21 '15 at 16:45
  • \$\begingroup\$ so if I went for a 18v that would be a better option ? \$\endgroup\$ – Chris May 21 '15 at 16:52
2
\$\begingroup\$

As the comments to your question say, you need to find out a suitable capacitor value. You can then see what the valley voltage of your ripple is, add in the voltage drop caused by the rectifier diodes, then arrive at the MINIMUM transformer secondary voltage. You will then select the next highest available transformer secondary voltage, re-do all of the calculations, and finally find out how much heat that you need to get rid of.

Note that you need to design for the minimum available AC Mains voltage available when choosing the minimum secondary voltage. You need to design for the MAXIMUM AC Mains voltage when calculating how much heat will be generated.

Engineering is all about trade-offs. If the available transformer secondary voltage is higher than you like, you can choose a smaller reservoir capacitor such that the worst-case ripple valley voltage is still above the regulator drop-out voltage. This extra ripple translates into less heat wasted in the regulator.

Do note that you must NOT allow the ripple valley voltage to drop below the regulator drop-out voltage or the resulting regulated DC supply will have AC Mains ripple present in the output. That can cause all manner of problems, especially in audio circuits.

Finally, pay attention to the ripple current rating of the reservoir capacitor when choosing a suitable part. Note that the peak ripple current is substantially higher than the DC current output of the supply - the capacitor has to charge in only a small portion of the incoming AC sine-wave waveform.

\$\endgroup\$
  • \$\begingroup\$ thank you for this , the peak volatge for the 12v transformer would be 1.414 x 12 = 16.96 taking the bridge rectifier into account -2v = 14.96v so an 18v transformer would be suitable minimum (18-10%) x 1.414 - 2 = 21.15v maximum =28.8v ripple voltage 21.15- 14 = 7.15 for a suitable capacitor do i 1 / (2 x 50 x 7.15 ) = 1.398 (10-3) ?? ? could i have some guidance on the calculations \$\endgroup\$ – Chris May 21 '15 at 23:41

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.