0
\$\begingroup\$

I am new to Electronics. I am trying the Full wave rectifier.

I use a transformer and 4 1n5408 diodes (Vdrop is 1.2V) and try to get 24VDC output from 220VAC 50Hz. I measure the AC output (of the transformer) and it is 24VAC. I measure the output after these 4 diodes, it is about 23VDC.

Everything goes fine until I put the smoothing capacitor, no load yet. It is a 3300uF 25V electrolytic capacitor. When I turn on the power, the capacitor gets hot really fast, changes its shape and is going to explode.

I check the polarity, it is correct. So my question is:

  • Is it because my capacitor has a "too close" maximum voltage? If it is how should I choose a capacitor voltage for this circuit?

  • Another question though, I have seen this formula around while searching for my problem: C = I(load)/2.f.V(ripple). So if I need Vr = 2V and I(load) = 2A I need to use a 2/(2.50.2) = 0.01F = 10000uF Capacitor right?

Thank you all, really sorry for my English.

Improve this question
\$\endgroup\$
9
  • \$\begingroup\$ Welcome to EE.SE! You have a transformer in there, yes? If you get 23 Vdc with no capacitor, 35 Vdc rating on you capacitor will not be enough (boarderline for a bench test, but not continuous use). Try 50 Vdc rating. Is the polarity correct? Your English is fine, but please draw a schematic. \$\endgroup\$ – winny Feb 25 '18 at 14:51
  • 2
    \$\begingroup\$ Tell me there is a transformer between that 240V and the rectifier.... \$\endgroup\$ – Trevor_G Feb 25 '18 at 14:54
  • \$\begingroup\$ Yes, the polarity correct, I am trying the schematic here "Full-wave Rectifier with Smoothing Capacitor" electronics-tutorials.ws/diode/diode_6.html \$\endgroup\$ – Kalaradin Feb 25 '18 at 15:02
  • \$\begingroup\$ Yes, there is a transformer between that 240VAC and the rectifier. \$\endgroup\$ – Kalaradin Feb 25 '18 at 15:03
  • 1
    \$\begingroup\$ As @winny points out the CAP needs to rated for at least double the DC measurement without it. And needs to be the right way around. \$\endgroup\$ – Trevor_G Feb 25 '18 at 15:05
6
\$\begingroup\$

The voltage you are measuring without the capacitor is the average DC level. However the capacitor must withstand the maximum AC level, which is 1.414× the average.

enter image description here

As such your capacitor needs to be rated at least double your DC voltage, i.e. 50V. However, to account for tolerances and aging 75 or 100V would be even better.

The cap also needs to be in the right way around.

enter image description here

The capacitor is charged to the peak level every half cycle of the AC and the ripple is caused by the capacitor discharging through the load. Notice, in the image above, the average DC level is now higher than without the capacitor.

Note: With no load to drain the capacitor it will charge up to the peak level and stay there with virtually zero ripple. As such you can expect to measure close to 32.5V across the capacitor with your multimeter with no load.

Improve this answer
\$\endgroup\$
5
  • \$\begingroup\$ Ok, so if I put a 50V cap it will be enough. And the VDC still about 23V right? \$\endgroup\$ – Kalaradin Feb 25 '18 at 15:20
  • 1
    \$\begingroup\$ @Kalaradin after you add the cap the average voltage will go up to 32.5Vish if nothing is attached to the output. Assuming the cap is not really leaky. \$\endgroup\$ – Trevor_G Feb 25 '18 at 15:22
  • \$\begingroup\$ I just measure the AC output (of the transformer) and it's 24VAC, so it is an average voltage too right? \$\endgroup\$ – Kalaradin Feb 25 '18 at 15:30
  • \$\begingroup\$ @Kalaradin Its 24V RMS so yes... \$\endgroup\$ – Trevor_G Feb 25 '18 at 15:31
  • 2
    \$\begingroup\$ Thank you very much, now my capacitor can rest in peace. \$\endgroup\$ – Kalaradin Feb 25 '18 at 15:39
4
\$\begingroup\$

If you measure about 23v without the capacitor, that means you have an average rectified DC level of about 23v.

Unfortunately for your capacitor, the peak value of rectified AC is about 1.57 times higher than the average, or about 36v. This is what the capacitor will try to charge to.

With a capacitor, the average value delivered to the load will depend somewhat on the current the load draws, but will be nearly 36v.

Use a 50v (or more) rated capacitor.

Improve this answer
\$\endgroup\$
3
  • \$\begingroup\$ Thank you, I have another question, my average rectified DC level is about 23v, so after I put capacitor, is the output voltage (to the load) goes up above 23V? Sorry if it is something dumb. \$\endgroup\$ – Kalaradin Feb 25 '18 at 15:25
  • \$\begingroup\$ "...the peak value of rectified AC is 1.414 times higher than the average..." I am afraid it should be 1.571 instead \$\endgroup\$ – carloc Feb 25 '18 at 16:32
  • \$\begingroup\$ @carloc I had a nagging doubt as I entered that figure, but didn't bother to check. You are of course right. \$\endgroup\$ – Neil_UK Feb 25 '18 at 17:01
1
\$\begingroup\$

The 24V transformer produces 24V only with its rated load current. The windings on the transformer have resistance so its voltage will be higher, maybe 26VAC with no load current. Then the peak voltage is 26V x 1.414= 36.8V which is reduced to 35.6V by the rectifier diodes and the capacitor will try to charge to the 35.6v. But explode if it is rated for only 25V.

What about the higher electricity voltage that happens when everybody in town turns off their air conditioners? Then the capacitor might try to charge to 40V or higher.

Improve this answer
\$\endgroup\$

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.