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I tried to build a circuit that could convert 220 volts AC into 12 volts DC:

I tried to achieve a circuit that could convert 220 volts AC into 12 volts DC

My voltage output is as I wanted but the measured current is always blank.

Can anyone explain why this is happening and what should I do to get the (12V / 5A) output (I want to do it with real components.)

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  • \$\begingroup\$ your load resistor is broken? \$\endgroup\$
    – user253751
    Mar 22, 2021 at 10:23
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    \$\begingroup\$ @Hissoka: The ampere meter is clearly showing 0.5A, so it isn't "blank." \$\endgroup\$
    – JRE
    Mar 22, 2021 at 10:28
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    \$\begingroup\$ as a note, your 500uF cap is upside down. the striped side, is the negative side but you have connected it to positive output of rectifier. \$\endgroup\$
    – Tirdad Sadri Nejad
    Mar 22, 2021 at 11:43
  • \$\begingroup\$ Load resistor not broken. Because the current is closly correct for the voltage and resistor value. 2200 uF capacitor may give more current. \$\endgroup\$
    – upali
    Mar 23, 2021 at 11:40

3 Answers 3

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12.4V / 24.9ohm = 0.5A which is what your Amp meter shows.

If you want more current, ex 5 Amps, you should decrease your R Load to about 2.4ohm.

Note as you pull more current you will have more voltage drop on the transformer so it's not exactly linear.

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    \$\begingroup\$ Also, with only 500 uF, the ripple voltage will be huge with a 5 A load. \$\endgroup\$
    – Dave Tweed
    Mar 22, 2021 at 10:31
  • \$\begingroup\$ @DaveTweed Surely, that's probably some basic classroom simulation task to understand U=R*I. \$\endgroup\$
    – Damien
    Mar 22, 2021 at 10:36
  • \$\begingroup\$ No simulation required; the formula is \$\Delta V=\frac{I\Delta t}{C}\$, where \$\Delta t\$ = 10 ms in a 50-Hz full-wave rectifier. \$\endgroup\$
    – Dave Tweed
    Mar 22, 2021 at 10:42
  • \$\begingroup\$ @DaveTweed what I meant was, given how basic is the question, it's probably some basic classroom exercise and ripple is probably not relevant at this stage. \$\endgroup\$
    – Damien
    Mar 22, 2021 at 10:55
  • \$\begingroup\$ Well, you mentioned voltage drop in the transformer. Excessive ripple will cause the average output voltage to drop faster than resistive losses in the transformer will. Why do you think the simulation is already showing only 12 .4 V at the output? \$\endgroup\$
    – Dave Tweed
    Mar 22, 2021 at 11:05
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For a reasonable amount of ripple at 5A you want more like 50mF than 500uF (~100x more).

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    \$\begingroup\$ 100 mF would actually be rather high. Think about what that means in terms of the conduction angle and the peak current through the rectifier. \$\endgroup\$
    – Dave Tweed
    Mar 22, 2021 at 10:50
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    \$\begingroup\$ @DaveTweed Based on 0.5V p-p ripple. Yes, that's a bit high. If we go with 10% (1.2V p-p ripple), then it's more like 40,000uF \$\endgroup\$
    – Spehro Pefhany
    Mar 22, 2021 at 10:51
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It seems to me that you are confused with the power supply rating. you wrote it be 12 V 5A. This simply means that it will supply a maximum of 5 amperes. If it is a somewhat constant voltage supply. Current will flow according to the load.

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  • \$\begingroup\$ In practical 5 A out put gives by a transformer of 20v open. \$\endgroup\$
    – upali
    Mar 23, 2021 at 11:44

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