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Well, my question is: "Is this calculator wrong?" It calculates the voltage after rectifier as

Vdc = Vac * 1.41

It seems it forgets the voltage drop on diodes (0.7V + 0.7V).

Is the site accurate or we should we take into consideration the voltage drop on diodes?


Update:
For 12Vac I get 15.9Vdc after the rectifier (4x 1n4007 & 470uF capacitor, no load).

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    \$\begingroup\$ It's correct if no current is drawn. Please edit your question so it still makes sense if that link dies. \$\endgroup\$ – Transistor Aug 5 '16 at 21:38
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    \$\begingroup\$ It also is not accounting for ripple in the DC. A capacitor is needed to filter out the ripple to get a smooth DC current. \$\endgroup\$ – user105652 Aug 5 '16 at 21:54
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    \$\begingroup\$ What value is the capacitor? 470uF to 2,200uF is common. For clean DC for audio amps the 'loose' rule is 2,000uF per amp of current consumed. The 'static' ripple is equal to delta V = 1/FC, where F is frequency and C is capacitance. \$\endgroup\$ – user105652 Aug 5 '16 at 22:22
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    \$\begingroup\$ It's not wrong, so much as imprecise. It posits an incomplete model useful for approximations. \$\endgroup\$ – Stephen Collings Aug 5 '16 at 22:23
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    \$\begingroup\$ You are missing the point of the formula if you think it is wrong. The formula tells you the maximum dc voltage you can achieve from a given rms ac voltage. That value is very important when it is time to specify the voltage ratings of your components, such as the filter capacitor. The formula is very useful if you understand what it means. \$\endgroup\$ – Elliot Alderson Jun 4 at 16:46
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  1. The sites formula is correct, but only under ideal conditions. The AC input must be a clean sine-wave. Some UPS's and DC-AC inverters put out a choppy sine wave that would make the 1.414 ratio of RMS value to peak value not true.

  2. This peak voltage assumes no load, whether a single diode is used or a bridge rectifier, plus capacitor of sufficient value to remove any AC ripple. Even a tiny load of .1% of capacity will drop the voltage by the amount the diodes dropped. So subtract .7 volts or 1.4 volts from the expected peak, and the numbers should match better.

  3. With heavier loads a bridge or full-wave rectifier will provide the most current. At high current levels >10 amps the Vdrop across each diode can be 1 volt. The output voltage will drop as the load increases until a full safe load is reached.

  4. By now the peaking effect is gone and the DC voltage is more like the AC-RMS value. The RMS-to-Peak formula is correct but only under ideal conditions, and as you might have guessed by now, real-world conditions dominate once you apply a load or use a inverter/UPS for AC power, making the RMS to Peak formula useless, especially under heavy loads.

  5. @ElliotAlderson made an important observation;

You are missing the point of the formula if you think it is wrong. The formula tells you the maximum dc voltage you can achieve from a given rms ac voltage. That value is very important when it is time to specify the voltage ratings of your components, such as the filter capacitor. The formula is very useful if you understand what it means.

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  • \$\begingroup\$ I have done so. I hoped it helped with your understanding of the issue. \$\endgroup\$ – user105652 Jun 6 at 22:42
  • \$\begingroup\$ @sprky - Many thanks. now the response is more clear \$\endgroup\$ – Ultralisk Jun 7 at 11:45
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Have a look at the following image:

enter image description here

That's a rough idea of what the voltage at the capacitor looks like in a full wave bridge rectified system. (Once it reaches an equilibrium state.) The grey curve is supposed to show the rectified DC out of the bridge, but this will actually be about two diode drops lower and there will be a tiny gap around 180 degrees and 360 degrees, and so on. But it's close. The main point here is that the thick black line shows you what the capacitor voltage roughly looks like when there is a real load applied and the capacitor is designed within some range of reason for the load.

As the rectified voltage gets past the bridge and is rising, at first it does nothing much since the capacitor voltage is higher. But the capacitor is still supplying current to the load and drooping, so eventually the drooping capacitor voltage and the rising rectified voltage cross over sufficiently to forward bias the diodes in the bridge and the capacitor voltage follows the rising voltage (or what remains of it, this first half of the first half cycle.) For this very short time before the bridge voltage peaks, some few degrees before 90 degrees, the transformer/bridge system is supplying current to the load and the capacitor.

As the rectified voltage rapidly declines and falls away from its peak at 90 degrees, it also falls away from the capacitor voltage and the capacitor is then supplying all of the current to the load. It must continue to do this until the next half cycle, usually not much but somewhere before 270 degrees when the transformer/bridge system supplies all the current again.

That lowest point in the droop of voltage must still be sufficiently high for the following voltage regulator system (if there is one.) Note that if the load draws more current than before, then the slope of this droop will steepen and it will dip still further down before the rising voltage from the bridge rectifier catches back up. Also, if you use a smaller capacitor even if the load current stays the same, the slope also steepens. So you need to make sure your capacitor and expected worst case load match up with the needs of the minimum input voltage for your following regulator system.

As you might guess, by now, there is no simple, linear, one-equation-solves-all-problems here. Some thinking is required.

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  • \$\begingroup\$ This is a nice answer too (different approach). I accepted only Sparky's because he was the first to answer. Thanks! \$\endgroup\$ – Ultralisk Jun 5 at 8:43

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