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. 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. 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.Īs 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. 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.Īs the rectified voltage gets past the bridge and is rising, at first it does nothing much since the capacitor voltage is higher. (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.
That's a rough idea of what the voltage at the capacitor looks like in a full wave bridge rectified system. The formula is very useful if you understand what it means. Specify the voltage ratings of your components, such as the filterĬapacitor. That value is very important when it is time to Theįormula tells you the maximum dc voltage you can achieve from a given
You are missing the point of the formula if you think it is wrong. 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 made an important observation The output voltage will drop as the load increases until a full safe load is reached.īy now the peaking effect is gone and the DC voltage is more like the AC-RMS value. At high current levels >10 amps the Vdrop across each diode can be 1 volt. With heavier loads a bridge or full-wave rectifier will provide the most current. 7 volts or 1.4 volts from the expected peak, and the numbers should match better. 1% of capacity will drop the voltage by the amount the diodes dropped.
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. 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. The sites formula is correct, but only under ideal conditions.