Dear All, whilst writing about the “Symmetrical Full Bridge Doubler” – I kind of silently assumed that you know the details related to the “maths” associated with “what to expect”. But as the old saying goes: “Never “ass|u|me” ….
So allow me to present to you just a quick reminder as to what might be expected of some typical scenarios.
A “traditional” full wave rectifier bridge, following a transformer secondary, followed by a C filter ( or C-R-C …)
The transformer that you may have purchased will have a label stating that the secondary voltage is XX Volts AC. For arguments sake, let us focus on an example. We have a transformer, with a label, stating: ”…. 10 V AC, 2 Amperes …. ”
So … What does this actually “mean” ?
Does this mean that if I plug it into the mains, rated accordingly, and take a multimeter, I shall measure “10 V AC” on the stated secondary ?
The simple answer is …. NO !
The reading will actually be a tad higher. It may be, depending on the type of transformer, be even 10% to 15% than the voltage rating as can be read off the label. Why is this so? The transformer is designed to deliver 10 V AC, whilst delivering the RATED LOAD CURRENT. In our example case, the 2 A. But now you need to take into consideration that the wire used to create the secondary winding has some resistance of it’s own. If the resistance of the wire would be, say, 0.5 ohms, that would imply that 0.5ohms x 2 A = 1 Volt of “voltage loss” on the internals of the winding itself. Therefore, the secondary, as produced by the manufacturer, needs to be made for a slightly higher voltage, i.e. 11 Volts, so that after substracting the internal voltage drop of 0.5R x 2 A = 1V, we end up with 11V – 1V = 10V, as “promissed” on the label.
Why am I telling you all this? Just to be sure that your fliter capacitors and rectifier diodes to not explode. Why should they? Because they will not withstand the reverse polarity voltage peaks. It is good to know what we may expect and what we are dealing with.
We started off with 10V AC on the label of the transformer. But now we know that at “idle” conditions, when no significant current is drawn, this will pop up as a full 11V on the output terminals of the transformer (for examples sake). Now, we allow this voltage to forward bias a diode and load up a capacitor within the filter bank.
Oh, did I mention that the 10 V as stated on the transformer label is an RMS value, meaning “Root Mean Square”? It is actually stated as “power-averaged” value of the voltage, which means that it is a sine, but one which shall deliver the same power to the load as an equivalent DC voltage of the same value. But a sine has it’s up’s and down’s, it’s high’s and low’s, and the RMS value is actually only a fraction of the peak value of the sine wave. The RMS Voltage value for a sinewave is equal to 0.7071 of the peak value of the wave [more specifically, V_peak * 1/sqrt(2) ].
Similarly, The “peak” value of a sine wave, the “crest” of the wave is 1.4142 times higher than the RMS rated voltage [more specifically, V_rms * sqrt(2) ].
Hence, our 11 V AC RMS actually creates occasional peaks, with a value of 1.4142 V, i.e.
11 * 1.4142 = 15,55 V_peak.
Now, the capacitor within the filter bank, will load up close to the full value of this peak voltage (less voltage drop on a forward bias diode). Now you can plainly see, that although the transformer is rated only but 10 V AC, it seems that the capacitor voltage rating for the filter bank needs to be at least 16V, … and we just barely make it, if you take into consideration occasional fluctuations and overvoltage conditions of the mains power supply. If you were to use a 10V rated capacitor, it would blow up, as it would not withstand the voltage stress.
But what about the diodes? How should the diodes be rated then?
The reverse voltage capability of the diodes is even more demanding. Just imagine: We have just charged the capacitor to a value of 10V * 1,1 * 1,4142 = 15,55 Volts positive.
But now comes the negative half cycle of the sine wave, and at the lowest possible point of the negative swing of the wave, we have MINUS 15,55 Volts on the other end of the diode. So now you can see that from the one side of the diode you have the cap “holding” the PLUS 15,55 Volts, and from the other side of the diode you have a negative peak of the wave, at MINUS 15,55 Volts.
Obviously, the diode needs to withstand a reverse polarity voltage of 31,1 volts. Intriguing. We started off with a transformer rated at 10 V AC RMS on the secondary winding, and it turns out that the reverse polarity capability of the rectifier diodes within our bridge needs to be over 3x that value.
Keep that in mind while designing even your most basic rectifiers an filter banks.
In case of the “voltage doubler” circuit, things get even more confusing. From the one side, it may be argued that two diodes in series should withstand twice the reverse polarity voltage than that of a single such diode, but unfortunately that is not quite exactly true, as the “division of voltage” on such reverse biased two diodes does not need to be a division into two equal valued halves. Indeed, if one of the diodes has a higher value of leakage current, then the gross value of voltage drop with burden the other diode, i.e. the one with less leakage. In extreme cases of gross mismatch of the diodes, one of them may actually need to take the burden of close to the whole value of the reverse voltage. This can be mitigated by an even valued resistor divider network, in parallel to the diodes, a network that “enforces” an even division of the reverse voltages burdening the diodes. There are some intricacies involved in the design of such divider networks.
So, for simplicity, if you do not use such resistor divider networks, assume that each diode, by itself, needs to withstand the full value of the reverse voltage. In the case of a voltage doubler, the diodes should at best be capable to cope with 62,2 Volts worth of reverse voltage (if no resistor divider networks are used).