Electrical – Full Bridge Rectifier with Filter Capacitor circuit

I can't tell if this is homework, or not. But if you actually are trying to figure out how to design a bridge rectifier system that yields a desired output and ripple specification, and you have complete freedom to build a transformer for it rather than being forced to buy one from a list of readily available units, then the process is something like this:

Specifications:

• \$V_{OUT_{REGULATED}}=5\:\textrm{V}\$ (the final regulated voltage)
• \$V_{DROP_{REGULATOR}}=1.5\:\textrm{V}\$ (the minimum overhead used by the linear regulator)
• \$I_{MAX}=1\:\textrm{A}\$ (the maximum load current to be supported)
• \$V_{RIPPLE}= 500\:\textrm{mV}\$ (the ripple voltage at the input of the linear regulator)
• \$f=60\:\textrm{Hz}\$ (the operating frequency of the mains supply to the transformer)

From that you can work out the rest. For example, let's use the above values to get:

$$\begin{align*} t_{charge} = \frac{\textrm{sin}^{-1}\left(\frac{V_{OUT_{REGULATED}}+V_{DROP_{REGULATOR}}}{V_{OUT_{REGULATED}}+V_{DROP_{REGULATOR}}+V_{RIPPLE}}\right)}{4 \pi f}&\approx 1.6\:\textrm{ms} \\ \\ C = \frac{I_{MAX}}{V_{RIPPLE}}\left(\frac{1}{2 f}-t_{charge}\right)&= 13.5\:\textrm{mF}\\ \\ I_{DIODE_{MAX}} = \frac{\sqrt{2}\cdot V_{RIPPLE}\cdot C}{t_{charge}}&\approx 6\:\textrm{A} \end{align*}$$

At this point, even though the capacitor isn't a standard value, you can make an estimate for the voltage drop of your rectifier diode. I generally take this to be about \$800\:\textrm{mV}\$ when supplying \$1\:\textrm{A}\$, so in this case I'd guess about:

$$V_{DIODE}=800\:\textrm{mV}+120\:\textrm{mV}\cdot \textrm{ln}\frac{6\:\textrm{A}}{1\:\textrm{A}}\approx 1\:\textrm{V}$$

Now, you can estimate the transformer's RMS value as:

$$V_{RMS} = \frac{V_{OUT_{REGULATED}}+V_{DROP_{REGULATOR}}+V_{RIPPLE}+2 V_{DIODE}}{\sqrt{2}}\approx 6.3\:\textrm{V}$$

Luckily, that happens to be a standard transformer value.

But note that this is under load. The capacitor chosen matters here.

I'm going to stop at this point and let you clarify your question some more. Perhaps the above process isn't what you wanted, at all. But it gives you an idea of some of the factors involved. Perhaps someone else understands you better and can provide a better response. But for now, I'm still not sure what you are doing.

^{simulate this circuit – Schematic created using CircuitLab}

Figure 1. (a) A single winding with full-wave bridge rectifier. (b) A centre-tapped transformer with rectifier diodes.

• I am assuming that the centre-tapped transformer has twice the number of turns as the single-winding transformer. e.g., If (a) is a 0 - 12 V transformer then (b) will be a 12 - 0 - 12 V transformer.
• If you trace the current flow in (a) when the dot end is positive you will see that the current will flow through D2, out to the load via V+, back on V- and through D3 back to the secondary. Each diode will create a voltage drop of 0.7 V so the voltage out will be down about 1.4 V.
• You can see in (b) that when the dot end is positive that D5 will conduct and the return will be straight to the centre-tap. The voltage drop will be 0.7 V approx.

The centre-tapped version is more efficient but probably more expensive to make.

Note that the XFMR2 wire gauge can be lighter than XFMR1 as it only carries current on alternate half-cycles. Total weight of copper will be about the same.