Rectifier

Overview

All active circuits require DC power to operate properly. Batteries can already supply DC voltages directly but can be sometimes impractical for your circuits. Instead, you can design your device to be compatible with household outlets to draw power from the grid. Standard outlets are energized with a single-phase AC, meaning that we have to design an AC-DC converter.

Power Outlets

The current is carried by two conductors: the hot and neutral. Neutral is supposed to be held at ground potential (would not recommend touching it, though), while the hot devilers AC relative to ground. For this reason, we must place our disconnection elements on the hot conductor.

In United States and Canada, the outlets give us a nominal 120 Vrms at 60 Hz. Other regions of the world will have different standards, although the fundamental concepts still apply. The root mean square (Vrms) rating is useful for power calculations, but we are more interested in the center-to-peak (Vp) value when converting to DC.

$$V_p = V_{rms} \times \sqrt 2 \approx 170 \text{ V}$$

The circuit I will use to connect to mains will look like this:

I’ve added a fuse and power switch on the hot conductor, although they are not part of the analysis. The transformer can be used to step the AC up or down, but for now it will provide isolation.

Half-Wave Rectifier (Ideal)

The most basic AC-DC converter is the half-wave rectifier. It consists of a single diode, which blocks the negative part of the AC mains wave:

Clearly, the load is not seeing a true DC voltage, although there is a DC component which is transferred to the load. In order to get a better result, we need to store energy between the peaks of the AC wave using a capacitor in shunt with the load:

The load voltage is now pretty much DC, although there still is some oscillation, which is called ripple. It is impossible to eliminate ripple entirely (because it is impossible to build a perfect low-pass filter). Depending on your application, you will want to keep the ripple voltage below a certain level. A conservative estimate (see appendix for more info) can be obtained using:

$$\Delta V = \frac{I_{load}}{f_{ripple} C}$$

where

• $I_{load}$ is the average load current
• $f_{ripple}$ is the mains frequency
• $C$ is the filter capacitor value

For the circuit shown with a 10k load, the ripple voltage is:

$$\Delta V = \frac{170 / 10k}{(60)(10 \mu)} = 28.3 \text{ V}$$

In other words, the load voltage is always between 141.7 and 170 volts for this circuit.

Full-Wave Rectifier (Ideal)

The half-wave rectifier has a major flaw. Only the positive swing of the AC wave is used to produce DC power, while the negative swing is completelly unused, despite being equally as good.

In order to use the negative phase we need somehow to extract a positive voltage from the negative swing. We start by adding another another coil and attaching a half-wave rectifier to the bottom instead:

Since we flipped the bottom secondary, it will see the reverse voltage across it. Plotting the transient plots shows that the rectifiers are effectively out-of-phase:

This is exactly what we need, although we still need to combine the two circuits together. Both the secondary coils are floating, therefore we can combine a signle node without any issues. Unifying GND1 and GND2 into a single GND is reasonable. Another observation from the transient plot is that the diodes will never conduct at the same time, meaning we can safely connect the cathodes together:

Some final touches for making the circuit more concise:

• Convert the secondary coils into a single center-tapped coil
• The loads are now in parallel, so we can combine them

Lastly, adding the filter capacitor we can notice that the ripple is considerably smaller, since the period between peaks has been cut in half.

In fact, when computing the ripple, we will double our ripple frequency to 120 Hz, which cuts the ripple voltage in half:

$$\Delta V = \frac{170 / 10k}{(120)(10 \mu)} = 14.2 \text{ V}$$

This circuit was extensively used in old tube power supplies, because it only uses two diodes to achieve full-wave rectification, which could be bought as a single tube package [1, pp. 5, 37]. This configuration is not really used nowadays because the secondary needs to be center-tapped and twice as big to get the same output voltage. With cheap semiconductor diodes, we can spare the extra 2 diodes to make a bridge rectifier and eliminate the hassle.

Full-Bridge Rectifier (Ideal)

An alternative path for achieving full-wave rectification involves switching the entire secondary current path when the AC polarity flips. Effectively, we want to create a circuit which switches the current path between state 1 and state 2 shown below:

On each side of the coil, we add two diodes. The first diode is forward-biased and goes to the positive side of the load, just like in a half-wave rectifier. The second diode is reverse-biased and goes to ground.

When the winding voltage is positive, the forward diode conducts and connects to the load input. Meanwhile, the other winding is connected to ground via its reverse diode:

And the opposite happens when the polarity flips:

Drawing the circuit more consicely gives it a bridge appearance, hence its name. This is the standard way to draw a full-bridge rectifier. The transient plot and ripple are identical to the full-wave rectifier (at least with ideal diodes), so I will omit it here.

This is the most common rectifier circuit in the present day. You can get exteremly cheap (cents) integrated FBR packages with various ratings at most distributors.

Sources