Since a diode conducts only in one direction, if it is placed in an AC circuit (this may be a transformer output, then it will be sinusoidal current, or a DCC output, the current being then rectangular), it will modify the shape of the current applied to the load.
Example: a load made up of an LED with its protective resistor is powered by a 15 V AC transformer output via a diode. During the positive half-wave, the current passes. During the negative half-wave, it doesn’t. The voltage at the terminals of the load (LED + resistance) will have the appearance shown if viewed with an oscilloscope.
What will be the visual effect? Well, the LED will illuminate less than if it was powered at 15 V DC, because the eye doesn’t have time to perceive the succession of turns on and off (50 times per second). Instead of perceiving it 50% of the time on and 50% of the time off, we’ll have the impression that its luminosity is 50% lower.
We need four diodes to obtain this rectification, in which the negative half-wave will no longer be suppressed, but will be made positive. These four diodes can be included in an integrated bridge, more convenient to wire, but not necessarily less bulky.
The schematic representation can take two apparently different forms, but representing exactly the same thing. The traditional shape in electronics is the diamond, but the arrangement of the diodes in this diamond shape may be difficult to apprehend for an uninitiated guy.
Personally, I prefer a representation that comes rather from electrotechnics, but which seems to me simpler, since all the diodes are drawn in the same direction.
Note: I have indicated by red circles — not to be confused with standardized symbols — wires that cross but are not connected. Drawing small bridges in these areas has long been out of the industry. If you want the wires to be connected, you put a big black dot on the crossing.
Now, we’d like, if the source current is alternative, getting something closer to the direct current. The current shape of the rectified voltage is far from it: it has a strong ripple. In the DCC case, since the source wave is rectangular, this problem is less important (however, there will be small gaps because the sides of the rectangles are not perfectly vertical). But if we use this diagram for coach lighting, we are at the mercy of bad wheel / rail contacts that will make the light flicker. We will therefore place a capacitor whose role will be to store electrical energy (it is a real reservoir, moreover, we speak of its capacity) when the input voltage is present and large, then to restore this energy to the load when this voltage is absent or low.
If you are using real strong rectifier diodes, the previous diagram is suitable. But even the smallest ones are relatively bulky and of unnecessarily great power for the currents involved, which are expressed in milliamperes. 1N4148 type switching diodes, very small, very cheap and very widespread, are suitable. Except… at power up!
Be aware that a discharged capacitor has by definition 0 V at its terminals, exactly like a simple piece of wire. This is called a short circuit. Of course, when it is getting charged, its voltage will increase, and the short circuit will disappear. Nevertheless, because of this, the charging current can be very large at the beginning, being limited only by the internal resistance of the source and wires. The poor diodes could suffer the consequences (it happened to me). A circuit composed of a diode V and a resistor R will therefore be added.
The circuit operates as follows: when the voltage VSOURCE is applied, the diode V is in reverse and cannot conduct. Thus, the capacitor C charging current in red passes through the resistor R which will limit it to a suitable value (here, with 15 V and 100 Ω, the peak current would be 15 V / 100 Ω = 0.15 A or 150 mA, perfectly permissible for 1N4148 diodes).
On the other hand, if the power supply is interrupted, the voltage of the charged capacitor is higher than that of the load; the diode V is forward and the current supplied to the load, in blue, passes through this diode, not through the resistor R.
Note that this circuit is perfectly valid if you replace the diode bridge with a decoder auxiliary output, which also has a limited current. The decoders are supposed to be short-circuit protected, but let’s not tempt fate…