Electronic – What role does this diode play

Well, I'm studying electrical engineering right now and I can tell you that such jumps as you described take around two years of lectures at my university.

First thing which is important is to know which elements are passive and which are active. Then you need to know which elements are linear and which aren't. Next step is to get equivalent schematics for elements which you have and to see how they behave.

For example, let's take the switch. In off state, it functions as an open circuit, while in on state it functions as short circuit. Next, if you have sensitive equipment, you'll be able to notice that the switch isn't actually short circuit because it has some resistance, but that it's very low. Now let's take a look at the diode. Diode isn't linear component, so it doesn't have resistance in the classical sense in which for example resistors have. Instead there's the V-I curve of the diode. On a resistor, it's a linear function and we can use resistance as its characteristic, but on diode, it looks exponential.

As you can see from the image, certain voltage is needed for diode to start working properly and when you trigger the switch, that voltage disappears. That means, that the "resistance" of the diode just became huge. To get a feeling for this, use the parallel resistor calculation for say 1 mΩ resistor and 1MΩ resistor and take a look how much current goes through each of them. This is the way the circuit you mentioned behaves.

Ordinary semiconductor diodes are a junction of N and P semiconductor material. It turns out you can make a diode from what is sort of half of a semiconductor junction.

Schottky diodes are a junction with one side a P or N semiconductor, but the other side just metal. The result still functions like a diode, but has the following differences relative to circuit design:

1. The forward drop is about half. This is very useful in high current applications since the diode will dissipate less power. It also helps with efficiency in switching power supply applications.

2. The reverse leakage is significantly higher, especially at high temperatures. This is something you have to watch out for and design for accordingly. Take a look at the datasheet of an ordinary Schottky diode, like 1N5818. You may be surprised how much it can leak backwards, especially at high temperature.

3. The reverse recovery time is much faster, essentially instantaneous for most applications. This is very useful in switching power supplies that run in continuous mode. In that case, there is forward current thru the diode when the switch is turned on, reverse biasing the diode. Silicon diodes can be a problem in this application because for the first few 10s or 100s of ns the switch is essentially shorted by the diode still conducting even though it is reverse biased. This leads to inefficiency and a lot of stress on both the switch and the diode.

4. Schottky diodes are not available with as high reverse voltages as silicon diodes are. Above about 100 V they get difficult to find or expensive when you do.