You are a beginner with Croduino. Or electronics? A specific module caught your eye, but you do not know how to use it? Do not worry, HUM is here for you! How to Use Module (HUM) is a blog tutorials series by e-radionica where you will find all you need in order to begin working with your favorite module. Tutorials include: technical characteristics, work principle, instructions on how to connect module with Croduino and the basic code. Everything else is left to your imagination.
In this tutorial, we will see how to adjust the voltage necessary for our projects. You must have had a situation when some part of your project required a certain voltage, of any value..
But, what to do if we only have a source of much higher voltage than we need, what is more, if that voltage can damage the electronics? Do not worry, the so-called buck converters are here to help us. They are DC-DC converters, i.e. they convert the DC voltage of certain values to DC voltage of lower values. There are also boost converters and you can read more about them on the link.
• Input 3V - 40V
• Output 1.5V - 35V
• Max. current on the output 3A
HOW DOES IT WORK?
So, buck converters or step-down converters, as their name suggests, can only lower the voltage, and therefore, it is necessary to ensure an input voltage that is at least 0.7V higher than the desired output voltage (in the case of LM2596 module), because of failures in the module itself.
First, let's look at the simplified buck converter scheme in general because it is best to learn the basic work principle of these converters first.
In this image, it is seen that buck converters consist of a few basic components: coil, capacitor, diode, and switch (usually mosfet). To work, it also needs a signal that will control the switch. The most common case is the one where we have a control chip that produces the PWM signal and thus, controls the mosfet. For an easier understanding of these converters, we can observe two cases.
1. FIRST CASE:
In the first case, the switch is closed and current flows through the coil, loads it with magnetic energy and the voltage drops on the coil. The voltage on the output will be lower just because the coil behaves differently in AC conditions, e.g. when some changes, such as switching, occur in the network. In this case, the coil will not be a short circuit as in DC conditions; rather, it will have a certain impedance which causes the voltage to drop. However, if we keep the switch closed for too long, the coil will get saturated and voltage on the output will be equal to the source voltage. That is why we have to open the switch quickly in order to get a lower voltage. If you are not interested in what is the use of each component, but you only want to understand the basic work principle, skip to the 2. SECOND CASE.
You certainly have a few questions like: "Why do we even have the coil, why don't we put the resistor to its place?"
It is true that, on the output, we can get lower voltage this way, but this is a very inefficient way. Lower voltage gain is obtained by dissipating energy in the form of heat energy, and this is how linear regulators work. The coil increases this efficiency by converting the input energy into the magnetic energy (rather than thermal), then accumulating the accumulated magnetic energy back into the circuit to ensure constant current flow.
Another question you might ask yourself: "Why not exclude the coil and put the resistor we thought about in its place? If we only pulse the input voltage, we have lower voltage on the output, because the middle value is lower?" In that case, we would get a capacitive loop in which the voltage load (capacitor) is connected to a voltage source.
At the moment of turning on the switch, a voltage spike on the capacitor will occur. The capacitor, at this point, is in a short circuit to DC current if it is empty. It is going to be empty because it is drained by the resistor (load) connected in parallel, which means very large current from the source (in theory infinite). In practice, this is a bit different because there are different resistances of the elements themselves: internal voltage source, serial capacitor resistance, resistors' resistance, etc. Therefore, the capacitive loop looks milder in practice but is certainly not desirable and it is likely that something will be destroyed. By adding the coil, we limit the current and branches, and then we have a current load (not the voltage).
Furthermore, can we exclude the capacitor too? If we pulse the input power now, we will have a lower voltage on the output (middle value), and we have avoided the capacitive loop?
For some applications it may be good, e.g. LED light control and the like, but on the output, we no longer have a stable DC voltage, but a pulsating signal and then we are no longer talking about buck converters. In digital electronics, we would encounter big problems if the voltage of the source went down to zero, our microcontroller would be reset. Also, the maximum voltage values would then be equal to the source voltage, if we only dispose of higher voltage than needed, the microcontroller will certainly not be satisfied.
Then we can return the coil? It will accumulate energy, and will not allow the current/voltage to drop to zero? Do we need a capacitor?
Actually, without the capacitor, in the end, with a sufficiently high frequency of the switch control, the current/voltage at the output would not drop to zero, but the output signal would be something like a triangular signal ... the capacitor is there to "iron out" the voltage.
It is interesting that increasing the frequency can reduce the physical dimensions of the coil. LM2596 module, for example, uses a fixed frequency of the 150 kHz oscillator.
We have now concluded why we need a coil for the buck converter, why the capacitor and why it does not work without them. Let's look at the second case and find out what the diode is for.
2. SECOND CASE
In the second case, the switch is off, the source is off the circuit. We know the coil resists the change of current, so it will continue to deliver the accumulated energy to the load, i.e. it has a source function.
We see that without a diode the circuit would not be closed, and the coil contains loads of magnetic energy that it wants to convert to electric, i.e. throw it out of itself. On the switch, there would be a short impulse of high negative voltage which could easily damage the switch (mosfet).
That is why we connect the diode in such a way that it will not lead in the first case, and the other will enable the coil to close the circuit.
Finally, let's check the datasheet of the LM2596 module. A typical topology of the LM2596 converter is provided. We can see all the previous components with one simple, but very important addition. These are the feedback resistors, which monitor the output voltage and as a voltage divider, provide information to the module on how to control the switch. They are very important because the output voltage will always be the same if the input voltage changes or the load is changed. The datasheet also states that the efficiency of this module reaches 73%.
HOW TO CONNECT IT?
Connecting this module is very simple, all four connectors are indicated on the board. Anyway, we must watch out not to change the polarity since the module does not have a protection diode. In this example, we have connected a 12V battery, and a voltmeter which we used to observe the output voltage, on the output. Voltage is adjusted using the variable resistor on the board.
IN+ ===> +12V
IN- ===> minus (gnd)