In this tutorial, you will learn how to use a step-up converter to power equipment that requires higher voltage values, which your available voltage source cannot meet. In this case, a 12V fan is going to be used as a load.
- XL6009 step-up converter;
- 12V fan;
- Voltage source (9V battery);
- Male to male jumper wires;
- 1 breadboard (optional);
The Step-up Converter
The device we are working with today is an XL6009 step-up (or boost) converter, as shown in Figure 1. Power converters have an important role in electronic systems. They can be devices of great complexity, but their function is actually quite simple: to serve as an interface between two electronic systems with different voltage or current levels that cannot be directly connected. They are used in various applications such as CD players, cellphones, PC power supplies and hybrid or electric vehicles.
Figure 2 shows the internal circuit of a step-up converter. It is composed of an inductor (L), a switching element (S, usually a transistor), a diode (D), a capacitor (C) and the load (Ro).
In order for the output voltage to be greater than the input voltage, the circuit needs to have a storage element in between, which in this case is the inductor. While the switch is on (Figure 3(a)), the diode is reversed biased and the current only flows through the inductor, charging it with electromagnetic energy. Since it is charged, the capacitor insures that the load voltage remains the same, while it is cut off from the input. When the switch is off, however, the current flows throughout the entire circuit and the load is supplied with energy from both the input and the inductor, hence making the output voltage greater than the input voltage and charging the capacitor (Figure 3(b)) .
The amount of voltage that the output will increase depends upon a number of factors including the value of the inductor, the maximum voltage storage of the capacitor, the switching speed, and the input voltage of the circuit.
Suppose that the switching element S is on from t=nT to t=nT+ton, and off for the remaining time, i.e. with a duty cycle δ=ton/T. The circuit waveforms can be analysed in Figure 4.
The step-up converter used has 4 pins, as seen in Figure 1:
- IN+ Input voltage (in this case, 9V);
- IN- Input Ground (0V);
- OUT+ Output voltage (in this case, 12V);
- OUT- Output Ground (0V);
Regarding the fan, I am using an F9025B model fan, 12V, 0.45A, as seen in Figure 5. This module has 3 pins:
- Red – Positive pin
- Black – Negative pin;
- White – Control pin (not used in this tutorial).
From the XL6009 step-up converter’s main specifications (Table 1), one can conclude that this converter is capable of powering the fan.
|Input Voltage||3 – 32 V|
|Output Voltage||5 – 35 V|
|Maximum Output Current||4 A|
The circuit is pretty straightforward. Start by connecting the positive and negative terminal of the 9V battery to the IN+ and IN- pins of the converter, respectively. Then, measure the voltage between the output pins of the converter. Make sure you measure at least 12.7V (Figure 6): 12V to power the fan and 0.7V to surpass the diode’s forward-biased voltage. You can adjust the voltage by rotating the potentiometer present in the module, which will change the system’s duty cycle.
The use of the diode is to prevent the fan rotation from creating an opposite current and damaging the circuit. Hence, connect the OUT+ pin to the diode’s anode, its cathode to the positive terminal of the fan and, finally, the OUT- pin of the converter to the negative terminal of the fan. In the end, your circuit should be set as shown in Figure 7.
That’s it! The next step will be to try to control the fan’s speed, but I’ll leave that for another tutorial.
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Towards the Future !!! 😉