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Tiny MCU hosts dual dc/dc-boost converters

23 Mar 2016  | Dhananjay Gadre

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Batteries are the usual power sources for portable-system applications, and it is not atypical these days to find microcontroller-based portable systems. A variety of microcontrollers operates at low power-supply voltages, such as 1.8V. Thus you can employ two AA or AAA cells to power the circuit. However, if the circuit requires higher voltage—for LED backlighting for an LCD, for example, which requires approximately 7.5V dc—you must employ a suitable dc/dc converter to boost the power-supply voltage from, for example, 3V to the required voltage. However, you can also employ a microcontroller to develop a suitable dc/dc-boost-voltage converter (Reference 1) with the help of a few additional discrete components.

Figure 1: The output voltage in a boost switching regulator is more than the input voltage. The boost switching regulator operates in either CCM (continuous-conduction mode) or DCM (discontinuous-conduction mode)

This Design Idea shows how to create not just one, but two dc/dc converters with just a tiny eight-pin microcontroller and a few discrete components. The design is scalable, and you can adapt it for a wide range of output-voltage requirements just by changing the control software for the microcontroller. You can even program the microcontroller to generate any required output-voltage start-up rate. Figure 1 shows the basic topology of a boost switching regulator. The output voltage in such a regulator is more than the input voltage. The boost switching regulator operates in either CCM (continuous-conduction mode) or DCM (discontinuous-conduction mode). It is easier to set up a circuit for DCM operation (Reference 2). The name comes from the fact that the inductor current falls to 0A for some time during each PWM period in DCM; in CCM, the inductor current is never 0A. The maximum current passes through the inductor at the end of high period of the PWM output (when the switch is on) and is:

Equation 1

where VDC is the input voltage, D is the duty cycle, T is the total cycle time, and L is the inductance of the inductor. The current through the diode falls to zero in time TR.

Equation 2

Equation 2

The load current is the average diode current,

Equation 2

from equation 1 and equation 2 and simplifies to:

The output voltage, VOUT, is:

Equation 5

The value of the output capacitor, which determines the ripple voltage, is:

Equation 6

where dV/dt represents the drop in the output voltage during the period of the PWM signal, I is the load current, and C is the required output capacitor.

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