# How a boost DC-DC converter works

There are some electrical systems applications in which we have a low voltage source (e.g. 24 V) and we need to power a component which requires a higher voltage (e.g. 120 V). There are several ways of raising a lower voltage to a higher voltage. When the voltage ratio of the output voltage compared to the input voltage is between 1 and 5, a boost DC-DC converter is used.

A boost DC-DC converter is a class of switching-mode power supply, which contains at least two semiconductor switches (a diode and a transistor) and at least two energy storage components (capacitor and inductor). The circuit of the PWM boost DC-DC converter is shown in the image below.

where:

VIN [V] – input voltage (lower, e.g. 24)
VOUT [V] – output voltage (higher, e.g. 120)
L [H] – inductance of the load inductor
C [F] – capacitance of the filter capacitor
R [Ω] – resistance of the resistor (load)
D – diode
S – switch (transistor, usually power MOSFET)
iL [A] – current through inductor
iD [A] – current through diode
iOUT [A] – output current

It’s called a PWM boost DC-DC converter because the switch (S) is controlled through a pulse width modulated (PWM) signal. Its output voltage VOUT [V] is always higher than the input voltage VIN [V] during steady-state operation. With other words, it “boosts” the voltage to a higher level.

How does it work ? First of all we need to know that the switch (S) is turned ON and OFF at the switching frequency fs = 1/T [Hz] and has the ON duty ratio D = tON/T [-], where:

T [s] – period (duration) of the PWM signal
tON [s] – time interval for which the switch (S) is ON

Also, there are two modes in which a boost DC-DC converter works:

• continuous conduction mode (CCM): means that the current in the inductor never goes to zero between switching cycles;
• discontinuous conduction mode (DCM): means that the current goes to zero during part of the switching cycle.

In this article we are going to cover only the boost DC-DC converter operating in CCM. To simplify the equations defining the circuit, we are going to make the following assumptions:

• the transistor (power MOSFET) and the diode will be considered ideal switches;
• the transistor output capacitance, the diode capacitance, and lead inductances (and thereby switching losses) are considered to be zero.

There are basically two states of the boost DC-DC converter:

• switch (S) ON (closed), when 0 < t ≤ D·T
• switch (S) OFF (open), when D·T < t ≤ T

Let’s make an analysis of the circuit for each state.

### Switch (S) ON (closed)

In this state the switch (S) is ON and the diode (D) is reversed biased (OFF). The current and magnetic flux in the inductor starts to increase. Since the diode (D) is OFF, it separates the load resistor from the low-resistance part of the circuit (collector–emitter of the transistor). The load current is maintained by the energy stored in the capacitor.

The voltage across the diode is vD = −VOUT , causing the diode to be reverse biased. The voltage across the inductor is vL = VIN . As a result, the inductor current increases linearly with a slope of VIN ∕L. The voltage across the switch is vS = 0. Consequently, the magnetic energy in the inductor also increases. The switch current is equal to the inductor current and the diode current is zero.

The voltage across the inductor vL is:

$v_{L}(t) = V_{IN} = L \frac{di_{L}}{dt} \tag{1}$

From equation (1) we can find the mathematical expression of the current through the inductor and the switch:

$i_{L} = i_{S} = \frac{1}{L} \int_{0}^{t}{v_{L}dt} + i_{L}(0) = \frac{1}{L} \int_{0}^{t}{V_{IN}dt} + i_{L}(0) = \frac{V_{IN}}{L}t + i_{L}(0) \tag{2}$

where iL(0) is the initial inductor current at time t = 0.

During this stage, the inductor current keeps rising and it reaches maximum value at the end of the time interval when t = D·T. From equation (2) we can find the maximum (peak) inductor current as:

$i_{L}(DT) = \frac{V_{IN} \cdot D \cdot T}{L} + i_{L}(0) \tag{3}$

The peak-to-peak value of the inductor current is expressed as:

$\Delta i_{L} = i_{L}(DT) – i_{L}(0) = \frac{V_{IN} \cdot D \cdot T}{L} + i_{L}(0) – i_{L}(0) = \frac{V_{IN} \cdot D \cdot T}{L} \tag{4}$

We know that the period T [s] is equal with the inverse of the switching frequency fs [Hz]:

$T = \frac{1}{f_{s}} \tag{5}$

Replacing (5) in (4), gives the expression of the peak-to-peak inductor current of the boost DC-DC converter when the switch (S) is closed as:

$\Delta i_{L} = \frac{V_{IN} \cdot D}{f_{s} \cdot L} \tag{6}$

### Switch (S) OFF (open)

In this state the switch (S) is OFF and the magnetic flux in the inductor starts to decrease. The electromotive forces in the winding are pulling in the direction opposite to that while the transistor was ON. As a result, the diode is conducting and the output voltage VOUT [V] is the sum of the input voltage VIN [V] and the voltage across the inductor vL [V]. The current through the inductor decreases and reaches its minimum at the end of the period T [s]. The switch current iS [A] and the diode voltage drop vD [V] are both zero.

The voltage across the inductor can be written as:

$v_{L} = V_{IN} – V_{OUT} = L \frac{di_{L}}{dt} \tag{7}$

The inductor current can be found by integration of equation (7):

$i_{L} = i_{D} = \frac{1}{L} \int_{DT}^{t} v_{L} dt + i_{L}(DT) \tag{7.1}$

Replacing first part of eqtion (7) in (7.1) gives:

$i_{L} = \frac{1}{L} \int_{DT}^{t} (V_{IN} – V_{OUT}) dt + i_{L}(DT) = \frac{V_{IN} – V_{OUT}}{L} (t – DT) + i_{L}(DT) \tag{8}$

The minimum inductor current is found at the end of the cycle (period) when t = T:

$i_{L}(T) = \frac{V_{IN} – V_{OUT}}{L} (T – DT) + i_{L}(DT) \tag{9}$

The peak-to-peak value of the inductor ripple current is:

$\Delta i_{L} = i_{L}(DT) – i_{L}(T) \tag{10}$

Replacing (10) in (9) gives the expression of the peak-to-peak inductor current of the boost DC-DC converter when the switch (S) is open as:

$\Delta i_{L} = i_{L}(DT) – \frac{V_{IN} – V_{OUT}}{L} (T – DT) – i_{L}(DT) = \frac{(V_{OUT} – V_{IN})(1 – D)T}{L} \tag{11}$

Taking into account that the change in inductor current during the ON period (equation (4)) and OFF period (equations (11)) are equal or the average inductor current over one switching cycle is zero, the following equations can be derived:

$\frac{V_{IN} \cdot D \cdot T}{L} + \frac{(V_{OUT} – V_{IN})(1 – D)T}{L} = 0 \tag{12}$

Simplifying the common terms of equations (12) gives the expression of the output voltage function of the input voltage and duty cycle:

$\bbox[#FFFF9D]{V_{OUT} = \frac{V_{IN}}{1-D}} \tag{13}$

### Boost DC-DC converter idealized waveforms

Idealized waveforms of the currents and voltages that explain the principle of operation of the boost converter are depicted in the image below.

To summarize the behaviour of the boost DC-DC converter, when the switch (S) is ON (closed), the inductor current increases linearly. When the switch (S) is OFF (open), the inductor current starts to decrease linearly, but never reaches zero throughout the operating range. The average of the inductor curent is equal with the load current.

Usually the duty cycle is 0 < D < 0.8, which means that the output voltage can be controlled within limits VIN < VOUT< 5VIN . Essentially, the limitation on the ratio VOUT /VIN is imposed by the real parameters of the elements of the converter.

In the following articles we are going to calculate the parameters of a boost DC-DC converter for a given output voltage and also simulate the behaviour of the converter using Scilab/Xcos environment.