# Basic Power Conversion Examples

Let’s look at a few examples of power electronic systems.

Consider the circuit shown below. It contains an AC source, a switch, and a resistive load. It is therefore a simple but complete power electronic system.

A very simple power electronic system described in the first example.

Just to get an idea of how power conversion might take place, let us assign some kind of control to the switch. What if the switch is turned on whenever $V_{ac}>0$, and turned off otherwise? The input and output voltage waveforms are illustrated below. The input has a time average of 0, and an RMS value equal to $V_{peak}/\sqrt{2}$. The output has a nonzero average value given by

\begin{aligned} \langle v_{out}(t)\rangle &= \frac{1}{2 \pi} (\int_{-\pi/2}^{3\pi/2} V_{peak}cos\theta d\theta + \int_{\pi/2}^{3\pi/2} 0 d\theta)\\ &= \frac{V_{peak}}{\pi}\\ &= 0.3183V_{peak} \end{aligned}

and an RMS value equal to $V_{peak}/2$. The output has some DC voltage content. The current can be thought of as an AC-DC converter

Input and output waveforms for the example circuit.

The circuit example above is a half-wave rectifier with a resistive load. A diode can be substituted for the switch. The example shows that a simple switching current can perform power conversion functions. But, notice that a diode is not, in general, the same as the switch. A diode places restrictions on the current direction, while a true switch would not. An ideal switch allows control over whether it is on or off, while a diode’s operation is constrained by circuit variables. Consider a second half-wave circuit, now with a series L-R (inductor-resistor) load, shown below.

Half-wave rectifier with R-L load described in second example.

A series D-L-R (diode-inductor-resistor) circuit has AC voltage-source input. This circuit operates much differently than the half-wave rectifier with resistive load. Remember that a diode will be on if forward biased, and of if reversed biased. In this circuit, an off diode will give $i=0$. Whenever the diode is on, the circuit is the AC source with R-L load. Let the AC voltage be %latex V_0 cos(\omega t)\$. From Kirchoff’s Voltage Law,

$V_o cos(\omega t) = L \frac{di}{dt} + Ri$

Let us assume that the diode is initially off (this assumption is arbitrary, and we will check it out as the example is solved). If the diode is off, $i=0$, and the voltage across the diode $v_d = v_{ac}$. The diode will become forward-biased when $v_{ac}$ becomes positive. The diode will turn on when the input voltage makes a zero-crossing in the positive direction. This allows us to establish initial conditions for the circuit: $i(t_0) = 0, t_0 = -\pi/(2\omega)$. The differential equation can be solved with conventional methods. Though one is able to solve such equations in closed form by hand, it is generally much simpler to evaluate with modern symbolic computer programs such as Mathematica. Other, more complex problems, must be solved numerically. In this case, the somewhat complex result is

$i(t) = V_0 [\frac{\omega L}{R^2 + \omega^2L^2} exp (\frac{-t}{\tau}-\frac{\pi}{2\omega\tau}) + \frac{R}{R^2+\omega^2L^2}cos(\omega t) + \frac{\omega L}{R^2+\omega^2L^2}sin{\omega t} ]$

where $\tau$ is the time constant $L/R$. What about diode turn-off? One first guess might be that the diode turns off when the voltage becomes negative, but this is not correct. We notice from the solution that the current is not zero when the voltage first becomes negative (check this for yourself!). If the switch attempts to turn off, it must drop the inductor current to zero instantly. The derivative of current in the inductor, $di/dt$, would become negative infinite. The inductor voltage $L(di/dt)$ similarly becomes negative infinite – and the devices are destroyed. What really happens is that the falling current allows the inductor to maintain forward bias on the diode. The diode will turn off only when the current reaches zero. A diode has very definite properties that determine the circuit action, and both the voltage and current are relevant.