Step-up chopper average and rms output

A step-up chopper produces an average output voltage higher than the source. An inductor in series with the source stores energy while the switch is ON and pushes it into the load, in series with $V_s$, while the switch is OFF.

Circuit and operation

      L        D
 +---mmm---+---|>|---+------+
 |   iL    |  vx     |      |
 Vs       CH        ===C   Load  V0
 |         |         |      |
 +---------+---------+------+

1. CH ON ($0 \le t \le T_{on}$): node $x$ is shorted to the negative rail ($v_x = 0$). D is reverse biased by $V_0$. $V_s$ appears across L, so $i_L$ rises linearly: $L\,di/dt = V_s$. The capacitor feeds the load.
2. CH OFF ($T_{on} \le t \le T$): the inductor EMF reverses and adds to $V_s$; D conducts, $v_x = V_0$, and $i_L$ falls: $L\,di/dt = V_s - V_0$ (negative).

 vx  V0 |    ____      ____
        |   |    |    |    |
      0 |___|    |____|    |___ t
 iL  I2 |   /\        /\
        |  /  \      /  \
     I1 | /    \____/    \___ t
        0 Ton  T

Average output voltage

Rise of current in $T_{on}$ equals the fall in $T_{off}$ (steady state):

$$\begin{aligned} \Delta I &= \frac{V_s}{L}T_{on} = \frac{V_0 - V_s}{L}T_{off} \\ V_0\,T_{off} &= V_s (T_{on} + T_{off}) = V_s T \\ V_0 &= \frac{V_s\,T}{T - T_{on}} = \frac{V_s}{1-k} \end{aligned}$$

Equivalently, the switch-node voltage $v_x$ is 0 for $kT$ and $V_0$ for $(1-k)T$; its average must equal $V_s$ (no average voltage across L): $(1-k)V_0 = V_s$.

For $k = 0 \to 1$, $V_0$ varies from $V_s$ to a very large value; e.g. $k = 0.5$ gives $V_0 = 2V_s$.

RMS value of output voltage

• With a large filter capacitor the load voltage is nearly ripple-free, so $V_{0,rms} \approx V_0 = \dfrac{V_s}{1-k}$.
• The chopped voltage at the chopper terminals ($v_x$, the voltage the load sees before filtering) is a pulse of height $V_0$ for $(1-k)T$:

$$\begin{aligned} V_{x,rms} &= \left[\frac{1}{T}\int_{kT}^{T}V_0^2\,dt\right]^{1/2} = V_0\sqrt{1-k} \\ &= \frac{V_s}{1-k}\sqrt{1-k} = \frac{V_s}{\sqrt{1-k}} \end{aligned}$$

and its average is $V_{x,avg} = (1-k)V_0 = V_s$.

Example: $V_s = 100$ V, $k = 0.6$: $V_0 = 100/0.4 = 250$ V; $V_{x,rms} = 100/\sqrt{0.4} = 158.1$ V.

Uses: regenerative braking of dc motors, battery charging from a lower voltage, boost SMPS.