Step-up chopper and controllable power transfer

A step-up (boost) chopper gives an average output voltage higher than the input. It stores energy in an inductor while the switch is ON and releases it, added to the source, into the load while the switch is OFF.

Principle and circuit

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

• CH ON ($0 < t < T_{on}$): D is reverse biased. $V_s$ is across L, so $i_L$ rises linearly from $I_1$ to $I_2$; energy is stored in L. The capacitor supplies the load.
• CH OFF ($T_{on} < t < T$): the inductor voltage reverses ($L\,di/dt$ adds to $V_s$), D conducts and current flows to the load at voltage $V_0 = V_s + L\,di/dt$. $i_L$ falls from $I_2$ to $I_1$.

 vCH  V0|    ____      ____
        |   |    |    |    |
       0|___|    |____|    |___ t
 iL   I2|   /\        /\
        |  /  \      /  \
      I1| /    \____/    \___  t
         Ton Toff

Output voltage

In steady state, the volt-second balance on L (average inductor voltage = 0):

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

The same result follows from energy balance: energy put into L during $T_{on}$ is $V_s I T_{on}$, energy given out during $T_{off}$ is $(V_0 - V_s) I T_{off}$. As $k$ goes from 0 to 1, $V_0$ rises from $V_s$ towards infinity (in practice limited by losses).

Controllable power transfer (energy from a low voltage to a higher voltage)

The same circuit transfers energy from a low-voltage source $V_s$ (e.g. a dc motor's back EMF in regenerative braking) into a fixed higher-voltage dc source $E$ (supply or battery) through D. With resistance $R$ of the source/inductor:

$$\begin{aligned} \text{ON: } & V_s = L\frac{di_1}{dt} + Ri_1 \\ \text{OFF: } & V_s = L\frac{di_2}{dt} + Ri_2 + E \end{aligned}$$

For the current to be controllable:

1. During ON, current must rise: $di_1/dt > 0$, which needs $V_s > 0$.
2. During OFF, current must fall: $di_2/dt < 0$, which needs $V_s < E$ (neglecting $Ri$).

So the condition for controllable power transfer is

$$0 \le V_s \le E$$

Using the boost relation, average values give $V_s = (1-k)E$; as $k$ varies from 1 to 0, $V_s$ can be anywhere from 0 to $E$. If $V_s > E$, current would rise in both intervals and could not be controlled by the switch; if $V_s \le 0$, no energy would flow to $E$.