Step-up chopper operation and output voltage

A step-up (boost) chopper is a dc–dc converter whose average output voltage is greater than the input voltage. It uses an inductor to store energy while the switch is ON and releases it, added to the source voltage, to the load while the switch is OFF.

       L          D
 +---^^^^^---+---->|---+--------+
 |   iL ->   |         |        |
 |           |         |        |
(+) Vs       S        === C    Load  Vo
 |   (chopper switch)  |        |
 |           |         |        |
 +-----------+---------+--------+

Operation

Let the switch S (thyristor with commutation circuit, MOSFET, IGBT, etc.) be ON for $T_{on}$ and OFF for $T_{off}$; period $T = T_{on} + T_{off}$, duty cycle $D = T_{on}/T$.

Mode 1: S ON ($0 < t < T_{on}$)

• The inductor is connected directly across the source; diode D is reverse biased (blocked by the capacitor voltage).
• $v_L = V_s$, so $i_L$ rises linearly from $I_1$ to $I_2$; energy is stored in L.
• The capacitor supplies the load.

Mode 2: S OFF ($T_{on} < t < T$)

• The inductor current cannot change suddenly, so the inductor voltage reverses and adds to $V_s$; D becomes forward biased.
• Current flows from source + inductor to the load and capacitor: $v_L = V_s - V_o$ (negative), so $i_L$ falls from $I_2$ to $I_1$.
• Energy from both the source and the inductor is delivered to the load, so $V_o > V_s$.

 vL
 +Vs  +------+      +------+
      |      |      |      |
 0 ---+      |      |      |
             +------+      +---
 -(Vo-Vs)
 iL      /\      /\        I2 (max)
        /  \    /  \
       /    \  /    \      I1 (min)
      |<Ton>|<Toff>|

Derivation of average output voltage

In steady state the average voltage across an inductor over one period is zero (volt-second balance), i.e. the rise in current in Mode 1 equals the fall in Mode 2:

$$\begin{aligned} \Delta I &= \frac{V_s}{L}T_{on} = \frac{V_o - V_s}{L}T_{off} \\ V_sT_{on} &= (V_o - V_s)T_{off} \\ V_o T_{off} &= V_s(T_{on} + T_{off}) = V_sT \end{aligned}$$ $$V_o = V_s\frac{T}{T_{off}} = V_s\frac{T}{T - T_{on}} = \frac{V_s}{1 - D}$$

Alternative (energy method): energy taken by L in Mode 1, $V_sIT_{on}$, equals energy given in Mode 2, $(V_o - V_s)IT_{off}$, which leads to the same result (assuming ripple-free $I$).

Discussion

• As $D$ varies from 0 to 1, $V_o$ varies from $V_s$ to (theoretically) infinity; in practice losses limit the gain, and D is kept below about 0.8–0.9.
• Input current $I_s = I_o/(1-D)$ for a lossless converter, since $V_sI_s = V_oI_o$.
• Example: $V_s = 100$ V, $D = 0.6$: $V_o = 100/(1 - 0.6) = 250$ V.
• Applications: regenerative braking of dc motors (motor emf pumping energy back into a higher-voltage supply), battery-powered systems, PFC front ends and solar PV converters.