Why doesn't the polarity of the inductor reverse in this buck-boost?

Question

Given the buck-boost below

_{Fig. 3.33, page 62 from Fundamentals of Power Electronics 3rd edition by Robert W. Erickson and Dragan Maksimovic}

If we assume the transistor is in conduction mode and the inductor has parasitic resistance \$R_L\$, applying KVL clock-wise we would get

$$ V_{batt} - I_LR_L - V_L = 0 \Rightarrow V_L = V_{batt} - I_LR_L $$

When the transistor is in cut-off mode, we apply KVL in the second circuit, that is:

$$ -V_L + I_LR_L + V_D + V = 0 \Rightarrow V_L = I_LR_L + V_D + V $$

When we use volts-seconds balance, the part of the relation that contains the inductor current for interval \$DT\$ and \$(1-D)T\$ will yield

$$ -I_LR_LD+I_LR_L(1-D) =I_LR_L(1 - 2D) $$

This seems to be wrong because there is no \$(1 - 2D)\$ in the final relation \$V(D)\$. The final relation assumes that the polarity of the inductor doesn't reverse, which would result in:

$$ V_L + I_LR_L + V_D = V \Rightarrow V_L = -I_LR_L - V_D + V $$

The part which contains \$I_L\$ now being

$$ -I_LR_LD-I_LR_L(1-D) =-I_LR_L $$

Using the assumption that the inductor's polarity doesn't reverse, I can get to the right relation between \$V_{batt}\$ and \$V\$, but why do we assume that it doesn't reverse? The inductor has to be in series with the battery, such that

$$ V_{batt} + V_L - I_LR_L \ge V_x - V_D $$

Where \$V_x\$ is the voltage at the top rail, but using the assumption that the polarity doesn't reverse, then

$$ V_{batt} - V_L - I_LR_L\ge V_x - V_D $$

This can't be since the current through the inductor is decreasing, hence its polarity will reverse. Why is it assumed that it is not reversed? Is this assumption only valid in this context, that is an ideal world, because the battery is connected to the terminal of the inductor so it keeps its polarity fixed?

Answer 1

When the transistor is in cut-off mode, we apply KVL in the second circuit, that is: $$ -V_L + I_LR_L + V_D + V = 0 \Rightarrow V_L = I_LR_L + V_D + V $$

Knowing that the inductor voltage polarity will reverse, you explicitly reversed it in the KVL equation. Let KVL tell you that it is reversed. The assumed "KVL polarity" must remain the same as when the switch is open to observe the polarity change.

The correct KVL equation is: $$ V_L + I_LR_L + V_D + V = 0 \Rightarrow V_L = -(I_LR_L + V_D + V) $$ which then indicates the polarity reversal.

Answer 2

Too complex reasoning used in the question. A practical electrician cannot follow equations this long.

When switch Q1 starts to conduct the inductor current starts to grow. The inductor current is upwards in your drawing because the battery plus is the bottom side of the battery.

When the switch is turned OFF the inductor current is still upwards. Its only route without burning anything is through the diode and that current charges the output capacitor so that the plus pole of the output voltage is the marked V+.

I would claim that this reverses the polarity if the bottom horizontal line of the schematic is the GND.