DC-DC converters

The idea

A battery gives one voltage, but a robot needs several: one for the motors, one for the logic, one for the sensors. A DC-DC converter chops the input with a MOSFET (PWM), then smooths the result with an inductor (which stores energy in its current) and a capacitor (which stores energy in its voltage). The duty cycle $D$ sets the output.

Buck (step-down)

• Switch on: the battery drives current through L into the load. L stores energy.
• Switch off: L keeps the current flowing (inductors resist sudden changes in current), and the diode completes the loop. This is called freewheeling.

$\boxed{V_{out}=D{V}_{in}}$

Boost (step-up)

• Switch on: L is connected straight across the battery and charges up. The capacitor feeds the load.
• Switch off: the battery and L push together through the diode, so the output rises above $V_{in}$.

$\boxed{V_{out}=\frac{V_{in}}{1-D}}$

Deriving both: volt-second balance

In steady state, an inductor’s average voltage over one cycle must be zero. If it weren’t, its current would keep growing every cycle. So (voltage × time while on) + (voltage × time while off) = 0.

Buck. On for $DT$: the inductor sits between $V_{in}$ and $V_{out}$, so $V_L=V_{in}-V_{out}$. Off for $(1-D)T$: the diode connects it to ground, so $V_L=-V_{out}$.

$(V_{in}-V_{out})D=V_{out}(1-D)\Rightarrow V_{out}=D{V}_{in}$

Boost. On: $V_L=V_{in}$. Off: $V_L=V_{in}-V_{out}$, which is negative.

$V_{in}D+(V_{in}-V_{out})(1-D)=0\Rightarrow V_{out}=\frac{V_{in}}{1-D}$

Sizing the buck’s L and C

Inductor: $V=L\frac{\Delta I}{\Delta t}$. While off, $V_L=V_{out}$ for a time $(1-D)/F_{SW}$, and the current changes by the ripple $\Delta I_L$:

$\boxed{L=\frac{V_{out}(1-D)}{F_{SW}\Delta I_L}}$

Capacitor: it absorbs the inductor’s triangle-shaped ripple current:

$\boxed{C_{out}\ge \frac{\Delta I_L}{8F_{SW}\Delta V_{out}}}$

Higher frequency means a smaller L and C, which is why modern converters are tiny. The trade-off is more switching loss.

Worked example

Step a 48 V robot battery down to 24 V at 5 A, switching at 100 kHz, with 10% ripple on both current and voltage. Find D, L and C.

Answer

$D=24/48=0.5$. $\Delta I_L=0.1\times 5=0.5$ A. $\Delta V=0.1\times 24=2.4$ V. $L=\frac{24\times 0.5}{100000\times 0.5}=240\mu$H $C\ge \frac{0.5}{8\times 100000\times 2.4}=0.26\mu$F

Isolated or not?

	Transformer-less (buck, boost)	Transformer-based (isolated)
How	Switch, inductor, capacitor	DC → high-frequency AC → transformer → rectified back to DC
Galvanic isolation	No: a fault passes to the load	Yes: protects the user and load
Efficiency	90–98%	80–95%
Size and cost	Small, cheap	Bulkier, costs more
Voltage ratio	Low to moderate	Extreme step-up or step-down
Outputs	One	One or several

Buck-boost converters can go either way, which is useful when a battery sags below the output as it drains. Synchronous versions replace the diode with a second MOSFET to cut loss.

Check yourself

1. A boost converter takes 12 V and must output 36 V. What duty cycle?

$36=12/(1-D)$ → $1-D=1/3$ → $D=0.667$.

2. In the buck L formula, which voltage goes on top, $V_{in}$ or $V_{out}$?

$V_{out}$. During the off-time, the inductor has $V_{out}$ across it.