🔁 Converter Topologies

The previous power-tree lesson used a buck as a choice: efficient step-down instead of a hot linear drop. Here we open the converter and ask what the three common non-isolated topologies actually do.

By the end, you can

1. Predict ideal output voltage for buck, boost and inverting buck-boost converters
2. Connect duty cycle to the on-time and off-time energy path
3. Estimate inductor ripple from voltage across the inductor, switching frequency and inductance
4. Choose a topology from the relationship between input, output and polarity

Intuition first

An inductor resists changes in current. That one sentence is the heart of every switching converter. During one part of the cycle the circuit pushes energy into the inductor's magnetic field. During the other part, the inductor releases that energy into the load. The switch decides which direction the energy path points.

In a buck, the switch connects the input to the inductor only part of the time. The load sees the average of those packets, so the output is lower than the input:

$$V_\text{out} = D V_\text{in}$$

In a boost, the switch first stores energy in the inductor, then places the inductor in series with the input so their voltages add. The output rises above the input:

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

In an inverting buck-boost, the inductor is charged from the input, then flipped so it discharges into the load with opposite polarity:

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

These formulas are ideal. Real converters lose voltage across switches, diodes, inductors and copper, and their controllers limit duty cycle so they have enough time to recharge, discharge and sense.

Drive it

Choose a topology, move duty cycle and switching frequency, and watch the ideal output and inductor ripple change. The ripple estimate is the inductor-voltage slope multiplied by on-time.

The important pattern is not the exact number. It is which knob changes which burden:

• More duty raises buck output, raises boost output sharply, and drives buck-boost deeper negative.
• More inductance lowers ripple because the same voltage produces a slower current slope.
• Higher switching frequency lowers ripple because each energy packet is shorter.

?A buck converter runs from 24 V with 25% duty cycle. What is the ideal output?

• A 6 V
• B 12 V
• C 24 V
• D 96 V

?Which topology naturally makes an output above the input while keeping the same polarity?

• A Buck
• B Boost
• C Inverting buck-boost
• D LDO

↘ Continuous and discontinuous conduction

The simple formulas above assume continuous conduction mode: inductor current never falls to zero during a cycle. At light load, many converters enter discontinuous conduction mode, where the inductor empties before the next cycle begins. Then the output no longer follows the same simple duty-cycle expression; load current and inductance become part of the conversion ratio. This is why converter datasheets show mode boundaries and why light-load efficiency can look very different from full-load efficiency.

Key takeaways

• ▸A buck steps down: $V_\text{out} = D V_\text{in}$.
• ▸A boost steps up: $V_\text{out} = V_\text{in}/(1-D)$.
• ▸An inverting buck-boost can step up or down but flips polarity.
• ▸Ripple falls when inductance or switching frequency rises.
• ▸Topology choice comes from output voltage, polarity, current, noise and available duty-cycle margin.

Practice 1 core

You have a 12 V battery and need 36 V for a small actuator rail. Which topology do you choose, and what ideal duty cycle gives 36 V?

↘ Show worked solution

Choose a boost converter because the output must be above the input with the same polarity. For an ideal boost,

$36 = \frac{12}{1-D}$

So $1-D = 12/36 = 1/3$ and $D = 2/3$, about 67% duty. In a real design you would leave margin for losses and controller limits, so the actual duty would be a little higher.