Power electronics introduction
Introduced| English | Meaning |
|---|---|
| buck | A DC–DC converter that gives an output voltage lower than its input voltage. |
| duty cycle | duty cycle |
| volt-second balance | volt-second balance |
| boost | A DC–DC converter that gives an output voltage higher than its input voltage. |
| continuous conduction | continuous conduction |
| synchronous rectification | synchronous rectification |
Smaller than its own plug
- A 30 W phone charger is now smaller than its plug. Inside, a switch flips a hundred thousand times a second.
- Switching converters change a DC voltage with almost no loss. This lesson builds the two simplest ones.
The buck converter
- A buck 降压 converter is a switch, a diode, an inductor and a capacitor. With the switch on, the input feeds the inductor; with it off, the inductor keeps current flowing through the diode.
- The switch sets only the duty cycle 占空比 $D$, the fraction of each period it is on. The inductor and capacitor do the averaging.
Volt-second balance
- In steady state the inductor's average voltage over a period is zero: volt-second balance 伏秒平衡. For the buck, $(V_{in} - V_{out})D = V_{out}(1 - D)$, so $V_{out} = DV_{in}$.
- The boost 升压 converter rearranges the same parts and gives $V_{out} = \dfrac{V_{in}}{1 - D}$.
A buck converter makes 3.3 V from 12 V. What duty cycle does it need? Give 3 decimal places.
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$D = V_{out}/V_{in} = 3.3/12 = 0.275$.
A boost converter makes 12 V from 5 V. What duty cycle does it need? Give 3 decimal places.
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$1 - D = V_{in}/V_{out} = 5/12$, so $D = 0.583$.
Ripple
- In continuous conduction 连续导通 (CCM) the inductor current never reaches zero. It ramps up and down by $\Delta I_{L} = \dfrac{V_{out}(1 - D)}{Lf}$.
- The output voltage ripple is $\Delta V_{out} = \dfrac{\Delta I_{L}}{8Cf}$.
That 3.3 V buck switches at 500 kHz with L = 10 µH. What is the inductor ripple current, in A? Give 2 decimal places.
Difficulty 2/5 · 140 coins + a small bonus on first completion today
$\Delta I_{L} = 3.3 \times 0.725 / (10 \times 10^{-6} \times 500 \times 10^{3}) = 0.48$ A.
Worked example: 12 V to 5 V
- 5 V at 2 A from 12 V at 100 kHz with $L = 100$ µH: $D = 5/12 = 0.417$ and $\Delta I_{L} = \dfrac{5 \times 0.583}{100\text{ µH} \times 100\text{ kHz}} = 0.29$ A, 15% of the load.
- With $C = 220$ µF, $\Delta V_{out} = \dfrac{0.29}{8 \times 220\text{ µF} \times 100\text{ kHz}} = 1.7$ mV.
Worked example: 12 V to 24 V
- A boost for 24 V at 1.5 A from 12 V needs $D = 1 - 12/24 = 0.5$. The output is 36 W, so the input current is 3 A, not 1.5 A.
- Its output capacitor takes current only in pulses, so a boost needs a bigger, lower-resistance capacitor than a buck of the same power.
An ideal boost converter delivers 24 V at 2 A from a 12 V supply. What current does it draw, in A?
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Power is conserved: 48 W out means 48 W in, and $48/12 = 4$ A.
Three common slips
- Light load. Below $\Delta I_{L}/2$ the current reaches zero, CCM ends and $V_{out}$ rises above $DV_{in}$.
- Ignoring the diode drop. 0.4 V is 8% of a 5 V rail; synchronous rectification 同步整流 replaces the diode with a MOSFET.
- Equal currents in and out. Only power is conserved.
A buck designed for CCM has ΔI_L = 0.29 A. The load falls to 0.1 A. What happens?
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The boundary is $\Delta I_{L}/2 = 0.145$ A. Below it the inductor current reaches zero each cycle, and the simple formula no longer holds.
In a 1.8 V converter, a 0.4 V diode drop wastes a large share of the output power, so a MOSFET often replaces the diode.
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0.4 V against 1.8 V is about 22%. Synchronous rectification cuts that loss to a small resistive drop.
Recap
- Buck: $V_{out} = DV_{in}$. Boost: $V_{out} = V_{in}/(1 - D)$.
- Ripple: $\Delta I_{L} = V_{out}(1 - D)/(Lf)$ and $\Delta V_{out} = \Delta I_{L}/(8Cf)$.
- The switch sets the ratio; the inductor and capacitor set the quality.
Match each converter to its output voltage in CCM.
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Both formulas come from volt-second balance on the inductor.