[Circuit] Buck Converter

電路知識：Buck Converter

Tool：: Qucs

Example Circuits

簡介：Buck Converter 原理

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設計概念

Continuous mode

\begin{aligned} V_{o, a v} & = D V_{i} - \frac{1}{T} (\int_{0}^{t_{o n}} V_{s w} d t - \int_{t_{o n}}^{T} V_{D} d t) \\ Δ I_{L} & = \frac{D T V_{i}}{L} - \int_{0}^{T_{o n}} \frac{V_{s w}}{L} d t - \int_{0}^{T_{o n}} \frac{V_{o}}{L} d t = \int_{t_{o n}}^{T} \frac{V_{o} - V_{D}}{L} d t \\ I_{a v} & = \frac{V_{o, a v}}{R} \\ L & \geq \frac{R}{2 V_{o, a v}} \int_{t_{o n}}^{T} (V_{o} - V_{D}) d t \\ V_{r i p p l e} & = \frac{Δ I_{L} T}{8 C} i f C i s l a r g e \end{aligned}

By Buck_chronogram.svg: efadae derivative work: Efadae (talk) - Buck_chronogram.svg, CC BY-SA 3.0, Link

{\begin{cases} V_{o} = V_{i} - V_{s w} - V_{L} & if t i n T_{o n} = D T \\ V_{o} = - V_{L} + V_{D} & if t i n T_{o f f} = (1 - D) T \end{cases}

平均輸出電壓

\begin{aligned} V_{o, a v} & = \frac{1}{T} \int_{0}^{t} V_{o} (t) d t \\ = \frac{1}{T} (\int_{0}^{t_{o n}} V_{o} (t) d t + \int_{t_{o n}}^{T} V_{o} (t) d t) \\ = \frac{1}{T} (\int_{0}^{t_{o n}} (V_{i} - V_{s w} - V_{L}) d t + \int_{t_{o n}}^{T} (- V_{L} + V_{D}) d t) \\ = \frac{1}{T} (\int_{0}^{t_{o n}} V_{i} d t - \int_{0}^{t_{o n}} V_{s w} d t - \int_{0}^{t_{o n}} V_{L} d t - \int_{t_{o n}}^{T} V_{L} d t + \int_{t_{o n}}^{T} V_{D} d t) \\ = \frac{1}{T} (\int_{0}^{t_{o n}} V_{i} d t - \int_{0}^{t_{o n}} V_{s w} d t - \int_{0}^{T} V_{L} d t + \int_{t_{o n}}^{T} V_{D} d t) \\ = \frac{1}{T} (D T V_{i} - \int_{0}^{t_{o n}} V_{s w} d t - 0 + \int_{t_{o n}}^{T} V_{D} d t) \\ = D V_{i} - \frac{1}{T} (\int_{0}^{t_{o n}} V_{s w} d t - \int_{t_{o n}}^{T} V_{D} d t) \end{aligned}

V_{L} = L \frac{d I_{L}}{d t}

\begin{aligned} Δ I_{L_{o n}} & = I_{m a x} - I_{m i n} \\ = \int_{0}^{T_{o n}} \frac{V_{L}}{L} d t \\ = \int_{0}^{T_{o n}} \frac{V_{i} - V_{s w} - V_{o}}{L} d t \\ = \int_{0}^{T_{o n}} \frac{V_{i}}{L} d t - \int_{0}^{T_{o n}} \frac{V_{s w}}{L} d t - \int_{0}^{T_{o n}} \frac{V_{o}}{L} d t \\ = \frac{D T V_{i}}{L} - \int_{0}^{T_{o n}} \frac{V_{s w}}{L} d t - \int_{0}^{T_{o n}} \frac{V_{o}}{L} d t \\ Δ I_{L_{o f f}} & = I_{m i n} - I_{m a x} \\ = \int_{t_{o n}}^{T} \frac{V_{L}}{L} d t \\ = \int_{t_{o n}}^{T} \frac{- V_{o} + V_{D}}{L} d t \end{aligned}

\begin{aligned} Δ I_{L_{o n}} = I_{m a x} - I_{m i n} & = \frac{D T V_{i}}{L} - \int_{0}^{T_{o n}} \frac{V_{s w}}{L} d t - \int_{0}^{T_{o n}} \frac{V_{o}}{L} d t \\ Δ I_{L_{o f f}} = I_{m i n} - I_{m a x} & = \int_{t_{o n}}^{T} \frac{- V_{o} + V_{D}}{L} d t \end{aligned}

從另一方面解

V_{o, a v}

，因電感電流穩定

\begin{aligned} 0 & = Δ I_{L_{o n}} + Δ I_{L_{o f f}} \\ 0 & = \frac{D T V_{i}}{L} - \int_{0}^{T_{o n}} \frac{V_{s w}}{L} d t - \int_{0}^{T_{o n}} \frac{V_{o}}{L} d t + \int_{t_{o n}}^{T} \frac{- V_{o} + V_{D}}{L} d t \\ 0 & = \frac{D T V_{i}}{L} - \int_{0}^{T_{o n}} \frac{V_{s w}}{L} d t - \int_{0}^{T} \frac{V_{o}}{L} d t + \int_{t_{o n}}^{T} \frac{V_{D}}{L} d t \\ \int_{0}^{T} \frac{V_{o}}{L} d t & = \frac{D T V_{i}}{L} - \int_{0}^{T_{o n}} \frac{V_{s w}}{L} d t + \int_{t_{o n}}^{T} \frac{V_{D}}{L} d t \\ \int_{0}^{T} V_{o} d t & = D T V_{i} - \int_{0}^{T_{o n}} V_{s w} d t + \int_{t_{o n}}^{T} V_{D} d t \\ \frac{1}{T} \int_{0}^{T} V_{o} d t & = D V_{i} - \frac{1}{T} \int_{0}^{T_{o n}} V_{s w} d t + \frac{1}{T} \int_{t_{o n}}^{T} V_{D} d t \\ V_{o, a v} & = D V_{i} - \frac{1}{T} (\int_{0}^{T_{o n}} V_{s w} d t - \int_{t_{o n}}^{T} V_{D} d t) \end{aligned}

因電感平均電流位於最大值與最小值的中間
又根據之前推導的

Δ I_{L_{o n}}

可得下面兩式

{\begin{matrix} I_{m a x} + I_{m i n} & = & 2 * I_{a v} \\ I_{m a x} - I_{m i n} & = & \frac{D T V_{i}}{L} - \int_{0}^{T_{o n}} \frac{V_{s w}}{L} d t - \int_{0}^{T_{o n}} \frac{V_{o}}{L} d t \end{matrix} {\begin{matrix} 2 \cdot I_{m a x} & = & 2 * I_{a v} + \frac{D T V_{i}}{L} - \int_{0}^{T_{o n}} \frac{V_{s w}}{L} d t - \int_{0}^{T_{o n}} \frac{V_{o}}{L} d t \\ 2 \cdot I_{m i n} & = & 2 * I_{a v} - \frac{D T V_{i}}{L} + \int_{0}^{T_{o n}} \frac{V_{s w}}{L} d t + \int_{0}^{T_{o n}} \frac{V_{o}}{L} d t \end{matrix} {\begin{matrix} I_{m a x} & = & I_{a v} + \frac{D T V_{i}}{2 L} - \int_{0}^{T_{o n}} \frac{V_{s w}}{2 L} d t - \int_{0}^{T_{o n}} \frac{V_{o}}{2 L} d t \\ I_{m i n} & = & I_{a v} - \frac{D T V_{i}}{2 L} + \int_{0}^{T_{o n}} \frac{V_{s w}}{2 L} d t + \int_{0}^{T_{o n}} \frac{V_{o}}{2 L} d t \end{matrix}

或者根據之前推導的

Δ I_{L_{o f f}}

可得下面兩式

{\begin{matrix} I_{m a x} + I_{m i n} & = & 2 * I_{a v} \\ I_{m i n} - I_{m a x} & = & \int_{t_{o n}}^{T} \frac{- V_{o} + V_{D}}{L} d t \end{matrix} {\begin{matrix} I_{m a x} + I_{m i n} & = & 2 * I_{a v} \\ I_{m a x} - I_{m i n} & = & \int_{t_{o n}}^{T} \frac{V_{o} - V_{D}}{L} d t \end{matrix} {\begin{matrix} 2 \cdot I_{m a x} & = & 2 * I_{a v} + \int_{t_{o n}}^{T} \frac{V_{o} - V_{D}}{L} d t \\ 2 \cdot I_{m i n} & = & 2 * I_{a v} - \int_{t_{o n}}^{T} \frac{V_{o} - V_{D}}{L} d t \end{matrix} {\begin{matrix} I_{m a x} & = & I_{a v} + \int_{t_{o n}}^{T} \frac{V_{o} - V_{D}}{2 L} d t \\ I_{m i n} & = & I_{a v} - \int_{t_{o n}}^{T} \frac{V_{o} - V_{D}}{2 L} d t \end{matrix}

保持 continuous mode 的條件，就是

I_{L} \geq 0

\begin{aligned} I_{m i n} & = I_{a v} - \int_{t_{o n}}^{T} \frac{V_{o} - V_{D}}{2 L} d t \geq 0 \\ I_{a v} & \geq \int_{t_{o n}}^{T} \frac{V_{o} - V_{D}}{2 L} d t \\ \frac{V_{o, a v}}{R} & \geq \int_{t_{o n}}^{T} \frac{V_{o} - V_{D}}{2 L} d t \\ L & \geq \frac{R}{2 V_{o, a v}} \int_{t_{o n}}^{T} (V_{o} - V_{D}) d t \end{aligned}

前提假設，電容很大，所以負載

R

的電流趨近於直流
故

I_{a v}

為

R

的電流

\begin{aligned} Q & = C V_{C} \\ I_{C} & = C \frac{d V_{C}}{d t} \\ I_{L} & = I_{C} + I_{R} \\ I_{a v} + Δ I_{L} & = C \frac{d V_{C}}{d t} + I_{a v} \\ Δ I_{L} & = C \frac{d V_{C}}{d t} \\ \int_{t_{1}}^{t_{2}} \frac{Δ I_{L}}{C} d t & = Δ V_{C} \end{aligned}

取

I_{L} > I_{a v}

的

Δ I_{L}

可得最大 Voltage Ripple
直接求三角形面積

\begin{aligned} Δ V_{C} & = \int_{t_{1}}^{t_{2}} \frac{Δ I_{L}}{C} d t \\ = \frac{1}{C} \cdot \frac{1}{2} \cdot \frac{Δ I_{L}}{2} \cdot \frac{T}{2} \\ = \frac{Δ I_{L} T}{8 C} \end{aligned}

Discontinuous mode

\begin{aligned} V_{o, a v} & = D V_{i} - \frac{1}{T} (\int_{0}^{t_{o n}} V_{s w} d t - \int_{t_{o n}}^{t_{o n} + δ T} V_{D} d t - \int_{t_{o n} - δ T}^{T} V_{C} d t) \\ Δ I_{L} & = \frac{D T V_{i}}{L} - \int_{0}^{T_{o n}} \frac{V_{s w}}{L} d t - \int_{0}^{T_{o n}} \frac{V_{o}}{L} d t = \int_{t_{o n}}^{t_{o n} + δ T} \frac{V_{o} - V_{D}}{L} d t \end{aligned}

By No machine-readable author provided. CyrilB~commonswiki assumed (based on copyright claims). - No machine-readable source provided. Own work assumed (based on copyright claims)., CC BY-SA 3.0, Link

{\begin{cases} V_{o} = V_{i} - V_{s w} - V_{L} & if t i n T_{o n} = D T \\ V_{o} = - V_{L} + V_{D} & if t i n δ T \\ V_{o} = V_{C} & if t i n o t h e r s \end{cases}

平均輸出電壓

\begin{aligned} V_{o, a v} & = \frac{1}{T} \int_{0}^{t} V_{o} (t) d t \\ = \frac{1}{T} (\int_{0}^{t_{o n}} V_{o} (t) d t + \int_{t_{o n}}^{t_{o n} + δ T} V_{o} (t) d t + \int_{t_{o n} + δ T}^{T} V_{o} (t) d t) \\ = \frac{1}{T} (\int_{0}^{t_{o n}} (V_{i} - V_{s w} - V_{L}) d t + \int_{t_{o n}}^{t_{o n} + δ T} (- V_{L} + V_{D}) d t + \int_{t_{o n} + δ T}^{T} V_{C} d t) \\ = \frac{1}{T} (\int_{0}^{t_{o n}} V_{i} d t - \int_{0}^{t_{o n}} V_{s w} d t - \int_{0}^{t_{o n}} V_{L} d t - \int_{t_{o n}}^{t_{o n} + δ T} V_{L} d t + \int_{t_{o n}}^{t_{o n} + δ T} V_{D} d t + \int_{t_{o n} + δ T}^{T} V_{C} d t) \\ = \frac{1}{T} (\int_{0}^{t_{o n}} V_{i} d t - \int_{0}^{t_{o n}} V_{s w} d t - \int_{0}^{t_{o n} + δ T} V_{L} d t + \int_{t_{o n}}^{t_{o n} + δ T} V_{D} d t + \int_{t_{o n} + δ T}^{T} V_{C} d t) \\ = \frac{1}{T} (D T V_{i} - \int_{0}^{t_{o n}} V_{s w} d t - 0 + \int_{t_{o n}}^{t_{o n} + δ T} V_{D} d t + \int_{t_{o n} + δ T}^{T} V_{C} d t) \\ = \frac{1}{T} (D T V_{i} - \int_{0}^{t_{o n}} V_{s w} d t + \int_{t_{o n}}^{t_{o n} + δ T} V_{D} d t + \int_{t_{o n} + δ T}^{T} V_{C} d t) \\ = D V_{i} - \frac{1}{T} (\int_{0}^{t_{o n}} V_{s w} d t - \int_{t_{o n}}^{t_{o n} + δ T} V_{D} d t - \int_{t_{o n} - δ T}^{T} V_{C} d t) \end{aligned}

V_{L} = L \frac{d I_{L}}{d t}

\begin{aligned} Δ I_{L_{o n}} & = I_{m a x} - I_{m i n} \\ = \int_{0}^{T_{o n}} \frac{V_{L}}{L} d t \\ = \int_{0}^{T_{o n}} \frac{V_{i} - V_{s w} - V_{o}}{L} d t \\ = \int_{0}^{T_{o n}} \frac{V_{i}}{L} d t - \int_{0}^{T_{o n}} \frac{V_{s w}}{L} d t - \int_{0}^{T_{o n}} \frac{V_{o}}{L} d t \\ = \frac{D T V_{i}}{L} - \int_{0}^{T_{o n}} \frac{V_{s w}}{L} d t - \int_{0}^{T_{o n}} \frac{V_{o}}{L} d t \\ Δ I_{L_{o f f}} & = I_{m i n} - I_{m a x} \\ = \int_{t_{o n}}^{t_{o n} + δ T} \frac{V_{L}}{L} d t \\ = \int_{t_{o n}}^{t_{o n} + δ T} \frac{- V_{o} + V_{D}}{L} d t \end{aligned}

Simulation GitHub

參考

Buck converter
Inductor Calculation for Buck Converter IC
STUDY AND DESIGN OF BUCK CONVERTER

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