[Circuit] Snubber 平均功率推導

電路知識：Snubber 平均功率推導
工具：Qucs
線上模擬

簡介：Snubber 平均功率推導

$$\begin{array}{rl}\mathbf{Z}& =R+\frac{1}{jwC}\\ & =R-\frac{j}{wC}\\ & =\sqrt{R^2+(\frac{1}{wC}{)}^2}\mathrm{\angle }-ta{n}^{-1}\frac{1}{wRC}\\ \mathbf{V}& =V\mathrm{\angle }0\\ \mathbf{I}& =\frac{\mathbf{V}}{\mathbf{Z}}\\ & =\frac{V}{\sqrt{R^2+(\frac{1}{wC}{)}^2}}\mathrm{\angle }ta{n}^{-1}\frac{1}{wRC}\\ P& =\frac{1}{2}|\mathbf{I}{|}^{2}R=\frac{1}{2}\frac{V^2}{R^2+(\frac{1}{wC}{)}^2}R\end{array}$$

若電壓來源非 Sinusoidal，可參考此篇計算之
[Circuit] 平均功率

範例

弦波

從上圖得知平均功率為 $0.4876W$，用圖中的 $I_{rms}$ 計算 $|I_{rms}{|}^{2}R$，也是同樣的結果
但若用 $P=C{V}^{2}f=1W$ 則完全超估，所以並不能如此計算

推導過程

$$\begin{array}{rl}v(t)& =sin(wt)=cos(wt-\frac{\pi }{2})\\ \mathbf{V}& =1\mathrm{\angle }-\frac{\pi }{2}\\ \mathbf{Z}& =1+\frac{1}{j2\pi \ast 1M\ast 1u}\\ & =1-j\frac{1}{2\pi }\\ & \approx 1.013\mathrm{\angle }-0.158\\ P& =\frac{1}{2}{\left|\frac{\mathbf{V}}{\mathbf{Z}}\right|}^{2}R\\ & =\frac{1}{2}(\frac{1}{1.013}{)}^2\\ & =0.4872\end{array}$$

程式模擬結果

import cmath

n = 1000
f = 1e6
C = 1e-6
p = 0

# Z=R+jwC
Z = 1+1/(1j*2*cmath.pi*f*C)

v = cmath.rect(1, -cmath.pi/2)
i = v/Z
p += (1/2 * abs(i)**2 * Z.real)
    
print(p)
# 0.4876477384840711

方波 I

從上圖得知平均功率為 $0.97942W$，用圖中的 $I_{rms}$ 計算 $|I_{rms}{|}^{2}R$，也是同樣的結果
但若用 $P=C{V}^{2}f=1W$ 看似可以，但實際還是不行，請看接下來的例子

推導過程

$$\begin{array}{rl}{v}_{square}(t)& =\frac{4}{\pi }\sum _{k=1}^{\mathrm{\infty }}\frac{sin(w(2k-1)t)}{2k-1}\\ & =\frac{4}{\pi }\sum _{k=1}^{\mathrm{\infty }}\frac{cos(w(2k-1)t-\frac{\pi }{2})}{2k-1}\\ {\mathbf{V}}_{\mathbf{square}}& =\sum _{k=1}^{\mathrm{\infty }}{\mathbf{V}}_{\mathbf{k}\mathbf{,}\mathbf{w}\mathbf{(}\mathbf{2}\mathbf{k}\mathbf{-}\mathbf{1}\mathbf{)}}\\ {\mathbf{V}}_{\mathbf{k}\mathbf{,}\mathbf{w}\mathbf{(}\mathbf{2}\mathbf{k}\mathbf{-}\mathbf{1}\mathbf{)}}& =\frac{4}{\pi }\sum _{k=1}^{\mathrm{\infty }}\frac{1\phase{-\frac{\pi }{2}}}{2k-1}\\ \\ {\mathbf{Z}}_{\mathbf{k}}& =1+\frac{1}{j2\pi \ast f\ast (2k-1)\ast C}\\ & =1+\frac{1}{j2\pi \ast 1e^6\ast (2k-1)\ast 1e^{-6}}\\ & =1+\frac{1}{j2\pi (2k-1)}\\ & =1+j\frac{-1}{2\pi (2k-1)}\\ & =\sqrt{1+(\frac{-1}{2\pi \ast (2k-1)}{)}^2}\phase{ta{n}^{-1}\frac{-1}{2\pi \ast (2k-1)}}\\ P& =\frac{1}{2}\sum _{k=1}^{\mathrm{\infty }}{\left|\frac{{\mathbf{V}}_{\mathbf{k}}}{{\mathbf{Z}}_{\mathbf{k}}}\right|}^{2}R\\ & =\frac{1}{2}\sum _{k=1}^{\mathrm{\infty }}(\frac{\frac{4}{\pi }\frac{1}{2k-1}}{\sqrt{1+(\frac{-1}{2\pi (2k-1)}{)}^2}}{)}^2\\ & =\frac{1}{2}\sum _{k=1}^{\mathrm{\infty }}\frac{(\frac{4}{\pi }\frac{1}{2k-1}{)}^2}{1+(\frac{-1}{2\pi (2k-1)}{)}^2}\\ & =\frac{1}{2}\sum _{k=1}^{\mathrm{\infty }}\frac{\frac{16}{{\pi }^2}\frac{1}{(2k-1{)}^2}}{1+\frac{1}{4{\pi }^2(2k-1{)}^2}}\\ & =\frac{1}{2}\sum _{k=1}^{\mathrm{\infty }}\frac{64}{4{\pi }^2(2k-1{)}^2+1}\\ & \approx \frac{1}{2}\times 1.959345\\ & =0.9796725\end{array}$$

程式模擬結果

import cmath

n = 100000
f = 1e6
C = 1e-6
p_i2R = 0
p_V2R = 0

for k in range(1,n):
    # Z=R+jwC
    Z = 1+1/(1j*2*cmath.pi*f*(2*k-1)*C)
    v = 4/cmath.pi * cmath.rect(1, -cmath.pi/2)/(2*k-1)
    i = v/Z
    p_i2R += (1/2 * abs(i)**2 * Z.real)
    
    v = (4/cmath.pi * cmath.rect(1, -cmath.pi/2)/(2*k-1)) / Z * Z.real
    
    p_V2R += (1/2 * abs(v)**2/Z.real)
    
print('i^2R=', p_i2R)
print('V^2/R=', p_V2R)
# 0.9796726231709146

方波 II

將電容大小改為 0.1uF，平均功率為 $0.39382W$
但 $P=C{V}^{2}f=0.1W$ 完全低估，所以並不能如此計算
推導過程 與 程式模擬 與上面的例子相同，就不再贅述