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$\text { Assume that the numerical values for the signal voltage shown in Fig. are } V_{m}=50 \mathrm{mV} \text { and } t_{1}=1 \mathrm{s}.$

$\begin{array}{l}{\text { This signal voltage is applied to integrating-amplifier circuit as input. }} \\ {\text { The circuit parameters of the amplifier are } R_{s}=100 \mathrm{k} \Omega, C_{f}=0.1 \mu \mathrm{F} \text { , }} \\ {\text { and } V_{c c}=6 \mathrm{V} \text { . The initial voltage on the capacitor is zero. }} \\ {\text { a) Calculate } V_{o}(t) .} \\ {\text { b) Plot } V_{o(t)} \text { versus } t}\end{array}$

$t_{1}=1 \sec ، V_{m}=50 \mathrm{mV}$

$V_{0}(t)=\frac{-1}{R_{s} C_{f}} \int_{0}^{t} V_{s} d t$

$V_{0}(t)=\frac{-1}{100 * 10^{3} * 0.1 *10^{-6}}\int_{0}^{t}$

(1) $0 \leq t \leq 1 \text { sec }$                                        $V_{s}=50 m v$

$V_{0}(t)=-100+\int_{0}^{t} 50 \times 10^{-3} d t$

$V_{0}(t)=-5 t \Rightarrow 0 \leq t \leq 1 \sec$

(2) $1 \leq t \leq 2 \text { sec }$                        $V_{s}=-50 \mathrm{mv}$

$V_{0}(t)=\int^{t}_{t_0}V_s dt*\frac{-1}{R_sC_f}+V(t_0)$

$V_{0}(t)=\int_{1}^{t}-50 * 10^{-3} *-100+-5)$

$V_0(t)=5(t-1)-5$

$V_{0}(t)=\left\{\begin{array}{cc}{-5 t} & {0 \leq t \leq 1} \\ {5(t-1)-5} & {1 \leq t \leq 2}\end{array}\right.$

$V_{0}=-5 t$

$t=0 \quad \longrightarrow V_{0}=0 \quad(0,0)$

$t=1 \quad \longrightarrow V_{0}=-5 \quad(1,-5)$

$V_{0}=5(t-1)-5$

$t=1$                                $V_{0}=-5(1,-5)$

$t=2$                                $V_{0}=0(2,0)$