Ideal OPAMP Solution For the circuit shown below, the OPAMP is ideal and \(V_{REF}\) is fixed. Find the value of: \[ \frac{R_F}{R_{IN}} \] Given: \(V_{OUT}=1V\) when \(V_{IN}=0.1V\) \(V_{OUT}=6V\) when \(V_{IN}=1V\) Solution Ideal OPAMP Rules For an ideal op-amp: 1. Input current is zero: \[ i_+=i_-=0 \] 2. With negative feedback: \[ V_+=V_- \] Since \(V_{REF}\) is fixed, the positive input voltage is constant. Step 1: Write the output equation The circuit is an inverting amplifier with a reference voltage. The output equation is: \[ V_{OUT} = V_+ - \frac{R_F}{R_{IN}} (V_{IN}-V_+) \] Let: \[ K=\frac{R_F}{R_{IN}} \] Therefore: \[ V_{OUT}=V_+-K(V_{IN}-V_+) \] Step 2: Apply first condition Given: \[ V_{IN}=0.1V \] \[ V_{OUT}=1V \] Therefore: \[ 1=V_+-K(0.1-V_+) \] Step 3: Apply second condition Given: \[ V_{IN}=1V \] \[ V_{OUT}=6V \] Therefore: \[ 6=V...

MOS Amplifier Step-by-Step Solution The given circuit contains two NMOS transistors \(M_1\) and \(M_2\). Find the voltage gain: \[ A_v=\frac{v_{out}}{v_{in}} \] Solution Step 1: Identify the amplifier The input signal is applied to the gate of \(M_2\). Therefore \(M_2\) is the amplifying transistor. A MOS transistor converts voltage into current: \[ i_d=g_m v_{gs} \] For \(M_2\): \[ i_2=g_{m2}v_{in} \] Step 2: Output resistance of the circuit For small signal analysis, the power supply \(V_{DD}\) becomes AC ground. AC Ground       |       \(r_{o2}\)       | Vout       |       \(r_{o1}\)       | Ground Both \(r_{o1}\) and \(r_{o2}\) are connected between the same two nodes. Therefore they are parallel. \[ R_{out}...

Semiconductor Band Diagram Solution Question A non-uniformly doped p-type semiconductor has length L . The energy band diagram is shown below. Given: Energy difference = Δ Length = L Find: Electric field inside the semiconductor. Solution Step 1: The semiconductor is at equilibrium , so the Fermi level remains constant throughout the material. \[ E_F=\text{constant} \] A non-uniform acceptor concentration causes the valence band to slope, producing an internal electric field. Step 2: For a uniform electric field, \[ E=V/L \] \[ \Delta V=EL \] where \(E\) = electric field \(L\) = semiconductor length Step 3: The band diagram shows an energy difference (eV) \[ \Delta \] Energy and voltage are related by \[ \Delta=q\Delta V \] Substitute \[ \Delta V=EL \] to obtain \[ \Delta=qEL \] Step 4: Rearrange to obtain the electric field: \[ \boxed{E=\frac{\Delta}{qL}} \] Final Answer \[ \boxed{E=\frac{\Delta}{qL}} \] ...

  A bar of silicon is doped with boron concentration of $10^{16}\,\text{cm}^{-3}$ and assumed to be fully ionized. It is exposed to light such that electron-hole pairs are generated throughout the volume of the bar at the rate of $10^{20}\,\text{cm}^{-3}\text{s}^{-1}$. If the recombination lifetime is $100\,\mu s$, intrinsic carrier concentration of silicon is $10^{10}\,\text{cm}^{-3}$, and assuming 100% ionization of boron, then the approximate product of steady-state electron and hole concentrations due to this light exposure is (A) $10^{20}\,\text{cm}^{-6}$ (B) $2\times10^{20}\,\text{cm}^{-6}$ (C) $10^{32}\,\text{cm}^{-6}$ (D) $2\times10^{32}\,\text{cm}^{-6}$ Step-by-Step Solution Step 1 : Identify the Type of Semiconductor The silicon is doped with boron. Boron is an acceptor impurity , therefore the semiconductor is p-type . Since boron is fully ionized, \[ p_0=N_A=10^{16}\,\text{cm}^{-3} \] This is the equilibrium hole ...

Let the linear convolution of two discrete-time finite-length sequences $x[n]$ and $h[n]$, each of length 16, be denoted by $y[n]$. Let $z[n]$ denote the 16-point inverse discrete Fourier transform (IDFT) of the product of the 16-point discrete Fourier transforms (DFTs) of $x[n]$ and $h[n]$. The value of $k$ for which $z[k]=y[k]$ is (A) 0 (B) 1 (C) 15 (D) 16 Solution Step 1: Length of Linear Convolution Both sequences have length 16. Length of linear convolution: $ 16+16-1=31 $ Hence the linear convolution contains the samples $ y[0],y[1],\ldots,y[30] $ Notice that there is no sample $y[31]$. Step 2: What does the IDFT of the Product of DFTs Give? A standard DFT property states: $ \text{IDFT}\{X[k]H[k]\} = \text{Circular Convolution} $ Therefore, $ z[n] = \text{16-point circular convolution of }x[n]\text{ and }h[n]. $ So, $y[n]$ = Linear Convolution $z[n]$ = 16-point Circular Convolution Step 3: Folding (Wrap-A...

Covariance vs Correlation: Difference, Formula, Examples and Comparison Introduction Covariance and correlation are two important statistical concepts used to measure the relationship between two variables. They help answer questions such as: Does advertising increase sales? Does study time affect exam marks? Are height and weight related? Although both measure relationships between variables, covariance and correlation have important differences in interpretation and usage. What is Covariance? Covariance measures the direction in which two variables change together. $$ Cov(X,Y)= \frac{\sum(X_i-\bar X)(Y_i-\bar Y)} {n} $$ Interpretation of Covariance Positive Covariance When: $$ Cov(X,Y)>0 $$ Both variables increase or decrease together. Negative Covariance $$ Cov(X,Y) One variable increases while the other decreases. Zero Covariance $$ Cov(X,Y)=0 $$ There is no linear relationship between v...

Question Find the integrating factor of the differential equation $$ \frac{dy}{dx} + \frac{x}{1-x^2}y = x\sqrt{y} $$ Solution The given differential equation is $$ \frac{dy}{dx} + \frac{x}{1-x^2}y = x\sqrt{y} $$ Since the equation contains the term $\sqrt{y}$, it is not linear. Let $$ v=\sqrt{y} $$ Then, $$ y=v^2 $$ Differentiating, $$ \frac{dy}{dx} = 2v\frac{dv}{dx} $$ Substitute these values into the given equation: $$ 2v\frac{dv}{dx} + \frac{x}{1-x^2}v^2 = xv $$ Divide both sides by $2v$: $$ \frac{dv}{dx} + \frac{x}{2(1-x^2)}v = \frac{x}{2} $$ This is now a linear differential equation of the form $$ \frac{dv}{dx}+P(x)v=Q(x) $$ where $$ P(x)= \frac{x}{2(1-x^2)}. $$ The integrating factor is $$ \text{I.F.} = e^{\int P(x)\,dx} = e^{\int \frac{x}{2(1-x^2)}dx} $$ Let $$ u=1-x^2 $$ Then, $$ du=-2x\,dx $$ Therefore, $$ \int \frac{x}{2(1-x^2)}dx = -\frac14 \int \frac{du}{u} $$ Hence, $$ = -\frac14\ln|1-x^2| $$ ...

  Line Integral Solution Solution $ \mathbf{F}(x,y)=-x\hat{i}+y\hat{j} $ $$ \int_C \mathbf{F}\cdot d\mathbf{r} $$ $$ =\int_C(-x\,dx+y\,dy) $$ Parameterization: $$ x=\cos t,\qquad y=\sin t $$ $$ dx=-\sin t\,dt,\qquad dy=\cos t\,dt $$ $$ \int_{0}^{\pi/4} \left[ (-\cos t)(-\sin t) + (\sin t)(\cos t) \right]dt $$ $$ = \int_{0}^{\pi/4} 2\sin t\cos t\,dt $$ where $$ 0\le t\le\frac{\pi}{4} $$ $$ 2\sin t\cos t=\sin2t $$ $$ =\int_{0}^{\pi/4}\sin2t\,dt $$ $$ =\left[-\frac12\cos2t\right]_{0}^{\pi/4} $$ $$ =-\frac12\cos\frac{\pi}{2} +\frac12\cos0 $$ $$ =0+\frac12 =\frac12 $$ $$ \boxed{\int_C \mathbf{F}\cdot d\mathbf{r}=\frac12} $$ Option A GATE EC Previous Year Papers with Solutions → Electronics and Communication Study Material →