Semiconductor Electronics: Materials, Devices and

Voltage gain = $\frac{I_C{R}_0}{I_B{R}_i}$  

R₀ ->Output Resistance

R_i ->Input Resistance

I_C ->Collector current

I_{B ->}Base current

Voltage gain = $\frac{I_C}{I_B}\frac{R_0}{R_i}=\frac{\left(5*10^{-3}\right)}{\left(100*10^{-6}\right)}*\frac{60}{200}=15$  

Truth table for Input & Output

Now, Truth Table for option B.

V-l characteristics graph is plotted in 4th quadrant for solar cell.

$i_{\left(zener-max\right)}=25mA$                                                                

 20 – i_max R – 8 = 0

i_max R = 12At minimum zener current $\left(\mu A\right):-$  

$20-i_{min}R-i_{min}{R}_L=0$  

  $\frac{R}{R_L}=\frac{12}{8}=\frac{3}{2}$

${\mathcal{l}}_{min}R=12$

  $i_{min}{R}_L=8$

At max^m zener current –

$20-i_{max}R-8=0$  

$i_L=O\left\{as{i}_{z}ma{x}^m=25mA\right\}$             

i_maxR = 12v

25 * 10^-3 R = 12

$R=\frac{12*10^3}{25}=12*40=480\Omega$

Here diode-D₁ is forced biased and diode-D₂ is reversed biased, so

$R_{AB}=\left(20//20\right)+15=25\Omega$

$E_9=\frac{hc}{\lambda }=\frac{1242}{\lambda \left(nm\right)}=\frac{1242}{400}=3.105$

 $i_{\left(zener-max\right)}=25mA$

20 – i_max R – 8 = 0

i_max R = 12

At minimum zener current $\left(\mu A\right):-$

$20-i_{min}R-i_{min}{R}_L=0$

$\frac{R}{R_L}=\frac{12}{8}=\frac{3}{2}$

${\mathcal{l}}_{min}R=12$

$i_{min}{R}_L=8$

At maxm zener current –

$20-i_{max}R-8=0$

$i_L=O\left\{as{i}_{z}ma{x}^m=25mA\right\}$

i_maxR = 12v

25 * 103 R = 12

$R=\frac{12*10^3}{25}=12*40=480\Omega$

g_eff = g – $\left(\frac{{\rho }_w}{{\rho }_b}\right)g$

$T=2\pi \sqrt[]{\frac{\mathcal{l}}{g}}$

$T\text{'}=2\pi \sqrt[]{\frac{\mathcal{l}}{g_{eff}}}$

$\Rightarrow T\text{'}=\sqrt[]{\frac{5}{4}}T$

= $\sqrt[]{\frac{5}{4}}*10$

= $5\sqrt[]{5}sec$

So, x = 5