Semiconductor Electronics: Materials, Devices and

Fermi level of p-type semiconductors will go downward

Using truth table of logic gates

$Y=\overline{\left(A\overline{B}\right)+\left(\overline{A}B\right)}=\overline{\left(\left(A\overline{B}\right)\right)}\overline{\left(\left(\overline{A}B\right)\right)}$

$Y=\left(\overline{A}+\overline{B}\right)\left(\overline{A}+\overline{B}\right)=\left(\overline{A}+B\right)\left(A+\overline{B}\right)=AB+\overline{A}\overline{B}$

$Y_1=\overline{A·A}$

$=\overline{A}$

$Y_2=\overline{B+B}$

$=\overline{B}$

$Y=\overline{Y_1+Y_2}$

$=\overline{\overline{A}+\overline{B}}$

$=\overline{\overline{A}}·\overline{\overline{B}}$

$=A·B is similar to output of AND Gate$

According to transformer ratio,

$\frac{V_S}{V_P}=\frac{N_S}{N_P}=2:1$

$l=\frac{120-60}{4000}=0.015A$

Thus l₂ = l - L_L

= 0.015 – 0.006

= 0.009 = 9mA

For 90 mA in zener diode, current in R should be greater than 90 mA.

$\frac{45}{R}=90*10^{-3}\Rightarrow R=\frac{45}{90*10^{-3}}=500\Omega$

$E=\frac{hc}{\lambda }=\frac{1242eV-nm}{621nm}=2eV$

minimum 2eV is required to emit.

$P_z=V_z{l}_z\Rightarrow 0.5=8l_z\Rightarrow l_z=l_P=\frac{1}{16}Amp$

When zener is connected across a potential divider arranged with maximum potential drop across zener diode, then

$$ $V_P=V-V_z=20-8=12$  volt Potential difference across protective resistance RP

$R_P=\frac{V_P}{I_P}=\frac{12}{\frac{1}{6}}=192Amp.$

$R=\frac{\Delta V}{\Delta l}=\frac{0.75-0.70}{\left(5-3\right)*10^{-3}}=\frac{0.05*1000}{2}=25\Omega$

In give diagram Diode 1 and 2 are in forward bias with R = $30\Omega$ and Diode 3 is reverse bias with R = infinite l₁ current is flowing through 20 $\Omega$  

So $\frac{l_1}{2}and\frac{l_1}{2}$ current will flow through Diode 1 and 2. As resistance is same applying KCl in ABCD Loop $-\frac{l_1}{2}*130-\frac{l_1}{2}*130-l_1*20+200=0$

-100l₁ + 200 = 0

l₁ = 2