Semiconductor Electronics: Materials, Devices and

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

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

Capacitor makes potential difference constant.

Using truth table

A            B            V₀

0            0

0            1            1

1            0            1

1            1

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

 0.02i * R_G = 0.98i * 5

[in parallel, pot. Diff is same]

or, R_G = 49 * 5

R_G = 245 $\Omega$

Maximum current through zener diode,

  $i=\frac{2}{10}=0.2A$             

When unregulated voltage is 14V,

Potential difference across Rs,

              V_S = 14 – 10 = 4V

$R_S=\frac{V_S}{i}=\frac{4}{0.2}=20\Omega$

β $=\frac{I_C}{I_E}=\frac{I_C}{I_E-I_C}=\frac{I_C/I_E}{1-l_C/I_E}=\frac{\alpha }{1-\alpha }$