**Charger Circuit Operation**\(\xrightarrow[\text{110 V}]{\text{ac voltage}}\) \(\xrightarrow[\text{4 V}]{\text{Transformer}}\) \(\xrightarrow[\text{3.5 V}]{\text{dc voltage using LPF}}\)

Output of TF using black box exhibits a zero dc content as -ve and +ve half cycles enclose equal areas, leading to a zero average

Resistor \(\rightarrow\)

**linear device**\(\rightarrow\) current Vs voltage is a straight line.Diode \(\rightarrow\)

**nonlinear device**\(\rightarrow\) \(I\) Vs \(V\) is not a straight line.The reason is the barrier potential.

\[\boxed{V_d < V_b \rightarrow I_d \downarrow} \qquad \boxed{V_d > V_b \rightarrow I_d \uparrow}\]

**Ideal
Diode:** acts like a perfect conductor (zero
resistance) when forward biased and like a perfect insulator (\(\infty\) resistance) when reverse
biased.

\[\begin{aligned} & V_{\text{anode}} > V_{\text{cathode}} \leftarrow \text{on} \\ & V_{\text{anode}} < V_{\text{cathode}} \leftarrow \text{off} \\ & V_{\text{anode}} - V_{\text{cathode}} = V_D \end{aligned}\]

**Real
diode**

**Ideal
diode**

**Si Vs
Ge**

Above the knee voltage, the diode current increases rapidly.

Small increase in the diode voltage cause large increases in diode current.

**Bulk Resistance :**After barrier potential is overcome, all that impedes the current is the ohmic resistance of the p and n regions. \[\text{Bulk resistance}~R_B = R_p + R_n\] \(R_B\) depends on the size of the p and n regions and how heavily doped they are. Often, \(R_B\) is less than 1 \(\Omega\).**Maximum DC Forward Current :**If the current in a diode is too large, the excessive heat can destroy the diode. The \(I_{F(max)}\) is one of the maximum ratings given on a data sheet.**Power Dissipation :**similar to resistor \[\begin{aligned} P_D & = V_D \cdot I_D \\ P_{max} & = V_{max} \cdot I_{max} \end{aligned}\] The power rating is the maximum power the diode can safely dissipate without shortening its life or degrading its properties

The diode current equation relating the voltage \(V\) and current \(I\) \[I=I_o\left[\mathrm{e}^{\left(V / \eta V_T\right)}-1\right]\] where \[\begin{aligned} I & = \text{diode current} \\ I_o & = \text{diode reverse saturation current at room temperature}\\ V & = \text{external voltage applied to the diode} \\ \eta &=\text{ a constant, 1 for germanium and 2 for silicon} \\ V_T&=k T / q=T / 11600, \text{volt-equivalent of temperature, i.e., thermal voltage} \\ k & = \text{Boltzmann's constant}~\left(1.38066 \times 10^{-23} \mathrm{~J} / \mathrm{K}\right)\\ q & = \text{charge of the electron}~ \left(1.60219 \times 10^{-19} \mathrm{C}\right)\\ T & = \text{temperature of the diode junction}~ (\mathrm{K})=\left({ }^{\circ} \mathrm{C}+273^{\circ}\right) \end{aligned}\]

\[\boxed{I=I_o\left[\mathrm{e}^{\left(V / \eta V_T\right)}-1\right]}\]

When the diode is reverse biased, its current equation may be obtained by changing the sign of the applied voltage \(V\).

Thus, the diode current with reverse bias is \[I=I_o\left[\mathrm{e}^{\left(-V / \eta V_T\right)}-1\right]\]

If \(V \gg V_T\), then the term \(\mathrm{e}^{\left(-V / \eta v_T\right)} \ll 1\), therefore \(I \approx-I_o\), termed as reverse saturation current, which is valid as long as the external voltage is below the breakdown value.

**Effect of
temperature:**

T \(\uparrow~\Rightarrow\) generation of e-p pairs increases, which increase the conductivities

If T \(\uparrow\) at fixed V \(\Rightarrow\) I increases

To bring I to normal \(\Rightarrow\) V \(\downarrow\)

**Load line**is a tool to find the exact value of the diode current and voltage \[I_D=\frac{V_S-V_D}{R_s}\]

If \(V_s = 2~\mathrm{V}\), \(R = 100~\Omega\)

\(V_D = 0 \Rightarrow I_D = 20~\mathrm{mA}\)

\(I_D = 0 \Rightarrow V_D = V_s = 2~\mathrm{V}\)

The straight line is called the load line

\(Q\) is an abbreviation for

*quiescent*, which means “at rest.”