Distinguish between breakdown voltage and PIV for a diode, and draw the graph showing the effect of temperature on its I-V characteristic.

1) Distinguish between Breakdown Voltage and Peak Inverse Voltage (PIV) as they relate to a diode.

• Breakdown Voltage: This is the reverse voltage at which a diode begins to conduct significantly in the reverse direction, often due to Zener or avalanche effects. If the current is not limited, this can lead to permanent damage to the diode.
• Peak Inverse Voltage (PIV): This is the maximum reverse voltage that a diode can withstand across its terminals without breaking down. It is a critical rating for diodes used in rectifier circuits, ensuring the diode can block the peak reverse voltage of the AC input.

2) Draw the graph showing the effect of temperature on the I-V characteristic of a diode.

The I-V characteristic curve of a diode shows current (I) versus voltage (V).

• Forward Bias: As temperature increases, the cut-in voltage (or threshold voltage) decreases. This means the diode starts conducting at a lower forward voltage for a given current.
• Reverse Bias: As temperature increases, the reverse saturation current ($I_s$) increases significantly, leading to a higher leakage current. The breakdown voltage also slightly decreases with increasing temperature.

$$\begin{array}{c|c} \text{I (mA)} & \text{} \\ \uparrow & \text{Higher Temp} \\ \text{} & \text{Lower Temp} \\ \text{} & \text{Forward Bias} \\ \text{} & \text{Current increases exponentially} \\ \text{} & \text{Cut-in voltage shifts left with higher temp} \\ \text{} & \text{----------------------------------------------------} \rightarrow \text{V (V)} \\ \text{} & \text{Reverse Bias} \\ \text{} & \text{Leakage current increases with higher temp} \\ \text{} & \text{Breakdown voltage shifts right (less negative) with higher temp} \\ \downarrow & \text{} \\ \end{array}$$

(Imagine two curves: one for lower temperature and one for higher temperature. In forward bias, the higher temperature curve would be to the left of the lower temperature curve, indicating a lower cut-in voltage. In reverse bias, the higher temperature curve would show a larger negative current (leakage) and a breakdown point closer to 0V (less negative breakdown voltage) compared to the lower temperature curve.)

3) With the aid of graphs only

i) Compare the reverse characteristics of Ge and Si diodes.

• Silicon (Si) Diode: Exhibits a very low reverse leakage current (typically nanoamperes or picoamperes) until it reaches its breakdown voltage, which is generally higher than that of Germanium.
• Germanium (Ge) Diode: Exhibits a significantly higher reverse leakage current (typically microamperes) compared to Silicon diodes at the same temperature, and generally has a lower breakdown voltage.

$$\begin{array}{c|c} \text{I (mA)} & \text{} \\ \uparrow & \text{} \\ \text{} & \text{----------------------------------------------------} \rightarrow \text{V (V)} \\ \text{} & \text{Reverse Bias} \\ \text{} & \text{Si Diode (very low leakage)} \\ \text{} & \text{Ge Diode (higher leakage)} \\ \downarrow & \text{} \\ \end{array}$$

(Imagine two curves in the third quadrant: both show a small negative current that increases sharply at breakdown. The Ge diode curve would show a larger negative current (leakage) before breakdown and a breakdown voltage closer to 0V compared to the Si diode.)

ii) Compare the forward characteristics of Ge and Si diodes.

• Silicon (Si) Diode: Has a higher cut-in voltage (approximately $0.7 \text{ V}$) before significant forward current flows.
• Germanium (Ge) Diode: Has a lower cut-in voltage (approximately $0.3 \text{ V}$) before significant forward current flows.

$$\begin{array}{c|c} \text{I (mA)} & \text{} \\ \uparrow & \text{Ge Diode (lower cut-in voltage)} \\ \text{} & \text{Si Diode (higher cut-in voltage)} \\ \text{} & \text{Forward Bias} \\ \text{} & \text{Current increases exponentially} \\ \text{} & \text{----------------------------------------------------} \rightarrow \text{V (V)} \\ \text{} & \text{} \\ \downarrow & \text{} \\ \end{array}$$

(Imagine two curves in the first quadrant: both show an exponential rise in current after a certain voltage. The Ge diode curve would start rising significantly at around $0.3 \text{ V}$, while the Si diode curve would start rising at around $0.7 \text{ V}$.)

iii) Distinguish between transition and diffusion capacitances of a diode and express each mathematically.

• Transition Capacitance ($C_T$ or $C_j$): Also known as junction capacitance or depletion capacitance. It arises from the depletion region of a reverse-biased p-n junction, which acts like a dielectric between the p and n regions (the capacitor plates). Its value depends on the reverse bias voltage.

  • Mathematical expression: $$C_T = \frac{\epsilon A}{W}$$ where $\epsilon$ is the permittivity of the semiconductor, $A$ is the junction area, and $W$ is the width of the depletion region. A more detailed expression for an abrupt junction is: $$C_T = \frac{C_{j0}}{\left(1 + \frac{V_R}{V_0}\right)^{1/2}}$$ where $C_{j0}$ is the zero-bias junction capacitance, $V_R$ is the reverse bias voltage, and $V_0$ is the built-in potential.

• Diffusion Capacitance ($C_D$): Also known as storage capacitance. It arises from the storage of minority carriers (electrons in p-type, holes in n-type) in the semiconductor material when the diode is forward-biased. When the forward bias voltage changes, these stored charges must be added or removed, which takes time and behaves like a capacitance.

  • Mathematical expression: $$C_D = \frac{\tau_p I_F}{V_T}$$ where $\tau_p$ is the minority carrier lifetime (e.g., for holes in n-type material), $I_F$ is the forward current, and $V_T = \frac{kT}{q}$ is the thermal voltage.