Temperature Effect on a Diode Current MCQ Quiz - Objective Question with Answer for Temperature Effect on a Diode Current - Download Free PDF

Concept:

1) A diode is a two-terminal non-linear device.

2) It is passive because there is no built-in source of power.

Note: It is not an exponential device, it’s a logarithmic device.[∵ Current controlling]

The current equation is defined as:

$I_d=I_S{e}^{\frac{V_d}{\eta V_t}}$

V_d = voltage across the diode

V_t = temperature equivalent

V_t ≈ 26 mV at room temperature.

$V_t=\frac{T}{11600}inkelvin$

I_s = Reverse saturation current.

$V_d=\eta V_t\ln \left(\frac{I_d}{I_S}\right)$

If ‘η’ is not given the default value will be 1

Calculation:

Given forward bias characteristics are

From the current equation using  data, we get two equations

$V_1=\eta V_t\ln \left(\frac{I_1}{I_S}\right)$ ⋯ (i)

$V_2=\eta V_t\ln \left(\frac{I_2}{I_S}\right)$ ⋯ (ii)

Considering η = 1

$V_1-V_2=V_t\ln \left(\frac{I_1}{I_S}\right)-V_t\ln \left(\frac{I_2}{I_S}\right)$

$V_1-V_2=V_t\ln \left(\frac{I_1}{I_2}\right)$

$0.6-0.65=\frac{T}{11600}\ln \left(\frac{1}{5}\right)$

$-0.05=-1.609\times \frac{T}{11600}$

$\frac{0.05\times 11600}{1.609}=T$

$T=\frac{580}{1.609}$

T = 360.47 Kelvin

T ≈ 360 Kelvin

Extra concept:

Diode small-signal equivalent is shown below for both ideal and practical cases.

Concept:

1) A diode is a two-terminal non-linear device.

2) It is passive because there is no built-in source of power.

Note: It is not an exponential device, it’s a logarithmic device.[∵ Current controlling]

The current equation is defined as:

$I_d=I_S{e}^{\frac{V_d}{\eta V_t}}$

V_d = voltage across the diode

V_t = temperature equivalent

V_t ≈ 26 mV at room temperature.

$V_t=\frac{T}{11600}inkelvin$

I_s = Reverse saturation current.

$V_d=\eta V_t\ln \left(\frac{I_d}{I_S}\right)$

If ‘η’ is not given the default value will be 1

Calculation:

Given forward bias characteristics are

From the current equation using  data, we get two equations

$V_1=\eta V_t\ln \left(\frac{I_1}{I_S}\right)$ ⋯ (i)

$V_2=\eta V_t\ln \left(\frac{I_2}{I_S}\right)$ ⋯ (ii)

Considering η = 1

$V_1-V_2=V_t\ln \left(\frac{I_1}{I_S}\right)-V_t\ln \left(\frac{I_2}{I_S}\right)$

$V_1-V_2=V_t\ln \left(\frac{I_1}{I_2}\right)$

$0.6-0.65=\frac{T}{11600}\ln \left(\frac{1}{5}\right)$

$-0.05=-1.609\times \frac{T}{11600}$

$\frac{0.05\times 11600}{1.609}=T$

$T=\frac{580}{1.609}$

T = 360.47 Kelvin

T ≈ 360 Kelvin

Extra concept:

Diode small-signal equivalent is shown below for both ideal and practical cases.

Concept:

1) A diode is a two-terminal non-linear device.

2) It is passive because there is no built-in source of power.

Note: It is not an exponential device, it’s a logarithmic device.[∵ Current controlling]

The current equation is defined as:

$I_d=I_S{e}^{\frac{V_d}{\eta V_t}}$

V_d = voltage across the diode

V_t = temperature equivalent

V_t ≈ 26 mV at room temperature.

$V_t=\frac{T}{11600}inkelvin$

I_s = Reverse saturation current.

$V_d=\eta V_t\ln \left(\frac{I_d}{I_S}\right)$

If ‘η’ is not given the default value will be 1

Calculation:

Given forward bias characteristics are

From the current equation using  data, we get two equations

$V_1=\eta V_t\ln \left(\frac{I_1}{I_S}\right)$ ⋯ (i)

$V_2=\eta V_t\ln \left(\frac{I_2}{I_S}\right)$ ⋯ (ii)

Considering η = 1

$V_1-V_2=V_t\ln \left(\frac{I_1}{I_S}\right)-V_t\ln \left(\frac{I_2}{I_S}\right)$

$V_1-V_2=V_t\ln \left(\frac{I_1}{I_2}\right)$

$0.6-0.65=\frac{T}{11600}\ln \left(\frac{1}{5}\right)$

$-0.05=-1.609\times \frac{T}{11600}$

$\frac{0.05\times 11600}{1.609}=T$

$T=\frac{580}{1.609}$

T = 360.47 Kelvin

T ≈ 360 Kelvin

Extra concept:

Diode small-signal equivalent is shown below for both ideal and practical cases.

Concept:

The diode current equation is given by:

$I_D=I_s{e}^{V_D/\eta V_T}$

η = Ideality factor (usually taken as 1)

VD = Applied bias voltage

VT = Thermal voltage

Calculation:

V_Ge = 0.143V

V_Si = 0.178V 

$\mathrm{I}={\mathrm{I}}_{{\mathrm{s}}_{\mathrm{G}\mathrm{e}}}\exp \left(\frac{\mathrm{e}{\mathrm{V}}_{\mathrm{G}\mathrm{e}}}{\mathrm{k}\mathrm{T}}\right)={\mathrm{I}}_{{\mathrm{s}}_{\mathrm{S}\mathrm{i}}}\exp \left(\frac{\mathrm{e}{\mathrm{V}}_{\mathrm{S}\mathrm{i}}}{\mathrm{k}\mathrm{T}}\right)$

$\mathrm{o}\mathrm{r},\frac{{\mathrm{I}}_{{\mathrm{s}}_{\mathrm{G}\mathrm{e}}}}{{\mathrm{I}}_{{\mathrm{s}}_{\mathrm{S}\mathrm{i}}}}=\exp \left[\frac{\mathrm{e}\left({\mathrm{V}}_{\mathrm{G}\mathrm{e}}-{\mathrm{V}}_{\mathrm{S}\mathrm{i}}\right)}{\mathrm{k}\mathrm{T}}\right]$

$=\exp \left[\frac{1.6\times {10}^{-19}\times \left(0.143-0.178\right)}{1.38\times {10}^{-23}\times 300}\right]$

= 0.25

${\mathrm{C}}_{\mathrm{o}\mathrm{x}}$ and ${\mathrm{C}}_{\mathrm{D}\mathrm{R}}$ are in series. Thus, the total depletion capacitance ${\mathrm{C}}_{\mathrm{D}}={\mathrm{C}}_{\mathrm{o}\mathrm{x}}\parallel {\mathrm{C}}_{\mathrm{D}\mathrm{R}}$.

$\begin{array}{l}\Rightarrow {\mathrm{C}}_{\mathrm{D}}=\frac{{\mathrm{C}}_{\mathrm{o}\mathrm{x}}{\mathrm{C}}_{\mathrm{D}\mathrm{R}}}{{\mathrm{C}}_{\mathrm{o}\mathrm{x}}+{\mathrm{C}}_{\mathrm{D}\mathrm{R}}}\\ \Rightarrow {\mathrm{C}}_{\mathrm{D}}=\frac{0.5\times 0.125}{0.5+0.125}\mathrm{n}\mathrm{F}/\mathrm{c}{\mathrm{m}}^2\end{array}$