Question

If a diode at 300°K with a constant bias current of 100μA has a forward voltage of 700mV across it, what will the voltage drop across this same diode be if the bias current is increased to 1mA? g

Answer 1

Answer:

the voltage drop across this same diode will be 760 mV

Explanation:

Given that:

Temperature T = 300°K

current $I_1$ = 100 μA

current $I_2$ = 1 mA

forward voltage $V_r$ = 700 mV = 0.7 V

To objective is to find the voltage drop across this same diode  if the bias current is increased to 1mA.

Using the formula:

$I = I_o \begin {pmatrix} e^{\dfrac{V_r}{nv_T}-1} \end {pmatrix}$

$I_1 = I_o \begin {pmatrix} e^{\dfrac{V_r}{nv_T}-1} \end {pmatrix}$

where;

$V_r$ = 0.7

$I_1 = I_o \begin {pmatrix} e^{\dfrac{0.7}{nv_T}-1} \end {pmatrix}$

$I_2 = I_o \begin {pmatrix} e^{\dfrac{V_r'}{nv_T}-1} \end {pmatrix}$

$\dfrac{I_1}{I_2} = \dfrac{ I_o \begin {pmatrix} e^{\dfrac{0.7}{nv_T}-1} \end {pmatrix} }{ I_o \begin {pmatrix} e^{\dfrac{V_r'}{nv_T}-1} \end {pmatrix} }$

$\dfrac{100 \ \mu A}{1 \ mA} = \dfrac{ \begin {pmatrix} e^{\dfrac{0.7}{nv_T}-1} \end {pmatrix} }{ \begin {pmatrix} e^{\dfrac{V_r'}{nv_T}-1} \end {pmatrix} }$

Suppose n = 1

$V_T = \dfrac{T}{11600} \\ \\ V_T = \dfrac{300}{11600} \\ \\ V_T = 25. 86 \ mV$

Then;

$e^{\dfrac{V_r'}{nv_T}-1} = 10 \begin {pmatrix} e ^{\dfrac{ 0.7} { nV_T} -1} \end {pmatrix}$

$e^{\dfrac{V_r'}{nv_T}-1} = 10 \begin {pmatrix} e ^{\dfrac{ 0.7} { 25.86} -1} \end {pmatrix}$

$e^{\dfrac{V_r'}{nv_T}-1} = 5.699 \times 10^{12}$

${e^\dfrac{V_r'}{nv_T}} = 5.7 \times 10^{12}$

${\dfrac{V_r'}{nv_T}} =log_{e ^{5.7 \times 10^{12}}}$

${\dfrac{V_r'}{nv_T}} =29.37$

$V_r'=29.37 \times nV_T$

$V_r'=29.37 \times 25.86$

$V_r'=759.5 \ mV$

$Vr' \simeq$ 760 mV

Thus, the voltage drop across this same diode will be 760 mV