Question

A diode fed with a constant current I=1mA has a voltage V = 690 mV at 20°C. Find the diode voltage at −20°C and at +85°C.

Answer 1

Answer:

602.23mV at -20°C

852.88mV at +85°C

Explanation:

===> First, let's get the saturation current of the diode

The current (I) through a diode is given by;

I = $I_{S}$ x $e^{r}$   ---------------------(i)

Where;

$I_{S}$ = saturation current of the diode

r = $\frac{V}{V_{T} }$  

$V_{T}$ = thermal voltage

V = diode voltage

At 20°C, the thermal voltage ($V_{T}$) of a diode is 25 x 10⁻³V

From the question, the following are given;

At 20°C

I = 1mA = 1 x 10⁻³A

V = 690mV = 690 x 10⁻³V

Find the value of r by substituting the values of V and $V_{T}$ into the equation;

=> r = $\frac{V}{V_{T} }$

=> r = (690 x 10⁻³) / (25 x 10⁻³)

=> r = 27.6

Substitute the values of r and I into equation (i) to give;

1 x 10⁻³ = $I_{S}$ x e²⁷°⁶

1 x 10⁻³ = $I_{S}$  x 9.69 x 10¹¹

Solve for $I_{S}$

$I_{S}$ = 1 x 10⁻³ / (9.69 x 10¹¹)

$I_{S}$ = 0.103 x 10⁻¹⁴A

$I_{S}$ = 1.03 x 10⁻¹⁵ A

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===> Second, let's get the diode voltage at -20°C using the diode voltage formula as follow;

V = $V_{T}$ x ln (I / $I_{S}$)              --------------------(ii)

Where;

V = diode voltage

$V_{T}$ = thermal voltage = k x T / q

k = Boltzmann's constant = 1.38 x 10⁻²³ Joules/kelvin

T = temperature = -20°C = 273 - 20 = 253K

q = absolute value of electron charge = 1.6 x 10⁻¹⁹C.

I = diode current = 1mA = 1 x 10⁻³A

$I_{S}$ = saturation current (calculated above) = 1.03 x 10⁻¹⁵A

Solve for $V_{T}$;

$V_{T}$ = kT/ q

$V_{T}$ = 1.38 x 10⁻²³ x 253 / (1.6 x 10⁻¹⁹)

$V_{T}$ = 218.2 x 10⁻⁴V

Substitute these values into equation (ii)

V = 218.2 x 10⁻⁴ x ln [(1 x 10⁻³) / (1.03 x 10⁻¹⁵)]

V = 218.2 x 10⁻⁴ x ln (0.97 x 10¹²)

V = 218.2 x 10⁻⁴ x 27.6

V = 6022.32 x 10⁻⁴V

V = 602.23 x 10⁻³V

V = 602.23mV

Therefore, the diode voltage at -20°C is 602.23mV

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===> Third, let's get the diode voltage at +85°C using the diode voltage formula as follow;

V = $V_{T}$ x ln (I / $I_{S}$)              --------------------(ii)

Where;

V = diode voltage

$V_{T}$ = thermal voltage = k x T / q

k = Boltzmann's constant = 1.38 x 10⁻²³ Joules/kelvin

T = temperature = 85°C = 273 + 85 = 358K

q = absolute value of electron charge = 1.6 x 10⁻¹⁹C.

I = diode current = 1mA = 1 x 10⁻³A

$I_{S}$ = saturation current (calculated above) = 1.03 x 10⁻¹⁵A

Solve for $V_{T}$;

$V_{T}$ = kT/ q

$V_{T}$ = 1.38 x 10⁻²³ x 358 / (1.6 x 10⁻¹⁹)

$V_{T}$ = 308.8 x 10⁻⁴V

Substitute these values into equation (ii)

V = 308.8 x 10⁻⁴ x ln [(1 x 10⁻³) / (1.03 x 10⁻¹⁵)]

V = 308.8 x 10⁻⁴ x ln (0.97 x 10¹²)

V = 308.8 x 10⁻⁴ x 27.6

V = 8522.88 x 10⁻⁴V

V = 852.88 x 10⁻³V

V = 852.88mV

Therefore, the diode voltage at +85°C is 852.88mV