Respuesta :
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 = [tex]I_{S}[/tex] x [tex]e^{r}[/tex] ---------------------(i)
Where;
[tex]I_{S}[/tex] = saturation current of the diode
r = [tex]\frac{V}{V_{T} }[/tex]
[tex]V_{T}[/tex] = thermal voltage
V = diode voltage
At 20°C, the thermal voltage ([tex]V_{T}[/tex]) 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 [tex]V_{T}[/tex] into the equation;
=> r = [tex]\frac{V}{V_{T} }[/tex]
=> 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 = [tex]I_{S}[/tex] x e²⁷°⁶
1 x 10⁻³ = [tex]I_{S}[/tex] x 9.69 x 10¹¹
Solve for [tex]I_{S}[/tex]
[tex]I_{S}[/tex] = 1 x 10⁻³ / (9.69 x 10¹¹)
[tex]I_{S}[/tex] = 0.103 x 10⁻¹⁴A
[tex]I_{S}[/tex] = 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 = [tex]V_{T}[/tex] x ln (I / [tex]I_{S}[/tex]) --------------------(ii)
Where;
V = diode voltage
[tex]V_{T}[/tex] = 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
[tex]I_{S}[/tex] = saturation current (calculated above) = 1.03 x 10⁻¹⁵A
Solve for [tex]V_{T}[/tex];
[tex]V_{T}[/tex] = kT/ q
[tex]V_{T}[/tex] = 1.38 x 10⁻²³ x 253 / (1.6 x 10⁻¹⁹)
[tex]V_{T}[/tex] = 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 = [tex]V_{T}[/tex] x ln (I / [tex]I_{S}[/tex]) --------------------(ii)
Where;
V = diode voltage
[tex]V_{T}[/tex] = 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
[tex]I_{S}[/tex] = saturation current (calculated above) = 1.03 x 10⁻¹⁵A
Solve for [tex]V_{T}[/tex];
[tex]V_{T}[/tex] = kT/ q
[tex]V_{T}[/tex] = 1.38 x 10⁻²³ x 358 / (1.6 x 10⁻¹⁹)
[tex]V_{T}[/tex] = 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
