<h2>
Answer:</h2><h2>
</h2>
602.23mV at -20°C
852.88mV at +85°C
<h2>
Explanation:</h2><h2>
</h2>
===> First, let's get the saturation current of the diode
The current (I) through a diode is given by;
I = x ---------------------(i)
Where;
= saturation current of the diode
r =
= thermal voltage
V = diode voltage
<em>At 20°C, the thermal voltage (</em><em>) of a diode is 25 x 10⁻³V</em>
<em />
<em>From the question, the following are given;</em>
<em>At 20°C</em>
I = 1mA = 1 x 10⁻³A
V = 690mV = 690 x 10⁻³V
<em>Find the value of r by substituting the values of V and </em><em> into the equation;</em>
=> r =
=> r = (690 x 10⁻³) / (25 x 10⁻³)
=> r = 27.6
<em>Substitute the values of r and I into equation (i) to give;</em>
1 x 10⁻³ = x e²⁷°⁶
1 x 10⁻³ = x 9.69 x 10¹¹
<em>Solve for </em><em />
= 1 x 10⁻³ / (9.69 x 10¹¹)
= 0.103 x 10⁻¹⁴A
= 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 = x ln (I / ) --------------------(ii)
Where;
V = diode voltage
= 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
= saturation current (calculated above) = 1.03 x 10⁻¹⁵A
Solve for ;
= kT/ q
= 1.38 x 10⁻²³ x 253 / (1.6 x 10⁻¹⁹)
= 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
<em>Therefore, the diode voltage at -20°C is 602.23mV</em>
<em></em>
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===> Third, let's get the diode voltage at +85°C using the diode voltage formula as follow;
V = x ln (I / ) --------------------(ii)
Where;
V = diode voltage
= 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
= saturation current (calculated above) = 1.03 x 10⁻¹⁵A
Solve for ;
= kT/ q
= 1.38 x 10⁻²³ x 358 / (1.6 x 10⁻¹⁹)
= 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
<em>Therefore, the diode voltage at +85°C is 852.88mV</em>