Final Answer:
The ratio of current at 100°C to -55°C will be approximately 0.00. This means the current at -55°C is significantly higher than that at 100°C.
Step-by-step explanation:
To determine the ratio of current at 100°C to -55°C, we need to consider the diode equation and how temperature affects it.
The diode equation relates the current through the diode (I) to the applied voltage (V) and temperature (T) as:
I = I_s * (exp(eV / (n * kT)) - 1)
where:
- I_s is the saturation current (a constant dependent on the diode material and doping)
- e is the electron charge (approximately 1.602 × 10^-19 C)
- n is the ideality factor (usually close to 1 for silicon diodes)
- k is the Boltzmann constant (approximately 1.38 × 10^-23 J/K)
- T is the absolute temperature in Kelvin
Given:
- Forward-bias voltage (V) = 0.6 V
- Temperature at 100°C (T1) = 100 + 273.15 = 373.15 K
- Temperature at -55°C (T2) = -55 + 273.15 = 218.15 K
Calculation:
1. Calculate the current at each temperature using the diode equation:
I1 (at 373.15 K): I1 = I_s * (exp(0.6 * 1.602e-19 / (1 * 1.38e-23 * 373.15)) - 1)
I2 (at 218.15 K): I2 = I_s * (exp(0.6 * 1.602e-19 / (1 * 1.38e-23 * 218.15)) - 1)
2. Calculate the ratio of currents:
Current ratio = I1 / I2
The ratio of current at 100°C to -55°C will be approximately 0.00
We need to know the value of the saturation current (I_s) to calculate the actual current values. However, we can still find the ratio of currents without it.
As temperature increases, the thermal voltage (kT) increases, causing the exponential term in the diode equation to dominate. This means that a smaller increase in voltage will result in a significantly larger increase in current at higher temperatures. This is why the current at -55°C is much higher than that at 100°C for the same forward-bias voltage.