Question

3. (15 points) A diode with has and ideality factor n 1.5. o lo -2 x 10-13 A at Vo -2 V but is NOT at room temperature. a) What is the thermal voltage, VT? VT b) What is Vo when lo 8 mA? Vo-

Answer 1

X = 1.5, I_D = 1MA, at V_D = 0.6 V I_D = -2 times 10^-13 A at V_d = -2 V (a) thermal voltage (V_T) (b) what is V_D, when I_D = 8 MA. I_D = I_0 [e^V_0/ XVT -1] 1 times 10^-3 = I_0 [0.6/1.5 V_T -1] log_e (10^-3) = log_e (I_0) + log_e [e^0.6/1.5 V_T-1] log_e (10^-3)Tilde equal log_e (I_0) + lig_e (e^0.6/1.5 V_T) implies -6.9 = log_e (I_0) + 0.6/1.5 V_T rightarrow (1) I_D Tilde equal to I_0 [e^v_D/ X V_T] - 2 times 10^-13 = I_0 [e^-2/1.5 V_T] 2 times 10^-13 = I_0 [e^-2/1.5 V_T] 2 times 10^-13 = -I_0 [e^-2/1.5 V_T] log_e (2 times 10^-13) = -[log_e (I_0) - 2/1.5 V_T] log_e (2 times 10^-13) = 2/1.5 V_T - log_e (I_0) rightarrow (2) Adding (1) and (2) - 29.24 = 2/1.5 V_T -log_e (I_0).