A non-ideal diode is biased with a voltage of -0.03 V, and a diode current of I1 is measured. The thermal voltage is 26 mV and the ideality factor for the diode is 15/13. The voltage, in V, at which the measured current increases to 1.5 I1 is closest to:
-0.09
The effective exponential voltage is \(\eta V_T=(15/13)(0.026)=0.03\,V\). Initially \(I_1/I_s=e^{-1}-1\). Requiring \(I_2=1.5I_1\) gives \(e^{V_2/0.03}=1+1.5(e^{-1}-1)=0.051819\). Thus \(V_2=0.03\ln(0.051819)\approx-0.0888\,V\), closest to −0.09 V. The negative current increases in magnitude.
Important Questions from Diodes and Its Applications
Which of the following statements is FALSE about LED lamps?
A Zener diode is used in _______ for voltage regulation.
Which of the following diodes can be used as Voltage Regulator if used in reverse bias?
What is it called when a diode exceeds the reverse voltage and then it fails and passes more current in the reverse direction?
Determine the resonant frequency of an IMPATT diode having the following specifications:
Carrier drift velocity: 2 × 107 cm/s
Drift-region length: 5 μm
Maximum operating voltage: 100 V
Maximum operating current: 250 mA
Efficiency: 20%
Breakdown voltage: 90 V
