Question

A particular diode is found to conduct 1mA with junction voltage of 0.7V. What current will flow in this diode if the junction voltage is raised to 0.71V? To 0.8V? If the junction voltage is lowered to 0.69V? To 0.6V? What change in junction voltage will increase the diode current by a factor of 10?

Answer 1

In the forward-biased configuration, the current through a diode increases exponentially with the voltage, while the voltage across the diode remains approximately 0.7V. Changing the junction voltage to 0.71V, 0.8V, 0.69V, or 0.6V will not significantly affect the diode current, which will still be around 1mA. To increase the diode current by a factor of 10, the junction voltage needs to change by approximately 0.0577V.

Step-by-step explanation:

In the forward-biased configuration of a diode, the current increases exponentially with the voltage. However, the voltage across the diode remains approximately 0.7 V. So, if the junction voltage is raised to 0.71V or 0.8V, the diode will still conduct with a similar current of 1mA. If the junction voltage is lowered to 0.69V or 0.6V, the diode will still conduct with a similar current of 1mA. To find the change in junction voltage that increases the diode current by a factor of 10, we can use the equation:

I_2 = I_1 * e^(deltaV / Vt)

Where I_2 is the new current, I_1 is the initial current, deltaV is the change in voltage, and Vt is the thermal voltage (approximately 0.025V). Solving for deltaV and substituting the given values:

1. I_2 = 10 * I_1
2. deltaV = Vt * ln(I_2 / I_1)

Substituting I_2 = 10mA and I_1 = 1mA:

1. deltaV = 0.025V * ln(10)

So, the change in junction voltage that increases the diode current by a factor of 10 is approximately 0.0577V.