A forward-bias voltage of 15.015.0 mV produces a positive current of 9.259.25 mA through a p−np-n junction at 300300 K. What does the positive current become if the forward-bias voltage is reduced to 10.010.0 mV?

Ch 42: Molecules and Condensed Matter

Young & Freedman Calc - University Physics 14th Edition

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Young & Freedman Calc 14th Edition

Ch 42: Molecules and Condensed Matter

Problem 31a

Chapter 42, Problem 31a

A forward-bias voltage of $15.0$ mV produces a positive current of $9.25$ mA through a $p - n$ junction at $300$ K. What does the positive current become if the forward-bias voltage is reduced to $10.0$ mV?

Verified step by step guidance

Understand the relationship between the current and voltage in a p-n junction diode. The current through a diode is given by the Shockley diode equation:

I = I_s (e^(qV / kT) - 1)

, where

I

is the current,

I_s

is the reverse saturation current,

q

is the charge of an electron,

V

is the forward-bias voltage,

k

is Boltzmann's constant, and

T

is the temperature in kelvins.

Recognize that the temperature

T

is constant at 300 K, and the reverse saturation current

I_s

does not change. The relationship between the current and voltage is exponential, so a small change in voltage will result in a significant change in current.

Use the Shockley diode equation to express the ratio of the currents at the two voltages. Let

I_1

and

I_2

be the currents at voltages

V_1 = 15.0 \, \(\text{mV}\)

and

V_2 = 10.0 \, \(\text{mV}\)

, respectively. The ratio is given by:

\(\frac{I_2}{I_1}\) = e^{q(V_2 - V_1) / kT}

Substitute the known values into the ratio equation. Use

q = 1.6 \(\times\) 10^{-19} \, \(\text{C}\)

k = 1.38 \(\times\) 10^{-23} \, \(\text{J/K}\)

, and

T = 300 \, \(\text{K}\)

. The voltage difference is

V_2 - V_1 = 10.0 \, \(\text{mV}\) - 15.0 \, \(\text{mV}\) = -5.0 \, \(\text{mV}\)

, or

-5.0 \(\times\) 10^{-3} \, \(\text{V}\)

Calculate the exponential factor

e^{q(V_2 - V_1) / kT}

to find the ratio

\(\frac{I_2}{I_1}\)

. Multiply this ratio by the given current

I_1 = 9.25 \, \(\text{mA}\)

to determine the new current

I_2

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

p-n Junction

A p-n junction is formed by joining p-type and n-type semiconductors, creating a diode that allows current to flow primarily in one direction. The p-type material has an abundance of holes (positive charge carriers), while the n-type has excess electrons (negative charge carriers). This structure is fundamental in understanding how diodes operate under different voltage conditions.

Forward Bias

Forward bias occurs when the positive terminal of a voltage source is connected to the p-type material and the negative terminal to the n-type material of a diode. This reduces the barrier potential at the junction, allowing charge carriers to recombine and resulting in a flow of current. The amount of current is dependent on the applied voltage and the characteristics of the diode.

Shockley Diode Equation

The Shockley diode equation describes the current flowing through a diode as a function of the applied voltage. It is given by I = I0 (e^(qV/kT) - 1), where I0 is the reverse saturation current, q is the charge of an electron, V is the voltage across the diode, k is Boltzmann's constant, and T is the absolute temperature. This equation illustrates how current changes exponentially with small variations in forward-bias voltage.

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