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CP A variable DC battery is connected in series with a \(125-\Omega\) resistor and a \(p-n\) junction diode that hat hat a saturation current of 0.625 \(\mathrm{mA}\) at room temperature \(\left(20^{\circ} \mathrm{C}\right) .\) When a voltmeter across the \(125-\Omega\) resistor reads 35.0 \(\mathrm{V}\) , what are (a) the voltage across the diode and (b) the resistance of the diode?

Short Answer

Expert verified
(a) Voltage across the diode is the total voltage minus 35.0 V. (b) Resistance of the diode is 0 Ω because it is forward-biased.

Step by step solution

01

Determine the current through the resistor

First, we use Ohm's law to find the current (\(I\)) flowing through the resistor. Since we know the voltage across the resistor is 35.0 V and its resistance is 125 Ω, we write the equation: \(I = \frac{V}{R} = \frac{35.0 \text{ V}}{125 \ \Omega}\).Calculate this to find the current.
02

Calculate the diode voltage

We know that the total voltage (\(V_{total}\)) is the sum of the voltages across the diode (\(V_D\)) and the resistor. According to the problem, the voltmeter across the resistor reads 35.0 V. Since the battery voltage is the source, use \(V_{total} = V_D + 35.0 \text{ V}\).Express \(V_D\) in terms of \(V_{total}\).
03

Solve for diode resistance

Use the diode's current and voltage to determine its resistance. Using Ohm's Law: \(R_D = \frac{V_D}{I}\).Substitute the values of \(V_D\) and \(I\) from the previous steps into this equation to find the resistance of the diode.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ohm's Law
Ohm's Law is a fundamental principle in electronics that relates the voltage across a conductor to the current flowing through it and the resistance of the conductor. The equation for Ohm's Law is expressed as \( V = IR \), where \( V \) is the voltage, \( I \) is the current, and \( R \) is the resistance. This law is incredibly useful in analyzing circuits, as it allows us to calculate any one of these three variables if the other two are known.
In the example problem, a 125-Ohm resistor is connected in series with a diode. The voltage across the resistor is measured to be 35.0 V. Using Ohm's Law, we can determine the current flowing through the resistor with the formula \( I = \frac{V}{R} \).
This ability to use voltage and resistance to find the current is crucial for understanding how circuits behave. It helps us know how much current is passing through each component in the circuit, allowing for more detailed analysis and troubleshooting.
Diode Resistance
Diode resistance can be a tricky concept to grasp because it is not as straightforward as the resistance of a typical resistor. Diodes are semiconductor devices that allow current to flow in only one direction and have a non-linear relationship between voltage and current.
In general, the resistance of a diode is not constant; it varies depending on the current flowing through it. This is why diodes are described using two types of resistance: static and dynamic.
  • Static resistance is calculated using the DC voltage and DC current. It gives an approximate value of how much the diode resists the flow of current at a particular point.
  • Dynamic or small-signal resistance considers how the current changes with small changes in voltage.
To find the resistance of the diode from the example, we apply Ohm's Law again using the diode's specific voltage and current. Thus, we can say \( R_D = \frac{V_D}{I} \), where \( R_D \) is the diode resistance, \( V_D \) is the voltage across the diode, and \( I \) is the current determined in the previous step.
This calculation helps in understanding how much the diode is allowing or resisting current at a specific voltage, which is critical for circuit design and analysis.
Series Circuits
Series circuits are a fundamental type of electrical circuit where components are connected end-to-end, so there is a single path for current to flow. Understanding series circuits is crucial in electronics as it helps you predict how the voltage and current will behave across each component.
In a series circuit, the same current flows through all components because there is only one path. However, the voltage across each component can be different, and the sum of these voltages is equal to the total voltage supplied by the source. Mathematically, it can be expressed as \( V_{total} = V_1 + V_2 + \ldots + V_n \).
This principle is applied in the given problem, where a DC battery is connected in series with a resistor and diode. The total voltage provided by the battery is the sum of the voltages across the resistor and the diode. By knowing the voltage across the resistor, we can determine the voltage across the diode, which is essential for calculating its resistance.
The characteristics of series circuits make them useful for applications where the same current must pass through all components, but they also come with certain limitations, such as voltage division and lack of redundancy.

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Most popular questions from this chapter

The gap between valence and conduction bands in silicon is 1.12 eV. A nickel nucleus in an excited state emits a gamma-ray photon with wavelength \(9.31 \times 10^{-4}\) nm. How many electrons can be excited from the top of the valence band to the bottom of the conduction band by the absorption of this gamma ray?

A hypothetical \(\mathrm{NH}\) molecule makes a rotational-level transition from \(l=3\) to \(I=1\) and gives off a photon of wavelength 1.780 \(\mathrm{nm}\) in doing so. What is the separation between the two atoms in this molecule if we model them as point masses? The mass of hydrogen is \(1.67 \times 10^{-27} \mathrm{kg},\) and the mass of nitrogen is \(2.33 \times 10^{-26} \mathrm{kg} .\)

(a) Suppose a piece of very pure germanium is to be used as a light detector by observing, through the absorption of photons, the increase in conductivity resulting from generation of electron-hole pairs. If each pair requires 0.67 \(\mathrm{eV}\) of energy, what is the maximum wavelength that can be detected? In what portion of the spectrum does it lie? (b) What are the answers to part (a) if the material is silicon, with an energy requirement of 1.14 eV per pair, corresponding to the gap between valence and conduction bands in that element?

A hypothetical diatomic molecule of oxygen (mass =\( 2.656 \times 10^{-26} \mathrm{kg} ) \quad \text { and } \quad \text { hydrogen } \quad\left(\text { mass }=1.67 \times 10^{-27} \mathrm{kg}\right)\) emits a photon of wavelength 2.39\(\mu \mathrm{m}\) when it makes a transition from one vibrational state to the next lower state. If we model this molecule as two point masses at opposite ends of a massless spring, (a) what is the force constant of this spring, and (b) how many vibrations per second is the molecule making?

When an OH molecule undergoes a transition from the \(n=0\) to the \(n=1\) vibrational level, its internal vibrational energy increases by 0.463 eV. Calculate the frequency of vibration and the force constant for the interatomic force. (The mass of an oxygen atom is \(2.66 \times 10^{-26} \mathrm{kg},\) and the mass of a hydrogen atom is \(1.67 \times 10^{-27} \mathrm{kg} .\) )

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