/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 40 In an \(N P N\) transistor the c... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 20: Problem 40

In an \(N P N\) transistor the collector current is \(24 \mathrm{~mA}\). If \(80 \%\) of electrons reach collector its base current in \(\mathrm{mA}\) is (A) 36 (B) 26 (C) 16 (D) 6

Short Answer

Expert verified
The base current in mA is 6 mA, which corresponds to the answer choice (D) 6.

Step by step solution

01

Calculate the Emitter Current

First, let's find the emitter current (I_E). Since 80% of the electrons reach the collector (I_C), the remaining 20% must be part of the base current (I_B). Therefore, we can express I_E as 100% of the transistor current. We are given I_C = 24 mA. Using the percentage information, we can calculate I_E: I_E = (100/80) * I_C I_E = (100/80) * 24 mA
02

Calculate the Emitter Current

Now, we can calculate the value of I_E: I_E = (5/4) * 24 mA I_E = 30 mA
03

Calculate the Base Current

Since we know the relationship between I_C, I_B, and I_E (I_E = I_C + I_B), we can now solve for I_B: I_B = I_E - I_C I_B = 30 mA - 24 mA I_B = 6 mA Thus, the base current in mA is 6 mA, which corresponds to the answer choice (D) 6.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The depletion layer in the \(P-N\) junction region is caused by (A) Drift of holes (B) Diffusion of charge carriers (C) Migration of impurity ions (D) Drift of electrons

For a transistor amplifier in common emitter configuration for load impedance of \(1 k \Omega\left(h_{f e}=50\right)\) and ( \(\left.h_{\text {oe }}=25 \mu \mathrm{A} / \mathrm{V}\right)\), the current gain is [2004] (A) \(-5.2\) (B) \(-15.7\) (C) \(-24.8\) (D) \(-48.78\)

In a common-base mode of transistor, the collector current is \(5.488 \mathrm{~mA}\) for an emitter current of \(5.60 \mathrm{~mA}\). The value of the base current amplification factor \((\beta)\) will be (A) 49 (B) 50 (C) 51 (D) 48

Then the value of \(\beta\) is (A) 100 (B) 50 (C) 200 (D) 150

Symbolic representation of photodiode is
See all solutions

Recommended explanations on Physics Textbooks

Further Mechanics and Thermal Physics

Read Explanation

Torque and Rotational Motion

Read Explanation

Rotational Dynamics

Read Explanation

Famous Physicists

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Magnetism

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.