Professional Courses
Industry-relevant training in Business, Technology, and Design to help professionals and graduates upskill for real-world careers.
Categories
Interactive Games
Fun, engaging games to boost memory, math fluency, typing speed, and English skillsβperfect for learners of all ages.
Typing
Memory
Math
English Adventures
Knowledge
Interactive Audio Lesson
Listen to a student-teacher conversation explaining the topic in a relatable way.
Understanding Common Emitter Amplifier Setup
Unlock Audio Lesson
Signup and Enroll to the course for listening the Audio Lesson
Today, we're going to delve into the common emitter amplifier. Can anyone remind me what a common emitter amplifier is?
It's an amplifier configuration where the emitter is common to both the input and output.
Exactly right! Now, when we use a current mirror in this setup, why do you think it's important for the transistors to be matched?
Because matching ensures that the currents through the transistors are equal, leading to consistent performance.
Good point! Remember: if transistors Q1 and Q2 are identical, their characteristics will align closely, allowing effective current mirroring. This will be crucial for your calculations moving forward.
Calculating Collector Currents
Unlock Audio Lesson
Signup and Enroll to the course for listening the Audio Lesson
Letβs calculate the collector current of transistor 1, starting with our assumption of I_C = 2 mA. Student_3, can you remind us how Ξ² affects the base current?
Base current, I_B, can be calculated as I_C divided by beta.
Correct! Therefore, what would our base current be when I_C is 2 mA and Ξ² is 100?
It would be 20 Β΅A.
Great! So to find the resistance R1 that biases our amplifier, can anyone tell me how we can calculate that?
We can use Ohm's law, R = V/I, substituting V as the supply voltage minus the V_BE drop.
Perfect! And what's the resistance we would calculate?
570 kβ¦.
Exactly! This resistance ensures we're maintaining a crucial current flow in our circuit.
Small Signal Output Resistance and Voltage Gain
Unlock Audio Lesson
Signup and Enroll to the course for listening the Audio Lesson
Now that we've established the values of our components, let's calculate the small signal output resistance, R_out. What do we recall about this in our circuit?
Itβs the equivalent resistance seen by the small signal at the output.
Right! If we have small signal resistances r_o1 and r_o4 both as 50 kβ¦, how do we find R_out?
R_out would be r_o1 in parallel with r_o4, so that would be 25 kβ¦.
Excellent! Now, can someone explain how we find the voltage gain of our amplifier?
The voltage gain is given by the formula: gain = -g_m * R_out, where g_m is the transconductance.
That's correct! And remembering our calculations, we found the voltage gain to be approximately 1923.
Output Voltage Determination
Unlock Audio Lesson
Signup and Enroll to the course for listening the Audio Lesson
Next, let's analyze the output voltage. If we assume an output voltage of 11.4 V, Student_1, how does this correspond with our earlier calculations?
It corresponds with our determination that the currents in transistors are equal, confirming stability in our circuit performance.
Exactly! But what happens if we consider base current losses or varying early voltages?
That could lead to a mismatch in currents and cause the output voltage to drift from our ideal value!
Absolutely! Understanding these intricacies allows us to appreciate the stability and operational conditions for our circuits.
Introduction & Overview
Read a summary of the section's main ideas. Choose from Basic, Medium, or Detailed.
Quick Overview
Standard
The section explains the process of finding collector currents in a common emitter amplifier using current mirrors and the significance of matching transistors. It also covers the concept of small signal output resistance, voltage gain calculations, and the effects of early voltage on the DC output voltage. The relationship between different currents and voltages in this configuration is thoroughly analyzed.
Detailed
Detailed Summary
In this section, the calculations for the common emitter amplifier using a current mirror are discussed. It begins by asserting the assumption that transistors Q1 and Q2 are identical, which helps in establishing that the collector current of transistor 1 (I_C1) must equal the collector current of transistor 4 (I_C4). The section works through finding the bias resistor R1 and its implications on the collector current (I_C = 2 mA) linked with a Ξ² value of 100. Subsequently, the small signal output resistance (R_out) and voltage gain calculations illustrate the high gain expected from the amplifier configuration. The output voltage is established as 11.4 V based on these conditions, leading into insights about how variations in early voltage or base current mismatches between transistors could result in significant output voltage fluctuations.
Youtube Videos
Audio Book
Dive deep into the subject with an immersive audiobook experience.
Understanding Current Relationships in Circuits
Unlock Audio Book
Signup and Enroll to the course for listening the Audio Book
As I said that the current flow current flow here and here should be equal and if you see it carefully the DC voltage here it is defined by this V β V drop. So, I should say the voltage here it is 12 V. So, the voltage here it is 12 β 0.6 so, that is 11.4 V.
Detailed Explanation
In this chunk, we discuss that for certain transistors in a circuit, the current flowing through them must be equal for proper functioning. If the DC voltage at a certain point in the circuit (V_BE) is defined and results from the voltage drop across components, we can calculate the specific voltage present at any node. For instance, a supply voltage of 12V and a base-emitter drop of 0.6V gives us a voltage of 11.4V.
Examples & Analogies
Think of a water pipeline where the pressure (voltage) must remain equal on both sides for water (current) to flow smoothly. If you have a pressure of 12 units at one end and a valve (like the base-emitter junction of a transistor) that requires a drop of 0.6 units to function correctly, the pressure after the valve will be 11.4 units, equalizing the flow on both sides.
Current Mirroring Concepts
Unlock Audio Book
Signup and Enroll to the course for listening the Audio Book
Now, with this 11.4 V here we can say that whatever the current we do have. So, that may be 2 m which is of course, 1 approximation that we are assuming ( ) = 1 please stop here.
Detailed Explanation
Here, the focus shifts to the concept of current mirroring. At the voltage of 11.4V, the expected current flowing through the circuit is estimated to be 2 mA. This is a common assumption in transistor circuits, where current mirrors facilitate the reproduction of this current in other similar components of the circuit.
Examples & Analogies
Imagine a school where a teacher (current) is teaching a group of students (transistors). The teacher's ability to educate (current level) can be mirrored in different classrooms. If the teacher successfully educates 2 students (2 mA) in one classroom, similarly, other classrooms can replicate this effort if the environment (voltage) is conducive and identical conditions (equal V_BE) are provided.
Effects of Base Current and Device Matching
Unlock Audio Book
Signup and Enroll to the course for listening the Audio Book
If this voltage here it is say 11.4, then you can say that this current and this current they will be equal. In intuitively also you can say that if this voltage it is 11.4. So, that makes the current here and current here, they are equal and also the current flow in transistor-3 and transistor-4 they are equal.
Detailed Explanation
This chunk emphasizes that if certain conditions (such as a fixed voltage of 11.4V) are met, equal currents will flow through respective transistors presuming they are identical in specifications (like Ξ² factors). Current mirroring relies on this principle, where two identical transistors (like transistor-3 and transistor-4) will share equal current under ideal conditions.
Examples & Analogies
Think of two identical scales (transistors) side by side. If you place the same weight (current) on both and ensure they are calibrated identically (matched specifications), both will show equal readings as long as the environment remains steady (constant voltage supply).
Impact of Early Voltage on DC Output
Unlock Audio Book
Signup and Enroll to the course for listening the Audio Book
In case if this current it is higher than the current flow through transistor-1 for 11.4 V DC, then the corresponding voltage here it would be higher +ve.
Detailed Explanation
This section discusses how the early voltage and the currents affect the DC output voltage. If the required current flowing exceeds what is being mirrored through the transistor (transistor-1), then the output voltage may change and deviate from the expected value. The circuit's precision is influenced by how closely the actual performance of the devices matches theoretical predictions.
Examples & Analogies
Recall the analogy of two water towers (transistors) regulating water flow. If one tower can handle more water flow than the other and tries to push more water than can be matched, the pressure in the pipelines (output voltage) can increase unexpectedly, leading to a need for adjustments to prevent overflow or system failure.
Definitions & Key Concepts
Learn essential terms and foundational ideas that form the basis of the topic.
Key Concepts
-
Common Emitter Amplifier: A configuration providing large voltage gain, using transistors for amplification.
-
Current Mirror Circuit: A configuration that replicates a current flowing in one transistor to another, ensuring stable current flow.
-
Voltage Gain: The ratio of output voltage to input voltage, vital for amplifying signals in electronic circuits.
Examples & Real-Life Applications
See how the concepts apply in real-world scenarios to understand their practical implications.
Examples
-
In a common emitter amplifier with an input voltage V_in of 1V, if the gain is 1923, the output voltage V_out would be approximately 1923V, allowing us to amplify weak signals significantly.
-
By using a current mirror, if a current of 2mA flows through transistor Q1, identical current flow will occur through Q2, maintaining consistency across the circuit.
Memory Aids
Use mnemonics, acronyms, or visual cues to help remember key information more easily.
π΅ Rhymes Time
-
To connect your amps with flair, ensure those currents are not rare.
π Fascinating Stories
-
Imagine a room with twin lamps illuminating equally; when one lamp shines brighter, the other dims, just like currents in a current mirror!
π§ Other Memory Gems
-
C.E.A for Common Emitter Amplifier: Current flowing Easily Amplified!
π― Super Acronyms
MIRROR
- Matching Identical Resistances Reflecting Output Resistance.
Flash Cards
Review key concepts with flashcards.
Glossary of Terms
Review the Definitions for terms.
-
Term: Common Emitter Amplifier
Definition:
An amplifier configuration where the emitter is common to both the input and output, typically providing significant voltage gain.
-
Term: Current Mirror
Definition:
A circuit configuration that clones the current flowing through one active device by controlling the current flow in another active device.
-
Term: Collector Current (I_C)
Definition:
The current flowing through the collector terminal of a bipolar junction transistor (BJT).
-
Term: Ξ² (Beta)
Definition:
The current gain of a transistor, a measure of how much the base current is amplified in terms of collector current.
-
Term: Early Voltage
Definition:
The voltage at which the output characteristic relationships of a transistor become linear, affecting the output impedance.