Loading the lesson
Loading the lesson
Search AllRounder.ai
Search the courses, subjects, tracks, games and features, or jump straight to a page.
Search the courses, subjects, tracks, games and features, or jump straight to a page.
Learn it your way
A short class with Sarah and her students. The other ways are under More.
Sign up free to listen. Transcripts are free to read.
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.
Let’s calculate the collector current of transistor 1, starting with our assumption of I_C = 2 mA. Student_3, can you remind us how β 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 β is 100?
It would be 20 µA.
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 kΩ.
Exactly! This resistance ensures we're maintaining a crucial current flow in our circuit.
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?
It’s 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 kΩ, how do we find R_out?
R_out would be r_o1 in parallel with r_o4, so that would be 25 kΩ.
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.
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.
A configuration providing large voltage gain, using transistors for amplification.
A configuration that replicates a current flowing in one transistor to another, ensuring stable current flow.
The ratio of output voltage to input voltage, vital for amplifying signals in electronic circuits.
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.
To connect your amps with flair, ensure those currents are not rare.
Imagine a room with twin lamps illuminating equally; when one lamp shines brighter, the other dims, just like currents in a current mirror!
C.E.A for Common Emitter Amplifier: Current flowing Easily Amplified!
MIRROR: Matching Identical Resistances Reflecting Output Resistance.
Term
What is a collector current?
Click to flip
Term
Define Early Voltage.
Click to flip
An amplifier configuration where the emitter is common to both the input and output, typically providing significant voltage gain.
A circuit configuration that clones the current flowing through one active device by controlling the current flow in another active device.
The current flowing through the collector terminal of a bipolar junction transistor (BJT).
The current gain of a transistor, a measure of how much the base current is amplified in terms of collector current.
The voltage at which the output characteristic relationships of a transistor become linear, affecting the output impedance.
This section discusses the calculation of currents and the output voltage in a current mirror setup, particularly focusing on a common emitter amplifier.
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.
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 β 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.