89.3.1 - Summary of Numerical Examples on Current Mirror
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Overview of Current Mirrors in Amplifiers
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Today, we will explore the use of current mirrors in differential amplifiers. Can someone tell me what a current mirror is?
Isn't it a way to copy current from one branch of a circuit to another?
Exactly! Current mirrors are essential for maintaining consistent current in different parts of a circuit. Now, why do you think this is important in amplifiers?
It helps keep the signal clean and stable, right?
Correct! A stable signal leads to improved performance in amplifiers. Let's now see how we calculate DC currents in these setups.
Calculating DC Currents
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Let's look at an example. If the load resistor is 11.4 k ohms and the current is 1mA, how do we determine the current through each transistor?
If the total current is 1 mA and it splits between two transistors, wouldn't each one carry 0.5 mA?
Exactly! And that mirrored current flows through both transistors in the mirror. Let’s calculate what happens when we adjust the input signal... What happens to the output?
The output current would mirror the input signal changes, right?
Yes! You're understanding the functionality perfectly.
Understanding Gain Calculations
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Now, with the DC currents established, how do we proceed to calculate the differential mode and common mode gains?
We use the transconductance and the output resistances, right?
That's right! A common mistake is overlooking how impedance affects gain. Can anyone summarize the gain calculation for differential mode?
We compute it using the formula: Gain = gm * (ro1 || ro2) for the differential mode.
Excellent! Now, how would the common mode gain differ?
It’s likely lower because of degeneration in the circuit, using the resistor to reduce common mode impact.
Absolutely correct! Time for some practice!
Practical Applications of Current Mirrors
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Let’s talk about where we can see current mirrors in real-world applications. Can someone give me an example?
I remember learning about them in transistor amplifiers, especially in active loads.
Great example! They also appear in voltage references and biasing circuits. Why do you think they are preferred?
Because they maintain a constant current, improving the stability over varying conditions.
Exactly! Stability is key in analog circuits. This is what enables efficient amplification and signal integrity.
Introduction & Overview
Read summaries of the section's main ideas at different levels of detail.
Quick Overview
Standard
In Section 3.1, we explore numerical examples related to current mirrors embedded in differential amplifiers. The section explains how to calculate the DC currents in transistors, establish gain measures, and analyzes the implications of these calculations on circuit performance, specifically focusing on differential and common mode gains.
Detailed
Detailed Overview of Current Mirror Numerical Examples
In this section, we elaborate on the applications of current mirrors within differential amplifiers. We begin by examining a differential amplifier that includes an active load, specifically emphasizing the role of current mirrors in ensuring matching currents through connected transistors. The primary calculations aimed at evaluating DC currents through the transistors demonstrate a clear relationship between theoretical values and practical applications.
The section also covers detailed examples of calculating differential and common mode gains. Furthermore, we analyze how the configuration of these circuits affects performance, leading to insights into designing more efficient analog electronic circuits. By the end of this section, the student will understand the operational principles behind current mirrors, appreciate their applications in amplifiers, and be proficient in executing relevant calculations.
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Introduction to Current Mirrors
Chapter 1 of 5
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Chapter Content
In the numerical examples, we will be reestablishing the same thing, to start with we do have to find the DC currents through all the transistors.
Detailed Explanation
This chunk introduces the concept of current mirrors within the context of example problems. The basic task is to determine the direct current (DC) flowing through each transistor in the circuit. Understanding the flow of current in transistors is essential, as it helps in analyzing the performance and behavior of current mirrors.
Examples & Analogies
Imagine a water system where pipes (transistors) carry water (current). To ensure that the right amount of water flows to each pipe, you need to calculate how much is needed at each point. In a circuit, similar calculations ensure each transistor receives the correct amount of current, allowing the circuit to function properly.
Understanding DC Current Calculation
Chapter 2 of 5
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Chapter Content
Again the value of this resistor it is 11.4 k, so that makes this current = 1 mA. So, the current flow here it is also 1 mA...
Detailed Explanation
This section provides specific values that illustrate how to calculate the DC currents through different components in the current mirror circuit. Initially, a resistor is given a value of 11.4 kOhm, which leads to a current of 1 mA. By applying the principles of circuit analysis, we see how this current splits into half (0.5 mA each) across two paths involving more transistors.
Examples & Analogies
Consider two garden hoses splitting water from a single source. If the total flow is 1 mL of water per second and the hoses are identical, each one will carry 0.5 mL per second. This is similar to how the current splits between the transistors in the circuit. Understanding this split helps to ensure that each component operates correctly.
Voltage Setting and Output Conditions
Chapter 3 of 5
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Chapter Content
So, since this is transistor-7 it is diode connected. So, the voltage here it is decided by the 12 V ‒ 0.6 V so the voltage here it is 11.4 V...
Detailed Explanation
The voltage across a diode-connected transistor is crucial for determining the output states within the circuit. By referencing a 12 V supply and the diode forward voltage drop of 0.6 V, a constant voltage of 11.4 V is established at this point. This voltage contributes to the stability required for the current mirror to function effectively.
Examples & Analogies
Think of this scenario like a water tower maintaining pressure by a drop in its spout. The pressure at the bottom (11.4 V) remains consistent regardless of the total amount of water (12 V) in the tank due to the outlet's resistance. This ensures every connected pipe receives a steady flow, just as transistors do in the current mirror design.
Differential and Common Mode Gain Calculations
Chapter 4 of 5
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Chapter Content
So, yeah so now our next calculation it is we need to find differential mode gain and common mode gain...
Detailed Explanation
This part explains how to calculate the differential mode gain and common mode gain, which are key performance metrics for amplifiers. The differential mode gain measures how well the circuit amplifies the difference in input signals, while the common mode gain focuses on signals that are common to both inputs. Understanding these concepts is vital for assessing amplifier performance in practical scenarios.
Examples & Analogies
Imagine you are measuring the height of two plants. If you want to know which plant is taller (differential gain), you compare their heights. But if you want to see if both plants reach a certain mark on a wall (common mode gain), you look at the height relative to that mark. Similarly, these gains in circuits help compare how well signals are amplified.
Conclusion of Numerical Examples
Chapter 5 of 5
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Chapter Content
Now, coming to the conclusion of this entire session it is what we have talked about basically we started with numerical examples of simple current mirror...
Detailed Explanation
In concluding the numerical examples section, the main takeaway is a summary of various current mirror implementations, calculations of quiescent currents, and an analysis of both differential and common mode gains. Competence in these calculations signifies a solid understanding of how current mirrors function within electronic circuits.
Examples & Analogies
Think of wrapping up a project presentation where you summarize your findings on how best to build a model. You outline what you did step by step, similar to retelling the calculations of the current mirrors in your circuit design, ensuring that your audience understands the importance of each part of your process.
Key Concepts
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Current Mirror: A circuit used to replicate a current across different transistors.
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Differential Gain: The amplification of the difference between two input voltages.
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Common Mode Gain: The response of the amplifier to common signals present at both inputs.
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Transconductance (gm): A measure of the performance of a transistor, indicating how well it converts input voltage to output current.
Examples & Applications
In a current mirror circuit, if a reference current of 1 mA is established, each mirrored branch should ideally also carry 1 mA.
When analyzing a differential amplifier, if a 12 V supply is used with a load resistor yielding 1 mA, the drop across the resistor confirms the expected outputs.
Memory Aids
Interactive tools to help you remember key concepts
Rhymes
A current mirror shows the flow, keeps it steady, helps us know.
Stories
Imagine a gardener (the current mirror) that waters two identical plants (transistors) equally from one hose.
Memory Tools
CAM - Copying Active Mirroring to remember how current mirrors operate.
Acronyms
DGAIN - Differential Gain Are Important Numbers, to remember the relevance of differential gains.
Flash Cards
Glossary
- Current Mirror
A circuit configuration used to provide a constant current to a load while replicating the current from a reference branch.
- Differential Amplifier
An amplifier that amplifies the difference between two input signals while rejecting any signals that are common to both inputs.
- Common Mode Gain
The gain of an amplifier when the same input signal is applied to both inputs.
- Differential Mode Gain
The gain of an amplifier when the difference between the input signals is amplified.
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