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Today, we will discuss the common emitter amplifier and its critical components, particularly focusing on the role of the current mirror. What do you think is the most vital aspect of an amplifier's design?
Is it the gain of the amplifier?
That's a great point! The gain is indeed crucial, but we also need to ensure that the transistors are biased correctly to work efficiently. How do you think we'll ensure that?
By balancing the current between them?
Exactly! We'll be using current mirrors for that purpose. Remember, a good acronym for that is B.I.G. - Biasing, Identical transistors, and Gain.
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Now, let's focus on calculating the DC output voltage. What relationships do we need to account for to derive this? Can anyone start?
I think we need to consider the collector current and the voltage drops at the base-emitter junction!
Yes, precisely! The voltage drop at the base-emitter junction (V_BE) is typically around 0.6V. So if our supply voltage is 12V, what would the DC output voltage be?
That would be 12V minus 0.6V, which gives us 11.4V.
Correct! So we have V_OUT = 11.4V. Remember this formula: V_OUT = V_CC - V_BE.
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Next, let's dive into the concept of early voltage. Why do we need to consider it when determining the output voltage?
Could it change the characteristics of the transistors?
That's right! The early voltage influences how much the collector current may vary with the output voltage. If we ignore early voltage, it can lead to significant discrepancies in our calculations!
So, if early voltage is not high, we could underestimate the output voltage fluctuations?
Exactly! It's crucial we consider it for a more accurate result.
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Let's now consider what happens if there's a mismatch in collector current. How might that affect the output voltage?
If one transistor carries more current than expected, we might get a lower output voltage?
That's correct! For instance, if Q_3 and Q_4 have differing beta values, we might see that effect, resulting in an adjustment needed in our DC output voltage. Letβs go back to our formula.
So a small current change could lead to a significant voltage variation?
Absolutely! A good rule of thumb is to keep track of variations and their influence on our amplification qualities.
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The section provides a detailed approach to calculating the output voltage in a common emitter amplifier configuration, focusing on equalizing current flow through various transistors and considering factors such as base current loss and early voltage. Key calculations related to current and voltage relationships are also illustrated.
In this section, we explore the concept of DC output voltage in a common emitter amplifier configured with a current mirror. The primary goal is to balance the current flow across various transistors involved in the amplification process. The transistors, specifically Q_1 and Q_2, are considered identical with a common beta (Ξ²) value, facilitating easier calculations. The section delves into the steps involved in determining the biasing resistances and establishing the collector currents.
Furthermore, the significance of the early voltage (denoted as V_A) is crucial, especially when assessing output voltage consistency. The calculations derive an output voltage of 11.4V, but this can vary based on non-idealities in the transistor characteristics, such as beta mismatches and early voltage discrepancies.
The section concludes with considerations for future analyses, hinting at the necessity for adjustments in output voltage owing to varying current conditions in subsequent examples.
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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 cc BE voltage here it is. So, this is 12 V. So, the voltage here it is 12 β 0.6 so, that is 11.4 V.
The DC output voltage (V_OUT) in a circuit is often influenced by the voltage drops across various components. In this scenario, the voltage drop across the base-emitter junction of a transistor is considered. The example shows a power supply voltage (V_CC) of 12 V, and after subtracting the base-emitter voltage (0.6 V), we arrive at a voltage of 11.4 V at a particular point in the circuit. This indicates how much voltage is available for the transistor's operation after accounting for this drop.
Imagine a water tank supplying water to a garden. The tank represents the power supply, and the water pipes represent the circuit paths. Just like some water will be lost due to friction in the pipes, some voltage is lost due to the base-emitter drop in the transistor.
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Now, with this 11.4 V here we can say that whatever the current we do have. So, that may be 2 mA which is of course, 1 approximation that we are assuming ( ) = 1 please stop here.
At the voltage level of 11.4 V, we assume that the current flowing through the circuit is approximately 2 mA. This reflects a typical behavior in current mirror circuits, where the current generated is mirrored (or replicated) across different nodes in the circuit. This assumption of equality is crucial for maintaining consistency in circuit behavior.
Think of it like a team relay race where one runner (the first transistor) passes off the baton (current) to the next runner (the second transistor). If the first runner has a consistent speed (current), the second runner will likely have to match that to keep the team competitive.
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So, whatever the current. Now if you see this current is getting mirrored here ignoring the base current loss here, because for transistor-3 and transistor-4, we have considered their Ξ²βs are very high. So, they are identical, this information it is also given.
The discussion emphasizes that due to the high gain (B2) of the transistors, the collector currents of transistors-3 and -4 will remain identical. In a current mirror configuration, each of these transistors replicates the current flowing through the first transistor. This simplifies the analysis since we assume minimal loss in base current due to the high beta.
Imagine two identical twins wearing the same outfit. If one twin gets ready first and has a specific style, the other is likely to replicate that style closely. Similarly, in this circuit, one transistor's output influences the other directly due to their identical configurations.
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So, that gives us the DC output voltage V = 11.4 V right. But of course, if the values of early voltage or in case we cannot ignore say this base current and then of course, the corresponding the current here and here there will be a mismatch and the DC voltage here it will deviate from here.
The calculated DC output voltage is 11.4 V, which presumes ideal conditions. However, real-world elements like Early voltage and base current loss might cause deviations from this calculated value. In scenarios where the gain of the transistors isn't high or there is a mismatch, the actual DC output voltage could change, leading to performance issues in the circuit.
Similar to a budget estimate for a party, where factors like unplanned guests (deviation) can lead to overspending. In electrical circuits, just as unexpected guests can throw off your budget, deviations due to varying transistors can change your expected voltage output.
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So, in the next numerical example it is continuation of this, the same example, but considering finite value of this Ξ² of transistor-3 and 4 will be giving us a situation where we need to consider mismatch of the 2 current and then we can try to find what will be the change of this output voltage right.
In the next step, the importance of considering transistor gain (Ξ²) in practical scenarios is highlighted. If the current between the two transistors does not match perfectly due to various factors, it will result in a change in the expected output voltage. Understanding this helps in designing circuits that are more robust and reliable, accounting for real-world imperfections.
This situation can be compared to a pair of friends trying to lift a heavy object together. If one friend is stronger (higher gain) than the other, they may not lift it evenly, which could cause unnecessary strain or misalignment. In circuits, ensuring that all components are well balanced is essential for proper functioning.
Learn essential terms and foundational ideas that form the basis of the topic.
Key Concepts
Current Mirror: A configuration that helps maintain constant current through a transistor.
Biasing: The process of setting a transistorβs DC operating voltage and current levels.
Voltage Gain: The ratio of output voltage to input voltage in an amplifier.
See how the concepts apply in real-world scenarios to understand their practical implications.
In an ideal case, if a common emitter amplifier has a collector current of 2mA and V_BE of 0.6V, the DC output voltage can be calculated as V_OUT = V_CC - V_BE = 12V - 0.6V = 11.4V.
If mismatches in transistors Q_3 and Q_4 lead to different collector currents, say 2.228mA vs. 2.210mA, the resultant effect on output voltage would tally with early voltage considerations, indicating decreased voltage.
Use mnemonics, acronyms, or visual cues to help remember key information more easily.
In biasing, transistors need to agree, to achieve the gain that we want to see.
Imagine two friends equally sharing their pocket money to ensure they both can buy their desired candy β that's how current mirrors work in circuits.
B.I.G: Biasing Identical transistors for Gain to remember the amplifier configuration.
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Review the Definitions for terms.
Term: DC Output Voltage
Definition:
The steady-state output voltage of an amplifier that delivers direct current.
Term: Early Voltage (V_A)
Definition:
The parameter that quantifies the change in output voltage with respect to the change in current for a transistor, indicating its output resistance.
Term: Common Emitter Amplifier
Definition:
A basic amplifier configuration where the emitter is common to both the input and output.
Term: Collector Current (I_C)
Definition:
The current flowing through the collector terminal of a transistor.
Term: Transistor Beta (Ξ²)
Definition:
The current gain of a transistor, defined as the ratio of collector current to base current.