89.2.2. Calculation of DC Currents through Transistors

In the context of this circuit, transistor-7 is expected to mirror its own DC current into transistor-8.

Transistor-7 acts as a current source that mirrors its current to transistor-8. This mirroring ensures that both transistors operate at the same current level, providing balance in the circuit. The goal is to avoid the need for external biasing circuits, as the internal current mirror handles the necessary current distribution.

Think of transistor-7 as a manager who ensures that both workers (transistors) under them are always doing the same amount of work. This helps in creating a balanced and efficient work environment without needing to hire more staff (external circuits) to supervise.

The value of the resistor is 11.4 k, making the current through it equal to 1 mA. Thus, half of this current, which is 0.5 mA, flows through each of transistor-7 and transistor-8.

The resistor value of 11.4 kΩ means that it creates a 1 mA current according to Ohm's law. When we have a balanced circuit with a common mode DC input, this current divides equally between the two paths, resulting in 0.5 mA through each of transistor-7 and transistor-8. This division is crucial for maintaining equal operating conditions for both transistors.

Imagine you have a water pipe (the resistor) that can push 1 liter of water per minute (1 mA). If you have two cups (transistors) placed at the end of this pipe, and you evenly split the flow, each cup will get 0.5 liters of water. This ensures that both cups are filled equally, just like how both transistors receive equal current.

The voltage at the output due to transistor-7 being diode connected is 11.4 V, assuming both Q7 and Q8 are identical.

In a diode-connected transistor configuration, the voltage drop across the junction is typically around 0.6 V. Given a supply voltage of 12 V, the output voltage is calculated by subtracting this drop from the supply voltage, resulting in 11.4 V. Since both transistors are identical, we can expect a similar output voltage at both points in the circuit.

Think of the 12 V supply as a water tank that can fill up a pool. The 0.6 V drop is like a drain that consumes some water. The remaining water level (11.4 V) represents how much water can still be used in the pool, which we want to be the same for both sides to have a balanced pool construction.

We will calculate the differential mode gain and common mode gain using small signal parameters and equivalent resistances.

The differential mode gain is determined by the ratio of the output voltage to the input voltage across the load resistances. It quantifies how effectively the circuit amplifies the difference between two input signals. The common mode gain, on the other hand, is much lower and reflects the circuit's ability to reject signals common to both inputs, ensuring that external noise does not affect the output significantly.

Think of differential gain like a microphone that effectively picks up only the voice of a speaker (the difference), while ignoring background noise (the common mode). If it works well, the speaker's voice is amplified loudly, but any irrelevant sounds are muted, demonstrating good audio quality.

In the small signal model, we consider various resistances, including output resistances of the transistors and how they interact with the signal current.

In this model, the resistances at the output of each transistor influence how the circuit responds to small signal changes. These resistances determine the impedance seen by the signal, which affects how the signal is amplified through the circuit. Calculating these resistances accurately is essential for predicting the gain.

Imagine a concert where the quality of speakers influences how well the music is conveyed. Just like high-resistance speakers might produce less sound volume when the same input signal is given, the output resistances in our circuit shape how signals are amplified, determining the overall sound quality.

The common mode gain is given by the expression A = g / (1 + g(2r)), indicating how the circuit handles common mode signals.

Common mode gain calculations show us how sensitive the circuit is to signals that affect both transistors equally. It’s typically lower than differential gain, which is desired to ensure that disturbances affecting both transistors don’t lead to significant output variations. This analysis is crucial for maintaining signal integrity in circuits used for amplifying small signals.

Think of the common mode gain like a noise-canceling headphone that selectively reduces sounds from both sides of your environment. While the headphones effectively reduce unwanted background noise (common mode signals), they allow your music (differential signals) to play clearly and loudly.

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