98.4. Numerical Example

In this circuit the value of this R it is given here it is 5 kΩ, R it is 940 kΩ and supply voltage it is 10 V. And let you consider V ≈ 0.6 V; early voltage of the device it is let say it is 100 V, β of the transistor current gain it is a 100.

This chunk introduces the key parameters of the circuit being analyzed in this numerical example, including the resistance values and supply voltage. The variable R denotes the feedback resistance (5 kΩ), while R refers to the base resistance (940 kΩ). The supply voltage of the circuit is given as 10 V. The forward bias voltage (V_BE(on)) is approximately 0.6 V and represents the voltage across the base-emitter junction of the transistor during operation. The Early voltage (V_A) is specified as 100 V, and the current gain (β) of the transistor is specified as 100. These parameters are fundamental for analyzing the performance of the amplifier circuit.

Think of the circuit as a small factory operation where R represents workers on the production line (5 kΩ workers) and R represents the total number of workers dedicated to quality control (940 kΩ). The factory runs on 10 V of electrical energy, which energizes the machines (the transistors). The 0.6 V voltage across the workers indicates the minimum energy required to motivate them, while the 100 V is like a maximum threshold that ensures the factory doesn’t overload. Having a current gain of 100 means that for every 1 unit of worker effort, the output (product) is 100 units stronger.

Now, with this information quickly we can say that the DC current here it is 10 µA and the β is a 100. So, we can say that the collector current it is 1 mA.

Using the provided parameters, we calculated the direct current (DC) in the circuit, which is 10 µA. Given the transistor current gain (β) of 100, this means that the collector current (I_C) is 1 mA because I_C = β * I_B, where I_B is the base current. Since the base current is 10 µA, multiplying it by the current gain gives us the collector current immediately. This calculation shows the amplification of the input current by the transistor.

Imagine pouring 10 gallons of paint (10 µA current) into a machine that produces detailed art. If this machine amplifies the effect of each gallon by 100 times (β = 100), you are effectively creating 1,000 gallons of output art (1 mA collector current). Like how the machine enhances the effort of each gallon, the transistor enhances the input current, showing how small inputs can lead to significant outputs.

And so, we can see that r = 2.6 kΩ. So, that is the input resistance of the main circuit the output resistance if I consider load affected. So, that is and in fact, with this value of early voltage we can say that r intrinsic output resistance it is = 100 kΩ.

Next, we calculate the input resistance (r) of the main circuit, which comes out to be 2.6 kΩ. This is a key parameter in amplifier design as it indicates how much resistance the input signal will encounter. The output resistance, taking into account the load effect, is calculated to be 100 kΩ, influenced by the Early voltage. The Early voltage affects the output resistance and helps in improving the linearity of the transistor operation.

In our factory analogy, if the initial effort (input resistance of 2.6 kΩ) shows how quickly the workers can respond to the tasks assigned, the output resistance (100 kΩ) reflects the control measures in place to ensure quality stays high, allowing more robust outcomes from the effort put in initially.