88.3. Next Numerical Example

As I said that the problem here is very similar, except that transistor-3 and transistor-4 they do have β = 250. Now, here we can try to find what will be the current I and I at V = 11.4.

In this portion, we are introducing a new example that builds upon the previous discussions. The focus is on two transistors (transistor-3 and transistor-4) that have a current gain (β) of 250. The goal of this example is to calculate the currents (I) under the condition that the voltage V is set at 11.4V. By changing the β value of transistors, we can explore how this impacts the circuit behavior. This serves as a transition to the actual calculation of the currents.

Imagine you are baking cookies, and you usually use a specific recipe with all the ingredients tightly controlled (like keeping β constant). Now, you decide to make a variation in one ingredient (like changing a type of flour) and want to see how it impacts the taste of your cookies. In a similar way, we are adjusting the β of the transistors here to see how it affects the current output.

As I say that is the voltage here it is 11.4 and supposes if we make this voltage also 11.4, then we know this current and this current they will be equal.

This chunk states that if we hold the voltage at 11.4V, we can assume that the currents through transistor-3 and transistor-4 will be equal. This is based on the principle that when two pathways are kept at the same voltage in a circuit, the current flowing through both should be the same under ideal conditions. Here, we're setting a reference voltage of 11.4V and assessing how it affects the operation of the transistors.

Think of two water pipes that are both being filled from the same reservoir. If the pressure (akin to voltage in our case) is the same in both pipes, the amount of water (current) flowing through each should also be the same. If we adjust the pressure to match (11.4V), we predict similar flow rates in both paths.

In fact, that is the base current 20 µA × β that is 100 × ( ). In fact, if you do the calculation here what you will get it is the 2 ×1.114.

This portion explains how to calculate the collector current in transistor-3 by multiplying the base current (20 µA) by the transistor's current gain (β). In this case, β is assumed to be 100 for calculations and when adjusted leads to a result of 2 times the value derived from the initial assumptions. This elucidates how input factors influence the output current through multiplication.

Consider a factory where each worker is multiplied by a certain productivity factor (β). If each worker produces a certain number of items, the total output is calculated by multiplying the number of workers by their individual productivity rates. Similarly, here we’re multiplying the base current (20 µA) by the efficiency factor (β) to find the overall current produced—illustrating how simple inputs can be scaled to show larger outputs.

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 β’s are very high.

The focus here is on the concept of current mirroring in transistors, particularly transistors-3 and -4, under the assumption that base current loss can be ignored because of the high β values. In a current mirror configuration, the output current ideally replicates the input current, with minimum loss. In practical scenarios, recognizing whether to consider losses helps in determining accurate current values.

Imagine a perfect copy machine that duplicates every document exactly each time. However, if the machine has a slight fault, it might miss copying some small details (base current losses). In this case, we assume our machine works perfectly (high β) for simplicity, enabling us to focus on overall performance rather than the tiny imperfections.

So, the voltage here at the output the DC voltage = V = 11.4 ‒ 0.45 = 10.95 V.

In this final segment, the actual output DC voltage is calculated. With the adjustments made to account for excess current demand and other variations, the resulting output voltage is found to be 10.95V. This value is significant because it represents how every incremental change in current flow can lead to variations in output voltage. We conclude that precision in current management substantially impacts overall circuit functionality.

Think of adjusting a garden hose to provide water to a plant. If too much water (current) is demanded by the plant, the pressure from the faucet (supply voltage) will lower, resulting in less water reaching the plant. In our example, as we assess the output voltage (supply), shifting parameters can lead to a noticeable drop (from 11.4V to 10.95V) showing just how sensitive our system can be!

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