86.6. Continuation of BJT Current Mirror Example

Now, let us consider the next part of the same question in the next slide. So, that will be continuation of this problem. But probably, we can consider some finite value of λ's.

In this segment, we are continuing the previous calculations while introducing a finite value for the channel length modulation parameter, λ. This parameter helps to understand the variations in the output current from the current mirror when transistors operate under non-ideal conditions.

Consider λ as a filter that allows only specific currents to pass through. By adjusting this filter, we can see how the output current changes under different conditions, similar to how changing a water filter can affect water flow and quality.

Now for this case, V_DS2 = 2.5, both this V_DS and this V_DS1 are the same. So, and also the lambdas are equal. So, we can say that this part of this equation it is becoming 1 for 2.5. As a result, the corresponding current here it is coming. So this ratio it is 4 and I_REF as I said it is 0.5. So, that gives us 2 mA.

In this chunk, we analyze how the drain-source voltage (V_DS) affects the current output in a current mirror configuration. By maintaining V_DS values as equal, we simplify the calculation of the output current, resulting in a calculated output of 2 mA based on earlier reference values.

Imagine adjusting the pressure in water pipes. If the pressure is equal throughout, the flow (current output) remains stable. Similarly, equal V_DS values help maintain consistent output current in the circuit.

So to keep this transistor in saturation, we know that the drain voltage it should be higher than the gate voltage minus V_th, the gate voltage we know, it is 2.5 V.

This chunk describes finding the minimum drain-source voltage (V_DS) necessary for maintaining a transistor in saturation mode, which is crucial for ensuring accurate and stable current mirroring. The calculations involve understanding the fundamental voltage relationships within the transistor.

Think of this as ensuring a car has enough fuel to keep running smoothly. Just as a car needs a minimum amount of fuel (energy) to operate efficiently, a transistor requires a certain minimum voltage to function effectively.

In the next slide we do have, so, what we have here it is all the other parameters remaining same. In fact, it is continuation of the same example, but we are considering λ = 0.01 V^-1.

After determining the minimum voltage requirements, the subsequent calculations involve the introduction of a finite channel length modulation parameter (λ). This adjustment allows for a more realistic evaluation of the output current based on practical transistor operation.

Imagine tuning a musical instrument. Initially, it sounds perfect, but adjusting its pitch (like changing λ) helps improve its overall sound quality based on real conditions.

So, if I say V_DS2 = 5.5, for this, I_DS2 is equal to 2 mA multiplied by this non-ideality factor.

This chunk discusses calculating the new output current when adjusting the drain-source voltage (V_DS2) to 5.5 V. It illustrates the relationship between V_DS and the calculated output current, demonstrating how variations affect transistor behavior.

Consider filling a container with water. Increasing the height from which you pour (analogous to increasing V_DS) raises the water flow into the container; likewise, adjusting V_DS directly impacts current output.

To calculate the small signal output resistance, we can get the calculate the slope of this line and reciprocal of that is the small signal output resistance.

In this section, we learn how to derive the small signal output resistance from the output current variations and the corresponding drain voltages. It's essential for analyzing the performance characteristics of the current mirror.

Think of it as measuring the stretch of a rubber band: the tighter you pull (changes in current), the more you can determine how stretchable it is (output resistance).

So, now we are going to BJT. So, we have considered this simple current mirror.

This chunk introduces the transition from MOSFET-based current mirrors to those using Bipolar Junction Transistors (BJTs). This step is significant since BJTs have different characteristics and parameters affecting current mirroring.

Imagine switching from using LED lights (MOSFETs) to incandescent bulbs (BJTs). While they both provide light (current), their power supplies, heat, and brightness control are unique, demanding distinct understandings and calculations.

In this case, just for a change, instead of giving a reference current, we are giving a resistor here, supply voltage is given to us 12 V.

This section covers the how-to of establishing a BJT current mirror by using a resistor to set the reference current rather than feeding a direct current value. It discusses the significance of setting up conditions for current flow with careful consideration of parameters like supply voltage.

Imagine trying to balance on a seesaw by applying weight (current) precisely. In this case, controlling current through a resistor functions like adding weight to the seesaw ensures a balanced operation.

Now, if we consider these finite β values and early voltage, the ratio instead of 1∶3, it is becoming different.

Here, we explore how non-ideality factors, including the finite beta (β) and early voltage, affect the current ratio within a BJT current mirror setup. Understanding these factors is crucial for achieving precision in applications.

It's akin to baking a cake with fluctuating oven temperatures. Different settings (non-ideality factors) can lead to varying degrees of success (output results).

So to take care of the base current loss, we can have a Beta-helper circuit here.

The chunk introduces the Beta-helper circuit, which mitigates base current losses in BJTs, enhancing the output current accuracy. This technique is especially useful for improving the performance of current mirrors in real-world applications.

Think of the Beta-helper like an assistant at a job; it helps reduce the workload (current loss), ensuring tasks are completed effectively and efficiently.

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