86.4. Continuation of Current Mirror Example

So here we do have the example circuit where M and M are forming current mirror. We do have a reference current here and then we do have the application circuit here. So in this example, the of transistor-1 and transistor-2 along with the K factor, it is given. For transistor-1, we do have 1 mA/V2. For transistor-2, we do have 4 mA/V2. And let me assume that both the transistors are having equal threshold voltage of 1.5 and then we do have the reference current = 0.5 mA, and then supply voltage it is 12 V.

In this initial segment, we're examining a circuit that includes two MOSFET transistors configured as a current mirror. The K factor represents the transconductance of the transistors, with one having a value of 1 mA/V² and the other 4 mA/V². A critical assumption made is that both transistors share the same threshold voltage (Vth) of 1.5 V, while the reference current is set to 0.5 mA and the supply voltage is 12 V. These parameters will influence the circuit’s performance and the current that flows through each transistor.

Think of the current mirror as a water faucet. The reference current is like adjusting how wide you want the faucet to open, and the supply voltage is akin to how much water pressure you have available in the pipes. The two different K factors represent two types of faucets that can dispense water at different rates, affecting how much water flows (or current) when you open them.

Let me start with the calculation of V or for I = 0.5 mA so, that is the I = its corresponding K which is 1 mA/V2 by 2 × ( ). So note that for this calculation, we are ignoring ( ).

Here, we calculate the value for the gate-source voltage (Vgs) for transistor-1, using the reference current of 0.5 mA and the given K factor for that transistor. The formula used involves relationships among the currents and the defined K values. It's important to note that we are initially ignoring the channel-length modulation effect (lambda), which simplifies our calculation. The result from this calculation provides us with the necessary Vgs to further explore how the current behaves in the circuit.

This step is like figuring out how much you need to open the faucet (Vgs) to get the desired amount of water (current). Ignoring the lambda effect can be compared to assuming the water pressure is constant, simplifying your calculations for now.

Now we can find next part it is that we need to find what is the minimum value of this V of transistor-2, particularly for transistor-2 for proper operation of the circuit.

In this section, we delve deeper into the operational requirement for transistor-2, specifying that it must remain in saturation for the current mirror to function properly. This necessitates a minimal voltage from the drain to the source (Vds) that must be greater than the gate voltage minus the threshold voltage. Calculating this value establishes the lower limit for Vds to ensure effective operation, which is set at 1 V in our scenario.

Imagine that for your faucet to properly dispense water, there has to be a certain amount of pressure in the pipes. If the pressure drops too low (in this case, the Vds being lower than the calculated minimum), the faucet won’t work correctly.

So, 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.

This next part focuses on re-evaluating our previous calculations by including a finite value for the channel-length modulation factor (λ). This adjustment takes into account the fact that when Vds changes, it can affect the output current slightly. We explore how different values for Vds lead to variations in the output current, and adjust our equations accordingly for two specific cases.

This is like realizing that the water pressure from the main supply line isn't constant, and calculating how slight changes in pressure affect the flow out of your faucet. By considering channel-length modulation, we're better modeling the real-world behavior of the circuit.

So, small signal output resistance at the output of the current mirror R . So, that is the voltage difference and we know that ΔV it is 3 V and the corresponding variation of this ΔI it is 0.06 mA.

The small signal output resistance is defined as the ratio of the change in voltage (ΔV) to the change in current (ΔI) resulting from small variations around the operating point. By measuring how the output current responds to variations in output voltage, we can calculate a resistance value that indicates how stable the current mirror is under small-signal conditions. In this specific case, we found it to be 50 kΩ.

This output resistance can be likened to how sensitive a faucet is to the pressure in the water pipes. A higher resistance means the water flow changes less with fluctuations in pressure, indicating a more stable faucet.

So now, if we continue this exercise for say, other condition. So, let we see in the next slide the third part of this, no now we are going to BJT.

Having covered the workings of a current mirror using MOSFETs, this transition introduces a current mirror built using BJTs (Bipolar Junction Transistors). We will follow similar concepts but must also consider the unique parameters and characteristics of BJTs, such as their different operational principles and parameters for calculating current flow.

Switching from MOSFETs to BJTs is like moving from one style of kitchen faucet to another; they're both designed to dispense water, but they operate on different mechanics and designs that need careful consideration in calculations.

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