87.3. MOSFET Version of Current Mirror

So in the next example, we are starting with MOSFET version and then after that we will go to the BJT version. And so, here transistor-3 it has been added on the main current mirror. So, the main current mirror it is constructed by M and M and whatever the 1 2 reference current it is coming, it is mirroring here.

In this part, the conversation introduces the MOSFET version of the current mirror circuit. The current mirror is constructed using transistors, and in this specific case, two transistors (M1 and M2) are used to mirror a reference current. The addition of a third transistor (M3) helps in improving the operation of the circuit by ensuring the correct functioning of the current mirror.

Imagine a group of friends where one friend (the reference current) always sets the pace while others (the mirroring transistors) try to keep up. The third friend (M3) acts as a motivator, ensuring that everyone stays in sync, even if the situation changes.

And then we do have this transistor M, to have meaningful operation of this circuit, we require its gate voltage V, denoted here. V it should be sufficiently high. So, that transistor-2 and maybe 3 also are in saturation region of operation.

For a current mirror to function correctly, transistors need to operate in the saturation region. This means the gate voltage (V) for transistors M2 and M3 must be set high enough so they can remain in this operating region. In saturation, the drain current remains constant, which is crucial for maintaining a stable mirrored output current.

Think of a bicycle race where some riders need specific conditions to maintain their speed. If they don't have enough push (high enough gate voltage), they will fall behind. Similarly, the MOSFETs need sufficient voltage to 'keep pace' and work effectively.

Now coming to the different sizes of different transistors given here, we do have for transistor-1, we do have the it is 1 mA/V2. On the other hand for transistor-2 and transistor-3, we do have = 4 mA/V2...We need to find the value of this V and the current flowing here.

The parameters for each transistor are defined using their transconductance factors, which relate gate voltage changes to output current changes. With these values, calculations are made to determine the gate-source voltage and the output currents through the transistors, ensuring that they will function as intended under specified conditions.

Consider these transistors as different engines in a car. Each engine has specific power ratings (transconductance) that dictate how much acceleration (current) you will get from a given throttle input (gate voltage). By understanding each engine's capabilities, you can determine how fast you will travel.

So, V = V ‒ I × R. And on the other hand, we know that expression of I = of transistor-1 and then ( ) .

Here, voltage drops across the bias resistor are taken into account when calculating the gate voltage for the transistors in the current mirror. The relationship expressed in Ohm's Law (V = I × R) helps in understanding how changes in current will affect the voltage at various points in the circuit.

Imagine filling a water tank (voltage) through a pipe (resistor). The amount of water flowing in (current) affects how quickly the tank fills up. If the pipe narrows (increasing resistance), less water gets through, necessitating adjustments in the faucet (gate voltage) to maintain the flow.

So, we need to find what will be the small signal output resistance. And R is g m3 r o3 + r o2 + r o2.

The small signal output resistance of the current mirror is calculated using the transconductance of the active transistor along with the output resistances of each of the other transistors in the circuit. This resistance indicates how well the current mirror can maintain output current despite voltage changes across it.

Imagine trying to maintain a constant speed while driving uphill (output resistance). If the car's engine (transconductance) is powerful and has reserves (output resistances of transistors), it can keep that speed regardless of the incline (voltage changes).

Now we can try to find the value of this I in same way and here now the ratio aspect ratio of transistor-3 channel and transistor-1 channel it is 2. So, the I = (I 1 mA) × (2/1).

Adjusting the ratio of aspect ratios between the channels of transistors allows for control over the output current. By understanding how these ratios influence the current (I), designers can effectively tune the current mirror for specific applications.

It's like adjusting the gears of a bicycle. If you change the gear ratio (aspect ratio), you can make pedaling easier or harder, impacting your speed and efficiency on different terrains. Similarly, adjusting transistor dimensions alters the current flow.

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