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**Contents**show

MOSFET current mirror is a popular current mirror circuit to be used on a MOSFET amplifier circuit. Basically, a MOSFET current mirror is used to act as a current source so we can use another approach to provide a “current source” to the transistor.

**What is a Current Mirror Circuit**

A current mirror circuit is an analog circuit which is able to sense a reference current (I_{REF}) and produces numbers of current with the same characteristic as that reference current.

The duplicated currents values can be:

- Equal to the reference current (I
_{o}= I_{REF}) - Multiplied from the reference current (I
_{o}= N x I_{REF}) - Divided from the reference current

For an easy illustration, you could see the image below.

**Why do We Need Current Mirror Circuit**

As we already know, an integrated circuit needs a current source for the biasing purpose of its amplifier transistor. Biasing the amplifier will produce a higher voltage gain and stability. Since an integrated circuit will be smaller overtime, an actual current source is not usable. That is why we will use a current mirror circuit instead.

A current source can be generated simply by a PMOS (MOSFET P-type) or MOS cascode configuration. Both can produce higher voltage gain for the amplifier.

The circuit below is the PMOS used to increase the gain.

While the circuit below is the cascode current mirror (also uses PMOS).

Another thing to consider, the cascode circuit is less stable if the biasing voltage or the temperature changed. An integrated circuit will have more than hundreds of amplifiers, biasing all of them with good precision will need a lot of work.

To make things much simpler, a current source is fabricated inside the IC and duplicated by the current mirror to produce multiple current sources for hundreds or thousands of amplifiers. Of course all the currents will be stable since the IC’s current source is stable.

**MOSFET Current Mirror Calculation**

Assume that we have an NMOS amplifier with a current source as drawn below.

We have:

- No resistors or capacitors are present in the circuit.
- This is a common source amplifier.
- The stability of I
_{D}(drain current) can be neglected.

Remember from our discussion about active load MOSFET? Every practical or real current source has their own inner or source resistance (r_{o}). Thus we can redraw the current source above into

In an ideal condition, a current source will have infinite resistance (r_{o} = ∞). Since there will be no ideal condition, a current source will have a very large output resistance (r_{o} = 100KΩ). For a simpler calculation, we will ignore this and write (r_{o} = ∞) but keep in mind that this resistance will make a difference when calculated thoroughly.

Our circuit will be more accurate if redrawn into:

If what we have is a small-signal circuit

The hybrid-pi model will be

Of course we will not finish this discussion since we need to learn about the current mirror. Replacing the current source at the drain with PMOS current mirror will result in:

The transistors Q_{2}, Q_{3}, and Q_{4} are used together as a current source. Keep in mind that Q_{4} is used as a resistor in the current mirror circuit since it acts as an enhancement load.

The amplifier circuit above only consists of NMOS and PMOS transistors and uses this as an integrated circuit.

How do we calculate the output resistance (r_{o})?

First we need to determine the small-signal circuit for this set of amplifiers. Our main focus here is the gate-source voltage (V_{GS}) of transistors Q_{2}, Q_{3}, and Q_{4} in DC values.

Thus the small signal voltages V_{GS} for each transistor is equal to zero.

The small-signal source v_{i}(t) mostly doesn’t produce small-signal voltages and currents throughout the amplifiers.

There will be small voltages v_{gs1}(t) and v_{ds1}(t) for transistor Q_{1}, together with its drain current i_{d1}(t). Also true for transistor Q_{2}, the small-signal voltage v_{ds2}(t) and current i_{d2}(t) will be found.

But take note that the rest of the voltages and currents in that circuit (V_{DS4}, V_{GS2}, I_{D3}), the small-signal elements are zero.

The gate-to-source voltage in Q_{2} has a DC value (no small-signal element), and we will have a small signal drain current i_{d2}(t) using KCL.

This is when we need to determine the MOSFET output resistance r_{o2}. The small-signal drain current for the PMOS transistor is

And v_{gs2} = 0, thus

The hybrid-pi model for the small-signal PMOS is

Since v_{gs2} = 0, the small-signal model will be

We can simplify it into

Simplifying all the calculations above, we can simplify the current mirror into the output resistance of MOSFET Q_{2} as drawn below.

We remember our previous circuit with a current source as shown below.

Comparing those two we will conclude that the output resistance of a current mirror circuit is equal to the output resistance of MOSFET Q_{2}.

We draw again our MOSFETs to act as a current mirror below.

The complex circuit above is equal to the

We draw the hybrid-pi model for the resulting small-signal circuit from this amplifier.

The open circuit voltage gain is

The voltage gain will be much higher if compared to the enhancement load of R_{D}.

Not only that, the output and input resistance of this amplifier is equal to the enhancement load: