What is MOSFET Current Mirror – Complete Calculation

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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 Q₂, Q₃, and Q₄ are used together as a current source. Keep in mind that Q₄ 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 Q₂, Q₃, and Q₄ 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 Q₁, together with its drain current i_d1(t). Also true for transistor Q₂, 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 Q₂ 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 Q₂ 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 Q₂.

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:

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