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MOSFET Tutorials
   Enhancement Type MOSFET Operation, P-Channel, and CMOS
   MOSFET Circuit Symbols, iD-vDS Characteristics
   MOSFET as an Amplifier. Small-Signal Equivalent Circuit Models
   MOSFET Small-Signal Amplifier Examples
   Biasing MOSFET Amplifiers. MOSFET Current Mirrors
   Biasing MOSFET Amplifiers. MOSFET Current Mirrors
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   Common Source Amplifier with Source Degeneration.
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Biasing MOSFET Amplifiers. MOSFET Current Mirrors

Biasing MOSFET Amplifiers. MOSFET Current Mirrors.
There are two different environments in which MOSFET amplifiers are found, (1) discrete circuits and (2) integrated circuits (ICs). The methods of biasing transistor amplifiers are different in these two environments.Why? Primarily because it’s “expensive” t o fabricate resistors (and large capacitors) on ICs. Of course, this is not a problem for discrete component circuits.We will discuss these two environments separately.
Biasing Discrete MOSFET Amplifier Circuits
The methods we can use here are virtually identical to the BJT amplifier circuits we saw in Chapter 5. A few of these biasing topologies are:


Example N30.1. Design the MOSFET amplifier below so that ID =1 mA and allow for a drain voltage swing of ±2 V. The amplifier is to present a 1-M beta input resistance to a capacitively coupled input signal. The transistor has kn W| L = 0.5 mA/V2 and Vt =2 V.

We can see directly from this circuit that at DC, VG = 0. Recall that for operation in the saturation mode VGD= Vt (with VGS >0 ). Now, for 2 ± -V swing in vo and large AC gain, we want RD to be large. Hence, let’s choose VD =0 (since Vt = 2V). Then for this ±2 -V swing in v0 V GD|max =0-2=-2t
VGD|max=0+2 V =Vt
Because of these results, the MOSFET stays in saturation.
Consequently,

For a saturated MOSFET

For R D=1 mA Þ (VGS -2 )2=4
Or
VGS = + 2 + 2 +4 V or 0 V With V G =0 and V GS =4 V then V 0 V

Lastly, for a 1-M. AC input resistance, then referring to the input portion of the small-signal model

we see that
Rin=RG Þ RG=1 M beta
Biasing IC MOSFET Amplifiers. Current Mirrors.
For MOSFET amplifier biasing in ICs, DC current sources are usually used. As discussed in Lecture 17, “golden currents” are produced using sophisticated multi-component circuits. Then current mirroring (aka current steering) circuits are used to replicate this golden current to provide DC biasing currents at different points in the IC. The basic MOSFET current mirror is shown in Fig. 4.33b for NMOS. (This is basically the same circuit we saw with the BJT current mirror in )

Q1 has the drain and gate terminals connected together. This forces Q1 to operate in the saturation mode in this particular circuit if In this mode

With a zero gate current,
I REF=I D1
where we can easily see from the above circuit that

Now, we’ll assume the two MOSFETs in the circuit have the same VGS. Consequently, the drain current in the second transistor is

If these two transistors are perfectly matched but perhaps fabricated with different channel dimensions, then 1 2 n n kn1 '= Kn2 , and Vt1= Vt2 so that we see by comparing (1) and (4) that

In this NMOS current mirror shown above, Q2 acts as a current sink since it pulls current I0 =ID2 from the load, which is the amplifier circuit of in this case.In PMOS this current mirror circuit is constructed as

Here Q2 acts as a current source since it pushes current I0I =ID2 into the load. Example N30.2. Design an NMOS current mirror with V DD= 5 V, VSS = 0, and IREF =100 uA. For the matched transistors L =10 um , W =100 um, Vt = 1V, and 20 n kn ' = uA/V2. Referring to the NMOS current mirror circuit in Fig. 4.33b above, notice that the drain of Q1 is connected to its gate so that VGD1 = 1, which is less than Vt. This means Q1 is operating in the saturation mode (or is possibly cutoff).
Assuming operation in saturation,

For I REF=100 uA Þ 100=10.10(VGS-1)or V GD=+1+1=2V or 0 V
Now , by KVL V DD=IREFR+VGS
With V GS=2V then

Here are a few additional questions based on this design:
• What is the lowest possible value for V0 =V2D and still have a functioning current mirror? As with Q1, the transistor Q2 must also operate in saturation if it’s going to supply a constant current.
Hence
V GD2£ V t Þ VG2-VD2£ V t \ V 0= V D2 ³ V G2- V t or V0 ³ V GS-Vt=2-1=1 V
Therefore,
V0|=1V
• Imagine that VA =107 L. (Notice that VA is proportional to the channel length, which is commonplace.) What is ro? VA =107.10*10-6=100 V

• What is change in the output current IO if VO changes by 3 V?



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