Solution to Mid-Semester Test (Problem Set 3 Question 3)

This exercise is to let you show by yourself how linearity of a balanced transconductor can be greatly improved by applying variable biasing. The transconductor we are dealing with here is a simple differential stage with emitter resistor. The emitter resistor is realized by a pair of MOS transistors operating in triode region.

We will compare two cases: (i) fixed biasing applied to the triode MOS (Q3 and Q4); (2) variable biasing applied to the triode MOS (Q3 and Q4). We then compare the linearity.

 

(a)

Before we begin, we do a quick calculation of the output currents for two values of v1, assuming perfect linearity. The ideal transconductance formula for this circuit is

With the values given, we have

k1 = 96*20 / 2 = 960 μA/V2
k3 = 96*3 / 2 = 144 μA/V2

Using the formula, we get
Gm = 116.2 μA/V

Thus, for v1 = 2.5 mV, we have

io1 = Gm (v1 - v2) = 0.2905 μA. (Remember v2 = 0.)

Similarly, for v1 = 250 mV, we get

io1 = 29.05 μA.

Therefore, if the transconductor is perfectly linear, the output current io1 is 0.2905 μA for v1 = 2.5 mV, and io1 is 29.05 μA for v1 = 250 mV.

(b)

We now consider the case where the bias of Q3 and Q4 is fixed at VG = 0. First, we know that the drain currents of Q1 and Q2 are

For convenience, we let Veff = Vgs - VTH. Thus, since iD1 + iD2 = 200 μA, we have

Veff12 + Veff22 = (5/24) V2 ............ (1)

Since Vgs2 = Vgs3 = Vgs4 and Vds3 = Vds4, the total current through Q3 and Q4 is

which must be equal to I1 - iD2. Hence, we have

960 Veff22 + 576 [ Veff2 Vds3 - Vds32 / 2 ] = 100 ................ (2)

Suppose all threshold voltages are the same. We have

vin = v1 - v2 = Veff1 - Veff2 + Vds3 ............ (3)

Solving equations (1), (2) and (3) numerically, we get

  • For vin = 2.5 mV, we have
    Veff1 = 0.32322 V
    Veff2 = 0.32228 V
    Vds3 = 0.0015648 V
    Thus, iD1 = 100.2898 μA and iD2 = 99.7104 μA. This gives io1 = iD1 - 100 = 0.2898 μA, which is same as in part (a).

  • For vin = 250 mV, we have
    Veff1 = 0.35452 V
    Veff2 = 0.28749 V
    Vds3 = 0.18297 V
    Thus, we get io1 = 20.66 μA, which is 30% lower than in (a). Clearly, we can see that the transconductor is not linear when the triode MOSs Q3 and Q4 are fixed biased.

(c)

Now, we consider a variable bias applied to Q3 and Q4. Our purpose is to show that linearity improves drastically.

Basically, we connect the gates of Q3 and Q4 to the input signals. The current through Q3 and Q4 will become

Hence, equation (2) becomes

960 Veff22 + 288 [ (2Veff2 + vin) Vds3 - Vds 32 ] = 100 ................ (2*)

The rest is exactly the same as before.

  • For v1 = 2.5 mV, we get io1 = 0.29 μA.

  • For v1 = 250 mV, we get
    Veff1 = 0.36620 V
    Veff2 = 0.27245 V
    Vds3 = 0.15625 V
    Hence, we get io1 = 28.74 μA, which is only 1% lower than the linear case found in (a).

So, we see the improvement in linearity of this transconductor.

Michael Tse; 16 October 2003.