1. Instrumentation Amplifier

A three Op-Amps instrumentation Amplifer is shown in Figure 1. ThreeOp-Amps instrumentation amplifiers are popular because they offer highinput resistance, adjustable differential gain, and high common mode re-jection ratio (CMRR).

Figure 1: An Instrumentation Amplifier

To decrease the number of unknowns, R1, R2, R3, R4, R5, and R6 areusually made equal R. The differential gain of the amplifier (Ad =

vovI2vI1 )

can be determined by Eq.1.

Ad = 1 +2R

RG(1)

The common mode gain of the instrumentation amplifier can be deter-mined by driving both vI2 and vI1 with the same source (e.g. vicm) andand calculating vo/vicm.

Ac =vovicm

=R4

R3 +R4

R1 +R2R1

R2R1

(2)

Ac is zero in theory since Ri = R for i = 1, 2...6. But in practice, Ac isnot zero because the resistors can not be perfectly matched.

The common rejection ratio (CMRR) in decibles can be calculated by

CMRR = 20 log10Ad|Ac| (3)

The probability that a resistor is within 1 percent of its nominal value is

P [0.99Rn < Rn 1.01Rn] = (0.01RnRn

)(0.01RnRn

) = 2(0.01RnRn

)1(4)

Rn is the nominal value of a resistor.0.01RnRn

is the standard normal

random variable (i.e. z).

Submission checklist:

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(a) (Extra credit, 1 point) Derive Eq. 1.

(b) (Extra credit, 1 point) Derive Eq. 2.

(c) (1 points) Let R = 10 K. What should be the value of RG in orderto obtain 60 dB of differential gain?

(d) (2 points) Model Ri, i = 1, 2...6 as a Gaussian random variable witha mean of Ri and a standard deviation of Ri . How is Ri relatedto Ri if the probability that Ri is within one percent of Ri is 0.99?

(e) (3 points)Generate a histogram of CMRR of one million instrumen-tation amplifiers in matlab. Please submit the plot along with thecode.

2. Common Rejection Ratio of a Differential Pair

The small signal equivalent circuit of a differential pair is shown in Fig-ure 2. RSS represents the resistance of the current source. If all of thetransistors and resistors are perfectly matched, then the common rejectionratio of a differential pair is . Otherwise, the common rejection ratio isfinite. We will explore the mismatch due to the transconductance of thetransistor in this problem. Let gm1 and gm2 be the transconductance ofQ1 and Q2 respectively.

Figure 2: A Differential Amplifier

The output voltage of vo1 is

vo1 = gm1RD1 + (gm1 + gm2)RSS

vicm (5)

The output voltage of vo2 is

vo2 = gm2RD1 + (gm1 + gm2)RSS

vicm (6)

The differential output voltage vod is then

vod =(gm1 gm2)RD

1 + (gm1 + gm2)RSSvicm (7)

2

The common-mode gain of the differential amplifier is

Ac =(gm1 gm2)RD

1 + (gm1 + gm2)RSS(8)

It can be shown that the differential gain of the diff pair is

Ad = gmRD (9)

The common mode rejection ratio (CMRR) is

CMRR =AdAc

(10)

CMRR can be expressed in dB by 20 log10(CMRR).Assume that RD=5 K and RSS=25 K. Assume that gm1 and gm2 can

be modeled as Gaussian random variable. Let gm (i.e. gm be 1 mA/V with astandard deviation of 0.1 mA/V. Generate a hisotram of one million differentialpairs. What is the average CMRR in dB? (5 points)

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