University of California, Berkeley Fall 2010

EE142, Problem Set #6 Solutions Amin Arbabian

Prof. Jan Rabaey

Problem 1

We have

Γ =z − 1

z + 1

We can also show that

z = −Γ + 1

Γ− 1

Let Γ = u+ jv, then

z = −u+ 1 + jv

u− 1 + jv

= − (u+ 1 + jv)(u− 1− jv)

(u− 1 + jv)(u− 1− jv)

= −u2 − 1 + v2 − j2v(u− 1)2 + v2

= r + jx

Therefore, r = 1−u2−v2(u−1)2+v2 . Since r < 0, we have 1−u2−v2

(u−1)2+v2 < 0, i.e.

u2 + v2 > 1

⇒ |Γ| > 1

Hence, r < 0 region falls outside the unit circle on the Smith chart.

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Problem 2

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From Smith Chart

Zin = 3.6− 0.3j

Problem 3

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Problem 4

If the currents through the MOSFETs are Id1 and Id2, then v0 = (Id1 − Id2)R where R = 500Ω.

Also, we dene the current factor β = kWL = 250× 10−6 × 500µ1µ = 0.125A/V

2.

Therefore,

Id1 =β

2(VGS1 − VT )2

⇒ VG1 = VT +

√2

β

√Id1 (1)

Id2 =β

2(VGS2 − VT )2

⇒ VGS2 = VT +

√2

β

√Id2 (2)

Also, we have

vin = VGS1 − VGS2 =

√2

β(√Id1 −

√Id2)

⇒√Id1 −

√Id2 =

√β

2vin (3)

Also,

Id1 + Id2 = 2I0 (4)

where I0 = 1.25mA.Substituting (4) in (3), and squaring we get

2I0 − 2√Id1(2I0 − Id1) =

β

2v2in

4

⇒ Id1(2I0 − Id1) = (I0 −β

4v2in)2

⇒ I2d1 − 2I0Id1 + (I0 −β

4v2in)2 = 0

⇒ Id1 = I0 +

√I20 − (I0 −

β

4v2in)2

⇒ Id1 = I0 +

√β

2I0v2in − (

β

4)2v4in

and similarly

Id2 = I0 −√β

2I0v2in − (

β

4)2v4in

Therefore,

v0 = (Id1 − Id2)R = 2

√β

2I0v2in − (

β

4)2v4inR

⇒ v0 =

√2βI0v2in −

β2

4v4inR

Hence,

v0 =√

2βI0Rvin

[1− β

8I0v2in

] 12

≈ gmRvin[1− β

16I0v2in]

= gmRvin −gmRβ

16I0v3in

a have β = 0.125A/V2and gm =

√2βI0 =

√2.5× 10−3 × 0.125 = 17.68mS.

Therefore,

a1 = gmR = 8.839

a2 = 0

a3 = −gmRβ16I0

= −55.243

which gives

v0 = 8.839vin − 55.243v3in

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Theoretical Calculations

For the untuned amplier,

HD2 =a22a1

Sin = 0

HD3 =

∣∣∣∣ a34a1

∣∣∣∣S2in

For 50mV,

HD3 =55.243

4× 8.839(0.05)2 = 3.906× 10−3

For 200mV,

HD3 = 0.0625

For the tuned amplier,

Q = ω0RC = 10

With R = 500Ω, ω0 = 2π109Hz, C = 2.183 pF and L = 7.958 nH.Since the circuit is fully dierential or in other words a2 = 0, we have

IIP2 =∞

With the two tones at frequency f1 and f2 , the 3rd order products 2f1 − f2 and 2f2 − f1 fall in band. Hence,for the tuned case also, we can treat the non-linearity as if the capacitor and inductor are not present as they tuneout close to 1GHz.

IIP3 =

√4

3

∣∣∣∣a1a3∣∣∣∣ = 461.88mV

P−1dB = IIP3

√0.11 = 153.19mV

Simulation Results

Figure 1 shows the distortion components at the output for A = 50mV and A = 200mV.For A = 50mV,

HD3 = −48.13dB ≡ 3.924× 10−3

For A = 200mV,

HD3 = −20.34dB ≡ 0.096

In both cases the second harmonic content is much lower as the circuit is dierential.For A = 50mV the HD3 matches well with calculations whereas for A = 200mV, the simulated value is much

higher. This is because the theoretical expression assumes a cubic non-linearity, whereas in the circuit we havehigher order non-linearities which come into play as A > P−1dB .

Figure 2 shows the voltage gain at the fundamental frequency as a function of the input amplitude. The 1dBcompression point occurs at 147mV matching well with the calculated value of 153.19mV.

Once the amplier is tuned, the response is shown in Figure 3. We observe that the Q of the network is 10.Since the circuit is dierential, the second order distortion components are so low (for simulation they will be at

the noise oor of computer precision) that we can neglect them and assume IIP2 =∞. To calculate IIP3, we havethe input as two tones and the simulated output third order products for three amplitudes is shown in Figure 4.

The extrapolation for IIP3 calculation is shown in Figure 5. The IIP3 is found to be −6.994 dBV ≡ 447mVmatching well with the theoretical value.

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Figure 1: HD2 and HD3 simulation

Figure 2: P−1dB simulation

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Figure 3: Frequency respones of tuned amplier

Figure 4: Simulated third order intermodulation distortion for A = 25mV,50mV, 75mV

8

−35 −30 −25 −20 −15 −10 −5−70

−60

−50

−40

−30

−20

−10

0

10

20

X: −6.994Y: 10.62

Input voltage (dBV)

Vol

tage

(dB

V)

Fundamental

Third order products

Figure 5: IIP3 calculation

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