6.1
Solution: For VG1= VG2= 0 V, ID1= ID2 = 0.4/2 = 0.2 mA
To obtain
VD1 = VD2 = 0.1 V VDD-ID1,2RD = 0.1 0.9-0.2RD=0.1 →RD=4KΩ
For Q1 and Q2: ID1,2 = 1 μ C (W) V2 2 a ox L 1,2 OV
→ 0.2 = 1 × 0.4(W)
2 L 1,2
× 0.152 → (W) = 44.4 L 1,2
For Q3:
0.4=1×0.4×(W)3×0.152 → (W)3 =88.8
Since Q3 and Q4 form a current mirror with ID3=4ID4:
(W) = 1(W) = 22.2 L4 4L3
VGS4 =VGS3 =Vtn +VOV =0.4+0.15=0.55V
R = 0.9−(−0.9)−0.55 = 12.5kΩ. 0.1
2LL
The lower limit on VCM is determined by the need to keep Q3 operating in saturation. For his to happen, the minimum value of VDS3 is VOV= 0.15 V. Thus:
VICMmin = -VSS + VOV3 + VGS1,2 = -0.9 + 0.15 + 0.4 + 0.15 = -0.2 v
The upper limit on VCM is determined by the need to keep Q1 and Q2 in saturation, thus
VICMmax=Vt+VDD -𝐼𝑅𝐷 =VD1,2+Vtn=0.1+0.4=0.5v 2
Thus,
-0.2v ≤ VICM ≤ +0.5 𝑣
6.2 Solution:
(a) The figure shows the differential half-circuit. Recalling that the incremental (small-signal) resistanceofadiode-connectedtransistorisgivenby(1 ||𝑟),theequivalentloadresistanceofQ1
𝑔𝑚
will be: RD= 1 ||𝑟 and the differential gain of the amplifier will be Ad =𝑉𝑜𝑑 = gm1 𝑔𝑚3 𝑜3 𝑉𝑖𝑑
[ 1 ||𝑟𝑜3||𝑟𝑜1] 𝑔𝑚3
Since both sides of the amplifier are matched, this expression can be written in a more general way as
Ad = gm1,2[ 1 ||𝑟 ||𝑟 ]
(b) Neglecting ro1,2 , 𝑟 (much larger that 1/gm3,4), 𝑜3,4
Ad≅ 𝑔𝑚1,2 = √2𝜇𝑛𝐶𝑜𝑥(𝑊/𝐿)1,2(𝐼/2) = √𝜇𝑛(𝑊/𝐿)1,2 𝑔𝑚3,4 √2𝜇𝑝𝐶𝑜𝑥(𝑊/𝐿)3,4(𝐼/2) 𝜇𝑝(𝑊/𝐿)3,4
(c)
𝜇𝑛 = 4𝜇𝑝 and all channel lengths are equal, Ad=2√ 1,2 ; 𝐴𝑑 = 10 → 10 = 2√ 1,2 → 1,2 = 25
𝑜
𝑔𝑚3,4
𝑜3,4 𝑜1,2
𝑊𝑊𝑊
6.3
Solution:
The value of R is found as follows: R=𝑉𝐺6−𝑉𝐺7 = 0.8−(−0.8) = 8 𝑘Ω
𝐼𝑅𝐸𝐹 0.2
Since I = IREF, Q3 and Q6 are matched and are operating at |𝑉 𝑂𝑉
| = 1.5 − 0.8 − 0.5 = 0.2 𝑉
𝑊𝑊𝑊 3,4 3,4 3,4
Thus,
0.2=1×0.1×(𝑊)6,3×0.22 →(𝑊)3 =(𝑊)6 =100
Each of Q4 and Q5 is conducting a ds current of (I/2) while Q7 is conducting a dc current IREF = I. Thus, Q4 and Q5 are matched and their W/L ratios are equal while Q7 has twice the (W/L) ration of Q4 and Q5. Thus,
𝐼 = 1 𝜇 𝐶 (𝑊) 𝑉2 ; where: 𝑉 = −0.8 − (−1.5) − 0.5 = 0.2 𝑉 2 2 𝑛 𝑜𝑥 𝐿 4,5 𝑂𝑉4,5 𝑂𝑉4,5
Thus,
0.1=1×0.25×(𝑊) ×0.04 →0.1=1×0.25×(𝑊) ×0.04 → (𝑊) =(𝑊) =20
2𝐿𝐿𝐿
2 𝐿 4,5
And (𝑊) = 40 𝐿7
2 𝐿 4,5
𝐿 4
𝐿 5
𝑟 =𝑟 = 𝐴𝑃 =
=100𝑘Ω
𝑜4 𝑜5
2
|𝑉 | 10
𝐼
|𝑉 | 10
0.1
𝑟 =𝑟 = 𝐴𝑃 =
=100𝑘Ω
𝐴 =𝑔 (𝑟 ||𝑟 ) →50=𝑔 (100||100)→𝑔 =1𝑚𝐴/𝑉
𝑜1 𝑜2
2
But
𝐼
𝑑 𝑚1,2 𝑜1,2 𝑜4,5
0.1
2(𝐼) 0.2
𝑚1,2
𝑚1,2
𝑔 = 2 →1=
𝑚1,2 |𝑉 | |𝑉 | 𝑂𝑉1,2
|=0.2𝑉
→ |𝑉
𝑂𝑉1,2 𝑂𝑉1,2
The 𝑊 ratio for Q1 and Q2 can now be determined from: 𝐿
0.1=1×0.1×(𝑊) ×0.22 → (𝑊) =(𝑊) =50 2 𝐿1,2 𝐿1𝐿2
A summary of the results is provided in the table below: