8.1
𝑉 25𝑚𝑉
𝑟= 𝑇= =250Ω
𝑎 × 𝑇𝑜𝑡𝑎𝑙 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑖𝑛 𝑐𝑜𝑙𝑙𝑒𝑐𝑡𝑜𝑟𝑠
𝑒
𝐼𝐸 0.1 𝑚𝐴
𝑉 𝑜=
0.99 × 25𝑘Ω 𝑉
8.2
𝑇𝑜𝑡𝑎𝑙 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑖𝑛 𝑒𝑚𝑚𝑖𝑡𝑒𝑟𝑠
0.99 × 25𝑘Ω
= = ≅25
𝑉 𝑖𝑒
𝑅 =(1+𝛽)(2𝑟 +500)=101×(2×250+500)=101𝑘Ω 𝑖𝑛 𝑒
A 2-mV input offset voltage corresponds to a different Δ𝑅𝐶 between the two collector resistances,
2=𝑉 Δ𝑅𝐶 =25×Δ𝑅𝐶 →Δ𝑅 =1.6𝑘Ω 𝑇𝑅𝐶 20 𝐶
Thus a 2-mV offset can be nulled out by adjusting one of the collector resistances by 1.6 𝑘Ω. If the adjustment mechanism raises one 𝑅𝐶 and lowers the other, then each need to be adjusted by
only 1.6 𝑘Ω = 0.8 𝑘Ω 2
If a potentiometer is used, the total resistance of the potentiometer must be as least 1.6 𝑘Ω.
2𝑟 + 500Ω 500 + 500 𝑉
8.3
Utilizing two matched transistors, Q5 and Q6, the value of R can be found from:
𝐼𝐼
𝐺𝑚 =𝑔𝑚1,2 = 2 →5= 2 →𝐼=0.25𝑚𝐴
𝑉𝑉 𝑇𝑇
𝐼=0−(−5)−0.7=0.25𝑚𝐴 → 𝑅=17.2𝑘Ω 𝑅
𝑅 =2𝑟 =2 𝛽 =2×100=40𝑘Ω
𝑖𝑑 𝜋
𝑔𝑚 5
|𝑉 | 100
𝑅 =𝑟 ||𝑟
𝑜 𝑜2 𝑜4
; 𝑟 =𝑟 = 𝐴 =
=800𝑘Ω
𝑜2 𝑜4
𝐼 2
0.125
Thus
1
𝑅𝑜 = 800||800 = 400 𝑘Ω
𝐴𝑑 =𝐺𝑚 ×𝑅𝑜 =5×400=2000𝑉 𝑉
𝑟 |𝑉 | 100 𝐴=−𝑜4 ;𝑟=𝐴= =800𝑘Ω
𝑐𝑚
𝛽3𝑅𝐸𝐸 𝑜4 𝐼 2
𝛽3 = 100
|𝑉 | 100 𝑅 =𝑟 = 𝐴 =
0.125
𝐸𝐸 𝑜5
𝐼 0.25
=400𝑘Ω
CMRR= |𝐴𝑑| = 2000 = 100000 𝑜𝑟 100 𝑑𝐵 |𝐴𝑐𝑚| 0.02
𝐴𝑐𝑚 =− 800
100 × 400 𝑉
=−0.02𝑉
The CMRR can be found as:
2
3