7.1
𝑉 𝛼 × 𝑇𝑜𝑡𝑎𝑙 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑖𝑛 𝑐𝑜𝑙𝑙𝑒𝑐𝑡𝑜𝑟𝑠 0.99 × 25 𝑜==
𝑉 𝑇𝑜𝑡𝑎𝑙𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒𝑖𝑛𝑒𝑚𝑖𝑡𝑡𝑒𝑟𝑠 2𝑟 +2×0.25 𝑖𝑒
𝑉 25𝑚𝑉
𝑟= 𝑇= =250Ω
𝑒
→ 𝑉𝑜 = 0.99×25
𝑉𝑖
2×0.25+2×0.25
= 25 𝑉/𝑉
𝐼𝐸 0.1 𝑚𝐴
𝑅 =(𝛽+1)(2𝑟 +2𝑅 )=2×101×(0.25+0.25)=101𝑘Ω 𝑖𝑛 𝑒𝑒
7.2
a) Assuming 𝑣𝑖𝑑 = 0 and the two sides of the differential amplifier are matched. Thus, 𝐼𝐷1 =𝐼𝐷2 =0.5𝑚𝐴
𝐼 =1𝜇 𝐶 (𝑊)𝑉2 → 0.5=1×2.5×𝑉2 → 𝑉 =0.632𝑉 𝐷1,22𝑛𝑜𝑥𝐿𝑂𝑉 2 𝑂𝑉 𝑂𝑉
𝑉 =𝑉 +1𝑚𝐴×𝑅𝑠𝑠=𝑉 +𝑉 +1×𝑅𝑠𝑠=0.7+0.632+1=2.332𝑉 𝐶𝑀 𝐺𝑆 𝑡 𝑂𝑉
b)𝑔𝑚 = 2𝐼𝐷 =2×0.5 =1.58𝑚𝐴/𝑉 𝑉𝑂𝑉 0.632
𝐴𝑑 =𝑔𝑚𝑅𝐷 → 8=1.38×𝑅𝐷 → 𝑅𝐷 =5.06𝑘Ω
c)𝑉 =𝑉 =𝑉 −𝐼 𝑅 =5−0.5×5.06=2.47𝑉 𝐷1 𝐷2 𝐷𝐷 𝐷𝐷
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