ECEN 3714 Lecture 8: Common Source Review of Class ver2
1 May 2017
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Document Summary
For the typical device vt0n equals 1v, kpn = 100ua/v2, cgs = 0. 1pf, va = 40v, kr = 2, and w/l = Cl s gds = id rds = v /id: device equations. From id observe that ( t0n)2(1 +/) 2 ( t0n)2 = = v + = 2id1 +1 = 2 100 0. 5um. For small signal - gm can be written as v, 2 id/ v, and 2 using . id1 = id2 = 100ua = id3 = Ma gate bias mirror mi equal 25x ibias = 25x 100ua = 2. 5ma, as aresult of vgs6= vgs4 =1. 25: finding gate voltages. M6 gate bias mirror m6 equal 16/4x = 8x ibias = 0. 8ma as aresult of vgs1= vgsi =1. 25. For vgma= =5 = v +1 = id1. In order for mi (amplifer bias current) and ma (the amplifer) to both be in saturation.
