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from H. Camenzind: "Designing Analog Chips", chap. 1-34 :
"• Bipolar transistors have lower offset voltages. Generally true, but offset voltage depends on size. Make a CMOS transistor larger than a bipolar one (or use trimming) to achieve equally low offset voltage."
from H. Camenzind: "Designing Analog Chips", chap. 4-3 :
"• for equal sizes, MOS transistor have a larger offset voltage (i.e. mismatch) than bipolar ones (about 2:1, but this depends greatly on the process)."
Thanks to erikl. I know the offset voltage of bipolar is smaller if they are used as input differential pairs. I have thought it was caused by the I/V equation difference. ΔV is related to logΔI for bipolar. So it leads to smaller ΔVbe for bipolar. Is it correct? What's the reason of lower offset voltage for bipolar? That's why I ask about current mirror.
Yes, correct. The actual reason is current mismatch. Hence the logarithmic Vbe-dependency of BJT's generates lower offset voltage than the square-root dependency of MOSFETs working in strong inversion. For MOSFETs working in weak inversion, the same logarithmic (Vgs) dependency is valid as for BJT's, so their offset behavior should be similar in this case.
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