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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)."
Right!... 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.