why the voltage at the drain of NM0 is 2Vbe? I think the current is decided by I0, Q0 and DZ17 are current mirror.
No and yes: BJT current mirrors need to have the same Vbe potential, i.e. have both their emitters and bases in common, which obviously is not the case in this circuit. It's a
buffered current mirror, however:
Consider the circuit on the top including R0 & R13 as an OpAmp, the node I2-Q0-R0 being the non-inverting input (Vbe of Q0), negative feedback from NM0 drain output via resistive divider R14/RZ18 to the inverting input (R13). Then the gain from the non-inverting input to the output (NM0 drain) is vu = 1+(R14/RZ18) ≈ 2 -- with the given values. Of course this is only an approximation, as the input impedance Rbe(DZ17) is in parallel to RZ18; due to the strong negative feedback via RZ8 this input resistance Rbe(DZ17) >> RZ18, however.
could you say some more about the independent current or provide some reference? does it generated by NM0 and the Vbe?
Right: With a temperature-
independent current I2, a temperature-dependent Vbe(Q0) is generated. Neglecting DZ17's input resistance, the
buffered voltage at DZ17's base is Vbe(DZ17) = (1+(R14/RZ18))*Vbe(Q0)*RZ18/(R14+RZ18) = Vbe(Q0) , so you could consider DZ17 as a "buffered current mirror transistor" of Q0. As DZ17 receives the same (but buffered) temperature-dependent Vbe as that one generated by Q0, it will reproduce the same temperature behavior as I2 -- good thermal coupling between the 2 BJTs implied.
You can find more about BJT current mirrors in any book about (analog) transistor electronics, e.g. in
Hans Camenzind's "Designing Analog Chips", chap. 3.
---------- Post added at 13:40 ---------- Previous post was at 13:29 ----------
R13=8*R R14=2R, R0=4R , i dont know how to determine the collector current of QZ17,because i dont know the drain voltage of the nmos NM0
... and we don't know all the other values (other resistors, transistor areas, currents, VDD, VPOW). May be you can make an estimation with my equation given above?