# Replacing Ideal current source in two-stage opamp

I am now investigating on replacing the ideal current source with the circuit similar to the one attached below.

Why would the mosfet M16 be connected to drain of M12 ?

• Presumably the picture came from somewhere so what does the source of the information have to say about M16? – Andy aka Mar 21 '16 at 8:34
• It is from page 14 of cad.knu.ac.kr/lecture/analvlsi/opa.pdf . But there is no further info. – kevin Mar 21 '16 at 8:39
• Maybe it is because its easier to fabricate another mosfet inside a CMOS chip than do a resistor .M16 is acting like a resistor as part of the internal compensation. – Autistic Mar 21 '16 at 10:09
• I understand it is part of the zero compensation, and it is acting as a voltage-controlled resistor (M16) operating in ohmic zone. What I wish to know more is why it is connected to the drain of M12. – kevin Mar 21 '16 at 14:15

M16 is connected to the M12 to ensure operation in the ohmic region. Recall that ohmic region operation in MOSFET is characterized by the following conditions:

VDS < VGS - VTH (1)

In M12 the Vds = Vgs = Vdd- Vds(M11) In M16 the Vds = Vds(M7)-Vgs(M7)

where Vds(M7) = Vdd - Vds(M6)

Therefore for Vds(M16) = Vdd - Vds(M6) - Vgs (M7)

If transistor sizes are identical, the Vds in (M12) should be larger than the Vds in (M16) and it ensures ohmic (voltage controlled resistor) region operation. Refer back to (1).

A secondary reason it takes M12 is that M(10) to M(15) are the bias circuit components. You don't want to take biases from the opamp itself. That is the purpose of the bias circuit, to provide reference currents and voltages were needed.