Jul 6, 2018 #1 M monglebest Junior Member level 1 Joined Apr 26, 2011 Messages 18 Helped 1 Reputation 2 Reaction score 1 Trophy points 1,283 Activity points 1,482 I see a circuit below and wonder why the left two branches are needed? Why not just allow the 100uA current mirror to source the M1 in diode connected? What's the purpose of M0/M3/M4/M7/M10? Size of M10 is same as M4.
I see a circuit below and wonder why the left two branches are needed? Why not just allow the 100uA current mirror to source the M1 in diode connected? What's the purpose of M0/M3/M4/M7/M10? Size of M10 is same as M4.
Jul 6, 2018 #2 d123 Advanced Member level 5 Joined Jun 7, 2015 Messages 2,505 Helped 494 Reputation 992 Reaction score 525 Trophy points 1,393 Location Spain Activity points 27,148 Hi, I'd guess that M0 is part of the M1 current sink, as is M2. M3 and M4 are a replica of M5 and M6 but for presumably/possibly "Vcmo_INN". I don't know the answer but in my opinion M1 is not the direct receiver of the 100uA source, M3 M4 and M0 are, aren't they? It would be nice to know why the circuit is not symmetrical on both halves, I know little about CMFB circuits.
Hi, I'd guess that M0 is part of the M1 current sink, as is M2. M3 and M4 are a replica of M5 and M6 but for presumably/possibly "Vcmo_INN". I don't know the answer but in my opinion M1 is not the direct receiver of the 100uA source, M3 M4 and M0 are, aren't they? It would be nice to know why the circuit is not symmetrical on both halves, I know little about CMFB circuits.