Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

What's the purpose the bias on the left two branches in this CMFB circuit

Status
Not open for further replies.

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.
Untitled.png
 

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.
 

Status
Not open for further replies.

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top