ACWWong
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Hi Aaron,
The current ratio should be close to W/L ratio even in deep submicron, and low overdrive voltage. I am using weak inversion 130nm devices a have no such problem.
I am a bit confused as to your comments of " My LNA RF transistors are 5 times the size of the current mirror transistors but the current is more than 10 times the value. One thing is that i'm using minimum device length since it is at RF, so i expect poor matching...but not that poor. This primitive diagram might help (see below)...its obviously not and exact design but basically while W/L of B is 10 times W/L of A, the current through A is more than 20 times that of B."
If W/L(A)=1, and W/L(B)=10, then the current ratio should be close to 10. The exact match will depend on vds of each device, but based upon your diagram, it looks fine assuming the top device B (cascode of the LNA) is not in triode due to IR drop across the load, and you don't degenerate the bottom tier (you can degenrate of course, only if the R in bottom A source is 1/10 that of B source). An error of 10% is quite feasible with not exact vds's, but 100% means there is something wrong.
As i initially answered your previous post, i recall you said you had a differential LNA, so you bias from a common reference with two bias resistors, one for each side of the LNA. Your (factor of 2 ?) error is not due to the fact you have two differential arms ?
ANyway, as an aside, it is not necessary for the top tier A device (cascode bias), the gate of top device B can go to Vdd, the current matching of the bottom tier should suffice. Of course you can add IR drop/diode in the refernce side to keep Vds of bottom tier A & B the same.
Hope this helps.
aw
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