Current Mirror OTA gain calculation

[Fabien]

Dear all,

I designed a current mirror. Some papers such as http://www.imec.be/esscirc/essderc-e...ons/063_50.pdf said the DC gain is

A0 = gm1 x B x ro3.

I computed it with mine, but, in the simulation, the B factor doesn't seems to have any impact on the DC gain, only on the GBW. Not sure if there is a reason, my inputs transistors are working in subthreshold region.

Any Idea, suggestions?

Thank you...

[erikl]

A0 = gm1 x B x ro3.

... the B factor doesn't seems to have any impact on the DC gain, only on the GBW.

Any Idea, suggestions?

The DC gain ~ B * ro3

Now read the 1^st paragraph of point 2.3. of your a.m. paper!

[Fabien]

OK, thank you for you answer. So the B factor doesn't have any impact as the ro will change with B...
But, if the gain is boosted B times and the output resistor is reduced B times, it means the DC gain is gm1 x ro3, doen't it?
If yes, the gm1 is 12.8µS (qID/nkT) and ro3 = 5.5MOhms... it gives only 70 so 37dB...

[erikl]

But, if the gain is boosted B times and the output resistor is reduced B times, it means the DC gain is gm1 x ro3, doen't it?

Fabien: No! DC gain is gm1 x B x ro3 , as you stated yourself above!

[Fabien]

OK, so I probably made a mistake when I calculate ro3...
For gm1, in subthreshold I've only this equation: gm1 = qId/nkT (from Martin's book)

[erikl]

OK, so I probably made a mistake when I calculate ro3...
For gm1, in subthreshold I've only this equation: gm1 = qId/nkT (from Martin's book)

This is quite ok, I think. The gain enhancement doesn't depend on B , but is based on the (1−k)*I_D1 current shunting, see and study equns. (5) & (6) of item 2.3. of your a.m. paper thoroughly!