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ontheverge Junior Member Offline Posts: 31	any clue on this circuit Oct 25^th, 2010, 7:24am Hi, below is part of a ckt, can anybody explain the function of M9(M10), M13 and M14? and the purpose of the whole ckt? thanks, Steve
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raja.cedt Senior Fellow Offline Posts: 1516 Germany	Re: op-amp circuit Reply #1 - Oct 25^th, 2010, 7:47am hi, I am guessing about this ckt. M13 is degeneration resister. M9 and M10 are the common mode feedback for first stage. Thanks.
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aaron_do

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Re: op-amp circuit
Reply #2 - Oct 25^th, 2010, 8:41am

Hi,

are you sure that it is an op-amp? I'm not an expert on op-amps, but it appears to be some kind of fixed gain cell.

The inner core of the amplifier seems to be a super source follower if you take the transfer function from the inputs to the voltage across M13. i.e, the voltage across M13 follows the input very closely. If M13 is considered a linear resistor, then the Gm of the inner core is quite linear and depends on M13.

The outer part of the amplifier seems to simply be copying the action of M9, M10 and M13. So the gain of the cell appears to be W7/W9.

I'm not an expert on op-amps so I might be way off Cheesy

. But I suggest you try simulating the circuit to see. I'm not sure, but it may need some form of CMFB.

cheers,
Aaron

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there is no energy in matter other than that received from the environment - Nikola Tesla

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HdrChopper

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Re: any clue on this circuit
Reply #3 - Oct 25^th, 2010, 7:43pm

I'm with Aaron,

M9/10 are super source followers that lower the output impendace of the input followers.
M13 is a degen resistor across which the input voltage is copied. The current flowing through it will flow also through M9 and M10 and that current is copied to a load resistor M14.

Bottom line: is an open loop voltage amplifier whose gain is Ron(load) / Ron (degen) * current copy factor.

Best
Tosei

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Keep it simple

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subgold

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Re: any clue on this circuit
Reply #4 - Oct 26^th, 2010, 7:10am

i might be wrong, but i dont think it is intended to be a super source follower, although it may look to be so.

normally the feedback loop of a super source follower should be constructed by a common-gate stage, not a common source stage like M9 and M10.

i think M9 and M10 are used to further improve the linearity of the transconductance of the input transistor pair. there must be some papers addressing such linearization techniques, but unfortunately i dont have the paper titles by heart right now.

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HdrChopper

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Re: any clue on this circuit
Reply #5 - Oct 26^th, 2010, 2:55pm

This is still a �super-buffer structure: M9 and M10 will rob the small signal current from the input followers, such current created as a consequence of a small voltage across the input degen resistor.
Such robbed current lowers the output impedance of the input buffers --> thus a super-buffer.
Additionally, since such small signal current is robbed, input buffer drain current does not change as a consequence of a small signal voltage and thus their drain voltage does not change either: input impendace is increased (much more noticeable if PNP used as input buffers instead --> another reason to think the input stage as a super-buffer.
Finally, as a consequence of having a virtually invariant drain current with the applied voltage, input buffers linearity is increased (as you correctly suggested). Higher linearity is another feature of a super-buffer circuit as compared to a simple source follower.

best
tosei

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« Last Edit: Oct 26^th, 2010, 6:33pm by HdrChopper »

Keep it simple

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Magnus Wiklund

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Re: any clue on this circuit
Reply #6 - Oct 28^th, 2010, 5:47pm

The voltage gain is W13/W9 x W7/W14.

M9 and M13 defines the feedback of the first stage and hence
the voltage gain of the first stage is 1/(Gm9/Gd13).
The last stage has a voltage gain of Gm7/Gd14.
Note that both Gd13 and Gd14 are operating at Vds=0
where Gds has almost the same value as Gm when operating in saturation.

Cheers,
Magnus

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Magnus Wiklund New Member Offline Posts: 2	Re: any clue on this circuit Reply #7 - Nov 2^nd, 2010, 12:46am I just wanted to add that the conclusion of Tosei is the same as in my post with current copy factor = Gm7/Gm9 = W7/W9 and Ron(load) / Ron (degen) = W13/W14. Cheers, Magnus
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haykp Community Member Offline Posts: 40	Re: any clue on this circuit Reply #8 - Nov 2^nd, 2010, 1:35am Hi All, This is my first comment... (please consider this as somehow test comment) So you can find more about your circuit in the Jacob Backer book.
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vp1953 Senior Member Offline Posts: 172	Re: any clue on this circuit Reply #9 - Nov 2^nd, 2010, 4:13pm Hi Magnus, I concur with your analysis. I get the same results for voltage gain (after making some reasonable assumptions regarding the input transistor, Rout of M9 etc) as what you posted.
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