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https://designers-guide.org/forum/YaBB.pl Design >> Analog Design >> any clue on this circuit https://designers-guide.org/forum/YaBB.pl?num=1288016670 Message started by ontheverge on Oct 25th, 2010, 7:24am |
Title: any clue on this circuit Post by ontheverge on Oct 25th, 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 |
Title: Re: op-amp circuit Post by raja.cedt on Oct 25th, 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. |
Title: Re: op-amp circuit Post by aaron_do on Oct 25th, 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 :D . But I suggest you try simulating the circuit to see. I'm not sure, but it may need some form of CMFB. cheers, Aaron |
Title: Re: any clue on this circuit Post by thechopper on Oct 25th, 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 |
Title: Re: any clue on this circuit Post by subgold on Oct 26th, 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. |
Title: Re: any clue on this circuit Post by thechopper on Oct 26th, 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 |
Title: Re: any clue on this circuit Post by Magnus Wiklund on Oct 28th, 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 |
Title: Re: any clue on this circuit Post by Magnus Wiklund on Nov 2nd, 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 |
Title: Re: any clue on this circuit Post by haykp on Nov 2nd, 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. |
Title: Re: any clue on this circuit Post by vp1953 on Nov 2nd, 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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