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Julian18

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constant gm biasing question
Jan 07^th, 2008, 5:45am

Hi, there.
I am kinda confused with the analysis of constant gm circuit . I am referring Tom Lee's book "The Design of CMOS Radio-Frequency Integrated Circuits", 1st edition and the derivation of constant gm circuit is as follows(see also the attachment):
Cutting the point at V1 and applying a test incremental voltage at the gate of M8, this voltage gets amplified at the diode connected MOS M9, and as Tom Lee says, "A self-consistent solution is possible only if that voltage equals the original gate drive." Thus:
gm9=gm8/(1+gm8xRs)
But, were the above equation to hold, the positive feedback will force the circuit to oscillate, which is not very good for a biasing circuit, furthermore, as I simulated the circuit, I found the above equation does not hold(I considered all the nonidealities as possible as I can). So if there is something wrong with this derivation or I have done something dumb?

BTW, in the 2ed Version of this book, the derivation has changed, and a more reasonable relation is given, which seems accurate for long channel devices.

Thanks.

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constant_gm.JPG

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joel

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Re: constant gm biasing question
Reply #1 - Jan 7^th, 2008, 2:42pm

I'm no guru and can't offer any deep insight, but I'll pitch in because I've just been looking at that very circuit.
In my case, it's page 127 of "Analog MOS Integrated Circuits for Signal Processing", Gregorian/Temes 1986.
I think I've seen it elsewhere as well. In my book, it's described as a voltage-independent current reference.
Process and resistor variance do show up in the current equation though, so I don't know how great its
utility is. I know that the projects I've worked on where bias current must be controlled to better than 5%
(e.g. for bandwidth control) all end up using an off-chip resistor being driven by a bandgap. Of course that's
a much more expensive system.

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HdrChopper

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Re: constant gm biasing question
Reply #2 - Jan 7^th, 2008, 5:44pm

Hi Julian,

Actually the circuit will not oscillate if the gain of the negative feedback comprised of the degen resistor at the source of M8 has a larger gain than the positive feedback of your pmos self-biasing current mirror. This circuit has actually two self-consistent solutions, one of which is zero current (under this condition the circuit is stable though useless) and the other one the solution provided by the book; which I guess is derived by simply equating the currents in M8 and M9 (assuming the PMOS mirror ratio is 1x).
Something that is certainly needed in order to make this work is M8 to have a multiplicity larger than M9, which should be factored in either in the gm or the beta for M8 when deriving the equation.

Hope this helps
tosei

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Keep it simple

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Julian18

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Re: constant gm biasing question
Reply #3 - Jan 7^th, 2008, 7:57pm

thechopper wrote on Jan 7^th, 2008, 5:44pm:

Hi Julian,

Actually the circuit will not oscillate if the gain of the negative feedback comprised of the degen resistor at the source of M8 has a larger gain than the positive feedback of your pmos self-biasing current mirror. This circuit has actually two self-consistent solutions, one of which is zero current (under this condition the circuit is stable though useless) and the other one the solution provided by the book; which I guess is derived by simply equating the currents in M8 and M9 (assuming the PMOS mirror ratio is 1x).
Something that is certainly needed in order to make this work is M8 to have a multiplicity larger than M9, which should be factored in either in the gm or the beta for M8 when deriving the equation.

Hope this helps
tosei

Hi Tosei:
I know there is a negative feedback due to the source degenerated resistor, but I think when i am calculating the positive feedback, I already take the negative one into account, In fact, if you do the KCLs et KVLs, you inherently include all feedbacks into your calculation. So under some conditions the oscillation criteria is satisfied. but In Tom Lee's book(1st edtion), it says that only on this condition, namely, gm9=gm8/(1+gm8*Rs)), does the gm-constant biasing circuit work (if I am not wrong)

Thanks

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doho

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Re: constant gm biasing question
Reply #4 - Jan 8^th, 2008, 11:38am

Maybe I can be of help regarding the stability question.

This circuit has both a negative and a positive feedback loop. The negative feedback loop (LN) consists of M9-R0 and the positive feedback loop (LP) consists of the loop around M2-M3-M8-M9.

What you need to do to evaluate stability is to first solve the large signal equations to get the possible states of the circuit. Then you need to evaluate the small-signal loop gain in the state you are interested in. This small-signal loop gain will tell you if a state is stable or not.

The small-signal analysis assuming that all output resistances are infinite gives the loop gain around the loop LP to be LP_gain=gm8/(gm9*(1+gm9*Rs)). If LP_gain > 1, the circuit will be unstable. LP_gain is indeed positive (>0) for the non-trivial state of the circuit but remember that a positive feedback loop can still be stable if the loop gain is less than one (which is the case for this circuit).

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Julian18

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Re: constant gm biasing question
Reply #5 - Jan 8^th, 2008, 9:14pm

doho wrote on Jan 8^th, 2008, 11:38am:

Maybe I can be of help regarding the stability question.

This circuit has both a negative and a positive feedback loop. The negative feedback loop (LN) consists of M9-R0 and the positive feedback loop (LP) consists of the loop around M2-M3-M8-M9.

What you need to do to evaluate stability is to first solve the large signal equations to get the possible states of the circuit. Then you need to evaluate the small-signal loop gain in the state you are interested in. This small-signal loop gain will tell you if a state is stable or not.

The small-signal analysis assuming that all output resistances are infinite gives the loop gain around the loop LP to be LP_gain=gm8/(gm9*(1+gm9*Rs)). If LP_gain > 1, the circuit will be unstable. LP_gain is indeed positive (>0) for the non-trivial state of the circuit but remember that a positive feedback loop can still be stable if the loop gain is less than one (which is the case for this circuit).

Hi, thanks
but the derivation in Tom Lee's book shows that only when the positive feedback loop gain is equal to 1, does the constant-gm circuit work, meaning it will provide a constant gm biasing.

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