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Gm=1/R circuit (Read 13972 times)
sharezhao
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Re: Gm=1/R circuit
Reply #15 - Jul 03rd, 2014, 2:43am
 
Loop gain smaller than 1 to ensure the stability of loop.
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raja.cedt
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Re: Gm=1/R circuit
Reply #16 - Jul 3rd, 2014, 5:21am
 
@aaron,

I didn't
gm ∝ (1+ΔT/1000)

it doesn't imply that there is any feedback...
understand this. What I am saying is if temp changes circuit will adjust current to compensate mobility degradation, assume for time being resister is constant.

I strongly believe current mirror also works because of proper -ve feedback at diode connection. Drain potential will be adjusted according to the reference current then matching concept comes. Any how I should thank for your many replies, I guess I should  stop now.

Thanks,
Raj.
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aaron_do
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Re: Gm=1/R circuit
Reply #17 - Jul 3rd, 2014, 6:13pm
 
Quote:
I didn't
gm ∝ (1+ΔT/1000)

it doesn't imply that there is any feedback... understand this.


I was just saying that the fact that a circuit is insensitive to a parameter (like in the equation) doesn't mean there must be -ve feedback.

I kind of see your point, but I feel that you are forcing the behavior of the circuit to fit a feedback model. Although mathematically it seems to work, it may not describe the real physical behavior of the circuit ...i.e. is there any signal that is actually being fed back from the output to the input?.

Anyway yeah not much point continuing since we're just repeating the same points...interesting topic though.

Aaron
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Lex
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Re: Gm=1/R circuit
Reply #18 - Jul 4th, 2014, 1:54am
 
Based on quadratic law:
I=4k(Vgs2-Vt)2 -> Vgs2=Vt+sqr(I/(4k))
I=k(Vgs1-Vt)2 -> Vgs1=Vt+sqr(I/(k))

Filling out
I*R=Vgs1-Vgs2
I*R=1/2*sqr(I/k)
-> I=(4*k*R2)-1

As for the loopgain, I guess you have to start with the partial derivatives of these equations, since they define the circuits operation. Because of the square root it might be a little tedious to do.
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raja.cedt
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Re: Gm=1/R circuit
Reply #19 - Jul 5th, 2014, 2:35am
 
Lex,
I have no problem with loopgain, I haven't done partial diff but still Gm&Gds were defined based on that manner. My question is "with out loop how come the above circuit producing stable Gm across PVT". I am sure this circuit might have invented by a clear idea rather than random placement. So just trying to understand this.

Thanks,
Raj.
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RobG
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Re: Gm=1/R circuit
Reply #20 - Jul 5th, 2014, 9:29pm
 
Hi Folks,
I'm not sure how to answer the loop gain question.  If you bias it so that R=1/gm then the output current will be independent of the input current to first order. This is because the Io/Iin transfer function is parabolic as mentioned earlier (see attachment). You don't have to bias it there - but you can. You may be thinking about it too hard. Wink

Anyway, nobody has mentioned it so I'll ramble a little bit. The bottom mirror is an old design called a "Peaking Mirror." I've seen a few articles on them but probably 70s vintage when two transistor circuits were still publishable. It is easier to analyze them using bipolars - but the behavior is the same with MOS.

I've sketched some principles... They may be helpful. Or incredibly boring.  If you were to bias the bootstrapped current source at the peak where gm=1/R it would be insensitive to PMOS mismatch - which actually isn't all that important since we know how to match transistors these days. I could see it having some startup issues if you injected it with high current (on the right side of the peak).

The more useful application is to reduce sensitivity of a current source to Vdd - see the sketch. You can cascade them and get even more power supply rejection... but it is usually just easier/more power efficient to build the old dVgs bootstrapped bias (aka beta multiplier). I never liked the fact that at high currents the output of the peaking mirror goes to zero so I now avoid it in bootstrapped sources, but I have used it to create startup currents with less sensitivity to Vdd.

I suppose you could use the mirror to clean up a noise input current - none of the small signal noise on the input current would pass through.
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« Last Edit: Jul 6th, 2014, 7:36am by RobG »  

Peaking_Mirror_2.jpg
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RobG
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Re: Gm=1/R circuit
Reply #21 - Jul 6th, 2014, 8:28am
 
I forgot to note that UT=kT/q.

Thinking about this a bit more... Consider a 1:1 PMOS mirror and the bootstrapped circuit of the original post. The bias current is
IOUT = (UT/R)ln(m).  

You are biased at the peak if
IOUT(peak) =(UT/R)(m/e).  

This is only possible if m=e. At least with bipolars it is impossible to bias on the right side of the peak. That is,
(UT/R)ln(m) ≤ (UT/R)(m/e).  

I expect IOUT is also to the left of the peak with MOSFETs.

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Re: Gm=1/R circuit
Reply #22 - Jul 8th, 2014, 11:39pm
 
raja.cedt wrote on Jul 5th, 2014, 2:35am:
Lex,
My question is "with out loop how come the above circuit producing stable Gm across PVT".

Thanks,
Raj.


Stability wrt PVT does not imply a negative feedback with a high loop gain. Another way to achieve it is open loop cancellation using matched nonlinearities. e.g. a CS amplifier with a diode connected load of the same type of transistor as the amplifier has a gain independent of PVT. OR a voltage divider with two identical nonlinear devices(e.g. diodes or diode connected transistors).
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Lex
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Re: Gm=1/R circuit
Reply #23 - Jul 9th, 2014, 5:13am
 
raja.cedt wrote on Jul 5th, 2014, 2:35am:
... My question is "with out loop how come the above circuit producing stable Gm across PVT"...


Going from I=(4*k*R2)-1, it is clear that there is no dependency on supply voltage. So it is a nice small circuit that will work fine over the 'V' in PVT.
For the rest, the current is based on k and R. So for NMOS mobility and resistor corners you'll probably have to account for, and their variation over temperature and process.

And there is definitely a loop, similar to a translinear loop. However this is with MOSFETs so its not exactly translinear because of the quadratic instead of exponential equations.
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