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Design >> Analog Design >> Gm=1/R circuit
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Message started by raja.cedt on Jun 21^st, 2014, 11:51am

Title: Gm=1/R circuit
Post by raja.cedt on Jun 21^st, 2014, 11:51am

Dear All,
Attached circuit is very well know to every one, but less frequent in usage(haven't seen quite frequently). Looks like it has loop gain of zero, if I assume gm=1/R. can any one coment on this, I mean 0 or less loop gain means more or less open loop. Any impact of this loop gain on normal operation?

Thanks,
Raj.

Title: Re: Gm=1/R circuit
Post by sheldon on Jun 22^nd, 2014, 5:09pm

Raja,

Why would you think that Gm=1/R? Shouldn't the open
loop gain be something like, gm_M4 * (go_M1 || go_M4)?
The output resistance of M1 and M4 is gds of a transistor
in saturation, not Gm. The output resistance of M1 should
be much larger than Rs so it can be ignored for small
signal analysis. Of course, you can't ignore it for large
signal analysis. So to me at first glance it seems like the
loop gain should be similar to the loop gain for a one pole
op-amp without a cascode. Certainly the open loop gain will
be low, but it should not be less than one, unless the power
supply voltage is so low the circuit does not setup properly.

Sheldon

Title: Re: Gm=1/R circuit
Post by ywguo on Jun 23^rd, 2014, 12:16am

Hi Raja,

The load resistance looked out of the drain of M4 is 1/g_m1(r_ds1+r_s)/r_s. So it is approximate 1/g_m1 assuming r_s is much less than r_ds1. So it is easy. The loop gain is around g_m3g_m1/g_m2g_m4.

Best Regards,
Yawei

Title: Re: Gm=1/R circuit
Post by raja.cedt on Jun 25^th, 2014, 1:39am

@ywguo: Here feedback taken from drain of M1 means there is drop trough Rs, I guess you miss this point.
@Shedlon: I have even included gds of M1 still same numerator, except minor difference in denominator.

Thanks,
Raj.

Title: Re: Gm=1/R circuit
Post by nrk1 on Jun 25^th, 2014, 5:54am

Interesting observation-the loop gain is indeed zero. If you evaluate the large signal transfer curve the loop by breaking at the gate of the 4x transistor, it turns out to be an inverted parabola and the operating point(when the loop is connected) is at the top of the parabola.

But this circuit is not much worse than the other circuit where the 1x transistor is diode connected and 4x transistor has R degenerating it. Both have the same operating point (ignoring body effect and assuming square law) and the other circuit has a loop gain of 1/3. The effect of errors will be different. If the 4x transistors threshold voltage changes by dVT, the bias current will change by dVT*gm4 in the circuit posted here and 3/4*dVT*gm4 in the other circuit with 4x transistor degenerated. gm4 is the transconductance of the 4x transistor at the op. point.

Title: Re: Gm=1/R circuit
Post by raja.cedt on Jun 25^th, 2014, 9:08am

Sir,
Thanks for your reply. if posb could you please explain about large signal transfer curve, I didn't understand the parabola concept.. One more thing the other circuit has +ve feedback with loop gain less than 1 so it is stable. My basic question is can +ve fb with loop gain less than 1 has de-sensitive properties like -ve fb??

Thanks,
raj.

Title: Re: Gm=1/R circuit
Post by ywguo on Jun 25^th, 2014, 7:09pm

Hi Raja,

You are right. I miss the point that the gate of M2 is taken from the drain of M1. But I am confused now. What are the sizes of each transistor in your schematic. And why do you think gm = 1/R?

Best Regards,
Yawei

Title: Re: Gm=1/R circuit
Post by aaron_do on Jun 25^th, 2014, 8:24pm

Hi,

Assume the current through the two branches is the same, but the RHS NMOS is 4x larger. Using simple device equations, since Ids ∝ (Vod)², Vod of RHS NMOS should be half of LHS NMOS for the same current. Assume 1xVod refers to RHS NMOS. So LHS NMOS has 2xVod, and voltage drop across Rs is 1xVod. Since gm1=2I/(2xVod) and I = Vod/Rs, gm1 = 1/Rs. gm2 = 2I/Vod = 2/Rs. So loop gain is 2/Rs*Rs = 2.

BTW, how do you get a loop gain of 0? Is that 0dB = 1?

I don't think weak positive or negative feedback has any desensitization property. So I assume your question is, why does the circuit stabilize to a fixed operating point? You are assuming there must be some feedback mechanism?

Personally, I wonder how well this circuit actually works, since I haven't seen it before. If those current mirrors are ideal, then I don't think this circuit has only 1 operating point. Seems it only requires that Vod,m1 = 2Vod,m2. So as long as the transistors follow square law behavior and the voltages never saturate, the circuit looks like it could take any current. The feedback mechanism looks like it could be a large signal one, something like in a VCO, where the saturating voltage swing causes the loop gain to drop. In this case, the drain voltage can't rise indefinitely...

Did I make a mistake somewhere? I assumed very simplistic equations...

regards,
Aaron

Title: Re: Gm=1/R circuit
Post by raja.cedt on Jun 26^th, 2014, 2:27am

@aaron_do,
1. Since gm1=2I/(2xVod) and I = Vod/Rs, gm1 = 1/Rs. gm2 = 2I/Vod = 2/Rs. So loop gain is 2/Rs*Rs = 2.. I didn't understand the bold section, how did you calculate. I guess here your mixing large signal and small signal, I normally don't use Vov in the loop gain however I could be wrong. Many people have been agreed for 0(no 0dB) loop gain.

2.I don't think weak positive or negative feedback has any desensitization property
. I guess week +ve feedback has, because if you check another traditional Gm=1/R circuit which is very robust from PVT point of view (forget about 2nd order effects) with less than 1 +ve fb loopgain.

3.Yes you are correct, this ckt strongly depends on square law, that's many people have tried to bi-pass this dependency and relaying on -ve feedback. check this reference.
http://www.ee.iitm.ac.in/~Shanthi/iscas2004.pdf

Thanks,
Raj.

Title: Re: Gm=1/R circuit
Post by aaron_do on Jun 26^th, 2014, 7:55pm

Hi Raj,

Quote:

1. Since gm1=2I/(2xVod) and I = Vod/Rs, gm1 = 1/Rs. gm2 = 2I/Vod = 2/Rs. So loop gain is 2/Rs*Rs = 2.. I didn't understand the bold section, how did you calculate. I guess here your mixing large signal and small signal, I normally don't use Vov in the loop gain however I could be wrong. Many people have been agreed for 0(no 0dB) loop gain.

So you agree that gm1 = I/V_od2 and gm2 = 2I/V_od2? Then for the loop gain we break the loop at the gate of M2 and go around the loop. It looks like thats what you did, but your impedance looking out from drain of M4 is (1/gm-Rs). I don't see how you arrived at this result. If I just look at M1 and Rs, when I apply a test voltage at gate of M1, the current = gmVtest, so impedance = 1/gm, not (1/gm-Rs), unless you are making some assumption about the output impedance of M1.

One mistake I see that I made though, since I = Vod/Rs, and I = 4K(Vod)², then Vod = 1/4KRs. Therefore I = 1/4KRs². So the current is indeed fixed.

EDIT:
OK I see my mistake, I forgot to take the voltage drop across Rs (same mistake ywguo made). So yes it becomes 2/Rs*(1/gm-Rs) = 0.

Quote:

I guess week +ve feedback has, because if you check another traditional Gm=1/R circuit which is very robust from PVT point of view (forget about 2nd order effects) with less than 1 +ve fb loopgain.

That's a really good question. So for a circuit with weak +ve feedback, what causes it to reach a stable operating point? Well, if you look at a VCO, if the +ve feedback is less than 1, then the output will not build up. So your original statement isn't always true. Perhaps it is a large signal effect (try calculating the large signal loop gain?). Or perhaps it is simply the result of the model parameters. For example suppose I have a series current source feeding a series resistor and diode, and the diode is ON.

---(->)----[diode]----[res]---GND
I

So the voltage will be I.R and the diode will have little effect. Even if the diode changes due to PVT, the voltage will still be I.R. That doesn't mean that there is some feedback mechanism working. Just that the circuit is insensitive to the diode.

I dunno, perhaps I am on the wrong track. what do you think?

regards,
Aaron

EDIT:
So I simulated this using ideal MOS devices and the loop gain is definitely 0...

Title: Re: Gm=1/R circuit
Post by raja.cedt on Jun 27^th, 2014, 2:44am

Hi Aaron,
I am also not an expert but have some idea. I strongly believe the following. "If you can define any parameter in a circuit as a ratio of like elements, definitely there is some -ve feedback, could be strong or week".
1. A simple potential divider, it has some feedback check the attached fig. for example if output voltage increased by small amount immediately feedback will decrease current and regains it's original operating point.
2.Current example, because outcome is (gm)/(1/r).

Coming to your current source example, I think you are slightly wrong in fact it has sensitive because after all it is current source pumping into a resister and diode small signal resistance(Vt/Ic). Most of the people use this insensitive assumption to make their life easy in DC calculation (However I could be wrong.). In-fact you can argue the following to show that there is no feedback, Increasing in the current will results increase in the voltage which doesn't change any parameter.

Thanks,
Raj.

Title: Re: Gm=1/R circuit
Post by aaron_do on Jun 27^th, 2014, 5:55pm

Hi Raj.,

I dunno. I'm sure some people would argue that you're stretching the definition of feedback here. As you say, any time the transfer function looks like x/(y+z) you can divide by y and it will become x/y/(1+z). Then you can model it as a negative feedback system. But that doesn't guarantee that any signal is fed back from the output to the input...

Its interesting to think about anyway.

For the diode and resistor, it depends on the size of the resistor and what you mean by sensitive. Insensitive IMO means it is not much affected, but doesn't mean that it is completely not affected. I'm simply pointing out that there are scenarios where a circuit has a fixed operating point and is insensitive to PVT, but there is no -ve feedback.

regards,
Aaron

Title: Re: Gm=1/R circuit
Post by nrk1 on Jun 30^th, 2014, 4:17pm

The parabola is the large signal transfer curve from the Gate of the larger transistor to the feedback point with the lip broken. Basically the same thing you evaluated, but in large signal. No there is no desensitisation here.

raja.cedt wrote on Jun 25^th, 2014, 9:08am:

Title: Re: Gm=1/R circuit
Post by raja.cedt on Jul 2^nd, 2014, 7:28am

Sir,
Thanks for your reply. I agree here there is desensitisation just because no loop gain. In case of increase in temp or PVT change current automatically change to adjust gm, I feel this is restoring mechanism or desensitisation. Could you please tell me where I am wrong here..

Thanks,
Raj.

Title: Re: Gm=1/R circuit
Post by aaron_do on Jul 2^nd, 2014, 5:45pm

Hi Raj.,

I'm not sure if you understood my point. I don't think its a restoring mechanism. Simply put, gm is insensitive to changes in PVT. For example (just an example), a current mirror. The current gain is insensitive to PVT because the devices are matched. But that doesn't mean that there's any feedback involved. In your case, maybe

gm ∝ (1+ΔT/1000)

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

regards,
Aaron

Title: Re: Gm=1/R circuit
Post by sharezhao on Jul 3^rd, 2014, 2:43am

Loop gain smaller than 1 to ensure the stability of loop.

Title: Re: Gm=1/R circuit
Post by raja.cedt on Jul 3^rd, 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.

Title: Re: Gm=1/R circuit
Post by aaron_do on Jul 3^rd, 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

Title: Re: Gm=1/R circuit
Post by Lex on Jul 4^th, 2014, 1:54am

Based on quadratic law:
I=4k(V_gs2-V_t)² -> V_gs2=V_t+sqr(I/(4k))
I=k(V_gs1-V_t)² -> V_gs1=V_t+sqr(I/(k))

Filling out
I*R=V_gs1-V_gs2
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.

Title: Re: Gm=1/R circuit
Post by raja.cedt on Jul 5^th, 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.

Title: Re: Gm=1/R circuit
Post by RobG on Jul 5^th, 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. ;)

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.

Title: Re: Gm=1/R circuit
Post by RobG on Jul 6^th, 2014, 8:28am

I forgot to note that U_T=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
I_OUT = (U_T/R)ln(m).

You are biased at the peak if
I_OUT(peak) =(U_T/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,
(U_T/R)ln(m) ≤ (U_T/R)(m/e).

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

Title: Re: Gm=1/R circuit
Post by nrk1 on Jul 8^th, 2014, 11:39pm

raja.cedt wrote on Jul 5^th, 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).

Title: Re: Gm=1/R circuit
Post by Lex on Jul 9^th, 2014, 5:13am

raja.cedt wrote on Jul 5^th, 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.