The Designer's Guide Community
Forum
Welcome, Guest. Please Login or Register. Please follow the Forum guidelines. Sep 23rd, 2017, 12:33am
  HomeHelpSearchLoginRegisterPM to admin  
 
Pages: 1
Send Topic Print
Deriving Gm of circuit (Read 1388 times)
blue111
Junior Member
**
Offline



Posts: 15

Deriving Gm of circuit
Jan 09th, 2017, 12:15am
 
For the following LC filter implementation,


I understand that this is a gyrator, but how to relate the two accuracy knobs (<7:0> and <2:0>) to the resonance frequency ?

Resonance frequency = 1/[sqrt(Leq*Cs)]
Leq = CL/(Gm1*Gm2)

In other words, how to derive equation for Gm2 in terms of the two accuracy knobs ?
Back to top
 
 
View Profile   IP Logged
subtr
Community Member
***
Offline

Analog Enthusiast

Posts: 48
USA
Re: Deriving Gm of circuit
Reply #1 - Jan 9th, 2017, 2:17am
 
Your amplifier is a source degenerated amplifier. So the actual Gm is gm/(1+gm.Rs). Now the gm is a function of current, W/L of your transistor. I hope you have the handles. If you want to increase Gm, increase current or reduce the Rs. As you see both vary Gm in different fashion and have different effects. Increasing current would look easy, but it will affect your current source's bias margin reducing CMRR. Increasing current will reduce the gain. But varying Rs will not give you a great range, but will not affect your dc operating much as it carries incremental current.
Back to top
 
 

Regards
Subtr
View Profile   IP Logged
deba
Community Member
***
Offline



Posts: 59

Re: Deriving Gm of circuit
Reply #2 - Jan 9th, 2017, 9:53pm
 
The Gm2 for the bottom transconductor is given by gm6/(1+gm6*Rs/2). Assuming that M6 and M5 are same. I have neglected gds and other contributions while deriving the expression. gm6 is a function of the bottom current source and Rs is the tunable resistance.
Back to top
 
 
View Profile   IP Logged
blue111
Junior Member
**
Offline



Posts: 15

Re: Deriving Gm of circuit
Reply #3 - Jan 16th, 2017, 3:43am
 
Hi,

1) I am not sure how you derive Gm2 without considering the impedance looking into the drain of M4 or M3.  http://whites.sdsmt.edu/classes/ee320/notes/320Lecture32.pdf

2) Besides, I was told that the tail current is a coarse accuracy knob, while the degeneration resistance (Rs) is a fine accuracy knob.

Any insights ? Thanks!
Back to top
 
 
View Profile   IP Logged
blue111
Junior Member
**
Offline



Posts: 15

Re: Deriving Gm of circuit
Reply #4 - Jun 24th, 2017, 6:31am
 
@subtr :
Quote:
Increasing current would look easy, but it will affect your current source's bias margin reducing CMRR. Increasing current will reduce the gain. But varying Rs will not give you a great range, but will not affect your dc operating much as it carries incremental current.


Could you elaborate more on your statement above possibly with some other references, diagrams or equations ?
Back to top
 
 
View Profile   IP Logged
blue111
Junior Member
**
Offline



Posts: 15

Re: Deriving Gm of circuit
Reply #5 - Sep 13th, 2017, 12:53am
 
For transconductance value of CMOS inverter, is it normal to be very small value ?

Back to top
 
 
View Profile   IP Logged
sheldon
Community Fellow
*****
Offline



Posts: 722

Re: Deriving Gm of circuit
Reply #6 - Sep 14th, 2017, 7:25pm
 
You should be able to turn the block into an oscillator, reverse the inputs
to generate positive feedback. Then run transient simulation and look at
the frequency of oscillation. You know the frequency and the value of C,
calculate Gm.
Back to top
 
 
View Profile   IP Logged
Horror Vacui
Community Member
***
Offline



Posts: 42

Re: Deriving Gm of circuit
Reply #7 - Sep 21st, 2017, 10:45am
 
gm is a small-signal parameter. It depends on your operation point. I have doubts whether the OP was set correctly.
If you have the DC transfer characteristic of the inverter, than you have the gm as well. gm is its derivative.
Sheldon's suggestion is also good, but oscillation is usually large signal. So you will get some kind of gm average of the "operation points" traveled through during oscillation.
Back to top
 
 
View Profile   IP Logged
sheldon
Community Fellow
*****
Offline



Posts: 722

Re: Deriving Gm of circuit
Reply #8 - Yesterday at 3:33pm
 
In general, if the gm stage is capacitively loaded, then the input voltage
should be Output Voltage Excursion/ open loop gain at frequency. So,
the effectively signal levels for the input devices should be small.

In general, the gm of a gm-C filter stage has some method of fixing the
gm to a constant value across the operating range. If it isn't then there
is distortion, that is, the gm changing as a function of input signal level
causes limits the dynamic range of the stage. The function of the source
degeneration resistors in the original schematic is region where the gm
in constant.
 
Back to top
 
 
View Profile   IP Logged
Pages: 1
Send Topic Print
Trouble viewing this site? Copyright © 2002-2014 Designer's Guide Consulting. 'Designer's Guide' is a registered trademark of Designer's Guide LLC. All rights reserved.

Our colleges are not as safe as they seem. Sexual assault is pervasive and the treatment of the victim by the adminstration is often as damaging as the assault: Campus Survivors, Campus Survivors Forum.

Some of our other sites that you might find useful: QuantiPhy.