Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronic Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Register Log in

[Cadence ADE XL] Optimize a Common Source Amplifier to 15dB gain via Corner Analysis

iSean

Newbie level 4
Joined
Nov 4, 2019
Messages
7
Helped
0
Reputation
0
Reaction score
0
Trophy points
1
Activity points
66
Hi there I'm a student intern, very new to ADE XL, with basic knowledge of Analog IC in Common Source Amplifier Characteristics.
I am seeking some help in some guide in using the Corner Analysis / Optimization tools in ADE XL.

Firstly, I need to optimize a Common Source Amplifier circuit as shown, with one single bias point, Vindc, maintain > gain of 15 dB @ saturation region [2]
I'm allowed to manipulate the W/L / R_D / Vindc. While keeping the V_DD constant @ 1.2V.

The circuit must work in -40, 27, 70 and 125'C at the corners 'tt', 'ss', 'ff', 'fnsp' and 'snfp'.

The problem arise caused by my corner and temperature settings, the Vout vs Vin generated is offset, and I couldn't determine the best Vindc point.
I've been struggling to manipulate to W/L and R_D / Vindc such that to minimumize this offset to all my circuit to provide a good consistent gain of > 15 dB ?

I've been struggling for 4 days with this question given by my supervisor...

WINWORD_aS7A6tNx9J.png

Corners.jpg
 

BigBoss

Advanced Member level 5
Joined
Nov 17, 2001
Messages
4,745
Helped
1,439
Reputation
2,876
Reaction score
1,301
Trophy points
1,393
Location
Turkey
Activity points
28,663
You cannot find Optimum Vdc(in) in such a way..Here, you're changing Vdc(in) and obviously Vout is changing..
Gain is another metric and specification.Look at Gain equation of this amplifier in a textbook then find another equation that relates this Gain to Vdc(in) then find optimum value ..
But I'm not sure you know what the Gain is..
 
  • Like
Reactions: iSean

    iSean

    points: 2
    Helpful Answer Positive Rating

iSean

Newbie level 4
Joined
Nov 4, 2019
Messages
7
Helped
0
Reputation
0
Reaction score
0
Trophy points
1
Activity points
66
You cannot find Optimum Vdc(in) in such a way..Here, you're changing Vdc(in) and obviously Vout is changing..
Gain is another metric and specification.Look at Gain equation of this amplifier in a textbook then find another equation that relates this Gain to Vdc(in) then find optimum value ..
But I'm not sure you know what the Gain is..
Well, the simplified form is just Av = - g_m * R_D ... ?
I was recommended to read Design of Analog CMOS Integrated Circuits: Behzad Razavi... but I think my background is too weak to understand most of the concepts mentions.

From what I realized from the Result Browser... The g_m is constantly changing as well with change in Vin.
How possible a proper method in obtaining a predictable calculable results...?

When the simulation will vary the values of g_m and 'betaeff' - u_n C_ox?

Any assisted guidance on materials to read is appreciated.
 

frankrose

Advanced Member level 3
Joined
Nov 27, 2013
Messages
772
Helped
216
Reputation
432
Reaction score
210
Trophy points
43
Activity points
5,369
If you need 15dB DC gain above circuit is pretty useless, only at the ouput transition you have gain (derivate of your curves), which is input gate threshold voltage dependent, thus corner dependent. You should use an OPAmp circuit rather, I suppose your supervisor won't be happy, but unfortunately I don't know better way.
If you need 15dB AC gain use a bias circuit (a current source + diode connected MOS transistor enough) and decouple it from your amplifier's gate with a big inductor or resistor. To couple your AC input signal to the amplifier's gate use a simple capacitor series with your AC source. It should handle corner variation and with proper output Rd load resistor you can cover most of the corners with relatively constant gain.
 

iSean

Newbie level 4
Joined
Nov 4, 2019
Messages
7
Helped
0
Reputation
0
Reaction score
0
Trophy points
1
Activity points
66
If you need 15dB DC gain above circuit is pretty useless, only at the ouput transition you have gain (derivate of your curves), which is input gate threshold voltage dependent, thus corner dependent. You should use an OPAmp circuit rather, I suppose your supervisor won't be happy, but unfortunately I don't know better way.

If you need 15dB AC gain use a bias circuit (a current source + diode connected MOS transistor enough) and decouple it from your amplifier's gate with a big inductor or resistor. To couple your AC input signal to the amplifier's gate use a simple capacitor series with your AC source. It should handle corner variation and with proper output Rd load resistor you can cover most of the corners with relatively constant gain.

Yes. I'm looking for 15dB AC Gain to allow AC Small Signal through. Hence I need to bias it at which Vindc should correspond to 0.5 VDD to allow maximum swing.
Are you proposing that I replace my R_D Load Resistor with a diode connected load, which acts as a current source?

https://i.stack.imgur.com/b6flD.png



As I'm not able what you are implying here.

Or are you proposing that I use two stages? A Common Source connected to a Source Follower?
A Circuit Diagram would be helpful.
 

FvM

Super Moderator
Staff member
Joined
Jan 22, 2008
Messages
46,978
Helped
13,934
Reputation
28,116
Reaction score
12,574
Trophy points
1,393
Location
Bochum, Germany
Activity points
273,927
Just read thoroughly. frankrose is suggesting a separate bias circuit to generate a variable gate voltage for the amplifier transistor.
 

frankrose

Advanced Member level 3
Joined
Nov 27, 2013
Messages
772
Helped
216
Reputation
432
Reaction score
210
Trophy points
43
Activity points
5,369
I recommended you the section a) of below picture. The bias circuit is M3(diode connected) + the current source, the DC coupling device is R, and the AC coupling device is C. You should replace (or not) diode connected M2 with a resistor.


The section b) of the picture is an other type of biasing, it is called self-biasing, where the M1 (amplifier device) sets its own operating Vgs by negative feedback through R. It also can handle corner variations.
 

iSean

Newbie level 4
Joined
Nov 4, 2019
Messages
7
Helped
0
Reputation
0
Reaction score
0
Trophy points
1
Activity points
66
I recommended you the section a) of below picture. The bias circuit is M3(diode connected) + the current source, the DC coupling device is R, and the AC coupling device is C. You should replace (or not) diode connected M2 with a resistor.


The section b) of the picture is an other type of biasing, it is called self-biasing, where the M1 (amplifier device) sets its own operating Vgs by negative feedback through R. It also can handle corner variations.
Thank you for your reply.
I try to see how the device works before coming back for you later.
how does one learn CMOS Transistor Level Circuits for its operations.
It seems very difficult as my university didn't really expose much indepth towards the operating principle of such devices.
 

iSean

Newbie level 4
Joined
Nov 4, 2019
Messages
7
Helped
0
Reputation
0
Reaction score
0
Trophy points
1
Activity points
66
I just realized I can't edit post here.
Supervisor asked me to 10dB AC gain instead.

He claims it is possible. I just need to analyze all the possible combinations...
Using this circuit. Got I feeling I might being trolled here...
 

iSean

Newbie level 4
Joined
Nov 4, 2019
Messages
7
Helped
0
Reputation
0
Reaction score
0
Trophy points
1
Activity points
66
after literally a week's of trial and error and playing around with 241 iterations.
... I reduced the 6kOhms to 1 kOhms it worked...
I didn't played with W. It is scary to see my stuff fail and readjusting the bias points over and over again.
 

Toggle Sidebar

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Top