Author Topic: Single Transistor Amplifier Design  (Read 1189 times)

0 Members and 1 Guest are viewing this topic.

Offline MattjdTopic starter

  • Regular Contributor
  • *
  • Posts: 230
  • Country: us
Single Transistor Amplifier Design
« on: March 04, 2019, 12:33:29 am »
Hi,

I'm trying to design an amplifier with the design specifications shown in the first image of the link.

Now I wanted to try a single common source amplifier. We're directed to use a specific nmos within cadence. Nmos4


I performed the simulations also shown in the imgur link to obtain the following


Considering the gain required is 10. I was going start by choosing a Q point from the graphs shown in the imgur link.

I chose, (11.112mA,2) @ Vgs = 2 as my Q point. This provides a transconductance gm of 10.621 mS. As well as...


Early voltage Va = 20.2 volts ... Va = ro*Id = 1818*11.112mA = 20.20
Channel Length modulation  lambda= 0.0495 ... lambda = 1/Va
conduction parameter Kn = 11.97 mA/V ... Kn = 2*Id/[(Vgs-Vt)^2(1+lambda*Vds)]


Additionally you can see from my Id vs Vgs graph that Vt is roughly .7 volts.


Since I chose a 4 resistor CS design, I was going to pick R1 = 300k Ohms and R2 = 200k Ohms. To calculate Rd I had planned on working backwards from this

-10 = -gm*(ro||RL||Rd)*(Rg/(Rsig+Rg)           (Rs isnt necessary bc i had planned on including by pass cap in parallel with it)...Rg = R1||R2

as I know gm, ro,RL

Once I calculated Rd, I was going to back calculate the necessary Rs value to give the proper Vgs and Vds as they're both known.

I get a negative resistance for Rd, im assuming something is wrong with my simulations OR i simply dont have a large enough gm to get the desired amplification.

I was hoping someone could point out my mistake.
« Last Edit: March 04, 2019, 07:45:34 am by Mattjd »
 

Offline Audioguru

  • Super Contributor
  • ***
  • Posts: 1507
  • Country: ca
Re: Single Transistor Amplifier Design
« Reply #1 on: March 04, 2019, 01:19:19 am »
What is the part number of your Mosfet that has a 'typical" threshold voltage of only 0.7V? or did you design the Mosfet?
Did you know that graphs in a datasheet are only for a "typical" one but the range from minimum to maximum are wide and are written in text?
 

Offline MattjdTopic starter

  • Regular Contributor
  • *
  • Posts: 230
  • Country: us
Re: Single Transistor Amplifier Design
« Reply #2 on: March 04, 2019, 01:33:18 am »
What is the part number of your Mosfet that has a 'typical" threshold voltage of only 0.7V? or did you design the Mosfet?
Did you know that graphs in a datasheet are only for a "typical" one but the range from minimum to maximum are wide and are written in text?

Its nmos4 in NCSU_Analog_Parts of cadence
 

Offline bson

  • Supporter
  • ****
  • Posts: 2497
  • Country: us
Re: Single Transistor Amplifier Design
« Reply #3 on: March 04, 2019, 01:48:18 am »
-10 = -gm*(ro||RL||Rd)*(Rg/(Rsig+Rg)           (Rs isnt necessary bc i had planned on including by pass cap in parallel with it)...Rg = R1||R2

as I know gm, ro,RL

Once I calculated Rd, I was going to back calculate the necessary Rs value to give the proper Vgs and Vds as they're both known.

I get a negative resistance for Rd,
Given,
-10 = -gm*(ro||RL||Rd)*(Rg/(Rsig+Rg)

I get:
-10/(-gm*Rg/(Rsig+Rg)) = ro||RL||Rd = 1/(go + gL + gd)
gm*Rg/10(Rsig+Rg) = go + gL + gd

gd = 1/Rd = gm*Rg/10(Rsig+Rg) - 1/Ro - 1/RL

Rg = 2/(2+3) = 0.4Ω = 200k||300k = 120kΩ
Rsig = 1kΩ
RL = 500Ω
Ro = 1818Ω

1/Rd = 10.621m*0.4/10(1k+0.4) - 1/1818 - 1/500 = 0.42246

Rd = 2.37Ω


EDIT: Oh wait, Rg is 200k||300k, not a divider.  Sorry.

Yup, corrected for that I get a negative Rd also; -668Ω.
« Last Edit: March 04, 2019, 01:59:04 am by bson »
 

Offline MattjdTopic starter

  • Regular Contributor
  • *
  • Posts: 230
  • Country: us
Re: Single Transistor Amplifier Design
« Reply #4 on: March 04, 2019, 01:58:55 am »
-10 = -gm*(ro||RL||Rd)*(Rg/(Rsig+Rg)           (Rs isnt necessary bc i had planned on including by pass cap in parallel with it)...Rg = R1||R2

as I know gm, ro,RL

Once I calculated Rd, I was going to back calculate the necessary Rs value to give the proper Vgs and Vds as they're both known.

I get a negative resistance for Rd,
Given,
-10 = -gm*(ro||RL||Rd)*(Rg/(Rsig+Rg)

I get:
-10/(-gm*Rg/(Rsig+Rg)) = ro||RL||Rd = 1/(go + gL + gd)
gm*Rg/10(Rsig+Rg) = go + gL + gd

gd = 1/Rd = gm*Rg/10(Rsig+Rg) - 1/Ro - 1/RL

Rg = 2/(2+3) = 0.4Ω
Rsig = 1kΩ
RL = 500Ω
Ro = 1818Ω

1/Rd = 10.621m*0.4/10(1k+0.4) - 1/1818 - 1/500 = 0.42246

Rd = 2.37Ω
your Rg is wrong

Rg is the parallel resistance of R1 and R2, which is 120k, so Rg/(Rsig+Rg) = 120k/121k ...approx 1, which is why that part of the gain is typically ignored when Rg is much much larger than Rsig
 


Share me

Digg  Facebook  SlashDot  Delicious  Technorati  Twitter  Google  Yahoo
Smf