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Designing Differential amplifier with differential output

Alex_bam

New Member
Hello!
I have done calculations for designing a differential amplifier (DA) with differential output. I simulate (using cadence ) the DA with the calculated values and perform the DC analysis of the circuit to inquire all the transistors are operating in saturation. However, I did not get all the transistors in the saturation region. I tried to tune the system (means to bring it to saturation region) but tuning one set of transistor results in the detuning of other transistors and vice versa.

if someone can review my calculations and suggest me correction. I will be thanked for my help.

Design Parameters:

GBW= 5MHz CL= 10pF, Vdd=5V

UnCox=151u A/V^2;

UpCox=26u A/V^2;

Av= 100

Vtn=0.85V Vtp=-1V

Slew Rate (SR) :5V/uSec

ICMR+= 4V ; ICMR- =1.5V (input Common mode range)

Step 1 _ Calculating Id:

Id=SR*CL 5V/uSec * 10pF= 50uA

Step 2 - Calculating (W/L) for M3 and M4:

Vds1>=Vmax-Vthn=4-0.85=3.15V (let consider it as 3.5V)

Vds3=Vdd-Vds1=5-3.5=1.5V

For M3 to be in saturation:

|Vds3|>=|Vgs3|-|Vthp|

1.5V >= |Vgs3| - 1V

|Vgs3|<= 2.5V (I take |Vgs3| : 2V )

Id3=Id/2 = 25uA

(W/L)3= 2* Id3 / UpCox * (|vgs3| - |Vthp| = 2

Step 3 - Calculating (W/L) for M1 and M2:

gm1=GBW * 2 * pi * CL

= 5M * 2 * 3.14 * 10pF

=314u

Id1=Id/2=25uA

(W/L)=gm1^2 / 2 * Id1 * unCox

(W/L) =13.06

Step 4 _ Calculating (W/L) for M5 and M6:

To keep M1 in saturation under ICMR- (1.5V):

Vin>=Vgs1+Vdsat5

Vgs1=1V (I calculate it as we know Id1 = 25uA , W1/L1=13.06 )

Putting Vin=1.5 (ICMR-) & Vgs1= 1V

Vdsat5<=0.5V

Id5=Id

(W/L) =2 * Id5 / UnCox * (Vdsat) ^ 2= 2.64

Finally, transistor dimensions are:

(W/L)1,2 = 1.9u / 1u

(W/L)3,4 = 13.06u / 1u

(W/L)5,6 = 2.64u / 1u
 

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unclejed613

Well-Known Member
Most Helpful Member
inquire all the transistors are operating in saturation. However, I did not get all the transistors in the saturation region. I tried to tune the system (means to bring it to saturation region) but tuning one set of transistor results in the detuning of other transistors and vice versa.
that's your problem right there... the transistors should be operating in the linear region, not the saturation region. preferably in the center of the linear region.
 

Alex_bam

New Member
that's your problem right there... the transistors should be operating in the linear region, not the saturation region. preferably in the center of the linear region.
Thanks for your suggestion.
My ultimate goal is to design a Fully Differential Amplifier. The first step is a differential amplifier with differential output, then CMFB design.
In a differential amplifier with differential output, all the FET's should be in saturation.
So my approach is to design & test the differential amp with differential output then will go for the CM detector and CMFB loop. Please comment, Is this approach is correct?
 

rjenkinsgb

Well-Known Member
Most Helpful Member
You need to explain the aims of the aim of the design, as what you post so far just makes no sense, at least without context.
 

tomizett

Active Member
the transistors should be operating in the linear region, not the saturation region
In this instance we're talking about MOSFETs, where the "saturation region" means where it's operating in a transconductance mode - it's the opposite of the terminology used for bipolar transistors, which is rather confusing. So yes - all the transistors should be in the "saturation region" for the amplifier to be operating linearly.

This sounds rather like university coursework or similar?
 

unclejed613

Well-Known Member
Most Helpful Member
most amplifiers are difficult to get working linearly without feedback. operated open loop, they tend to want to stick to one rail or the other.
 

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