Pls help in fully differential OTA??

[wael_wael]

hi,
i have this FD-OTA but i cant under stand why thy use the red color connection

regards

 

[hr_rezaee]

Help in fully differential OTA??

Hi
It is opamp tail.
if this red connection is removed then you have very small diff. gain.
regards

 

[wael_wael]

Re: Help in fully differential OTA??

thank you, could suggest me any artical related to this op amp, any way if i want to change the input mosfet to NMOS where this connection will be
regards

 

[sutapanaki]

Re: Help in fully differential OTA??

The input devices are already NMOS. And yes, the red connection is just the common source point of the differential pair. Think of the transistors below the red line as two cascoded current sources connected in parallel. Nothing more. If you want to use PMOS input, you'll have to flip everything upside-down i.e. Nmos transistors become pmos and vice versa. If you want to read about this circuit, look for boosted folded cascode opamp. There is also a common mode feedback at the output - this is done with the transistors connected with thick black lines.

 

[analog_prodigy]

Help in fully differential OTA??

You can find good opamp material in TI website

 

[wael_wael]

Re: Help in fully differential OTA??

thank you all, and really thank to mr. sutapanaki for his advise. i was mistaken about PMOS, you are right it is NMOS already
best regards

 

[safwatonline]

Help in fully differential OTA??

yep but y dividing the tail current source into two parallel , i cannot see any advantage however it decrease the Output impedance of the cascode mirror

 

[wael_wael]

Re: Help in fully differential OTA??

thanx all, could some one help me and suggest me article witch shows how to design the gain boosting amplifier(sup op amp in the figure), it is look like inverter.
regards

 

[bkt22]

Re: Help in fully differential OTA??

yep but y dividing the tail current source into two parallel , i cannot see any advantage however it decrease the Output impedance of the cascode mirror

No, it does not decrease the output impedance of the cascode mirror.

Imagine cascode mirror with current I. Let bottom transistor be M2 and cascode transistor be M1
Output impedance here is ~ gm1r01r02

Now if the same current is divided into 2 stages, ie, we have two cascodes but now they each carry I/2 current. Let bottom transistor be M4 and cascode transistor be M3.

The output impedance will be ~ (gm3r03r04)/2

Suppose current density in both are the same, ie, L is same and W is scaled according to current.

Then gm3 = (gm1/2) , r03 = 2*r01 r04=2*r02

Now, you see they have the same output impedance.

Why did you conclude that the output impedance is decreased? By the above calculation, output impedance is the same with the same current density.

 

[safwatonline]

Help in fully differential OTA??

bkt22,
i didn't calculate it so i thought it will decrease,but u seem to be right
thanks&regards

 

[wael_wael]

Re: Help in fully differential OTA??

hi, in this figure below what is the type of gain boosting op amp(sub OP-AMP), is it inverter? or class A OP AMP, or current source , the all circuit is folded cascode oP-AMP
regards

 

[liuyonggen_1]

why does the tail current use two branches, in my opinion, one branch is better in the espects of area and matching.

 

[hr_rezaee]

Re: Help in fully differential OTA??

hi, in this figure below what is the type of gain boosting op amp, is it inverter? or class A OP AMP, or current source
regards

Hi
it is simplest amplifier that I know.
as I remember, this is pseudo differential configuration.
regards

 

[starrinesss]

Just casode current source and the input transistors are already NMOS

 

[sutapanaki]

Re: Help in fully differential OTA??

hi, in this figure below what is the type of gain boosting op amp(sub OP-AMP), is it inverter? or class A OP AMP, or current source , the all circuit is folded cascode oP-AMP
regards

The type of the sub-amplifiers is shown on your circuit with the 2-transistor configurations - regular comon source amplifiers. Can't get simpler than that. The only problem is that it eats a bit of headroom, because there is a Vgs from the drain of the lower current source transistors in the real opamp to gnd.