a simple circuit question

[devop]

A Vdd independent biasing circuit
can I change B to A????why???

 

[pakitos]

can't see any image

 

[hr_rezaee]

Hi
As I know yes.
because we have the same equations.
regards

 

[JPR]

The two figures are DEFINITELY NOT THE SAME!

Figure A has positive feedback greater than unity and will tend to either collapse to the zero current state or increase to an extremely high current state.

Figure B has positive feedback less than unity and will tend to go to the expected current state, as long as there is a startup circuit preventing the zero current state.

 

[transbrother]

If you assume that p1 and p2 have same gm and n1 and n2 have the same gm,

from fig A, overall loop gain ~ (1 + gm * r) > 1 and figure B, has a loop gain of ~ 1/(1+gm * r) < 1. So do NOT use fig A as it will go to a non-steady state as mentioned above by JPR.

 

[abcyin]

figure A is a positive feedback, so it is instable, while the circuit in figure B is right, is can get a current source insensitive to the VDD supply.

 

[analog_prodigy]

how can we decide the type of feedback? please anybody can explain in detail?

 

[surno]

In the fig.1 if gate voltage of p1 is inreased source voltage of p2 is decreas. So source voltage of p1 is increase. That is positive feedback. OK~?!

 

[devop]

one more question
we know that the circuit B need a start up part.if the current of n2 is 10u when it is stable,will I give a start up current more than 10u to start it or just give it a small current and it will go to the stable state?

 

[Sadegh.j]

Could someone please explain why this is a positive feedback.

 

[liuyonggen_1]

Four transistors form a loop, it's not difficult to analyze the loop, two negative voltage gain form a positive feedback.

for start-up circuit, when the biasing operates normally, it often stop the start-up circuit, both large or small start-up current are ok!

 

[standup]

I think we can divided the circuits into two parts according to the book written by Razavi,generally,the device connected with the style of diodes must be drivered by an ideal current source.So we can analyse the circuits begin at the side of diode connection device.
A:I1p increases-->I2p incresase(now regards I2p as the ideal current of the right parts with N1、N2、R )-->VR increase--> to keep I2p rising,Vgn2 increase fast-->I1n increase.So current of the left side become bigger.
It can't be stable.
B:I2p increases-->I1p increases-->I1n increases -->
Vgn1 increases
I2p increases-->VR increases
Now,we should consider which one changes more.
VR=I×R,I∞(Vg)^2,so the varies rate of the VR must be faster than that of Vg
we can get that Vgsn2 decreases,the current flowed through n2 decreases.

 

[she_long]

I think P.R Gray's book Analysis and Design of Analog Integrated Circuits --Chapter 4: Current Mirrors and Current Source will be useful.
In figure A and B, p1 and p2 are current mirror, the gain is 1 ideally, n1 and n2 is current source, the gain is different, Figure A is large than 1 and Figure B is less than 1. If the loop gain is >1, it's not stable.

 

[northeast1]

A Vdd independent biasing circuit
can I change B to A????why???

I think it cant, because of the feedback is not same.

 

[Alan_Nesta]

you can give it start condition to get rid of zero current condition, so all current levels (inside your design and your device and stand) are OK

one more question
we know that the circuit B need a start up part.if the current of n2 is 10u when it is stable,will I give a start up current more than 10u to start it or just give it a small current and it will go to the stable state?

 

[shao]

b better