help:maybe not hard for you

[devop]

can someone tell me that how to calculate V1-V2,the equation bewteen V1-V2 and A.
thanks

 

[pankajdudulwar]

let me know whether the transistor are NPN or PNP or else their operating region.

 

[devop]

let me know whether the transistor are NPN or PNP or else their operating region.

it is a bandgap circuit for n well process,so it's a pnp,thanks
can you give me some suggestion?

 

[A.Anand Srinivasan]

is it a opamp that you have used or is it a amplifier......

because the output very much depends on the amplifiers gain and the amplifiers input current.....

 

[standup]

is the circuit a bandgap reference voltage?
where is the VDD?
thank you!

 

[devop]

is it a opamp that you have used or is it a amplifier......

because the output very much depends on the amplifiers gain and the amplifiers input current.....

I'm sorry I dont konw much about the difference between opamp and amplifiers that you mentioned.
it's a two stage opamp with pmos input.
I want to know the relationship between V1-V2 and the amplifiers gain .

to standup
it is the bandgap reference voltage circuit

vdd is connect to the amplifier

 

[libinphy]

idearly ,v1-v2=0

 

[rainman.cn]

why you want to know v1-v2, its a nonideal factor if not 0...

 

[devop]

why you want to know v1-v2, its a nonideal factor if not 0...

because if A is not big encough,v1-v2 will not small encough
so I want to konw how big the A encough to make the v1-v2 small than a given factor maybe 0.0001

 

[holddreams]

In Razavi's book,page 386,figure 11.11,it's the circuit you drawn.
And there are some equations about the circuit.

(V1-V2)*A=vout,
usually,A is large,and if the opamp is ideally,v1=v2,that is v1-v2=0,
so,0×∞=constant(vout value).

correct me if i'm wrong.

Thanks.

 

[transbrother]

V1 = Veb1
V2 = I2*R3 + Veb2

Also by KVL and assuming zero offset in opamp,
V1 = V2
Therefore, I2 = (Veb1-Veb2)/R3 = del.Veb/R3 (ptat current)

And,
I1*R1 = I2 * R2
That is, I1 = (del.Veb/R3) * R2/R1

Now, your bandgap voltage,
out = I1 * R1 + Veb1
= del.Veb R2/R3 + Veb1

 

[A.Anand Srinivasan]

in the bandgap reference circuit you have given the resistance R2 is chose to cancel out the temperature effect of the current equation of the transistor....

the circuit has a feedback and the amp used is an opamp.... opamps generally have a gain of about 10^5... due to feedback the input i.e V1-V2 would be in µV....

 

[rfzheng]

in the bandga reference ,the gain of two stages opamp is big
enough. and you do not need to care about the absolute gain of the opamp, if it is big enough(>60dB), because the opamp work in the closed loop. v1-v2 may equal to the offset voltage of the opamp, it may several mini-volts and this offset voltage will influence your output reference voltage ,because the offset voltage has changed delta Vbe and changed the PTAT current .

 

[FFFFFF]

look this :

**broken link removed**

http://www.eecs.umich.edu/~mpflynn/Design_Contest/Fall2003/Reports/kabir_bhaskar_tao.pdf


**broken link removed**

http://www.circuitsage.com/bandgap.html

**broken link removed**

6 * F

 

[devop]

in In Razavi's book,the homework 11.4,he gives a solution tha A(V1-V2)=Vdd-vgs(in the picture above,maybe A(v1-v2)=Vout),so V1-V2 inversely proportional to A
and I am puzzlede here.left part is a small signal and the right is a big signal?????????

 

[A.Anand Srinivasan]

it is due to the opamp.... opamp generally have gains of the range of 10^5.... opamps with much higher gains are also there....