Why a high gain opamp with negative feedback is needed in a bandgap reference source?

[sina25]

hi everybody

Why a high gain opamp with negative feedback is necesary in a bandgap reference source to ensure that the two inputs have the same potential?
I know that is neceary for getting the PTAT voltage,
but if the bandgap reference sources work with DC signal,
how does a high-gain AC and the negative feedback ensure that the two inputs have the same DC potential

please explain detailed

help me

thanks

 

[sunking]

OPAMP

 

[sina25]

Re: OPAMP

Please, somebody try to ask my questions

thanks sunking by the papers

HELP ME


hi everybody

Why a high gain opamp with negative feedback is necesary in a bandgap reference source to ensure that the two inputs have the same potential?
I know that is neceary for getting the PTAT voltage,
but if the bandgap reference sources work with DC signal,
how does a high-gain AC and the negative feedback ensure that the two inputs have the same DC potential

please explain detailed

help me

thanks

 

[surianova]

Re: OPAMP

Please, somebody try to ask my questions

thanks sunking by the papers

HELP ME


hi everybody

Why a high gain opamp with negative feedback is necesary in a bandgap reference source to ensure that the two inputs have the same potential?
I know that is neceary for getting the PTAT voltage,
but if the bandgap reference sources work with DC signal,
how does a high-gain AC and the negative feedback ensure that the two inputs have the same DC potential

please explain detailed

help me

thanks

in negative feedback, the transfer function is

H(s)= G/ (1+GH) , G=open loop gain, H=feedback gain

vout= G/ (1+GH) * vin

divide by G

vout= 1/ [1/G + H] * vin

When G is high , vout almost equal to vout=1/H

for example, test for 2 case

1. vin=1.2, G= 100,assume H=1 for buffer

vout= 1/ [1/G + H] * vin

what is vout ?


2. vin=1.2 , G=1000

what is vout?

From this 2 case, when the gain is high, the vout is more closer to 1.2 input


Hope this will help.

 

[sina25]

Re: OPAMP

Hi everybody

according to surianova G=open loop gain, and H=feedback gain

1. Are G y H DC gain?,
remember that the voltage reference sources work with DC voltages and DC
currents.

2. How should i obtain that DC gain?
With a large-signal analysis? or with a small-signal analysis without capacitors
and inductors? and WHY?

please
explain detailed

help me

 

[surianova]

Re: OPAMP

Hi everybody

according to surianova G=open loop gain, and H=feedback gain

1. Are G y H DC gain?,
remember that the voltage reference sources work with DC voltages and DC
currents.

2. How should i obtain that DC gain?
With a large-signal analysis? or with a small-signal analysis without capacitors
and inductors? and WHY?

please
explain detailed

help me

G and H is the DC gain. It must be measure in AC analysis. The DC gain is the gain before it drop 3 dB. Large-signal analysis just to make sure all the transistor in saturation.

 

[nandu_r]

Re: OPAMP

G and H are dc gain , that can be found in low frequency say 1HZ(in the ac analysis)
This is assuming that all capacitor are small enough they act as open circuit and all inductors as open circuit

 

[sina25]

Re: OPAMP

Hi everybody

surianova, nandu_r:

How could I do that AC analysis?
With a small-signal analysis? and why?
But a bandgap-voltage source doesn't work with small signals,
Could their voltages or currents be considered for a small-signal analysis? and WHY?

In a bandgap-voltage source the opamp-inputs voltages normally are 0.65Volts (300ºK), because this is the diode voltage (300ºK), then
HOW COULD THOSE VOLTAGES BE CONSIDERED SMALL-SIGNAL?
AND WHY?

PLEASE HELP ME

thanks

 

[huangjw]

OPAMP

you just consider the op-amp's ac characteristic.
I think the small signal in bandgap-voltage is the change of the voltage change with temperature or supply

 

[wanily1983]

Re: OPAMP

i think the opamp is used to get a precious PTAT voltage through negative feedback. it isn't necessery in all bandgap circuits.

 

[sina25]

Re: OPAMP

How does the OPAMP ensure that V+ and V- are equal?

is this equality due to OPAMP diferential input and its simetry?

HELP ME

THANKS

 

[rockycheng]

Re: OPAMP

How does the OPAMP ensure that V+ and V- are equal?

is this equality due to OPAMP diferential input and its simetry?

HELP ME

THANKS

Hi,

Maybe you need to read more about the feedback.

Put simply, if the product of feedback factor and the voltage gain of the opamp is significantly larger than unity, the voltage of the two input nodes should be almost equal, or in other words, the differential input is almost zero. If not, this differential input will make a large single-ended output voltage, due to the large opamp gain. And this large output will in turn reduce the differential input value, through the negative feedback path, until the differential input is so small that the ouput voltage is finally stable at a specific value. Since Vo=Vi*Av is still valid for the opamp, for a certain value of Vo, the larger the Av, the smaller the Vi. That's why large opamp gain helps make V+ and V- close, no matter what the value of Vo is.

I hope this may help.

 

[sina25]

Re: OPAMP

surianova says: G and H is the DC gain. It must be measure in AC analysis. The DC gain is the gain before it drop 3 dB. Large-signal analysis just to make sure all the transistor in saturation.

Hi everybody

surianova, nandu_r:

How could I do that AC analysis?
With a small-signal analysis? and why?
But a bandgap-voltage source doesn't work with small signals,
Could their voltages or currents be considered for a small-signal analysis? and WHY?

In a bandgap-voltage source the opamp-inputs voltages normally are 0.65Volts (300ºK), because this is the diode voltage (300ºK), then
HOW COULD THOSE VOLTAGES BE CONSIDERED SMALL-SIGNAL?
AND WHY?

PLEASE HELP ME

thanks

please, try to answer these questions

HELP ME

thanks

 

[zenisle]

OPAMP

good answers.

 

[sina25]

Re: OPAMP

Surianova said:

in negative feedback, the transfer function is

H(s)= G/ (1+GH) , G=open loop gain, H=feedback gain

vout= G/ (1+GH) * vin

divide by G

vout= 1/ [1/G + H] * vin

When G is high , vout almost equal to vout=1/H

for example, test for 2 case

1. vin=1.2, G= 100,assume H=1 for buffer

vout= 1/ [1/G + H] * vin

what is vout ?


2. vin=1.2 , G=1000

what is vout?

From this 2 case, when the gain is high, the vout is more closer to 1.2 input


Hope this will help.

***********************************************************************************************************************

IN THIS CIRCUIT PLEASE SHOW: vin, vout, G and H

THANKS