auto-zero comparator question?

[eegchen]

hi,
I read the auto-zero comparator, the first stage is to stage the offset error(top graph)
And the second stage is to compare Vin and Vref, which will eliminate the error( low graph)

I have question on the equation on second stage. Why (Vin+ - Vin-)- (Vc+ - Vc-)?
Could you anyone help to explain?

 

[BackerShu]

In the low graph:
Vin+ - Vc+ - Voffset=V1p (V1p is the positive input of P1)
Vin- - Vc- = V1n
So, the differential input of P1 is
V1p -V1n = (Vin+ - Vin-) - (Vc+ - Vc-) - Voffset

It seems that the sign before the Voffset is not correct.

 

[eegchen]

hi backerShu,

Thanks for your reply.
What i don't understand is why Vin+ - Vc+ not Vin+ + Vc+?

when i design charge pump, if you give a Vdd to Cap, it will boost the output.

best
Gang

In the low graph:
Vin+ - Vc+ - Voffset=V1p (V1p is the positive input of P1)
Vin- - Vc- = V1n
So, the differential input of P1 is
V1p -V1n = (Vin+ - Vin-) - (Vc+ - Vc-) - Voffset

It seems that the sign before the Voffset is not correct.

 

[BackerShu]

I think it's all about how to define the reference direction of Vc+/-.
Take the path with Voffset in the low graph for example. Assume that Vc+ has the same direction with Voffset, so Vc+=Vplus-Vminus.(Vplus is the terminal which is further away from P1 ). So we get
Vin+ - Vc+ - Voffset=V1p
But if you take the opposite defination.Than we get
Vin+ + Vc+ - Voffset=V1p
So it comes to
V1p -V1n = (Vin+ - Vin-) + (Vc+ - Vc-) - Voffset
In fact, they are the same when taking the reference direction into account.

 

[eegchen]

thanks.

But should be V1P and V1n equal to Vcm ?

best
gang

Added after 5 minutes:

thanks. I still have a little confusion.

If thing comes to
V1p -V1n = (Vin+ - Vin-) + (Vc+ - Vc-) - Voffset
Whey there are still the same?
if we assume reference direction is fixed
Best

I think it's all about how to define the reference direction of Vc+/-.
Take the path with Voffset in the low graph for example. Assume that Vc+ has the same direction with Voffset, so Vc+=Vplus-Vminus.(Vplus is the terminal which is further away from P1 ). So we get
Vin+ - Vc+ - Voffset=V1p
But if you take the opposite defination.Than we get
Vin+ + Vc+ - Voffset=V1p
So it comes to
V1p -V1n = (Vin+ - Vin-) + (Vc+ - Vc-) - Voffset
In fact, they are the same when taking the reference direction into account.