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Unclearity in the IC design book by Gray Hurst etc.

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radius2

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Hi

I was reading chapter 12 page 836 in Gray Hurst Lewis and Meyer book.
In figure 12.24a the DM half-circuit for the fully differential opamp on the
previous page is drawn. The input signal is -vid / 2 and the output signal is
vod / 2 which means that the half circuit is inverting. However, my opinion is
that the half circuit in fig. 12.24a cannot be inverting. The first common
source stage inverts and the second inverts back which results in a noninverting
operation. I understand that the fully differential amplifier (from vid to vod) is inverting but not why the half circuit in fig. 12.24a is.

Does anybody have an idea or am I missing something?

Thanks
 

Could you possibly post this figure?
 

Hi

Sure, fig 12.23 shows the fully differential op amp
and fig 12.24a shows its DM half circuit

BR



Added after 4 minutes:

Seems like the first figure wasn't uploaded.
Here i will try again.

 

I think you are right: the half-circuit doesn't invert, so the minus sign at -vid / 2 must be wrong.
The inversion of the fully differential amplifier just results from the connotation nomenclature: input to opposite output.
Rgds, erikl
 

You should not look at this half circuit as exact physical replica of the actual circuit. In he diff amplifier you close for example the feedback between one input and the output of the other half. However, you need to have some means of circuit analysis and the half circuit is good for that . Just think that there is an ideal inversion between the output and the input when you analyze it for feedback.
 

Yeah I see what you are saying, one can picture an ideal inversion.

It is also possible to derive the voltage gain from vi2 to vo2 under the assumption
that vi2 = -vid/2 and vi1 = vid/2 (the exact opposite). The gain of
the amplifier from vi2= -vid/2 to vo2 is then the same as for the half circuit pictured (This is only possible since the voltage at vi1 is exactly opposite otherwise
it is not true) but now you get the minus sign and thereby capture the inversion.
So if one wants to do it can be done in this way. Then it's possible to
regard the fully differential amplifier in feedback configuration as two
inverting single ended amplifiers.

BR
 

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