Hi,
all.
I am doing my project with AD623, there are two questions I don't understand.
1.On page 20 of datasheet of AD623, what's the '0.6V' stand for in the equations above Figure55? Why it adds 0.6V here? Does it mean the PNP junction voltage?

2.'For a bipolar input voltage with a common-mode voltage that is roughly half way between the rails, VDIFFMAX is half the value that the previous equations yield because the REF pin is at midsupply.' Why here it says that 'the REF pin is at midsupply'?
Thank you!

• Hello Keven,

You are correct, the 0.6V corresponds to the base-emitter junction voltage of the input transistors. The reason this voltage is included in the equations you have mentioned: The goal of this section of the data sheet is to explain the maximum input range of the amplifier. For this purpose, finding the output voltage of the input amlifiers is essential, since they will be limited to the supply rails. This is the main limiting factor at high gains.

Regarding your second quesiton: The calculations done previous to the paragraph you are referring, are done assuming the maximum input voltage for the maximum output voltage possible. Remember that, since the output voltage is always referenced to the REF pin, if this is at mid supply, the output voltage can only swing mid supply.

To help me explain this, here's an ideal example: assume no common-mode range and ideal rail-to-rail operation. If the supply rails are 5V and 0V, then the output can swing anywhere from 0V to 5V (again, this is idealized, there is no perfect rail-to-rail output). If the reference pin (REF) is at 0V, then the output can only amplify input signals that are positive. If there is no gain, then a differential input voltage of 5V will become 5V in respect to REF. However, if you want to amplify bipolar signals (both positive and negative voltages of about the same amplitude) then REF pin should be placed at mid supply, or 2.5V, so that the output can swing to 0V, or -2.5V below REF. In this case, if you take the same input of 5V as before, the output would need to become 7.5V. That's outside the rails, which would be impossible for the amplifier to do. This means that 2.5V is the maximum positive input voltage, which is one half as before. Note that the total swing is still the same, but now you've split the difference between positive and negative swings.

Regards,

Gustavo

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Thank you,