DC input random offset voltage for diff. amplifier with current-mirror load

AI Thread Summary
The discussion revolves around the derivation of equation 6.69 in "Analysis and Design of Analog Integrated Circuits" by Gray and Meyer, specifically regarding the input offset voltage (VOS) for a differential amplifier with a current-mirror load. The user expresses confusion over the absence of supply voltages VSS and VDD in the equation and questions whether the definitions provided in the book align with this derivation. It is noted that the supply voltages can be considered equal and opposite, which contributes to the circuit's symmetrical nature and independence from the differential input pair. Ultimately, the user discovers that the relevant derivation can actually be found on pages 332-333 of the same book. This highlights the importance of thorough reference checking in complex circuit analysis.
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TL;DR Summary
how the expression for DC input offset voltage for diff. amplifier with current mirror load come about
I am trying to figure out how the derivation of equation 6.69 come about on page 426 in the book Analysis and Design of Analog Integrated Circuits, 5th Edition by Gray and Meyer. They defined on page 424 under section 6.3.3 the input offset voltage (VOS) of op amps with differential inputs and single-ended outputs as the differential input voltage for which the dc output voltage is midway between the supplies. But I don’t see VSS and VDD in the equation so I am not sure if they are using this definition since they did mention another definition for VOS under 6.3.3 which is the differential input voltage for which the op-amp output voltage is zero but I try to follow the similar derivation on pg. 234-236 under 3.5.6.6 and 3.5.6.7 for diff. amp with resistors pair as loads but I still don’t get how do they come about equation 6.69?Analysis and Design of Analog Integrated Circuits, 5th Edition
http://fa.ee.sut.ac.ir/Downloads/AcademicStaff/24/Courses/73/[Gray___Meyer]_Analysis_and_Design_of_Analog_Integrated_Circuits_5th_ed.pdf
 
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eyeweyew said:
But I don’t see VSS and VDD in the equation so I am not sure if they are using this definition ...
The supply voltages can be assumed typical, equal and opposite, and sufficient for the circuit to operate.

Supply voltage is independent, since the circuit is symmetrical. The supplies are isolated from the differential input pair, by the collector or drain voltage of the other transistors employed.

Part of the elegance of employing a current mirror as the load, is supply voltage independence, while the gain of the differential input pair is effectively squared.
 
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eyeweyew said:
TL;DR Summary: how the expression for DC input offset voltage for diff. amplifier with current mirror load come about

I am trying to figure out how the derivation of equation 6.69 come about on page 426 in the book Analysis and Design of Analog Integrated Circuits, 5th Edition by Gray and Meyer. They defined on page 424 under section 6.3.3 the input offset voltage (VOS) of op amps with differential inputs and single-ended outputs as the differential input voltage for which the dc output voltage is midway between the supplies. But I don’t see VSS and VDD in the equation so I am not sure if they are using this definition since they did mention another definition for VOS under 6.3.3 which is the differential input voltage for which the op-amp output voltage is zero but I try to follow the similar derivation on pg. 234-236 under 3.5.6.6 and 3.5.6.7 for diff. amp with resistors pair as loads but I still don’t get how do they come about equation 6.69?Analysis and Design of Analog Integrated Circuits, 5th Edition
http://fa.ee.sut.ac.ir/Downloads/AcademicStaff/24/Courses/73/[Gray___Meyer]_Analysis_and_Design_of_Analog_Integrated_Circuits_5th_ed.pdf
Ok, Never mind. The derivation is actually on page 332-333 in the same book.
 

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