XXX of XXX circuit XX XX achieve the XXX XX XX,XX,v3 in amplified form at XXXXXX terminal.
XXX XX the Op-XXX XXX a XXXX XXXXX XXXXX XXXXXXXXX, XXXXX XXXX no XXXXXXX XXXXX XXXX XXX Op-amp.
XXXXXXXXX, i = XX+XX+i3 XX XXXX XXXXXXX at the output side. XXXX XXXXX XX a XXXXXXX short between XXX points X and B, there XXX voltage XX these two XXXXX will XX XXXXX. X point XX XX ground XXXXX the point A XXXX XXXX XX at X XXXXX. Considering XXXXX lets dive XXXX XXXXXXXX.
XXXXXXXX the currents XX XXXXX and output XXXX,
XX XXXXXXX XXXXXXX we will be XXXXXXX,
XXXX the XXXXXX XXXX be,
Now let's move XXXX XXX differential XXXXXXXXX circuit.
The circuit XXXXX XXXX,
XXXX current XXXX applies and the XXXXXXX XXXXX makes Voltage at point X (VA) will XX equal to XXXXXXX XX point B(XX).
XXX hold on! XXXX's VB XXXXXX be?
XX XX w XXX voltage XXXX XXXXXX XX resistor, XXXXX is,
XXX's get XXXX XXXXXXX XXXXXXXX,
The XXXXXXX XX XX XXXX XX also XXXXXXX at XXX output XXXX because XX hing input XXXXXXXXX XX XXX Op-amp,
equating XXX current XX XXXXX XXX output side,
XXXXXXX algebraically,
XX, XXXXXX XXXXXXX becomes,
Now to achieve XXX XXXXXXXXXXXX operation XX should do XXX XXXXXXXXX,
XXX XXX XXXXXX XXXXXXX,
Differential XXXXXXXXX achieved!!
XXX's XXXX onto XXX problem,
Looking XXX circuit, it's just a XXXXXXX amplifier and we XXXX put the required XXXXX given in XXX XXXXXXX XXXX the XXXXXXXX XX XXX,
where,
XXXXXXX XXXXX XXXXXX,
XXXXXX XXXX XX -XXX.