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 Differential summing amplifier for inputs having large common mode signals

Details
Inventors: Hedman, Robert L.;
Assignee: International Business Machines Corporation (Armonk, NY)
Primary Examiner: LaRoche; Eugene R.
Assistant Examiner: Mottola; Steven J.
Attorney, Agent or Firm: Anglin; J. Michael

Differential input voltages are converted into current unbalances in a parallel-branch circuit. A load means converts the sum of the current unbalances to a differential voltage representing the sum of the input voltages.

DETAILED DESCRIPTION OF A PREFERRED EMBODIMENT In the Figure, amplifier 100 has a summing stage 110, a buffer stage 120, and an output stage 130.
Input stage 110 receives two differential input voltages V1 and V2, and converts them to a voltage V3 by superposition of currents in its two branches, 111 and 112.
Current sources I1 and I2 are identical.
In the absence of any differential input voltages, equal currents flow between the supply potentials Vcc and ground through the two branches.
This remains true even in the presence of a large and variable common-mode voltage difference between V1 and V2.
(The common-mode voltage of V1 is defined as the voltage between ground potential and the average of the voltages on the two input terminals marked V1.
The common-mode voltages of V2 and V3 are defined similarly.
) A differential input voltage between the two terminals marked V1 appears between the bases of NPN transistors Q1 and Q2.
This produces a current in current-transfer resistor R1 proportional to the differential portion of V1, and thereby unbalances the branch currents by an amount V1/R1, in a direction determined by the polarity of that input voltage.
Differential input voltage V2 appears between the bases of another pair of NPN transistors, Q3 and Q4.
V2 produces a current in another current-transfer means R2, unbalancing the branch currents by V2/R2.
Thus, the total difference between the currents in branches 111 and 112 follows the weighted sum of the two input voltages.
The voltage across each branch load means Q5, R3 and Q6, R4 is proportional to the current through it.
Thus, the differential voltage V3 between the collectors of Q3 and Q4 is a sum signal V3=(V1/R1+V2/R2)*Z, where Z represents an effective impedance between the bases of transistors Q7 and Q8.
The value of Z is normally quite large, on the order of 1OK to 1OOK Ohms.
This large impedance provides high gain and a good compensation point, both attributes being useful when this circuit forms a part of a feedback amplifier



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