Analog Integrated Circuits

Summer and Subtractor OpAmp Circuits

25 questions By Tony R. Kuphaldt

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Question 22 of 25

Predict how the operation of this summer circuit will be affected as a result of the following faults. Consider each fault independently (i.e. one at a time, no multiple faults):

Resistor R₁ fails open:

Resistor R₂ fails open:

Solder bridge (short) across resistor R₃:

Resistor R₄ fails open:

Solder bridge (short) across resistor R₄:

For each of these conditions, explain why the resulting effects will occur.

Reveal answer

Resistor R₁ fails open: V_out becomes (inverted) sum of V₂ and V₃ only.

Resistor R₂ fails open: V_out becomes (inverted) sum of V₁ and V₃ only.

Solder bridge (short) across resistor R₃: V_out saturates in a negative direction.

Resistor R₄ fails open: V_out saturates in a negative direction.

Solder bridge (short) across resistor R₄: V_out goes to 0 volts.

Notes:
The purpose of this question is to approach the domain of circuit troubleshooting from a perspective of knowing what the fault is, rather than only knowing what the symptoms are. Although this is not necessarily a realistic perspective, it helps students build the foundational knowledge necessary to diagnose a faulted circuit from empirical data. Questions such as this should be followed (eventually) by other questions asking students to identify likely faults based on measurements.
Question 23 of 25

Predict how the operation of this difference amplifier circuit will be affected as a result of the following faults. Consider each fault independently (i.e. one at a time, no multiple faults):

Resistor R₁ fails open:

Resistor R₂ fails open:

Solder bridge (short) across resistor R₃:

Resistor R₄ fails open:

Solder bridge (short) across resistor R₄:

For each of these conditions, explain why the resulting effects will occur.

Reveal answer

Resistor R₁ fails open: V_out becomes equal to ¹/₂ V₂.

Resistor R₂ fails open: V_out saturates.

Solder bridge (short) across resistor R₃: V_out becomes equal to 2 V₂ − V₁ instead of V₂ − V₁.

Resistor R₄ fails open: V_out becomes equal to 2 V₂ − V₁ instead of V₂ − V₁.

Solder bridge (short) across resistor R₄: V_out becomes equal to −V₁.

Notes:
The purpose of this question is to approach the domain of circuit troubleshooting from a perspective of knowing what the fault is, rather than only knowing what the symptoms are. Although this is not necessarily a realistic perspective, it helps students build the foundational knowledge necessary to diagnose a faulted circuit from empirical data. Questions such as this should be followed (eventually) by other questions asking students to identify likely faults based on measurements.
Question 24 of 25

The instrumentation amplifier is a popular circuit configuration for analog signal conditioning in a wide variety of electronic measurement applications. One of the reasons it is so popular is that its differential gain may be set by changing the value of a single resistor, the value of which is represented in this schematic by a multiplier constant named m:

There is an equation describing the differential gain of an instrumentation amplifier, but it is easy enough to research so I’ll leave that detail up to you. What I’d like you to do here is algebraically derive that equation based on what you know of inverting and non-inverting operational amplifier circuits.

Suppose we apply 1 volt to the non-inverting input and ground the inverting input, giving a differential input voltage of 1 volt. Whatever voltage appears at the output of the instrumentation amplifier circuit, then, directly represents the voltage gain:

A hint for constructing an algebraic explanation for the circuit’s output voltage is to view the two “buffer” opamps separately, as inverting and non-inverting amplifiers:

Note which configuration (inverting or non-inverting) each of these circuits resemble, develop transfer functions for each (Output = … Input), then combine the two equations in a manner representing what the subtractor circuit will do. Your final result should be the gain equation for an instrumentation amplifier in terms of m.

Reveal answer

I won’t show you the complete answer, but here’s a start:

Equation for inverting side:

Output = − ( R
mR ) Input

Equation for non-inverting side:

Output = ( R
mR 1 ) Input

Notes:
This question actually originated from one of my students as he tried to figure out an algebraic explanation for the instrumentation amplifier’s gain! I thought the idea was so good that I decided to include it as a question in the Socratic Electronics project.

Astute students will note that the negative sign in the inverting amplifier equation becomes very important in this proof. As an instructor, I often avoid signs, choosing to figure out the polarity of the signal as a final step after all the other arithmetic has been completed for a circuit analysis. As such, I usually present the inverting amplifier equation as [(R_f)/(R_in)] with the caveat of inverted polarity from input to output. Here, though, the negative sign becomes a vital part of the solution!