Question 1

Operational Transconductance Amplifier (OTA) voltage gain For an OTA driving a resistive load RL (single-ended), the small-signal voltage gain can be expressed in terms of transconductance gm as:

Accepted Answer

Introduction / Context: An Operational Transconductance Amplifier converts an input voltage to an output current with proportionality gm (amps per volt). When that current is sent into a resistive load, it develops a voltage. Recognizing this relationship helps when designing voltage-controlled filters, oscillators, and automatic gain control circuits that use OTAs. Given Data / Assumptions: OTA modeled as io = gm * vi. Load is a linear resistor RL to ground (or to a reference node), producing vo = io * RL. Small-signal, linear operation around a bias point (gm may be set by a control current). Concept / Approach: By definition, the OTA emits an output current proportional to input voltage. That current flowing through RL generates an output voltage. Hence, voltage gain is the product of gm and RL. If the load is reactive or frequency dependent, the gain becomes frequency dependent, but the basic proportionality remains. Step-by-Step Solution: Start with io = gm * vi.Voltage across load: vo = io * RL = (gm * vi) * RL.Therefore, Av = vo / vi = gm * RL.Sign conventions depend on the particular OTA pinout and load connection; magnitude is gm * RL. Verification / Alternative check: If RL doubles, Av doubles, which matches intuition: more resistance creates more voltage from the same current. If gm increases via bias current, gain increases proportionally. Why Other Options Are Wrong: 1/gm or RL/gm: imply inverse dependence, incorrect for a simple resistive load.gm/CL: mixes dimensions (per farad), relating to a pole frequency rather than voltage gain. Common Pitfalls: Forgetting that real OTAs have limited output compliance and may need a buffer; high RL or large signals can lead to distortion or clipping. Final Answer: Av = gm * RL

Question 2

In the classic three-op-amp instrumentation amplifier topology used for precision differential measurements, which stage primarily provides the adjustable differential voltage gain (set by the gain resistor Rg), while the final op-amp mainly perform…

Accepted Answer

Introduction / Context: Instrumentation amplifiers (in-amps) are widely used to extract small differential signals in the presence of large common-mode voltages. The most common topology is the three-op-amp in-amp, valued for its high input impedance, excellent common-mode rejection ratio (CMRR), and easily programmable gain. Knowing which stage sets the gain is essential for proper design and troubleshooting. Given Data / Assumptions: Classic three-op-amp instrumentation amplifier. Two input buffer amplifiers (first stage) with a gain-setting resistor Rg between them. A third op-amp (second stage) configured as a differential amplifier with precision resistor ratios. Concept / Approach: In the classic in-amp, the first stage consists of two op-amps configured as noninverting amplifiers. Their gains are simultaneously set by Rg, which bridges their feedback networks. This stage establishes the bulk of the differential voltage gain while maintaining very high input impedance. The final (third) op-amp subtracts the two buffered signals using a resistor network to remove common-mode content and refine CMRR. Step-by-Step Solution: Identify stages: first stage = dual noninverting buffers with Rg; second stage = subtractor.Relate Rg to gain: decreasing Rg increases first-stage gain.Note function of last op-amp: implements subtraction with precise resistor ratios to maximize CMRR.Therefore, the differential voltage gain is primarily produced by the first stage. Verification / Alternative check: Manufacturers’ datasheets show gain equations like G ≈ 1 + (2R/Rg) for the first stage, confirming that Rg controls overall gain. The output stage gain is typically fixed by resistor ratios to enforce accurate subtraction. Why Other Options Are Wrong: Second stage/output op-amp: mainly subtracts; it is not the usual adjustable gain element. “Mismatched resistors” reduce accuracy and CMRR; they do not set correct gain. “Both stages equally” is incorrect; the programmable gain resides predominantly in the first stage. Common Pitfalls: Assuming the differential amplifier stage sets gain; in reality, its resistor matching is optimized for CMRR, not user-programmable gain. Final Answer: first stage

Question 3

In analog signal conditioning, what is the standard name for circuits whose primary purpose is to shift (raise or lower) the dc level of a waveform without significantly altering the waveform’s shape or peak-to-peak amplitude?

Accepted Answer

Introduction / Context: Many practical systems need to move a signal up or down on the voltage axis so it fits within an ADC input range, a coupling window, or a processing block that expects a certain dc bias. The circuits that perform this dc restoration or level shifting are commonly called clampers (or dc restorers). Given Data / Assumptions: We wish to change only the dc level of a signal. Waveform shape and peak-to-peak value should remain essentially unchanged. Standard passive/diode-capacitor or active variants may be used. Concept / Approach: A clamper uses a diode and capacitor (and sometimes a resistor) to force one reference point of the waveform (e.g., its peaks) to a chosen dc level. During one half-cycle, the capacitor charges to a peak-related voltage; during the other half, it adds or subtracts this stored voltage, effectively translating the entire waveform up or down. Step-by-Step Solution: Choose reference (e.g., clamp to 0 V or to a chosen bias).Orient diode so the capacitor charges to the appropriate peak value.During the opposite half-cycle, the capacitor’s voltage adds algebraically, shifting the waveform’s dc level.Select time constants so that discharge is minimal over a cycle to preserve peak-to-peak amplitude. Verification / Alternative check: Oscilloscope observation shows the same peak-to-peak swing but a different vertical offset versus ground. Simulations confirm that varying the reference (or adding a dc source) shifts the waveform accordingly. Why Other Options Are Wrong: Limiters clip amplitude; they do not merely shift dc. Peak detectors follow the envelope’s maxima; they do not translate the entire waveform. DC converters change voltage magnitude or polarity (regulation/translation), not simply the waveform’s offset. Slicers threshold signals rather than shift their dc level. Common Pitfalls: Choosing a capacitor too small (causing droop) or a resistor that discharges too quickly, both of which distort the waveform and defeat pure level shifting. Final Answer: clampers

Question 4

An instrumentation amplifier used for precision sensing is specifically designed to exhibit a very high value of which performance metric, enabling accurate extraction of small differential signals in the presence of large common-mode voltages?

Accepted Answer

Introduction / Context: Instrumentation amplifiers (in-amps) are the front-end choice in medical, industrial, and measurement systems precisely because they can reject large common-mode voltages while amplifying tiny differential signals. The figure of merit that quantifies this capability is the common-mode rejection ratio (CMRR). Given Data / Assumptions: Differential input signals can be microvolts to millivolts. Common-mode voltages may be tens of volts or more. Balanced resistor networks and topology are optimized for high CMRR. Concept / Approach: CMRR is the ratio of differential gain to common-mode gain. A high CMRR means the amplifier strongly suppresses any voltage common to both inputs (such as interference or ground shifts) while accurately amplifying the difference. Three-op-amp in-amps use matched resistor networks to achieve orders of magnitude better CMRR than simple differential amplifiers built from unmatched parts. Step-by-Step Solution: Recognize the desired attribute: reject common-mode, pass differential.Identify the metric: CMRR = Ad / Acm (often specified in dB).Relate to design: precision resistor matching and topology maximize CMRR.Conclude: an in-amp is characterized by very high CMRR. Verification / Alternative check: Datasheets list CMRR values (often 90–120 dB or higher), directly demonstrating the in-amp’s strength versus general-purpose op-amp stages. Why Other Options Are Wrong: Output impedance is deliberately low, not high, for driving loads. Power gain and supply voltage are application-dependent and not the defining features. Slew rate is important for speed but does not address common-mode error rejection. Common Pitfalls: Neglecting resistor tolerance and layout symmetry can degrade practical CMRR dramatically even when the IC specification is high. Final Answer: CMRR

Question 5

In an isolation amplifier that transfers information across an isolation barrier (for example, via a transformer or capacitive coupling), what is the primary role of the oscillator block at the input side of the system?

Accepted Answer

Introduction / Context: Isolation amplifiers provide galvanic isolation between input and output for safety and noise immunity. Because transformers and capacitors block dc, isolation systems typically translate the input signal into an ac form, send it across the barrier, and then recover a proportional output on the far side. Given Data / Assumptions: Isolation barrier uses a transformer or capacitors. Input of interest may be a slowly varying or dc signal. System contains modulator/oscillator, isolation element, and a demodulator/rectifier. Concept / Approach: The oscillator generates a carrier, generally at high frequency, onto which the input (even if dc) is modulated. High frequency is preferred to minimize transformer size, improve bandwidth, and ease filtering. After crossing the barrier, synchronous demodulation and filtering recreate a dc (or low-frequency) representation of the original signal. Step-by-Step Solution: Create carrier: oscillator produces high-frequency ac.Modulate input: encode the input amplitude onto the carrier.Isolate and transfer: carrier passes through transformer/capacitive barrier.Demodulate/rectify: recover proportional dc on the output side. Verification / Alternative check: Block diagrams in isolation-amplifier datasheets show a chopper/modulator feeding a transformer, followed by a demodulator and filter, confirming the oscillator’s role in converting dc to ac for transmission. Why Other Options Are Wrong: Low-frequency ac increases size and reduces bandwidth. Rectification is performed after the barrier, not by the oscillator. Producing dual-polarity rails is a power-supply function, not the oscillator’s primary role. Adjusting input common-mode range is handled by front-end design, not the oscillator. Common Pitfalls: Confusing isolation power converters with signal-path modulators; although both may use oscillators, the signal path’s primary goal is information transfer, not power delivery. Final Answer: convert dc to high-frequency ac

Question 6

When an operational transconductance amplifier (OTA) is configured as a Schmitt trigger, which control quantity sets and therefore determines the upper and lower trigger thresholds (by adjusting the device transconductance gm)?

Accepted Answer

Introduction / Context: An operational transconductance amplifier (OTA) converts an input voltage into an output current with transconductance gm. In many OTA implementations, gm is programmable via a bias current IBIAS. Using an OTA in a Schmitt-trigger (hysteretic comparator) arrangement exploits this controllability to set switching thresholds. Given Data / Assumptions: OTA exhibits gm proportional to a bias current IBIAS over its operating range. Schmitt-trigger topology adds positive feedback to create hysteresis (distinct upper and lower thresholds). Device is operated within its linear/controlled gm region for predictable thresholds. Concept / Approach: In an OTA-based Schmitt trigger, the effective loop gain and the translation between input voltage and output current depend on gm. Since gm ∝ IBIAS, changing IBIAS scales the currents in the feedback network, thereby moving the input voltages at which the circuit toggles. Iout is a result of gm * Vin; it does not independently set thresholds. Vout follows the comparator/regeneration behavior and likewise does not directly control thresholds; it is the outcome of the gm-controlled loop. Step-by-Step Solution: Express OTA current: Iout = gm * Vin.Relate gm to bias: gm = k * IBIAS (k depends on process and topology).Positive feedback creates two switching points proportional to gm.Adjust IBIAS to scale gm → thresholds shift accordingly. Verification / Alternative check: Device datasheets and application notes show threshold equations where IBIAS appears explicitly, confirming that altering IBIAS repositions the Schmitt trigger’s upper and lower trip points. Why Other Options Are Wrong: Iout is a dependent variable set by gm and input; it does not independently program thresholds. Vout is the end result of crossing the threshold; it does not set it. “Both Iout and IBIAS” overstates the role of Iout; the controllable parameter is IBIAS. Supply voltage alone does not determine hysteresis thresholds in an OTA-based design. Common Pitfalls: Assuming thresholds are fixed by passive resistors only; in OTA designs, gm (and therefore IBIAS) is a powerful knob for tuning hysteresis without changing resistors. Final Answer: the IBIAS

Question 7

Isolation amps are named as such because they are sealed for protection from outside environmental damage.

Accepted Answer

False

Question 8

Clampers are often called peak detectors.

Accepted Answer

False

Question 9

OTA stands for Operational Transistor Amplifier.

Accepted Answer

False

Question 10

Typical isolation amplifiers use optical coupling.

Accepted Answer

False

Question 11

One of the principal areas of application for an isolation amplifier is power plant instrumentation.

Accepted Answer

True

Question 12

The main purpose of an instrumentation amplifier is to amplify small signals riding on large common-mode voltages.

Accepted Answer

True

Question 13

An isolation amplifier provides ac isolation between input and output.

Accepted Answer

False

Question 14

An OTA is primarily a current-to-voltage amplifier.

Accepted Answer

False

Question 15

To construct an instrumentation amplifier, two op-amps and seven resistors are needed.

Accepted Answer

False

Question 16

The external resistor of a typical instrumentation amplifier is used to set the ________.

Accepted Answer

gain

Question 17

A voltage-to-current converter is used in applications where it's necessary to have an output load current that is controlled by _________.

Accepted Answer

input voltage

Question 18

A Schmitt trigger has _________.

Accepted Answer

three trigger points

Question 19

The ________ amplifier is ideal for amplitude modulation since its gain is current-controlled.

Accepted Answer

operational transconductance

Question 20

A(n) ________ amplifier is used if the application requires a high sensitivity to low-level signals and protection from high-voltage associated circuits.

Accepted Answer

isolation

Special-Purpose Op-Amp Circuits Questions