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Amplifier output headroom is the electric potential difference reserved between a usable output limit and its corresponding supply rail. If the supply potentials are $V_{-},V_{+}$ and the specified amplifier output range is $[V_{\mathrm{lo}},V_{\mathrm{hi}}]$, defineThese two margins can differ and depend on load and operating conditions. The switched-integrator datasheet lists supply conditions, output range and output-current limits separately.^[ref]

For an original chosen $0$–$5\mathrm{V}$ circuit with usable limits $0.25$ and $4.55\mathrm{V}$, the lower and upper margins are $0.25$ and $0.45\mathrm{V}$. A signal centered at $V_0$ has maximum symmetric peak excursionAt $V_0=2.50\mathrm{V}$ this is $2.05\mathrm{V}$. Centering instead at $(V_{\mathrm{lo}}+V_{\mathrm{hi}})/2=2.40\mathrm{V}$ permits $2.15\mathrm{V}$. Thus centering on the supply midpoint need not maximize the usable symmetric swing.

An output sweep into the actual load identifies where the specified gain or distortion tolerance fails. Changing the load tests whether the limit comes from output-current demand. These measurements concern static headroom; a fast pulse also tests frequency response and overload recovery. More final-stage headroom does not cure an internal violation of the amplifier input common-mode range.

High-side current sensing measures an electric current with a sensor between the supply and the load, rather than between the load and its return reference as in low-side current sensing. With a shunt resistor, the useful signal is the difference between two input potentials that may both lie far above the amplifier's local reference. The amplifier input common-mode range therefore matters independently of the differential signal size.^[ref]

For an illustrative12volt supply, a50milliohm shunt resistor carrying2amperes developsThe sense terminals are at12.0 and11.9volts relative to the supply return, so their common-mode voltage is11.95volts. An ideal differential amplifier of gain20 would convert the100millivolt difference to2volts. A3.3volt output or digitizer range does not by itself establish that the amplifier inputs can accept11.95volts: that requires a suitable input stage and operating specifications.

The LTC6115 demonstrates an actual high-side arrangement: an external shunt resistor generates a differential electric potential difference, and external resistors set its conversion to an output signal. Its separate current-sense and voltage-sense supply conditions must not be conflated.^[ref]

Keeping the load return connected directly to the reference avoids the reference shift introduced by a return-path shunt resistor. The supply-side drop still reduces the available load electric potential difference. Separate sense connections implement four-terminal sensing; measuring through shared high-current tracks instead adds their drops to the inferred electric current.^[ref]

An instrumentation amplifier amplifies a differential input while presenting high input impedance and rejecting a common component. A useful ideal relation isIts common-mode rejection ratio describes finite unwanted response to common-mode voltage. This is distinct from the amplifier input common-mode range over which the circuit can operate without internal saturation.^[ref]

For an original example, stipulate $v_{+}=2.501\mathrm{V}$, $v_{-}=2.499\mathrm{V}$, $G=100$ and $V_{\mathrm{ref}}=2.5\mathrm{V}$. The differential signal is $2\mathrm{mV}$ and the ideal output is $2.7\mathrm{V}$, assuming all input and output limits are satisfied.

Now suppose the differential-to-common gain ratio is $10^4$, or $80\mathrm{dB}$. The common-mode gain is then $100/10^4=0.01$. A $1\mathrm{V}$ common-input change contributes $10\mathrm{mV}$ at the output, five percent of the desired $0.20\mathrm{V}$ signal contribution. A high rejection ratio therefore does not mean zero error.

A source that changes both inputs together tests rejection; a source that changes their difference tests differential gain. Both tests must respect input common-mode limits, amplifier output range, load and frequency response. A plausible ideal output value alone does not prove that every internal stage remains linear.

Low-side current sensing places a sensor between a load and its return reference. For a shunt resistor, the electric potential difference across the shunt resistor is proportional to the returning electric current. The readout inputs can operate near the return potential, but the load's local return is raised by the burden voltage. High-side current sensing instead places the sensor toward the supply.^[ref]

For an illustrative12volt supply, a6ohm load and a50milliohm return-path shunt resistor, Ohm's law givesThe load return is about99millivolts above the source return, and the load itself receives about11.901volts. These potentials matter if a second instrument or communication cable connects the load return to another reference. A parallel return path can bypass the shunt resistor, so the measured branch electric current need not equal the total load electric current.

Pete Semig's reference design provides a concrete apparatus: a25ohm shunt resistor, an instrumentation amplifier and a single5volt supply measure a specified range from10microamperes to10milliamperes. Its circuit analysis and simulated responses separate the ideal current-to-output relation from practical offset and low-current errors.^[ref]

For a new setup, recording a zero-current output and several known electric currents tests offset and gain. Four-terminal sensing helps keep high-current connection drops out of the sensed electric potential difference; an amplifier input common-mode range extending to the required near-return potentials is still necessary. Neither a near-zero common-mode voltage nor a successful gain check proves that the return-path insertion leaves the load undisturbed.