Amplifier output range ($V_{\mathrm{lo}}\le v_o\le V_{\mathrm{hi}}$)

The amplifier output range is the interval over which an electronic amplifier can provide the specified output behavior under stated supply, load and operating conditions. Its limits need not coincide with the power-supply potentials. The historical integrator datasheet lists output range and output-current conditions separately from its supply conditions.^[ref]

For an original linear model, let $v_o=V_{\mathrm{ref}}+G{v}_i$ with positive gain. Avoiding static clipping requiresStipulate a $0$–$5\mathrm{V}$ supply, usable output limits $0.2$–$4.6\mathrm{V}$, $V_{\mathrm{ref}}=2.0\mathrm{V}$ and $G=200$. The permitted differential input interval is $-9$ to $13\mathrm{mV}$. A positive $12\mathrm{mV}$ input predicts $4.4\mathrm{V}$ and fits; a negative $12\mathrm{mV}$ input predicts $-0.4\mathrm{V}$ and does not.

These are stipulated limits, not a product guarantee. The amplifier output headroom relative to each supply can differ. Loading can change the available output swing, and input-stage saturation can occur even when the final ideal output lies inside its range. A slow input sweep identifies the static departure from linear gain; a pulse test additionally probes dynamic recovery. Amplifier clipping describes the waveform consequence of exceeding these limits.

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.