Question 12.16: Effect of Short-Circuit Current Limit on an Amplifier Determ...

Known Quantities: Short-circuit current limit, I_{\mathrm{Sc}}; amplitude of sinusoidal input voltage; amplifier closed-loop gain.

Find: Compute the maximum allowable load resistance value, R_{L \text { min }}, and sketch the theoretical and actual output voltage waveforms for resistances smaller than R_{L \text { min }}.

Schematics, Diagrams, Circuits, and Given Data: I_{\mathrm{SC}}=50 \mathrm{~mA} ; v_{S}(t)=0.05 \sin (\omega t) ; A_{V}=100.

Assumptions: Assume short-circuit-current-limited (otherwise ideal) op-amp.

Analysis: Given the closed-loop voltage gain of 100 , we compute the theoretical output voltage to be:

v_{\text {out }}(t)=-A_{V} v_{S}(t)=-5 \sin (\omega t)

To assess the effect of the short-circuit current limit, we calculate the peak value of the output voltage, since this is the condition that will require the maximum output current from the op-amp:

\begin{aligned}&v_{\text {out peak }}=5 \mathrm{~V} \\&I_{\mathrm{SC}}=50 \mathrm{~mA} \\&R_{L \text { min }}=\frac{v_{\text {out peak }}}{I_{\mathrm{SC}}}=\frac{5 \mathrm{~V}}{50 \mathrm{~mA}}=100  \Omega\end{aligned}

For any load resistance less than 100  \Omega, the required load current will be greater than I_{\mathrm{SC}}. For example, if we chose a 75-\Omega load resistor, we would find that

v_{\text {out peak }}=I_{\text {SC }} \times R_{L}=3.75 \mathrm{~V}

That is, the output voltage cannot reach the theoretically correct 5-\mathrm{V} peak, and would be “compressed” to reach a peak voltage of only 3.75 \mathrm{~V}. This effect is depicted in Figure 12.56.

Comments: The short-circuit current limit is listed in the device data sheets (see the data sheets in the accompanying CD-ROM, or in the device templates in the Electronics Workbench { }^{\mathrm{TM}} libraries, for examples). Typical values for a low-cost general-purpose amplifier (e.g., 741c) are in the tens of milliamperes.