An amplifier has a linear transfer characteristic passing through the origin (0, 0) and with output voltage saturation at L+ = 10 V and L– = –8 V. The amplifier gain is 100 V/V. What is the amplitude (in mV) of the largest sine-wave input having no dc component that can be applied without causing output voltage distortion?

I tried to take the middle point (L+ - L-)/2 = 9, dividing it by the gain, but this only gets me the sinusoidal voltage assuming there is a DC component.

(The answer to this problem is 80 mV, but I'm getting 90mV).

  • \$\begingroup\$ So what if you assumed that the DC component must be zero? And don't forget your units in the answer. Is the correct answer 80 parsecs or something else perhaps? \$\endgroup\$ – Bimpelrekkie Jun 25 at 13:48
  • \$\begingroup\$ My bad it is in mV* \$\endgroup\$ – O. Sinno Jun 25 at 13:51
  • \$\begingroup\$ Shouldn't it be the same process even if it was 0? \$\endgroup\$ – O. Sinno Jun 25 at 13:52
  • \$\begingroup\$ You calculate the midpoint but then a DC voltage is needed to "shift" the voltages such that the peaks fit nicely between +10 V and - 8 V. Now what if no "shift" was allowed. How would you fit the sinewave (without a DC offset so midpoint remains 0 V) in a +10 V to -8 V window? Perhaps you should make a drawing on a piece of paper. \$\endgroup\$ – Bimpelrekkie Jun 25 at 13:57

The greatest possible amplitude of an undistorted output sine is:

min(|L+|, |L−|) = 8 V

And the amplitude of the corresponding input sine is therefore:

min(|L+|, |L−|) ∕ 100 V/V = 8 V ∕ 100 V/V = 0.08 V = 80 mV

The minimum (min) value of the absolute values of the supply rail voltages needs to be taken in order not to exceed the supply range and so to avoid clipping:

  • \$\begingroup\$ Thank you, but why do we take the minimum between these two? \$\endgroup\$ – O. Sinno Jun 25 at 14:04
  • \$\begingroup\$ Please see my edited answer... \$\endgroup\$ – aschipfl Jun 25 at 14:26

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