0
\$\begingroup\$

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).

\$\endgroup\$
  • \$\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
1
\$\begingroup\$

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:
illustration

\$\endgroup\$
  • \$\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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.