Determining the output sine voltage using transfer characteristics

Question

Consider the following problem:

This question is tricky,

Here is my attempt:

$$v_o=3+12v_i \to \frac{v_o}{v_i}=\frac{3}{v_i}+12=A_v$$ where \$A_v\$ is the voltage gain in V/V.

We can essentially extract the saturation levels of \$v_i\$ from \$v_o\$ as follows:

$$-3=3+12L- \to L- = -0.5\text{ V}$$

$$4.2=3+12L+ \to L+ = 0.1\text{ V}$$

We know that the input voltage is bounded by $$L- / A_v \leq vi \leq L+/A_v$$

Therefore, if we replace \$v_i\$ by \$L+/A_v\$ in \$(3/v_i)+12=A_v\$ we get \$A_v=-0.413\$, yet if we replace \$v_i\$ by \$L-/A_v\$ we get \$A_v = 1.714\$.

Why did we get two different values for the voltage gain \$A_v\$? How can we proceed to solve this problem?

Answer 1

I think you way over-thought this!

The output voltage cannot be greater than +4.2V, or less than -3.0V.

That means that the largest sinusoid that can appear at the output must fit within those extremes. This corresponds to a peak-to-peak excursion of \$(+4.2V) - (-3.0V) = 7.2V\$.

That's peak-to-peak, which corresponds a sinusoid of amplitude \$\frac{7.2V}{2} = 3.6V\$. In other words, if \$D=3.6V\$, then the function

$$ v_O = C + D \space sin(t) $$

will produce a sinusoid which extends up to \$C+D\$ and down to \$C-D\$. You could find a value for C that fits \$v_O\$ exactly between +4.2V and -3.0V, but the question doesn't need you to find that! All you were asked for is a value for D which produces a sinusoid whose maximum-minus-minimum is 7.2V.