Asymetrical square wave or rectangle wave amp op
Intro

I hadn't really considered this before and now that I have thought a little further I think it's actually easier than I'd suggested in the comments. You could go with an added series resistance to the output, as I suggested. Or the approaches that have been also added here (an offshoot of what I was earlier thinking about.)

But it's not needed.



Basic idea

Here's the basic idea (I'm assuming you will use an rail-to-rail in/out opamp such as the LT1800, for example):







simulate this circuit &ndash; Schematic created using CircuitLab

To get a duty cycle of \$\frac13\$rd, you want the average (mean) current in \$R_4\$ to be twice as much when the output of the opamp is HIGH than when the output of the opamp is LOW. In effect, this means that the difference between the opamp's \$V_{\text{OUT}_\text{HIGH}}\$ and the average of \$V_{\text{H}_\text{LOW}}\$ and \$V_{\text{H}_\text{HIGH}}\$ must be twice as much as the difference between the opamp's \$V_{\text{OUT}_\text{LOW}}\$ and the average of \$V_{\text{H}_\text{LOW}}\$ and \$V_{\text{H}_\text{HIGH}}\$.

Or,

$$V_{\text{OUT}_\text{HIGH}}-\frac{V_{\text{H}_\text{LOW}}+V_{\text{H}_\text{HIGH}}}{2}=2\cdot\left(\frac{V_{\text{H}_\text{LOW}}+V_{\text{H}_\text{HIGH}}}{2}-V_{\text{OUT}_\text{LOW}}\right)$$

Since \$V_{\text{OUT}_\text{LOW}}=0\:\text{V}\$ and \$V_{\text{OUT}_\text{HIGH}}=V_\text{CC}\$ for any similar circuit, regardless of \$V_\text{CC}\$, it follows that:

$$V_{\text{H}_\text{LOW}}+V_{\text{H}_\text{HIGH}}=\frac23 \:V_\text{CC}$$

You also should be able to work out the following two equations:

$$\begin{align*}
V_{\text{H}_\text{LOW}} &amp;= V_\text{CC}\frac{R_2\:R_3}{R_1\:R_2+R_1\:R_3+R_2\:R_3}\\\\
V_{\text{H}_\text{HIGH}} &amp;= V_\text{CC}\frac{R_2\:\left(R_1+R_3\right)}{R_1\:R_2+R_1\:R_3+R_2\:R_3}
\end{align*}$$

Knowing all of the above, you could choose to specify \$R_3\$ and \$V_{\text{H}_\text{LOW}}\$ and then work out the details for the required values of \$R_1\$ and \$R_2\$ (and, obviously, the resulting \$V_{\text{H}_\text{HIGH}}\$, too.)



Addendum

As some time has passed now, I'll provide some equations:

$$\begin{align*}
R_1&amp;=2\:R_3\cdot\frac{V_\text{CC} - 3\:V_{\text{H}_\text{LOW}}}{3\:V_{\text{H}_\text{LOW}}}\\\\
R_2&amp;=2\:R_3\cdot\frac{V_\text{CC} - 3\:V_{\text{H}_\text{LOW}}}{V_\text{CC} + 3\:V_{\text{H}_\text{LOW}}}\\\\
V_{\text{H}_\text{HIGH}}&amp;=\frac23\:V_\text{CC} - V_{\text{H}_\text{LOW}}
\end{align*}$$

From these, it is easy to see that \$V_{\text{H}_\text{LOW}}\lt\frac{V_\text{CC}}{3}\$. So your choices for \$V_{\text{H}_\text{LOW}}\$ are constrained.