Electrical – Optocoupler 4N25 saturated CTR

Page 134 of your datasheet: "DC Current Transfer Ratio" and "N" for "Normalized". This is the ratio between collector current and LED current. If you vary the current through the LED, the current through the transistor will vary too:

\$\boldsymbol{NCTR} = \frac{I_C}{I_{LED}}\$

According to figure 2 - 5, measured under the following conditions: \$\boldsymbol{V_{CE}=10 \text{V}}, I_F= 10 \text{mA}\$ and at various ambient temperatures (\$T_A\$)

and

\$\boldsymbol{NCTR_{SAT}} = \frac{I_C}{I_{LED}}\$

According to figure 2 - 5, measured under the following conditions: \$\boldsymbol{V_{CE}=0.4 \text{V}}, I_F= 10 \text{mA}\$ and at various ambient temperatures (\$T_A\$)

In the latter case the transistor is in saturation, this means the collector-emitter voltage across it is so low that the transistor is unable to further increase the current by changing its conductivity (due to more light from the LED).

The key to maximum speed appears to be Figure 8, the frequency response, plotted at different values of RL (R2 in your circuit), and the fastest is with RL=100 ohms, where full amplitude is maintained up to 1 MHz corresponding to rise and fall times below 0.5 us, correlating reasonably well with Figure 7.

Now at the conditions for measuring Fig8, the LED current is 16mA, so R1 can be calculated to give 16mA from your supply voltage and the LED Vf. Not especially critical or interesting.

But the current transfer ratio (p.4) is guaranteed to be between 20% and 50%, giving somewhere between 3mA and 8 mA through RL/R2, or 0.3 to 0.8V pk-pk. On a 5V supply such a low value of RL keeps the phototransistor well out of saturation, but needs a gain stage to amplify the signal developed across RL to recreate logic levels.

If that's not fast enough, a different device or different approach may be necessary.