# Under what conditions does an optocoupler work fastest?

I want to use the optocoupler FODM452R2 in my design. I want this optocoupler to run as fast as it can (i.e.; work with minimum propagation delay and least rise/fall times). What is the optimum R1, R2 resistor values and magnitudes of V1, V2 voltages?

I know that when BJTs saturate they turn off very slowly, so I have to avoid saturation. I also know that BJTs don't respond fast when the resistor value is too high in common emitter configuration. So there is a trade off. The value of R2 must be chosen in a way to optimize the BJT's switching speed.
Is there an optimum value for R1 as well? Or, is the smaller the better without damaging the LED? On the other hand, does the value of V2 have any effect on the switching performance?

The datasheet doesn't directly answer my question, or the answers are encoded in it in a way I can't understand. It only gives the variation in propagation delay with ambient temperature.

simulate this circuit – Schematic created using CircuitLab

• It is also worth asking if a pulse transformer would be a better way of preserving the waveform characteristics you need. – Brian Drummond Nov 2 '14 at 21:54
• Phototransistor optocoupler tend to be sluggish, especially if you actually want some output voltage change, as Brian notes. I don't find 'fast as possible' to be a useful quantitative specification. – Spehro Pefhany Nov 2 '14 at 23:26

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.

I actually have some relevant experience on this very subject. Many, many years ago I grabbed a bunch of PS2505 optoisolators which turned out to be PS2506s. No big deal right? It turns out the PS2506s are INCREDIBLY slow compared to the PS2505s. My friend and mentor, Don Shepherd, gave me this sage advice.

Choose R1 so that about half of the available collector current (from the opto) flows through it. This ensures that the PNP transistor won't turn on from noise but still leaves enough current for the transistor to do its work. The available collector current for an optoisolator is found in the opto's datasheet as the Current Transfer Ratio (CTR). It defines the amount of collector current generated as a result of the LED current.

Since the voltage across the B-E junction of a transistor is about 0.7V, this means there will be 0.7V across R1.

This circuit speeds up the optoisolator by keeping the phototransistor collector voltage change to a minimum. Any collector voltage change comes with its a proportionate collector current change and this current is capacitively coupled back to the internal base of the optoisolator's transistor via parasitic capacitance present in all devices. The coupled collector current opposes the photocurrent coming from the diode and thus slows down the opto switching time due to "infighting." By using the PNP transistor we keep the opto's collector voltage change to a minimum and that way as much of the photocurrent as possible goes into turning the phototransistor on instead of fending off the collector current.

For my specific problem, the PS2506 datasheet states that the CTR is 80% to 600%, with 300% being typical. This is measured with If at 5mA and Vce at 5V. Plugging these values in, I see 3.52V across a 476 ohm resistor in series with the opto's LED, so If is 7mA. On the other side of the opto, Ic = If * CTR, so 300% of 7mA is is 21mA (typically).

With 0.7V across R1: if I want 10mA of current that makes R=V/I = 0.7 / 10.5 or 67 ohms.

I built this circuit and measured the pulse rise and fall times across the 100 ohm resistor at about 100ns. That's quite a speedup without this helper transistor.

BTW: The difference between the PS2505 and PS2506 is that the latter uses a photodarlington transistor. Since darlington transistors have such insanely high gain I am guessing that between the more complex structure of the darlington transistor (higher parasitic capacitance) and higher gain, the PS2506 spent much more of its available current fighting the parasitics.

• Interesting +1. I wonder if a cascode arrangement would be even better. – Spehro Pefhany Nov 3 '14 at 3:35