The attached graphs are from the datasheet of an optical isolator ( https://www.sparkfun.com/datasheets/Components/LTV-8x6.pdf ). Looking at the "response time" graph, the total on/off cycle time (tr + tf + ts + td) for a 10 kOhm load appears to be approximately 130 + 130 + 10 + 10 us, or 280 us. This equates to a frequency of 3.571 kHz. However, looking at the "frequency response" graph, the rolloff of a 10 kOhm load appears to start around 1 kHz, and frequency response is approximately -2 dB at 3.5 kHz. Meanwhile, a 1 kHz = 0 dB rating suggests to me an on/off period of closer to 1 ms. How to reconcile this?
If I am designing a circuit using one of these chips, and I was going to put a low-pass filter on the load side of the chip anyway, does the fact that high-frequency signals are being attenuated really matter? Or can I maybe even use that to my advantage and skip the filter entirely?
Graynomad:
ts and td look like propagation delays, as such they don't effect the freq just the phase and sometimes the duty cycle.
I wasn't sure what those were, but I looked up the terms and found another device's data sheet that gave tr+ts+td+tf as the total cycle time. They were the "curves" before and after the ramp up, before the plateau.
I would assume the freq spec was measured at the -3db point which gives ~4kHz.
Got it. That makes sense then. That also explains why when I was playing with a frequency-calculator, the cutoff frequency I was selecting was not 0 dB, but somewhere slightly less than that.
