# Interpreting optical isolator datasheet

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?

Thanks much.

ts and td look like propagation delays, as such they don’t effect the freq just the phase and sometimes the duty cycle.

I would assume the freq spec was measured at the -3db point which gives ~4kHz.

Rob

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.

Here we go. I found the datasheet: http://www.vishay.com/docs/83510/tcmt1100.pdf

t(d) is "delay time" and t(s) is "storage time". t(on) is t(d) + t(r), and t(off) is t(s) + t(fall).