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Comment by dragontamer | Hacker News Reader

What's the use-case here? That's the biggest optoisolator I've ever heard of.

Someone out there wants a lot of power pushed through electrically isolated islands. Normally, you'd have power on both sides but just send a signal (lower-power light + sensor to detect the light). It seems odd to use this technique to transmit power (Albeit a small 0.5W worth of power or so, but that's enough to run most microcontrollers and even some microprocessors).

My guess is some kind of optoisolator situation except you weren't allowed to run power on the other side for some strange reason. But I'm having difficulty thinking of a practical application where these requirements pop up.

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I've worked with these exact parts on a few designs for a large government institution. Exactly as you suspect: couldn't run power/conductive lines to the device on the receiving end for physics reasons. But, did work surprisingly well for the incredibly impractical use case.

Products like [1] an oscilloscope probe with ±60,000V isolation.

[1] https://www.tek.com/en/products/oscilloscopes/oscilloscope-p...

Why would an oscilloscope probe ever draw more than a few mW or even microwatts?

I'd expect a normal optoisolator module to work for a probe like that. No need for 180mA+ transferred between circuits.

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An oscillator probe is an example where electrical isolation + signal (aka: communications) are useful. But under no circumstances should 500mW of power be sent through such a setup in either direction.

The amplifiers in high-speed active probes can use nontrivial amounts of power. I don't have specific numbers as the probe/scope interfaces are proprietary, but I wouldn't be surprised if some of the wired (non-isolated) probes draw multiple watts.

> I'd expect a normal optoisolator module to work for a probe like that.

Absolutely not. "Standard" optoisolators tend to be slow and have analog characteristics which are unsuitable for an oscilloscope.

Active probes can draw several watts. Probably shouldn’t, but no one is optimizing them for efficiency

IIRC, leaker and worse-power BJTs handle higher-power and have lower noise. (Though Google says JFETs have even lower noise).

But the numbers discussed here is a lot more power than I was expecting. I'm guessing that at the GHz-range, everything uses more power though.

Opto-isolators don't transmit power per se. Since they use a phototransistor, it's more like they allow power to flow through from one pin to another, not provide a voltage across the pins.

Does fiber have significant transmission loss? I wonder how does that stand in comparison to typical power transmission over metal wires for the same distance.

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