Yeah, they've been doing that for a while; if you wanted to do it yourself, the NRE probably sucks, but evidently it comes down pretty well in quantity. Examples:
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RECOM's medical grade isolators are quite good as well, though I forget if they're quite 2pF good. May also be that I was using the 5W model, so of course the capacitance scales with that.
The best I've seen (probably also the best possible), is in the high-CMRR isolated scope made by, these guys I think it is:
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Dave Jones @eevblog caught them at a trade show and got a look inside the box, they're well engineered indeed. At least as good as anything comparable from the big brands, and at a fraction of the price!
The higher-voltage ones (for gate drivers) say 2.5pF.
I've been using SIP-style parts, less space. Mornsun QA01C, with +20, -4 volts out (I use it at +16, -3V). Their spec is 3.5pF, so I'd save 30%. IPD's IG120-20 is a good USA replacement. Murata's MGJ2 family has similar low capacitances.
Yikes, that'll leave a mark. How much current does it take to get it into regulation?
Cheers
Phil Hobbs
--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC / Hobbs ElectroOptics
Optics, Electro-optics, Photonics, Analog Electronics
Briarcliff Manor NY 10510
http://electrooptical.net
http://hobbs-eo.com
My board (all 0.9 square inches of it) was beautifully packed and routed when I measured the dc/dc thing just to be sure. Consternation. Adding a 1K 0603 resistor was tough.
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The 9V would have damaged things.
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John Larkin Highland Technology, Inc
lunatic fringe electronics
Yes, I thought so too. But for BJT's, the saturation voltage depends on many variables such as beta, drive level, temperature and so on. So it's a very unreliable test.
I think a much more precise test is to put a small resistor in the current path and measure the voltage across it. The patent claims this wastes too much power, but this is false.
For example, if you want to set the current limit to 150 mA, a 1 Ohm resistor will produce 0.15 Volt at the limit. This is 0.15 Watt. However, if the source supply is 24 Volt, the power delivered to the load is 3.6 Watts at the current limit.
The ratio is 4.16%. So the current monitoring resistor takes a negligible fraction of the total power, and Murata's claim is false. Not only that, it ignores the power dissipated in the transistors at maximum load, which is probably equal to or greater than the power lost in the current sense resistor.
Of course, you can always add foldback to the current limit and drop the power dissipation to low values.
Or you could use cycle-by-cycle limiting and shut off the drive as soon as the limit is reached.
So you have to take many of these patent claims with a grain of salt.
It short-cuts all that complexity and inferrs the transistor's power dissipation, so it accomplishes the goal of protecting the transistor.
Current doesn't kill transistors; power does. A saturated transistor doesn't dissipate much power.
The part I posted about doesn't seem to do this, and has not very good shorted-load protection. Luckily, my load is unlikely to short.
It's interesting that these sorts of dc/dc bricks still use discrete parts, especially those ones with discrete transformer windings. You'd think someone would spin an IC, which could include an oscillator and proper protections and other features.
--
John Larkin Highland Technology, Inc
lunatic fringe electronics
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