Hi, Thanks that Sebastien reminded me that those two USB host ports with total 2A specification which I indeed forgot this to consider. So that was 1.1 hold current fuse took.
> > Also I quite doubt that I can find a suitable Zerner to afford this > > time that diode doesn't fail within 10s. > > Use two of your previous zeners in parallel? > > some experimental results below on 2A PTC fuse and 2 * 5.1V Zener: http://en.qi-hardware.com/wiki/Protection_of_Reversed_Polarity_on_DC_plug-in#Sch._D 1, first table (No LOAD condition) - fuse and 2* 5.1V diodes in series. (without rc2 board load) ex. 12V inputs (12V - 5.1V) / 0.07 Ω = 98A @ transient, this won't be happened and limited by experimental power supply @ 2.85A max. drive out. When 2.85A ( > 2A holding current and < 4A trip current ) goes through PTC fuse, see p3 figure 4 of http://www.circuitprotection.com/catalog/fundamentals/PPTC_tech_brief.pdf "In region C, it is possible for the device to either trip or remain in the low-resistance state, depending on the individual device resistance and its environment.", after 11 seconds, the current descends to 0.09A. also the voltage on Zener two pads remains 5.127V. rc2 won't be booted. Safe though. With regards to why have 40mA I measured, pls see p.2 of http://www.onsemi.com/pub/Collateral/1N5333B-D.PDF be noticed that Zener acts as either linear under VZ or exponential curve under VF, so: a) any forwarding input voltage >= VF, Zener trys to keep normal diode action; < VF, said a '-5V' input to board, it must be have forwarding current 2.85A there. then this all huge current diverts through diode and fuse, a power dissipation on each Zener is (2.85/2) * 0.786 = 1.12 Watt which is better than only use 1 Zener(2.85A * 0.727A=2W). More safety to share dissipation. This won't damage original rc2 circuit design. Also it descends to 0.23A after 43 seconds. b) any forwarding input voltage >= VZ, Zener breaks down to try to keep a 5.1V stage; < VZ, said a '+5V' input, then there's few current( a 40mA got experimentally) as datasheet curve shown (Like IR). Of course this current acts by fuse thermal resistive factor too. There's no condition only "consume ZERO power. but instead of that you have 40mA through the diodes." :-) Surely that zener has probably been ageing enough to have such high 40mA reversed leakage current after my extremely high consumption(66.8°C) on it. :-) 2, second table (LOAD condition) - fuse and 2* 5.1V diodes in series. (with rc2 board) booted: @ 4.5 ~ 7V inputs unbooted: @ over 7.5V temperature measured on fuse & diode( one of two ) surface body: @ 4.75 ~ 5.25V inputs range fuse: 36.5 ~ 41.1 °C Zener: 39.9 ~ 46.3 °C ＠ 6 ~ 20V inputs range fuse: 42.9 ~ 57.2 °C, ------> 57.2°C @ 6V Zener: 50.6 ~ 66.8 °C, ------> 66.8°C @ 6V 3, from LOAD results, no matter what used a 1.1A fuse or 2A fuse, high temperature got on surface body on fuse and Zener especially @ 6V. This is caused by current stays while in Zener break down, as all experimental results shown any voltage inputs over VZ, an exponential curve works; so more watt consumed on zener. :-) Use a bigger [email protected] is not a bad idea though to lower consumption while 5.25V or 6V inputs. But don't forget that reviewing a 12V input result, there's 11 seconds remaining there when even using 5.1V. Of course we don't know if consumer usb device has over-protective parts on it even we insist on a [email protected]. Btw, for 5.6V Zeners, I've ordered today. :-) Will do test when I get. Next steps: * modify all solved h/w fixed in schematic, will send to list. * keep remaining parts sourcing. Thanks, -Adam
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