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I have a 6V input on my board. I need to put a load switch for USB 3.0 which needs 5V. I need to stepdown 6V to 5V using a series diode. 6V is going to other circuits also.

Can someone suggest the diode I can use & what parameters should I consider while selecting a diode.

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  • \$\begingroup\$ What is your minimum and maximum load current? \$\endgroup\$
    – EM Fields
    Commented Dec 21, 2015 at 9:15
  • \$\begingroup\$ Load current is for USB 3.0 i.e 900 mA. \$\endgroup\$
    – Oshi
    Commented Dec 21, 2015 at 11:17
  • \$\begingroup\$ You also need to know the minimum current your load might take. \$\endgroup\$
    – The Photon
    Commented Dec 21, 2015 at 15:59

3 Answers 3

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A diode won't do the job. The voltage drop will decrease as the current decreases leaving anything plugged in exposed to more than 5 V.

I suspect you need a LDO (low drop-out) voltage regulator with current limiting. The current limiting will protect your supply and anything else running on it.

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  • \$\begingroup\$ I agree. But in case I need to use a diode only, which one would it be? \$\endgroup\$
    – Oshi
    Commented Dec 21, 2015 at 11:17
  • \$\begingroup\$ I would use a couple of 1N4000 series diodes in series but that would drop 1.2 to 1.4 V at that current and might leave you on the low side of 5 V. They're rated at 1 A so you may wish for a higher safety margin too. Go through the common diode spec sheets - particularly the V-I charts and pay attention to the effect of temperature. \$\endgroup\$
    – Transistor
    Commented Dec 21, 2015 at 12:07
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We can also use Schottky diodes instead. That has forward voltage drop from 0.7 to 1V but be cautious it has a high leakage current.

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You can use a zener diode in parallel with your power source. You probably need a current limiting resistor so that you don't burn up your diode.

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