diodesstep-downvoltage
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.
That is not a diode but an thermistor, a resistor that has a it's resistance value dependent on temperature. This is common with Nickel Cadium battery where the temperature of the battery pack its used in the charging algorithm for this battery chemistry
Edit:
Here a link to digikey showing a NTC thermistor in the same glass package
Also, this is another example of the thermistor in the glass packaging also used for charging on NiCd batteries that I have on hand.
This is a classic saturated switching application. Your schematic is correct. In a saturated BJT switch there is a rule of thumb that the base current should be chosen to be around 1/10 of the collector current. This is sometimes referred to as using a forced beta of 10.
A forced beta of 10 is high enough to make sure that a typical BJT will be in saturation, and Vce will be low. Either of the suggested transistors will work. If memory serves, the 2222 is a bit better in saturated switching applications.
Since you know the collector current will be around 28 mA, you can pick R1 so that the base current will be around 2.8mA. Your formula for calculating the base resistor is correct. Just plug in 2.8mA instead of 15mA.
If you use one IO pin to supply multiple BJT's, make sure the total current is well under the limit. I would suggest that you use no more than 3 or 4 BJT's per IO pin in order to make sure the IO is not stressed. If you read the fine print, I am sure the IO pin cannot supply 22mA and stay close to VCC. Probably it is 22mA with Vout >= 2.4V or something like that.
If you use N-channel MOSFET's instead of BJT's, you can put all 8 on one IO pin. The BSS138 might be a good choice if you go that route.
Best Answer
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.