As we know, a USB 2.0 device, once it has properly negotiated with the host, is allowed to draw 500mA.
Suppose that you are designing such a device, and your input current limiter chip is tuned by a resistor. Let's say that the resistors available to you are ±1% tolerance.
Should you size your resistor to draw 500mA in the nominal case, knowing that in the maximum case you will draw 505mA? Or should you size your resistor to draw 495mA, so that you are guaranteed to respect the exact limit, but less power is available to your device?
Best Answer
The current limit detection by a host mode device that emits current to a USB device may or may not be accurate. Here are the types of host devices that you are likely to find:
No USB host mode device designer wants to make a device that nuisance trips and causes grief for their users. For this reason it will be rare to find a port that supports USB power source cut off at 501mA. Instead the cutoff, if any, will be much higher from say 600mA and possibly up to the current limit of the internal 5V supply.
An interesting point is to open up the typical powered USB four port hub. It is not uncommon to find the internal circuit board designed to use the 5V port power switch parts. However likely as not the parts may not even be installed and instead be replaced by a direct connection to the 5V supply. The manufacturer may then use the same circuit board in several models of their USB hubs that are offered at two different price points. As you can imagine the USB hub market place is an extremely competitive one and every extra part in the unit lowers the profit made on each sale.