Hello Sina S. ,

I am not sure I understand your question. A PC USB connector provides +5 VDC at a certain current. For example, the USB hub inside the computer might provide 500 mA total current to be shared amongst all the USB ports on the computer. If your computer had 5 USB ports with each connected to a USB device, then the total current drawn by these 5 USB devices cannot exceed 500 mA.

I don't know why you are referring to the USB port's charge.

Regards,

Tom Cuff

you can see the following links:

1. https://www.pcworld.com/article/420146/how-to-keep-charging-your-usb-devices-past-your-pcs-bedtime.html

2. https://www.cmd-ltd.com/advice-centre/usb-chargers-and-power-modules/usb-and-power-module-product-help/usb-charger-faqs/

Hi Sina S. ,

if you mean by "charge" voltage and current used to power the connected devices (whose purpose might include recharging of some accumulators) then please have a look at the examples starting at page 511 of the attached file.

As you can see in the examples, the usual way to control the programmable power supply is via requests by the attached devices. So, you had to implement the USB protocol or to modify the program of an existing device (if you can get hold of the source code).

Thomas Cuff It cannot exceed, but it can be reduced...

Hello Sina S. ,

You can exceed the current limit of the USB hub, but once you exceed the compliance level of the voltage source, the voltage source's nominal voltage will fall below +5 VDC due to the internal resistance of the voltage source. As far as reducing the current limit, I am not sure of what the point of that approach is. I mean, whatever load you attach to the individual USB port, the load will try to draw the current it needs, i.e., if the load needs 10 mA that is what it will try to draw from the voltage source. What would be the point of setting the current limit of the USB hub to 5 mA, if the load needs 10 mA? I don't understand what you are trying to accomplish by reducing the current limit. If you are trying to charge an accumulator, the accumulator (secondary battery) will automatically reduce its current demand as it begins to charge up. Can you explain your application, please?

Regards,

Tom Cuf

Hi Thomas Cuff ,

I'm joining you in being interested in Sina's application, too. (But if it is something basically new she might be reluctant to tell us at this stage.)

As for the meaningfulness of her approach, I think you are right as far as USB 1, 2, 3, BC, and C are concerned. But with USB-PD, the source can deliver up to 20 V, 5 A, and even the cables (i.e. ICs within the plugs) can join the communication (SOP'/SOP" communication, please see page 48 of the file I attached to my first answer). So, we could have a situation where a device is currently powered with 5 V, 0.5 A, now asks for 20 V, 5 A but the cable tells the source that it is limited to 3 A, thus the source rejects the request of the device, and the device's next request (based on choosing a less power hungry mode) is for 20 V, 1.5 A which will be granted.

It would be nonsensical, of course, if the source granted a request but reduced tacitly the voltage and/or the current limit to less than the granted values.

if you talking about using PC to charge external devices then USB port on PC has no control over the charging characteristics of the device attached to the USB port, all it does is provides +5V and 500mA minimum and in some cases as low as 100mA. It's an external device responsibility to adjust the current drawn from USB port so as not to exceed a maximum allowed by PC current or it will trip internal overcurrent protection. Many charge controllers adjust charge current by monitoring USB voltage drop. If the cable used is not to USB specification you can have a significant voltage drop under load so make sure USB cable can handle desired charge current.