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Tool/software:
Hi,
I believe that I have detected a mistake in the RM46L852 documentation. The following pictures show this mistake:
On page 1941 of User guide
On page 1942 of User guide
On page 23 of Datasheet
On page 17 of TMDXRM46HDK schematic
According to the documentation, it is necessary to add a signal level shifter (5 V -> 3.3 V) to avoid damaging the USB_VBUS pin. However, this recommendation it is not reflected in both schematics. So, I consider that there is a mistake. The 100 kohm resistor protects the pin of a high current, but I am not sure if a voltage of 3.3V can be obtained with this resistor. I would like to know your opinion about this, it is possible that I have misinterpretation all this.
Best regards,
Francisco.
Hi Francisco,
Apologies for the delay in the late response, i was stuck with other issues in this mean time.
Last time i didn't verify the schematic carefully, however now i found following things after schematic verification.
1. In actual schematic this pin is not directly connected to the controller pin.
In between there is a D-MUX "SN74CBTLV3257RGYR".
2. And after verifying its datasheet i found that it can easily support upto 4.6v
3. At the same time, it will take some input current on that pin right, so this will again create some voltage drop at 100K ohm resistor, i think this will be sufficient to protect this D-Mux and any way controller will not have any affect.
--
Thanks & regards,
Jagadish.
Hi Jagadish,
The information that you sent me is not safe if there is a risk of 5V voltage on this pin. It is true that there is a SN74CBTLV3257RGYR between MCU and USB connector. However, the V_BUS (ATTACH) signal is 5 V so is higher that the absolute maximum value of the SN74CBTLV3257RGYR (4.6 V). In addition, it is never safe to work with absolute maximum ratings of a device.
However, if we consider the pin such as a current source of 20 uA (as appears on page 23 of the datasheet), with just this resistor we can get a signal of 3 V. It is correct? Is this the reason why Texas Instrument implemented it like this?
On the other hand, I am not using this EVB without that I am developing my PCB and I detected this bug and searched information to verify this problem. Therefore, I think that a possible solution is use a resistor divider as it was commented in the post https://e2e.ti.com/support/microcontrollers/arm-based-microcontrollers-group/arm-based-microcontrollers/f/arm-based-microcontrollers-forum/306015/external-pull-down-resistor-on-usb-vbus/1066565?tisearch=e2e-sitesearch&keymatch=%22USB_VBUS%22%20AND%20%22external%20signal%20level%20shifter%22#1066565
, but the links not work. Or using a signal level shifter such as SN74LV1T32 to pass 5 V to 3.3 V.
Best regards,
Francisco.
Hi Francisco,
Apologies for the delay in the response, i was stuck with other issues.
5 V so is higher that the absolute maximum value of the SN74CBTLV3257RGYR (4.6 V). In addition, it is never safe to work with absolute maximum ratings of a device.
I agreed with you, and you are totally correct about this.
However, if we consider the pin such as a current source of 20 uA (as appears on page 23 of the datasheet), with just this resistor we can get a signal of 3 V. It is correct? Is this the reason why Texas Instrument implemented it like this?
I don't think they consider this because this current is sink from the SN74CBTLV3257RGYR, not from external USB right?
And the sink current for D-Mux is just 1uA only.
So, the voltage drop at 100k resistor could be 0.1v approximately. So, 4.9v(5-0.1) will get pass to the D-Mux and which is higher than its absolute max rating.
I think level shifter will be required here, i think it is missing in the original circuit. So, my suggestion would be better to consider level shifter if you are designing custom board.
--
Thanks & regards,
Jagadish.
Hi Francisco,
I mention this because I have tested the RC circuit (100 kohm and 0.1 uF) connected directly to the USB_DET signal (without SN74CBTLV3257RGYR, that is, using the LAUNCHXL2_RM46) and I get 3.1 V.
I am not aware of that you connected directly, i thought you are also using TI-board.
Then it is correct, i mean we will get ~3.1v only because:
Fixed 20 uA pulldown means it will take 20uA current in nominal conditions, please refer below thread once:
So that means the voltage drop across the 100K resistor will be 2v (100K * 20u) approximately that means the voltage drop at the controller pin will becomes ~3v approximately (5 - 2).
And microcontroller can tolerate upto 4.6v on input i/o pins:
So, in this case we don't get any issues.
--
Thanks & regards,
Jagadish.
