[PIC] Usb hid easypic7 pic18f4550

[salahd]

Hi,
According to Dogan Ibrahim book (Advanced PIC projects),i'm tryning to experiment the USB HID project. I used easypic7 to interface PIC18F4550 TO PC using USB.
The project run's ok in similation by using a GUI interface template EasyHid in VB6 and in VBexpress 2010, but in real hardware nothing changes in PORTB.
NB: the jumper J12 in right postion OF EASYPIC7
the sw3.2(PORTB) also in right position
I can see the device with HID Terminal.
My PC recognize the device.

Any help thanks a lot
Best regards
DAHOUATHI Salah.View attachment PROJECT_USB_DOGAN.rarView attachment PROJECT_USB_DOGAN.rar

 

[K33rg4t3]

/************************************************************************/
// When Recieved - Test Receive Buffer if bytes are in correct position
//  P = nT   P = 24T will switch PORTB Leds as 0b11000
//           P = 255T will switch all leds ON
/*************************************************************************/
             if(Read_buffer[0] == 'P' && Read_buffer[1] == '=' &&
             Read_buffer[3] == 'T' )
             {
             //LATC.B0 = 0;
             PORTB = Read_buffer[2];
             }
           // else
            // {
            //  LATC.B0 = 1;
            // }
            }  //end If (num > 0)

This is not valid.
"P=24T" will not switch PORTB leds, because "24" here is two characters.
See:
"P=24T" will be: Read_buffer[0] = 'P', Read_buffer[1] = '=', Read_buffer[2] = '2', Read_buffer[3] = '4', Read_buffer[4] = 'T'....

 

[salahd]

Hi,
Thanks for your reply
According to the book the code is correct:
For example, if bits 3 and 4 of PORTB are to be set, i.e., PORTB = “00011000,” then
the Visual Basic program sends packet P = 24T (number 24 is sent as a single binary
byte and not as two ASCII bytes) to the microcontroller over the USB link (24=16+8).
And the project run's ok in simulation
Best regards.