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GT911 Display Controller: Hardware + Firmware Needed for TouchGFX

KMew
Senior III

Hello,

We are looking to use the Rocktech RK070ER9427-CTG, which uses a GT911 display controller for an upcoming custom board we are designing. In the current evaluation board I have (STM32H7B3I-EVAL), they use RK070ER9427-CT, which uses a different display controller (FT5336).

We would like to still use the TouchGFX application to develop the display, but with a different controller and a custom board, I believe it would require significant driver changes and specific hardware. I would like some assistance to develop this as cleanly as possible. I have the following questions:

1) Are there developed drivers for the GT911 so I can use the display with the new controller? I will likely be using the same MCU from the evaluation board.

2) What are the core hardware pieces needed for the display? An MCU, SDRAM and QSPI is what I (think) I have identified so far, but is this all? Or would I need other things?

Thank you in advance.

1 ACCEPTED SOLUTION

Accepted Solutions

Requirments is based on your application, but minimum is one frame buffer RAM for L8 color mode 800x480 bytes , flash for your application plus graphics around 512k is ok for simple app. MCU need LTDC and pins for signals RGB... All MCU with this and high is ok.

QSPI and SDRAM is required only if app plan more as internal resources.

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8 REPLIES 8
MM..1
Chief III

You mix apple with ...

GT and FT is only touch hw no display.

GT911 is I2C device then read touch events is simple read some bytes from some register.

Hello MM!

Whoops, you are right. I apologize. I suppose I should reshape my question! I want to incorporate the Rocktech display, which uses the GT911 display controller. I want to integrate the entire package (Display + GT911 Controller) onto my custom board.

So what is the information I should be looking for to determine what all is needed to do this?

Thanks!

Requirments is based on your application, but minimum is one frame buffer RAM for L8 color mode 800x480 bytes , flash for your application plus graphics around 512k is ok for simple app. MCU need LTDC and pins for signals RGB... All MCU with this and high is ok.

QSPI and SDRAM is required only if app plan more as internal resources.

Hello MM,

Thank you for the reply! Our application is relatively simple. It is displaying data of a battery-based, portable charger that powers some 120Vac outlets, USB-C, USB-A, etc. We are looking for a nice presentation that has some basic screen animations (company logo on start-up that fades and options to switch screens to see different data).

For the MCU I'm looking at, it has:

  • 2 MBytes of Flash
  • 1376 kBytes of RAM
  • Sufficient pins for RBG signals
  • LTDC (I'm seeing references to it also being called LCD-TFT)

The only other data I'm storing is some CAN data received from peripherals and it's not a lot of data.

Based on that and your response, it sounds like QSPI and SDRAM won't be needed. Does that make sense?

Yes , MCU and LCD only is ok.

If you need 16bit RGB and double buffered animations then too SDRAM is required.

Thank you MM!

One follow-up question. Some videos I've found recommend some NOR Flash memory, but they don't really describe the requirements to know if it's necessary. Do you know what determines the necessity of additional flash?

(I apologize for the newbie question. Lots to learn :D )

simple you have 2M internal flash. Application without graphics for example 300kB, then 1,7M you have for font and images. 800x480 in L8 one full screen background 384k

and you need calculate all planned elements. If is more as 1,7M you need external.

Hello MM,

Again, thank you so much! I imagine the SDRAM could be sized with the same logic, but instead of static things (images, text, etc.), it would be based on the screen resolution and whether or not single or double buffering is used?

For example, if the screen resolution is 800 x 480 and I'm doing double buffering (for animations), the size I would need would be:

ResolutionWidth x ResolutionHeight x BitDepth x 3 (3 for the previous and next image for the animation). So for a 24 bit depth:

800 x 480 x 3 x 3 = 3,456 kBytes

Is that logic correct?