Asus N73 backlight enable circuit - explanation

[Adenitz]

Hello,

attached is shown part of LCD backlight circuit in Asus N73 laptop. Here, I'd like to clarify few things about how this circuit actually works.

First of all, some signals end with '#'. From the context I can conclude that this means that signal is present when actual voltage state is low, for example 0V.

In order to have backlight, the signal LED_BKLTEN must be present i.e. high voltage = 3.3V in this case.

This is possible only if the following signals are also on high voltage level:
BUF_PLT_RST# is 3.3V which means the signal is not present,
LCD_BACKEN is 3.3V, which means the signal is present,
LCD_BACKOFF# is 3.3V which means the signal is not present,
LID_SW# is 3.3V which means the signal is not present.

If for example, LCD_BACKOFF# is present, its voltage state goes to 0V (or belove certain voltage treshold) and the diode D4502 starts conducting from 3 to 1 and the signal LED_BKLTEN goes to low voltage state i.e. its not present anymore.

Am I reasoning this correctly?
Can you please confirm that # sign means signal is present when its voltage is low?

Thank you.

[mon2]

Re: Asus N73 backlight enable circuit - explanation

The use of '#' suffix in digital logic means 'ACTIVE LOW'.

So, when such a signal is LOW, the signal is ACTIVE or TRUE.

For example, LCD_EN# is considered to be ACTIVE LOW.

If this signal line is LOW (~0 volts) then the LCD is ENABLED.

If the signal line is HIGH (~3v3 volts) then the LCD is DISABLED.

Just some rough example discussion. In the real world, there are ranges where a logic LOW is consider LOW; ranges for the HIGH status.

Also, it varies if the max rail voltage is 1v8 vs. 3v3 vs. 5v0. Each have their thresholds for LOW and HIGH levels.

[Adenitz]

Re: Asus N73 backlight enable circuit - explanation

Thank you mon2, so I understood correctly. In this particular example 3.3V was the high voltage, so I used that as a reference.
So, when signal has #, it means it is ACTIVE or TRUE or PRESENT if it has low voltage.

I guess my explanation of how this LCD backlight enable circuit works is OK.

If the LID is closed, the signal LID_SW# will become TRUE and its voltage level will go toward zero.

[mon2]

Re: Asus N73 backlight enable circuit - explanation

In this particular example 3.3V was the high voltage, so I used that as a reference.
So, when signal has #, it means it is ACTIVE or TRUE or PRESENT if it has low voltage.

Yes.

If the LID is closed, the signal LID_SW# will become TRUE and its voltage level will go toward zero.

Yes, I would think the same for this active LOW signal. Certainly confirm it once you can probe this line. LID switches are often hall effect sensors that trigger by magnet - hidden somewhere in the frame. Once the magnet reaches the vicinity of the sensor, the lid is considered to be closed.

[reformatt]

I had an Asus X550LC motherboard with no backlight today. BL_EN was low at around 0.3V. This is derived from a dual diode (D4501) with inputs from the PCH (L_BKLT_EN) and the EC (LCD_BACKOFF#).

In this case, L_BKLT_EN was 3.3V (H) but LCD_BACKOFF# was 0V (active low). There was some strange gunk in the HDMI connector (so much so I couldn't even get a plug into it). This was also present on the board. When I went to check LID_SW#_IO at J6801, I could pin 5 and 6 covered in that gunk. So that line was being held low. Removing it and resoldering the pins solved the fault.

What is odd about that is that you have the screen enabled but backlight is off. So you don't immediately think LID_SW in this scenario as the screen was enabled.