0
\$\begingroup\$

Here you will see a schematic for an LCD panel. Header on the right, panel on the left, backlight power circuit on the bottom left and enable circuits in the middle. I am having difficult understanding some of the logic behind the enable circuits.

  1. The LCD_PWRE signal. When it goes high, T3 (npn transistor) activates creating a path from the power pins to ground, but I don't know what this accomplishes. How does it affect FET1 and FET2 (p-channel mosfets)? What purpose does the two channel diode D1 serve?

  2. The backlight enable circuit, I understand that when the base goes low on T1 (PNP transistor) that the backlight will turn on. What I don't understand is the signal connected the base of T1. What is happening with T2 (NPN transistor)? It would appear that the base of T2 is always high, causing the base of T1 to be equal to the LCD_BKL signal. I assume this is not the case as then why bother having T2 at all? Why not connect LCD_BKL directly to the base of T1 with a pulldown resistor?

\$\endgroup\$

1 Answer 1

2
\$\begingroup\$

You're on the right track, but missing an important step or 2.

  1. Yes, when LCD_PWRE goes high, T3 turns on, but this is not to "create a path from the power pins to ground".
    When T3 turns on, it sinks current through R6 and both legs of D1.
    This in turn has the effect of pulling the voltage at the gates of FET1 & FET2 down - which turns them on - they're P-channel FETs, so with their source pins held at the supply voltage , pulling their gates low turns them on.
    D1's purpose becomes clearer when you consider the other components around the FETs.
    Each one has a parallel R-C connected between source & gate. When T3 turns on to pull the gate low, the C's will charge up, and when T3 turns off, the C's will each discharge through the R next to it. But notice that although both resistors are 27k, C1 is 22uF while C2 is only 100nF - so C1 will obviously take much longer to discharge than C2.
    The effect of this is to have FET1 remain on for a little while longer than FET2 when T3 turns off. (Why this should be necessary I couldn't guess).

  2. Transistors T1 and T2 are used as a voltage level-shifter so that the 3.3V control signal can switch the 28V backlight supply.
    When LCD_BKL is high (3.3v) there is no voltage across the base-emitter junction of T2, therefore T2 stays off. With T2 off, no current can flow through T1's base, so T1 stays off.
    When LCD_BKL is pulled low, T2 turns on, allowing current through T1's base, so T1 turns on and supplies 28V to the backlight.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.