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iPhone 4 NO Backlight Repair

iPhone 4 LCD backlight is couple tiny LEDs. In order to light up these LEDs at the same time, a higher voltage drop is neccessary. When "VCC_MAIN" goes thru circuit components L18, Q1, C67, FL4, and J4 Pin 2 to LCD Panel, U48 Power IC Pin E1 "LX_LED" or "SW_BOOST" provides a chain of on-off rectangle pulses to Q1. Transistor Q1 will then be turned on and off accordingly. When Q1 is on, energy is charged on L18. When Q1 is off, energy goes thru internal diode which is integrated in Q1. Eneregy then piles up on C67. C67 finally has a enough voltage drop to light up all LEDs on LCD panel at the same time. Theory on Switching Boost Circuit is shown below. NO light, NO display or dim backlight is caused by the defective of one of the component in the circuit or by defective LCD panel.

Steps to troubleshoot:

1) check fuse coil FL4. If it is bad, missing or damaged, no beep or show more than 0.1 ohms, replace or jump it.

2) check trace is broken or not.

3) check resistor value on R225, R522 and R519, repalce if resistance value does not match.

3) check if C67 both legs are ground or same voltage, replace if yes.

4) check Q1 has any damage, otherwise repalce it.

5) check coil L18. If it is bad, missing or damaged, no beep or show more than 0.1 ohms, replace it.

6) if nothing solve above, U48 Power IC has to replaced.

Repair example 1: if there is no beep on the filter coil FL4 (fuse) or FL4 is damaged or missing, just simply put one jump from J4 Pin 2 to C67 and the repair is done.

Repair example 2: if on-off pulse signal at U48 Pin E1 is not found by an oscilloscope, U48 Power IC needs to be replaced and the repair is done.

















THEORY:



1) Switching Boost Circuit is a DC-DC converter which is by mean of ON-OFF rectangle pulse to pump up the voltage to desired level. Here is the brief illustration of this widely used circuit.

As the intial pulse is HIGH, transistor turns ON to allow current go thru coil L. Coil stores eneregy in the form of magnetic field.



2) As the pulse goes LOW, transistor turns OFF. This sudden drop of current causes coil L to produce a back e.m.f. in the opposite polarity. This results in two voltages, the input voltage U1 and the back e.m.f.(V=VL) across L in series.

Since the transistor is off, this combined voltage (U1 +VL) is no way out except that it goes thru the forward bias D and charges capacitor C.



3) As the pulse goes HIGH again, transistor turns ON and L stores energy accordingly. Due to the charge on C, the cathode of D is more postive than its anode. Diode is reverse bias and then turns off. C releases a voltage of U1+VL to output (U0).



4) The output voltage (U0) is determined by the input voltage (U1) and the duty cycle (D) which is ratio between pulse HIGH (TON) and period in a rectangle pulse.

For the voltage on a coil is V=L(di/dt) or di/dt=V/L. Integrate both side we have the current i=(V/L)xT where V/L is a constant for a rectangle pulse.

While the transistor is ON, current at node A is ramp up, i=(U1/L)xTON. While the transistor is OFF, current at node A is ramp down i=-(U0-U1)/LxTOFF. In a steady state, total current at this node is zero,
ie (U1/L)xTON + -(U0-U1)/LxTOFF = 0, simplify we have U0=U1(1+D/(1-D)), where D=duty cycle=TON/(TON+TOFF)=TON/Period

For example: if duty cycle is half, ie pulse HIGH and pulse Low are equal, then output voltage will be double.
U0 = 9V x (1+ 0.5/0.5) = 9V x (1+1) = 18V











Disclaimer: The content of this page is for reference only. You take your own risk to perform any test or repair on your device. We are not responsible to any damage or lose due to any false statement in this page.
Source of all circuit diagram are from Apple's schematic. iPad, iPhone, iPod, Macbook are trademarks of Apple company.