Why is my low dropout voltage regulator Diodes AP7370 burning up?

Question

I'm building a LED driver using Diodes AL8871Q IC to power a XHP70 LED powered from a 4 series protected Lithium battery pack, integrated on the board it has an Attiny85.

I plan to use the Attiny85 to control the PWM pin on the AL8871Q, read voltage from an NTC thermistor. I am using a Diodes AP7370 SOT23 regulator to power the Attiny85 from the incoming 14v VIN.

Using a bench top power supply to emulate a battery pack, I set to 14V 1A and hook everything up including LED load. Powering on the PSU, all works, I can control the PWM of the Attiny85 which in turns control LED brightness.

Problem I have is when I tried with a battery the LDO regulator burns out. So I tried with a new LDO regulator, turning on the PSU before connecting to my circuit, it burns the LDO regulator.

I figured that the PSU has a soft start that stops the in rush of current vs battery or connect an already on PSU.

Is this what is happening? How can I prevent it from happening with adding the least amount of components.

Scenerios

PSU set to 14v 10A

1. Works Circuit is connected to powered down PSU, LED/load is connected. Power on PSU, Attiny boots up fine, triggers PWM that controls AL8871Q IC that powers the LED. All this happens almost instantaneously. Scopes shows PWM duty cycle at 40% as programmed on Attiny85. PSU displays shows 25W power drawn, voltage reads 14v and 1.8A+-.

2. Burns PSU powered on NOT connected to circuit. LED is connected to circuit. PSU is then connected to circuit. AP7370 burns up and breaks.

3. Burns 14.8V battery pack is connect to circuit, AP7370 burns up

4. Works Circuit is connected to powered down PSU, NO LED/load connected. Power on PSU, Attiny boots up. Scopes shows PWM.

5. Works Attiny85 & AP7370 removed from circuit, PSU powered ON then connected to circuit with LED connected. PSU displays shows 50W power drawn, voltage displays 6+V and about 8+A.

Schematic has 7805 labelled wrongly, it should be AP7370

Answer 1

If the LED is drawing its power through the 7805, the regulator will require a large heatsink. If the bench supply has an adjustable current limit restricting the output to a max of 1 A, this might be why the regulator survives bench operation. A lithium battery has no such limiting.

BTW, the 7805 is not a low-dropout regulator. It is a second-generation single-chip regulator from the 60's/early 70's, and needs over two volts differential between its input and output to work properly.

Answer 2

There is only 100nF capacitor at the output of the LDO, so perhaps it is oscillating. This should not result in its destruction, but who knows.

The datasheet isn't very helpful as it contains no advice on capacitor selection, which is a red flag, but all the examples use a 1µF cap on the output, so perhaps replacing the 100nF cap with 1µF could help.

Answer 3

You have an oscilloscope as mentioned in the comments.

1. Check the battery voltage - most likely higher than 14V by a good bit.
2. Hook up your AL8871Q with the LED to the battery, but with the AP7370 and the processor removed.
3. Check the battery voltage with the oscilloscope while the LED is lit.

I think you will find that there are peaks on the battery voltage going up to more than 20V. This will be caused by the switching action on the inductor. It won't take much of that to kill the AP7370.

Verify that as the cause of the failure. If that's the cause, then the solution is to keep the peaks away from the regulator.

That could be as simple as a resistor with a Zener diode or a resistor with a capacitor filtering the power to the regulator.

Answer 4

The destruction of the regulator may be caused by a transient overvoltage you get when you connect your circuit to either the battery or the PSU. I think the story is the following:

1. when you connect the circuit, there is a large inrush current in capacitors C1 and C2
2. because of the stray inductance between the voltage source (battery or PSU) and the capacitor, a resonance appears, and capacitors C1 and C2 can be charged at a voltage clearly above 14V. If there where no losses (resistance), the voltage could reach twice the voltage of the battery or PSU, that is 28V
3. this voltage exceeds the maximum input voltage of the regulator (20V) and the device fails.

In order to test this hypothesis, you can connect a scope at the input of the circuit, in parallel to C1. You don't even have to put a new regulator. Then configure the oscilloscope in mode "single", and connect the voltage source. You should obtain the voltage at the entrance of your circuit. If it exceeds 20V, you have the explanation.

To avoid this phenomenon, you can:

1. put a resistor in series with the circuit, and a relay in parallel with the resistor. The resistor limits the inrush current, and then the voltage peak
2. the relay must driven by your arduino. It has to be closed after a delay sufficient for the capacitors to be charged
3. only after the relay is closed (which short circuits the resistor) you can start the PWM

You don't kill the regulator when you power the PSU, while it is already connected, because the voltage increases more slowly.

[EDIT] Actually there in a simplier solution, because the only component which is at risk is the regulator: only put a resistor in series with the input of the regulator, and you don't even have to add a relay, given the small current that will flow.

^{simulate this circuit – Schematic created using CircuitLab}

The actual values of Rld and Cld can be adjusted to avoid a large voltage drop across Rld (depends on the current drawn be the arduino) and to provide a time constant Rld*Cld large enough compared to the voltage transient across C1

Answer 5

The current sense and output circuit should be like this. This is why the inductor is causing a reverse spike in the circuit when the input voltage sags below the MOSFET output, causing the fault. :

^{simulate this circuit – Schematic created using CircuitLab}

Answer 6

1 - You are applying about 15 volts and drawing 1.8 amps.

2 - Your regulator output is 5 volts. That means that the regulator is dropping 10 volts across it.

3 - Since the device is passing 1.8 amps, it is dissipating 1.8 x 10, or 18 watts.

4 - You are using an SOT23. From the data sheet, P 2, the maximum thermal resistance from chip to ambient is 203 deg C/watt. If the chip could handle the situation, the interior temperature of the IC would be (203 x 18), or 3,654 degrees. Since ambient will be somewhere around 20 degrees C, the expected chip temperature is expected to be something like 3,674 degrees C. Do you see why it is failing? And let me guess - it's failing instantly upon connection. Do you understand why?

5 - You need a much beefier regulator/heatsink combination. Assuming you're willing to live with a 50 degree C rise, or a final chip temperature of 70 to 80 degrees C, you'll need a heatsink with a thermal resistance of no more than about 2.5 degrees/watt.

Here is the data sheet for a suitable heatsink, the Ohmite BGA425-125E, although it's intended to be clamped to a BGA IC, so mounting could be a problem. It's 42 mm square, and about 1/2 in high.

Perhaps more usefully, this sort of situation cries out for a switching regulator rather than a linear.