0
\$\begingroup\$

Inputs: BRAKE-IN, TURN-IN, REVERSE-IN, PARK-IN Outputs: BRAKE-PARK-LED-GND, TURN-ON, REVERSE-LED-GND, BRAKE-PARK-LED-GND

Application We have chosen the Texas Instruments TPL7407LA to drive groups of LEDs for an automotive lighting solution. The TPL7407LA inputs are connected to +12 VDC input signals to trigger the TPL7407LA’s outputs to turn on. These outputs are sinking current from groupings of automotive LEDs through the TPL7407LA.

INPUTS

BRAKE-IN +12VDC signal from vehicle's mechanical brake switch

TURN-IN +12VDC signal from vehicle's electronic flasher switch OR thermal hazard switch (depending which is enabled)

REVERSE-IN +12VDC signal from vehicle's mechanical reverse switch

PARK-IN +12VDC signal from vehicle's mechanical headlight switch

GND Vehicle's ground

COM Connected to 9V rail. 9V rail is also providing power to LED groupings.

OUTPUTS

BRAKE-PARK-LED-GND (O1/O2) LED common ground from brake light LED grouping

TURN-ON Connected to micrcontroller pin. Pullup resistor to 5V rail installed to preventing floating. 5V rail also connected to microcontroller power supply pin. REVERSE-LED-GND LED common ground from reverse light LED grouping

BRAKE-PARK-LED-GND (O6) LED common ground from brake light LED grouping. Resistor inline used to dissipate power and ultimately lower LED grouping's brightness.

Application We have chosen the Texas Instruments TPL7407LA to drive LEDs for an automotive lighting solution. The TPL7407LA inputs are connected to +12 VDC input signals to trigger the TPL7407LA’s outputs to turn on. These outputs are sinking current from groupings of automotive LEDs through the TPL7407LA.

Issue When applying power to the TPL7407LA on the test bench, the TPL7407LA inputs and outputs work just as expected. Power from the test bench is supplied by a generic adjustable power supply set at 12.5 VDC in this scenario. The TPL7407LA sinks a maximum of approximately 1.0 A total when all inputs into the TPL7407LA are activated with the given +12 VDC inputs.

When installing our product on a vehicle, testing goes mostly as expected except for the scenario where more than one of the +12 VDC inputs into the TPL7407LA goes high. The vehicle provides the switched +12 VDC power input triggers via mechanical OEM vehicle switches (brake light push button switch, turn signal electronic flasher, hazard thermal flasher, etc).

We notice that when multiple transistor are activated (typically from both the brake and the turn +12 VDC signal inputs), then after the lines are deactivated - upon the next activation of the lines, all outputs are activated no matter which singular transistor is activated. I believe this is a symptom of a damaged transistor array and we are unsure why this condition is occurring.

We have since ordered an automotive grade ULQ2003AQDRQ1 equivalent to see if this part is more resilient than the TPL7407LA for this application, but it is still very bizarre to us why this transistor array would be damaged in this type of environment.

TPL7407LA Datasheet: http://www.ti.com/lit/ds/symlink/tpl7407la.pdf

Update

We were able to purchase and make use of an oscilloscope and here were our findings:

TURN-IN Active (electronic flasher), No Diode On Input

TURN-IN Active (electronic flasher), No Diode On Input

TURN-IN Active (thermal flasher), No Diode On Input

TURN-IN Active (thermal flasher), No Diode On Input

TURN-IN Active (electronic flasher), Diode Added On Input

TURN-IN Active (electronic flasher), Diode Added On Input

TURN-IN Active (thermal flasher), Diode Added On Input

TURN-IN Active (thermal flasher), Diode Added On Input

TURN-IN Active (thermal flasher), BRAKE-IN Active, Diode Added On Input

TURN-IN Active (thermal flasher), BRAKE-IN Active, Diode Added On Input

PARK Active (Headlight Switch), Diode Added On Input

PARK Active (Headlight Switch), Diode Added On Input

\$\endgroup\$
10
  • \$\begingroup\$ So, I see 9 V on your diagram and I see 12 V in your text. What about protection diodes and conduction via those? Does it have such diodes? If so, is it possible the behavior is due to them? (Just a random question without fetching the datasheet as I'm feeling a little lazy, right now.) \$\endgroup\$
    – jonk
    Commented Sep 13, 2018 at 20:11
  • \$\begingroup\$ @jonk See updated photo - I originally uploaded the incorrect photo. \$\endgroup\$ Commented Sep 13, 2018 at 20:12
  • \$\begingroup\$ I'm still thinking the same questions. Of course, now I see a 5 V via a resistor, too. With all these rails, I'm imagining even more questions. \$\endgroup\$
    – jonk
    Commented Sep 13, 2018 at 20:16
  • \$\begingroup\$ Why does only one of the LED outputs have a limiting resistor? Why does one have a 7.5k pullup (to +5 of all things)? If that's not an LED (requiring a lot of current) why have you doubled up that line, and not the REVERSE channel? \$\endgroup\$ Commented Sep 13, 2018 at 20:34
  • \$\begingroup\$ The vehicles provides +12v from the battery, our regulators step that down to both 9V for the LEDs and 5v for the microcontroller. We do not have protection diodes associated with this device. Only one of the LED outputs has a resistor on that line because we need to dissipate power to reduce the brightness of the LEDs when the PARK-IN input is active (this is the same grouping of LEDs that the BRAKE-IN activates). The doubled up lines are LEDs for higher current requirements. \$\endgroup\$ Commented Sep 13, 2018 at 20:44

2 Answers 2

2
\$\begingroup\$

Unlike bipolar transistors, mosfet input stages are high impedance, which is usually an advantage but sometimes carries disadvantages as well. In noisy environments like automotive applications, there are often energetic transients coupled into lines, and the higher the impedance terminating the line, the more voltage produced by the energy of a transient.

What is needed is a means of dissipating short bursts of energy, and clamping diodes are excellent at the job. Diodes are quite robust and it's difficult to damage one in short spaces of time.

I suggest you try clamping each of your inputs with at least one diode, cathode to input and anode to ground. That will limit the negative voltage excursion to a diode drop below ground and allow the energy of any transient to dissipate in the diode.

As you found, trying to use a diode to block a transient presents it with a high impedance which cause high voltage spikes until the the transient finds a way to dissipate its energy.

If you want to be quite sure, you can also use another diode, anode to input and cathode to a positive rail to clamp the the inputs' positive excursions to one diode drop above that rail, although the TPL7407 datasheet indicates it has over-voltage protection on its inputs. In any case, you are usually better off allowing spikes to be dissipated in cheap, robust components like diodes, rather than inside an IC.

I understand that adding extra components partially defeats the point of using an IC for your low-side driver, but a few diodes in the right places can save you a lot of grief.

\$\endgroup\$
1
\$\begingroup\$

I suspect your problem is one of layout. If the chip ground bounces below MCU ground, you can easily exceed the -300mV absolute maximum input voltage. That can cause latchup with your relatively large currents.

I suggest series resistors on each input in the 1K range.

You may be a bit close on maximum total current, with all 7 outputs activated and a relatively modest 70°C Ta you're limited to a bit over 150mA per output in the PW package, but I don't think that's what's happening in your bench tests.

\$\endgroup\$
1
  • \$\begingroup\$ Thanks for the suggestion. We have not tried this yet, but we were able to get the oscilloscope on the inputs to the transistor array and have found that the reverse voltage upwards of -44 VDC on one of the lines is definitely causing the damage to the array. We have since added 1N4001 diodes in series on those inputs to resolve this, but it still pushes through past the diode breakdown voltage and ultimately reduces the reverse voltage spike to approximately -6 VDC. Will add the aforementioned images once I figure out how to in this thread. \$\endgroup\$ Commented Sep 23, 2018 at 19:07

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.