I'll try and keep this as organized as possible so everyone can understand my thought process as easy as possible.

  1. I will be using a 12V 5A switching power supply (available on eBay) to convert my wall outlet power down to a workable voltage. That will connect to a DC socket on my board

  2. From there I need two different voltages: 9V at up to 3A for special LED's, and 5V at probably less than 1A for normal SMD LED's and a couple MCU's

  3. I was thinking of using TI's LM1804 (http://www.ti.com/lit/ds/symlink/lm1084.pdf) LDO to regulate the 12V DC input to 9V for the special LED's, and then the ON Semi NCP1117 to get it down to 5V for the MCU's and LED's.

  4. One of the other reasons I need an 8.5V+ supply is for the TI TPL7407LDR (http://www.ti.com/product/tpl7407l/description) to use as an LED driver for the special LED's. The IC datasheet makes it sound like it requires an input of 8.5V+ on the COM pin to be able to function at optimal speed and sink max current.

So here's what I'm curious about:

  1. Is using an LDO like this a smart idea?
  2. Would I be better off getting a 9V 4A switching DC adapter instead of a 12V 5A one (efficiency wise). I do need an excess of 3A as these LED's draw about 100mA's a pop, and I'll be running about 3A's worth currently. But I would prefer as much available current as possible due to future expansions that will actually draw up to 4-5A.
  3. Can I use the NCP1117 5V LDO in series after the LM1084 9V LDO?
  4. Should I need to use a rectifier diode before the inputs of the LDO's (for instance, like the Arduino does)?

I understand LDO's work by dumping the extra voltage and such as heat. I intend on using thermal vias and heatsinks. But in the case of the Arduino, for example, it takes a 9V to 12V 0.25A to 1A adapter and can safely and efficiently drop that down to 5V. So I feel like I've done a good job selecting components to stay as efficient as possible, and for not being a power expert, I feel confident this will work. I'd just like an outside opinion and some thoughts.

However, that TPL7407LDR is a tad confusing. If someone could clarify the COM pin requirements, I'd greatly appreciate it.

Thank you guys!


2 Answers 2


As I read the datasheet, the TPL7407LDR requires at least 8.5 volts on the COM pin - higher voltages are fine.

I would suggest a 9 volt switching regulator (or 12V->9V DC-DC converter) rather than a linear LDO. With an LDO, you will be dissipating about 12 watts in the 9 volt regulator if you start with a 12 volt power supply.

  • \$\begingroup\$ Yikes. Okay, so I just looked and found a 9V 5A switching reg (so same thing I ordered now, just 9V, not 12V). The VDO on the LDO is ~1.3V.... so theoretically I could get that ~8.5V output. I'm assuming that would yield a far better outcome. But what about the +5V LDO? That's a 3.5V drop, so is that still going to be feasible, or should I use a step down there too? Thanks. \$\endgroup\$ Mar 30, 2015 at 2:17
  • \$\begingroup\$ The TPL7407 doesn't need exactly 8.5 volts on the COM pin - it just wants 8.5 volts or more, so you can connect the COM pin to your 9 volts. If you need 1 Amp at 5 volts, the 5 volt regulator will dissipate 4 watts if powered from 9 volts. A DC-DC converter would be more efficient, but 4 watts wouldn't need too big a heatsink. \$\endgroup\$ Mar 30, 2015 at 2:30
  • \$\begingroup\$ Right. I know it has an onboard LDO too. I was more using the offboard LDO just because the special LED's are very, very touchy, and therefore I figured it was important to ensure the line is as stable and conditioned as possible. Since I don't do much power related stuff, I just went straight to the LDO's because of the general simplicity. However, I found a couple TI DC>DC converters so far that I like (The TPS5450 and LMR14050) that end up with a smaller footprint and better efficiency, so I could swap the LDO's with switching regs and probably better off, yea? \$\endgroup\$ Mar 30, 2015 at 2:39

Start with a 9V 5 or 6 Amp PSU. This will provide proper switching and handle all of the power circuitry, allowing you to focus on your end product instead. It also prevents needing an extra regulating circuit. As for the 5V needs, use a 2 Amp USB car charger. These typically have a very wide input range and 9V will be more than enough. Again, this allows you to focus on your end product instead of the power circuitry.

As for the TPL7407LDR, the pin function description on page 3 says (emphasis mine):

Supply pin that should be tied to 8.5V or higher for proper operation (see Power Supply Recommendations for further instruction)

For further note, not applicable to you since you have a 9V source voltage for the specialty leds and the COM pin, but:

An integrated LDO reduces the COM voltage of 8.5 V to 40 V to a regulated voltage of 7 V. Though 8.5 V minimum is recommended for Vcom, the part will still function with a reduced COM voltage with a reduced gate drive voltage and a resulting higher Rdson.

8.5V is just the recommended minimum supply voltage at the specified response values in the datasheet, but it can work lower than that.

Stick to 9V on the COM pin and the LED supplies, and you will be fine.

  • \$\begingroup\$ Thanks for the input. So then, you think I should just use the 9V adapter straight then for the IC and LED's, no conditioning or anything? \$\endgroup\$ Mar 30, 2015 at 5:17
  • \$\begingroup\$ Yep. The IC has a built in LDO for itself, and the LEDs arn't going to be that picky. \$\endgroup\$
    – Passerby
    Mar 30, 2015 at 5:47
  • \$\begingroup\$ Yeah, I knew about the built in LDO. It was more for the LED's. They're experimental LEDs, and the max safe values are still being determined. But they require a very stable environment because slight changes in voltage have exponential repercussions on their current, and so on. \$\endgroup\$ Mar 30, 2015 at 16:38

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