I am reviewing my prototype PCB to iron out previous issues. The PCB is a Bluetooth data capture unit for liquid levels and flow. I am a software programmer by trade and I have some formal digital electronics training, the rest I learn as I go along, I would really appreciate it if some of the buffs over here could point me in the right direction.

Circuit breakdown:


10 - 15Vdc -> Buck/Boost -> 12VDC -> 5VDC -> 3.3VDC

Load: +-500mA

  1. two 5V MCUs
  2. 5V RDM6300 RFID reader.
  3. 3.3V HC-06 Bluetooth module.
  4. two 12V 4-20ma sensors
  5. four 5V LEDs
  6. two transistors


Reliability. Currently I drive a LD1117V33 3.3V regulator from a 7805, which I drive from the 12v DC/DC converter. As you might guess the 7805 gets hot even with a heat sink. I love the boards and would like them to last a long time out in the hot field. So due to the affordability and seeming simplicity of regulators as opposed to DC/DC regulators and the fallout of cascading ripples on my analog circuits, not to mention the questions about shared grounds and isolated grounds etc, I want to add an additional 7809. So that I have 12v -> 9V -> 5V -> 3.3V to share the thermal dissipation load.

My problem is how do I go about calculating the input and output capacitors for each of these regulators? Particularly, how to address that the output capacitor of one regulator is the input capacitor of the next regulator? The current 7805 -> LD1117V33 seems to work fine with 1uF input and 100nF output on both.

  • \$\begingroup\$ Is there a reason you aren't bucking down to 5V from the 12V rail, or the input supply for that matter? \$\endgroup\$ Commented Sep 10, 2016 at 17:32
  • \$\begingroup\$ Forget it and add a heatsink. \$\endgroup\$
    – user207421
    Commented Sep 11, 2016 at 22:28
  • \$\begingroup\$ @ThreePhaseEel To be brutally honest I started out with a breadboard Arduino setup and just added on, it wasn't until I started losing chips that I realized I might need a better power circuit. You see the water pump is run from the same battery/charger system which causes some severe power spikes. I have been thinking about replacing the 7805 with an OKI-78SR , drive the two 12V 4-20mA sensors from the fluctuating 12V since the current loop should be immune and use a 2.5V Zener diode for analog reference to mitigate the reported ripple in the 78SR. \$\endgroup\$ Commented Sep 12, 2016 at 11:22
  • \$\begingroup\$ @EJP That is what I have been doing, and it has been working well, but I am always trying to walk the line between the 'better way' and affordability. \$\endgroup\$ Commented Sep 12, 2016 at 11:27
  • \$\begingroup\$ @BinaryWizzard -- the OKI-78SR for the 5V supply is a good idea. Zeners are lousy reference diodes though, even the venerable TL431 is going to be a better analog reference, never mind an actual precision reference IC... \$\endgroup\$ Commented Sep 12, 2016 at 11:43

1 Answer 1


Yes, you could add an entire extra linear voltage regulator and capacitors for a voltage rail you don't even need, just to spread out power dissipation.

Or you could just use a resistor.

You say your problem is reliability? Then let me tell you about the most reliable electrical component ever made: the resistor. They're the only component that can, albeit unreliably, continue to function even after catching fire. Heck, I've seen those beefy sandstone power resistors explode a little and still work, at least until they explode a lot.

But I digress.

Just put some resistors between the 12V buck-boost converter and your 7805. You can parallel several if you like to really spread out the dissipation and give yourself lots of head room. This will be the most reliable, simple, and inexpensive solution.

And, as a bonus, you will almost certainly get better ripple reduction than cascading an additional 780x (7809 in your case).

The 780x series regulators are fairly slow. They have ~60-70dBV of ripple rejection but only if that ripple is 120Hz. I don't know what frequency your dc/dc converter is operating at, but it is almost certainly 100kHz or more, and will have switching harmonics into the MHz. Those will shoot through a slow linear regulator like the 7809 like a bullet shoots through air.

Ok, it's not quite that bad, but I would be surprised if you got more than 40dBV of rejection at 100kHz, and if your buck-boost converter is faster, it could be 20dBV or even worse.

But do you know what you get if you have a resistor in series with a capacitor? A low pass filter! YAY! And if you placed a voltage dropping resistor before your LM7805, you very conveniently have just such a capacitor - the input capacitor for the linear regulator. Sure, the 7805 needs a nice low impedance input, but that's what the capacitor is for. And since you know your maximum load current, you its a trivial matter to size your voltage dropping resistor. Its really just a matter of how much heat you want to dissipate in the resistor(s) vs. in the linear regulator. The nice thing about, say, 1W resistors, is that they don't mind getting hot and will dump that heat with natural convection alone, so I would recommend favoring the resistors when it comes to shedding watts.

I'll just do a quick and dirty example. At 500mA load current, let's drop the voltage to exactly what you'd get with a 7809. To do that, we need a 6Ω resistor. If you want to be really safe, perhaps use 3 18Ω 1W resistors in parallel, then each one will only have to dissipate 500mW. They will drop the voltage to 9V before it even reaches the 7805. If you put a nice fat ceramic directly on the input of the 7805, perhaps a 47µF one or even a fancy 100µF one, in all its 0805 sized glory, you'll solve your problem in a very reliable way, AND have a beefy low pass filter that will give you better ripple rejection than an extra linear regulator ever could, especially at higher frequencies.

And if you really want to clean up the voltage even more, toss in a ferrite bead in series between the resistors and the input capacitor. Ferrite beads are wonderful little critters. Unlike inductors, which might reduce ripple but also radiate some of that energy out as EMI and almost certainly make the situation worse, ferrite beads take high frequency ripple and dissipate it as heat due to core losses. They're best thought of as frequency dependent resistors that only have resistance above certain frequencies. They will aid you a great deal when it comes to cleaning up the analog sections of circuits - now is the best time to start using them!

If you still really have your heart set on adding a 7809, then you can still solve your problem this way. Add a smaller series resistor between the 7809's output and the 7805's input. This will dissipate a bit of power and spread things out further, and will decouple the 7809's output capacitor from the 7805's input capacitor. Just use the values you would if the regulators were being used by themselves. The input capacitor is what provides the low impedance power to the voltage regulator, so it is perfectly fine to have some resistance in series with the input, as long as it is before the input capacitor. Never put it between the input capacitor and the regulators actual input, obviously.


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.