We've currently got a PCB that needs to power a modem and a microcontroller. The PCB is powered via a battery (12V - 30V range). The modem and the microcontroller have separate power lines. The microcontroller + peripherals consumes about 200mA, while the modem can consume about 300mA (with spikes of 2A during transmission bursts).

During the design phase it was obvious that a step down switching regulator would be the preferred solution, as an LDO would need to dissipate too much heat. For an initial revision however, the PCB was designed in such a way so that we had the option to use either

(The reason being that the LDO was a lot cheaper to design / manufacter) and that the modem was only used a couple of times per day for really short periods of times (sending about 5 x 1kb per day).

The reason being that it was unclear what the cost would be of a decent switching regulator design.

We've tested with both and indeed found that the LDO needs cooling when heavily used. Obviously using the PTN78000WAH buck is not possible from a pricing perspective (14 eur).

We're wondering what the uptick would be compared to the LM317 LDO if we were to go for a switching regulator. Our PCB designer isn't clear on this, stating that a switching regulator design is very complicate, requires additional research and a proof of concept that we need to pay without have any idea what the impact on the final cost will be.

We just want to know what the pricing impact would be if we opt for a switching regulator on the PCB.

The reference design of the modem proposed the following:

enter image description here

We already have the fuse, diodes, caps, ferrite bead in place on our PCB. The only thing we don't have is the L101 (a inductor/ choke), as this is currently on the Texas Instruments breakout board.

Is it as easy as looking at the price of the 2 core components:

And conclude that for about 3EUR (as opposed to the current 0,3EUR) we would be able to put a switching regulator circuit on our PCB ? Or is there still something else I'm missing. And are there other alternatives for the 2 components above that might be suitable for our use-case (powering a SIM900 modem, 12-30V input, 2A current during transmission bursts, 300mA normal operations) ?

  • 2
    \$\begingroup\$ LM2596 is somewhat dated. For the same price, TPS54340 will work better. It has higher switching frequency which means smaller inductor and capacitor. Also it has good light-load efficiency. \$\endgroup\$
    – venny
    Commented Aug 31, 2014 at 21:22
  • \$\begingroup\$ Thx for the comment and the tip... is it safe to assume that the actual switching regulator chip + inductor will be the biggest cost in the complete switching circuit ? And is the LM2596 / TPS54340 the expected price-range for a switching regulator chip ?. I also noticed someone mentioning the MC34063ADG that is about 1/10 of the cost of the above mentioned chips. \$\endgroup\$
    – ddewaele
    Commented Aug 31, 2014 at 22:02
  • \$\begingroup\$ MC34063 is dirt cheap, but for this application it is underpowered. It has bipolar switch with saturation voltage of \$1\, \mathrm{V}\$, maximum current only \$1.5\, \mathrm{A}\$ and does not come in thermally enhanced package. Moreover, because of its low frequency (\$100\, \mathrm{kHz}\$) it needs large inductor with large inductance. \$\endgroup\$
    – venny
    Commented Aug 31, 2014 at 22:23
  • \$\begingroup\$ There are thousands of different switching regulators you can use that are better than the 2596. We service LG TVs and they seem to be fond of a company called Monolithic Power. I've recovered many of their switching regulators from dead boards and put them to good use in my own projects. MP1423 is one that comes to mind. Indestructible little IC. 3A full load and doesn't even heat up. I had problems with getting 100% performance from 2596. The MP1423 Just Worked. \$\endgroup\$
    – hjf
    Commented Sep 1, 2014 at 5:06
  • \$\begingroup\$ What DC voltage range is on the input of the LDO regulator and how much current is it requiring to pass on heaviest load? \$\endgroup\$
    – Andy aka
    Commented Sep 1, 2014 at 7:33

3 Answers 3


Switching regulator design is complicated indeed, and it even requires a proper PCB layout thanks to the high frequencies and EMI involved. But it's by no means impossible when building it around a controller IC and following its datasheet's recommendations.

But consider that for about the same 3 euros, you could buy a ready-made switching regulator module on its own little PCB from eBay.

And you don't necessarily even need a switching regulator circuit capable of 2A or more, if those spikes are short enough. You could buck down to let's say 8V and put that into a suitably large capacitor, and further regulate to 5V using a linear regulator. The 2A spikes would make the capacitor voltage dip down, but if it doesn't dip too much (thanks to a big capacitor), the output will remain a steady 5V.


I agree with Rennex answer and I want to add something.

Using LM317 or other linear regulator in your case is very bad idea. You will loose a lot of precious battery energy on linear regulator.

At 200mA ("normal operation"), with dropping from 24V to 5V regulator will waste:

P = I * U = 0.2A * (24V-5V) = 3.8W

  • \$\begingroup\$ It's running of a car battery that is constantly being charged when it needs to supply power to the board. But it's clear that a switching regulator is a lot more efficient. It was just a cost-based measure. \$\endgroup\$
    – ddewaele
    Commented Sep 1, 2014 at 16:34

your design will have much higher power loss than the LDO.

  1. the zener circuit is incomplete, you need a series resistor with the source(try googling) to limit the current the zener will get. P=VI, if your source has 20V and your zener is 5V then the 15V trimmed voltage will just dissipate heat through the wire. Let say the wire has 1 ohm and the voltage across it is 15V, then you get 15 watts of power through the wire, meaning you have 15A of current then the power through the zener is Pz=5*15=75watts. check the shunt or series zener regulator on google to see the circuit.

  2. why do you have the external pwm source other than your IC? meaning you will have 2 IC, switching regulator can be done and usually done with only 1 IC. the other IC will just consume power also you need the zener regulator for that IC too, so it is rather inconvenient.

  3. Lastly, your circuit for regulating voltage is wrong, but not very wrong. I mean if your IC can take the 2A switching current then it is ok. but usually, you will include a transistor bjt or mosfet as a switch because IC will tend to break if pass too much current through it. so basically you will use ICs for PWM and feeback voltage only.

  • \$\begingroup\$ Errm, (1) there is no Zener in the circuit - the thing at the input is a TVS which is a high voltage spike suppression device. (2) There is no external PWM source in that circuit - there is a power on/off input for turning off the power if needed. (3) The LM2596 circuit shown is wired up exactly as it should be, and is capable of supplying a 3A load. \$\endgroup\$ Commented Mar 13, 2016 at 5:51

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