I'm designing a basic table-top toy based around a dsPIC33 MCU. It'll be battery powered and I'm using a Micrel MIC2940A-3.3 regulator to produce the 3.3v for the chip. (I've fitted a PP3 clip, with the intention that my device can run from a 9v battery or 3x or 4x AA batteries in a cage with a PP3-clip.)

I'm currently using it in the basic configuration with a couple of capacitors to make the MCU happy. (The typical applications from the datasheet just give it used like this.)

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That works, but I've read elsewhere that the 3-terminal regulator can be used as a voltage reference in a feedback loop to a power transistor, so that the power current comes from the supply rather than the regulator.

Is it common practise to use these regulators bare in a simple application?

And is the more complex configuration only necessary when a large amount of power is drawn? (And would using it as a reference be more efficient/stable than the regulator by itself?)

Edit - More info: With the MCU running in its 40 MIPS mode and LEDs and other outputs all on this 3.3v output, it comes to 85mA max draw.

  • \$\begingroup\$ Are you referring to outboard boost? If so, see electronics.stackexchange.com/questions/180191/… \$\endgroup\$ Sep 3, 2016 at 14:30
  • \$\begingroup\$ Yes, I think so. I shouldn't need anything that complicated for a simple read buttons/blinky-lights device, I think? \$\endgroup\$
    – TempCat
    Sep 3, 2016 at 14:33
  • 1
    \$\begingroup\$ I think this LDO will not be efficient for your application since most of the power will be dissipated by the LDO. Thus, 2/3 of batteries energy will be wasted (assuming 9V to 3.3V conversion). Maybe using switching voltage regulator? \$\endgroup\$
    – Nazar
    Sep 3, 2016 at 14:50
  • \$\begingroup\$ I haven't accepted any answers yet, because nobody's answered the questions I asked. \$\endgroup\$
    – TempCat
    Sep 5, 2016 at 2:46

3 Answers 3


Your circuit draws 85mA maximum, so the maximum power dissipation is 490mW. That is acceptable for a TO-220 package with no heatsink.

However, you could save your users quite a bit of money by using something like 3-4 AA cells and an LDO. There is also more choice of LDOs if the input voltage is lower, and some have low enough quiescent current that you could avoid a hard on/off switch. Or you could use a switching regulator, but the cost per kWh of 9V batteries is higher than that from AA cells, so it would be even better to use a SMPS from AA cells.

Suggest you calculate the cost per hour of operation each way.

By the way, the capacitors are not just to keep the MCU happy- this LDO, like most, will oscillate if you don't provide the appropriate capacitance using the appropriate type of capacitors on the output. Not paying attention to this stuff is a shortcut to a lot of headaches:

External Capacitors A 10μF (or greater) capacitor is required between the MIC2940A output and ground to prevent oscillations due to instability. Most types of tantalum or aluminum electrolytics will be adequate; film types will work, but are costly and therefore not recommended. Many aluminum electrolytics have electrolytes that freeze at about –30°C, so solid tantalums are recommended for operation below –25°C. The important parameters of the capacitor are an effective series resistance of about 5Ω or less and a resonant frequency above 500kHz. The value of this capacitor may be increased without limit. At lower values of output current, less output capacitance is required for output stability. The capacitor can be reduced to 3.3μF for current below 100mA or 2.2μF for currents below 10mA. Adjusting the MIC2941A to voltages below 5V runs the error amplifier at lower gains so that more output capacitance is needed. For the worst-case situation of a 1.25A load at 1.23V output (Output shorted to a 22μF (or greater) capacitor should be used.

They also recommend at least a 0.22uF capacitor at the input and you only have 100nF. I suggest a 100uF/10V electrolytic in parallel with at least 100nF. Batteries greatly increase in internal resistance as they deplete.


Do the math: How much current does your device draw, and what is the voltage drop across this regulator? This determines how much power the regulator must dissipate. Does this fall within the ratings in its datasheet?

  • \$\begingroup\$ I'm not sure exactly what you mean by voltage drop. Is that the 'dropout voltage' listed in the datasheet (200mV for this current), or the drop the regulator makes from the incoming supply to its regulated voltage (9v/4.5v to 3.3v)? For this type of regulator, is that V*I power simply dissipated since this isn't a switching type? \$\endgroup\$
    – TempCat
    Sep 3, 2016 at 15:05
  • \$\begingroup\$ Yes, "voltage drop" is the difference between the actual input voltage and the output voltage. This drop, multiplied by the load current, gets turned into heat. \$\endgroup\$
    – Dave Tweed
    Sep 3, 2016 at 15:18

You say the regulator gets 9 V in and produces 3.3 V at 85 mA out. The voltage it drops, 9.0 V - 3.3 V = 5.7 V, times the 85 mA is what it will burn up in heat. (5.7 V)(85 mA) = 485 mW, nearly half a Watt. That's more than a SOT-23 or probably a SOT-89 package can handle. A TO-3 would be OK in free air, but would get noticeably warm.

Since this is a battery powered device, presumably power drain is important because it directly effects battery life.

Using a 3.3 V linear regulator powered from a 9 V battery is quite wasteful, as shown above. And, the typical 9 V batteries with the two clips on one end have low energy density to begin with.

(3.3 V)/(9 V) = 37%, which is how efficiently the battery energy is getting delivered to the where it's used. Even a simple buck regulator can do much better. Depending on how much power you need, one that does PWM but can fall back to PFM mode at low output currents could be useful. Such chips are available off the shelf. At 85% efficiency, you get over twice the battery life for doing the same thing.


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