Technical Article

A Low-Cost Method of Powering a Microcontroller

October 30, 2020 by Robert Keim

This article proposes a simple microcontroller power-supply circuit that is worth considering when you’re developing a (very) cost-sensitive system.

The Perils of Overdesign

I admit that this article is, in part, a reaction to my bad habit of overdesigning everything. Engineering is not about finding the highest performance, most sophisticated, most bulletproof solution to every problem. Engineers are expected to balance technical specifications with budgetary and schedule constraints, such that the final system is optimized with respect to all relevant design factors.

Furthermore, overdesign is sometimes a symptom of laziness. I can more easily find a ready-made IC than design a customized circuit, and this approach becomes even easier when I can tout the IC’s “superior” performance specs and reassure myself that I am making the system more robust. But what if the IC costs twice as much as a customized circuit that is adequate, i.e., that meets system requirements? Is this good engineering?

 

The Diode Voltage Regulator

The use case that I’m imagining here is a 3.3 V microcontroller on a PCB that already has a 5 V rail. Maybe the product’s primary supply (e.g., from a wall transformer) is 5 V, or maybe a switcher generates 5 V for other components on the board. In any event, you have 5 V, and you need 3.3 V for a small, low-power MCU that consistently performs basic monitoring, control, and communication tasks.

Let’s imagine that we’re working in the consumer electronics space—the product doesn’t need to endure extreme temperature variations, no one’s life is depending on it, and minimizing cost is a central design goal.

An overdesigner like me would instinctively reach for a linear-regulator IC, probably with five times more current capacity than I need and a collection of phenomenal specs that are irrelevant to my application—2% VOUT accuracy over the entire temperature range, input voltages up to 20 V, 0.001 %/V line regulation, etc.

You might think that I’m heading toward a Zener-diode-based regulator, but what I have in mind is even more basic than that: it’s just a series connection of three ordinary diodes.

 

    

Make sure you read the “Dealing with Low-Current MCU States” section below! You may need to include a resistor to ensure that this circuit is safe for your microcontroller.

 

All we’re doing here is using the diode voltage drops to bring the 5 V rail into the microcontroller’s acceptable supply-voltage range. Will this actually work? Well, first let’s look at the basic voltage and current considerations.

 

Supply Voltage and Diode Drop

I’m most familiar with processors made by Silicon Labs, so I’ll use their specs as representative examples. Older 3.3 V devices in their lineup have a VDD range of 2.7 V to 3.6 V, and newer ones tolerate 2.2 V to 3.6 V. We observe, then, that 

  • Three diodes dropping 600 mV a piece would put us at 3.2 V; 
  • We’ll stay within the 2.7–3.6 V range even if all three go as low as 470 mV or as high as 750 mV; and
  • With the 2.2–3.6 V range of the newer MCUs, the diode forward voltage can be anywhere from 470 mV to 930 mV.

 

MCU Current Consumption

The three-diode voltage “regulator” would be easy to implement if we could rely on the constant-voltage-drop model. But in an application like this one, we need a more precise analytical approach.

The exact voltage dropped by a diode is a smoothly varying function of current. Thus, the supply voltage remaining after the three diode drops will change significantly depending on how much current the MCU is consuming at any given moment.

What we have to do is carefully estimate the MCU’s current consumption, and then the plot of forward current vs. forward voltage in a diode’s datasheet helps us to determine if the diode drops will be within the acceptable range.

Here’s an example:

 

Plot of the BAS16GW diode's forward current as a function of forward voltage. Image used courtesy of Nexperia

 

Let’s say we’re trying to stay in the 2.7 V–3.6 V supply-voltage range. If we’re operating at room temperature and the microcontroller needs 1 mA, we’ll have diode drops in the 600 mV region. This puts us near the middle of the VDD range.

The lower and upper limits for diode voltage, as mentioned above, are 470 mV and 750 mV, and these correspond to a current-consumption range of about 80 μA to 10 mA (the 80 μA is an approximation because the plot doesn’t extend to such low currents).

For an application like the one we’re considering in this article, 10 mA would be fairly high, unless the MCU has to provide serious I/O current for driving LEDs or some such. The EFM8 Busy Bee, for example, draws only about 4.5 mA when operating at 24.5 MHz. The current consumption drops to 0.9 mA at 1.53 MHz.

When you look at these numbers, the three-diode regulator seems like a feasible option and could probably provide adequate performance in quite a few applications. However, before we move on to the cost analysis, we need to discuss an important limitation.

 

Dealing with Low-Current MCU States

As the MCU’s current consumption decreases, diode drop also decreases, and this leads to a higher voltage delivered to the device’s VDD pin. If the VDD voltage is too high, the MCU could be damaged. This would occur when the device enters some sort of low-power sleep or standby mode.

However, another possibility is that a normal reset would result in temporarily low-current consumption. This would be a transient event and consequently, the bypass capacitor might protect the MCU, but I’ve never used the three-diode regulator in a real circuit, so I’m not sure.

The solution to these problems is to include additional circuitry that prevents the current from ever becoming dangerously low. This might be accomplished automatically if you have other components drawing current through the diodes. Otherwise, you can include a resistor that is sized according to the minimum acceptable diode current:

 

    

 

Does This Really Reduce Cost?

The most inexpensive linear-regulator IC that I found on Digi-Key costs 4.3 cents (all of the prices mentioned in this section refer to large-quantity orders, meaning thousands of units).

If the MCU’s bypass cap doesn’t adequately stabilize the regulator, you will also need an output capacitor. (I always have separate capacitors for the regulator and the MCU, but if you think that they can share a cap, let us know in the comments.) A 1 μF ceramic cap will add 0.3–0.4 cents to the BOM cost.

I saw a three-diode array (i.e., three independent diodes in one surface-mount package) priced at 2 cents. Individual surface-mount diodes go as low as about 1.2 cents, for a total cost of 3.6 cents. A resistor would be a bit cheaper than a capacitor, maybe 0.2 cents.

These figures indicate that modest cost reduction is possible. However, this is a simplistic analysis, and my intuition tells me that when all purchasing factors are considered, the three-diode solution would look even better from an economic standpoint.

 

Conclusion

Am I crazy for proposing a primitive power-supply circuit that saves me a few cents while offering mediocre performance and complicating the design task? Or is this the sort of outside-the-box thinking that is an essential aspect of electrical engineering?

Have you ever used a circuit like this instead of a linear regulator? Feel free to share your thoughts and experiences in the comments section below.

21 Comments
  • G
    Gordonnp3 October 30, 2020

    I think that part of your assumption is that the MCU will act as a constant load and that is not the case. Any active device such as an MCU will cause your Vout to be unstable.

    Like. Reply
    • RK37 October 31, 2020
      Actually, two sections in the article address this exact topic: "MCU Current Consumption" and "Dealing with Low-Current MCU States." Here are some relevant excerpts: "The exact voltage dropped by a diode is a smoothly varying function of current. Thus, the supply voltage remaining after the three diode drops will change significantly depending on how much current the MCU is consuming at any given moment." "The lower and upper limits for diode voltage, as mentioned above, are 470 mV and 750 mV, and these correspond to a current-consumption range of about 80 μA to 10 mA." "As the MCU’s current consumption decreases, diode drop also decreases, and this leads to a higher voltage delivered to the device’s VDD pin. If the VDD voltage is too high, the MCU could be damaged. This would occur when the device enters some sort of low-power sleep or standby mode."
      Like. Reply
    • Chakit20 November 01, 2020
      Why will Voit be unstable Can you please explain ??
      Like. Reply
    • S
      splud November 24, 2020
      Congrats, this thread made it into a link in an AAC newsletter. How many circuits has anyone here made which consisted solely of a µC? What about the other components in the circuit driven by said µC, such as indicator LEDs, triacs, discrete transistors, other ICs? If your µC never goes into sleep mode and itself has fairly constant current demands, yay - but the instant it drives an LED, does TX on a radio module, etc, the current swings considerably from that steady state. Now, if you're driving an EN pin on a motor controller which itself is supplied by the supply rail, this might not add up to much, but you're also operating at the whim of a voltage dip when the motor switches on and your FIXED voltage drop leads to a brownout condition for your µC - something which it may not do if it had a proper Vreg and a couple of capacitors (all depending on the input voltage and how much of a dip happens of course). No offense, this may be a workable solution for someone bodging something together on their workbench when they don't have a suitable Vreg (shame!), but this approach should be avoided anywhere you want something working reliably under real-world conditions. $0.05 x 10M units adds up to real money, but it is a false cost savings if the design results in device failures because it is only stable if the circuit is always steady state. In the real world, these sorts of failures lead either to RMA costs (if you offer a warranty), or a reputation for junk product, which impacts future sales. Either way, the savings isn't realized. Saving a few cents is better done by selecting alternate sources for parts, or opting for a part that provides the characteristics you actually need rather than significantly over-rated ones (such as using a Vreg rated at 200mA instead of one rated at 1.5A, when your circuit tops out at 50mA). Those are savings which improve your bottom line without subjecting the design to failure.
      Like. Reply
  • Peter Antypas October 31, 2020

    In the real world, components are not sourced from the likes of DigiKey and Mouser. A truly low-cost LDO from a typical Chinese vendor would be under 1.5 cents for qty 1K and would still occupy the same or less board space. Some parts from Natlinear don’t even require an input capacitor. This is not a good solution. None should implement this in any production circuit, ever.

    Like. Reply
    • RK37 October 31, 2020
      I agree that it's not a good solution. As I explicitly stated in the article, I've never used it. However, that reflects my personal approach to circuit design. It's interesting to note that I formulated the idea for this article when I saw this circuit in an app note published by a leading manufacturer of microcontrollers. I don't pretend to understand all the factors that influence design choices in the countless engineering companies operating in countries all over the world. In my life as an engineer, this circuit doesn't make sense. But maybe it does make sense for someone else. And I believe in the value of thinking critically about possible alternatives.
      Like. Reply
  • Goodi October 31, 2020

    I stumbled upon a design which used just one resistor and one capacitor as a voltage regulator for a small MCU. If you assume that this MCU is doing the same simple stuff over and over again in one infinite loop the average power consumption will be very consistent. As far as I can tell this design was working perfectly fine.

    Like. Reply
    • B
      bhawk November 03, 2020
      Microchip makes some PICs with an internal shunt regulator - these parts can operate from a high voltage with a series dropping resistor to dissipate the difference between the voltage supplied and 5v. The parts are PIC16HVxxx.
      Like. Reply
  • A
    atait October 31, 2020

    This is a nice realization. Linear voltage regulators can be thought of as variable resistors, and diodes as well can be thought of as variable resistors. In a prototyping situation, if someone were to find themself without the correct voltage regulator, I could see them reaching for a few diodes to get a fast solution; although, they would also probably place an order for the correct part and not worry about the fractions of cents.

    In a production scenario, you also have to consider yield and customer base. The whole circuit depends on the voltage regulator, but it is not the cost bottleneck. Bottleneck calculations should always precede over optimization.

    MCUs are minimum 20¢. Presumably, that MCU is connected to other things, and there is also at least the cost of a PCB. Let’s call it $1 invested before you can do QC. Going with a home grown regulator that has not been tested, QC’ed, and certified by a manufacturer increases the probability that you’d have to throw out that $1. Even simply increasing the number of solder joints hurts your yield. With that 4.3¢, remember that you are contributing to the engineering, logistics, and verification efforts of the manufacturer, and that stuff is truly valuable.

    Also, some potential customers will require different capabilities in terms of operating temperature range and power demand bursts. A designer can’t know that ahead of time. A good strategy is to look at the specs of the most expensive component and design the support circuitry to slightly exceed that spec.

    That’s my 2 cents (or 4.3 cents). Thank you for the post to get us thinking.

    Like. Reply
  • YounusAAC October 31, 2020

    Hi
    Yes I have used 5 diodes to step down from 11.9 v to 8v for my arduino UNO.
    I was working with a project in which I had a microcontroller and a 12v relay board of 8 relays. The challenge was on one side I had microcontroller working with 5 v but I know that it has a regulator ic on its PCB to step down it, but to be on the safe end, I used five diodes I series and I checked voltage it was around 8 v which I thought was safe.
    And I connected 12v relay board directly to my supply I.e., 12v approx. which replaces your Resistor in this thread above to continue current through diodes but when relays are off there is no current drawn by relay module but current is drawn by arduino pins as my relay module is active low so my circuit is self checking for the continuity of current through diodes and this is by coincidence and luck not by my intelligence as I had not even considered what will happen if no current flow through diodes. My circuit worked as expected and smoothly I did not use any resistors that would have dissipated power and consumed my time for looking different wattage resistors.

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  • mm3 November 01, 2020

    Interesting.
    When the circuit starts each diode will have a 1.6V drop with high inrush current for a short time.
    I’m not sure if this could become a problem. It may be better to add a small resistor in series as well.
    I’m just an hobbyist, but few cents really seem irrelevant to me, but more components to solder seems a cons

    Like. Reply
  • YounusAAC November 01, 2020

    Oh , ho
    I forgot to mention here that when relays are energized, arduino pins are at high state meaning NO current is supplied only arduino may be running in low power mode or default current mode and I do not know what will the drops in diodes, however, my circuit worked smoothly.

    Like. Reply
  • the78mole November 01, 2020

    I once used a 12V zener diode to raise the operating voltage of some low-cost LDO (max. 15V) to be operated with 22-26V for providing 3V3 to some microcontroller.

    Like. Reply
  • A
    adx November 02, 2020

    I have seen done similar things, although the 3 diode circuit is getting a bit crusty. Something with a transistor might be just as simple and cheap, or even cheaper if there is already something on the board. The simplest I’ve used on the power supply pins of a micro is nothing more than a zener (maybe an optional cap). In my experience, considerations of reliability, yield, instability, specs and cost all go out the window when compared to whether it is a “good idea or not” (in hindsight, ie whether it works). Most often, this kind of “clever” design is not. But less often, mindless use of a regulator is also not a good idea (for example if the voltage goes negative unexpectedly, the example circuit will survive, while the regulator circuit will blow up in the field). Good design cannot entirely compensate for reality, which is something that occurs after the product is sold. You probably wouldn’t set out to use a passive regulator, but in some situations you might be told to do away with the monolithic one, and end up with a better solution. How did the micro with literally nothing more than a zener on its power pins get its power? 2 port pins used as a bridge rectifier (rated for the current without latching up) while also sensing the polarity of the line powering it.

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  • B
    bumtras November 03, 2020

    In a hobby project I used a single diode to power a 3.3V (3.6V - 2.7V) Bluetooth module from a li-ion battery (4.2V max). The 0.6V diode drop was perfect and it worked fine, but it feels dodgy.

    Like. Reply
  • J
    jamesmknox November 03, 2020

    It’s not wrong (especially for for one of those late night “gotta try this right now” prototypes.  Further, you did say VERY LOW COST, meaning it is likely never going to be repaired but simply tossed.  Otherwise, I would be very reluctant to use a component in a manner in which it was not designed for. 

    I recall many years ago a frustrated professor (not me) that tried to repair a piece of expensive electronics.  Initially, the problem seemed simple, in spite of not having a schematic or any other information.  Opening the device up, it was obvious that a capacitor had shorted, taking with it an SN7404 hex inverter.  Replaced both, turned it on, and it went berserk.  Finally, after days of manually tracing the schematic, he figured it out.

    At some point the designer needed a delay line to remove signal glitches and allow the states for the decision tree to line up.  So he wired in the hex inverter - all six of the inverter gates - in series.  No impact, except that it provided a nice brief but adequate delay for the signal.  Well, guess what?  In the years since the equipment was built, ICs have gotten MUCH faster.  And the datasheets list a MAXIMUM delay through the gate, not a MINIMUM.  So the new and perfectly good IC was not providing anywhere near the needed delay.  [Of course, we all had a good laugh - at his expense.  But he was a bit less amused.]

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  • oliverb November 03, 2020

    I’ve used a red LED for convenience when modifying a 5V design to take a 3.3V part since I could just cut a track and drop it across.

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  • B
    bgennette November 03, 2020

    I have done this ‘double sided’ with a 1A diode and 100R in both +V and Ground to isolate a decisions & comms uC from a motor control uC in an automotive environment.  1k resistors in the data interconnections soaked up voltage variations.  So the first uC got a well regulated 5V and the second a nice 3.5V with mild suppression that cleared up all the glitches that were occuring when both shared the ‘full’ supply.
    Not the same situation or solution as yours, but a cheap use of diodes to separate supply in a simmilar way.
    bye.

    Like. Reply
    • RK37 November 12, 2020
      Thanks for contributing to the discussion, everyone. I enjoyed reading your comments.
      Like. Reply
  • M
    mikewax November 24, 2020

    There’s no such thing as ALWAYS or NEVER, no assumptions that will ALWAYS apply, no such thing as a datasheet that’s absolutely comprehensive and covers every possible contingency. There WILL be situations, however rare, where solutions like this one will be the most appropriate. And any engineer who thinks he can do his job without EVER extrapolating from the curves on a data sheet WILL sometimes end up with extra components and cost that he doesn’t really need.

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  • D
    du00000001 November 25, 2020

    I’d just go for a resistive divider plus an NPN emitter follower: this way the variation in output voltage is just the variation of 1 diode (instead of 3), and could easily start at a drop of 0 V (for 0 µA). From my experience, a small-signal NPN transistor (up to 100 mA) comes for twice the price of a simple diode, so the emitter follower would come cheaper than the 3-diodes solution. (Resistors in e.g. 2 % accuracy come cheap as dirt these days.)
    And yes - I did similar tricks before - including powering a sub-circuit from a power OpAmp. (In this case, costs obviously weren’t the issue.)

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