Need portable power solution: 9V, 40mA for at least one month - Schematic UPDATE - Page 1
Forum section: electronics / projects

author: sentry7
author_rank: Regular Contributor
timestamp: March 06, 2017, 12:13:12 pm
content: Hi guys,Title says it all. I'm looking for a battery or battery pack solution. Is it realistic?

author: bktemp
author_rank: Super Contributor
timestamp: March 06, 2017, 12:20:04 pm
content: 9V 40mA = 360mW = 260Wh for one month.A typical car battery has 12V 40Ah = 480Wh, so that is roughly the size you need.

author: madires
author_rank: Super Contributor
timestamp: March 06, 2017, 12:41:11 pm
content: Or a large Li-Ion pack with a similar rating. Smaller, less weight, but much more expensive.

author: sentry7
author_rank: Regular Contributor
timestamp: March 06, 2017, 12:53:44 pm
content: Hmmm...I see. I guess I need to start trimming the wattage. A car battery is not as portable as I would like, and yes, li-ion packs are expensive.

author: mjs
author_rank: Regular Contributor
timestamp: March 06, 2017, 12:54:23 pm
content: One-off not cheap: 3series3parallel D-size lithium thionyl battery can make this in relatively small space. Depending on source, it can also make some space in you wallet.

author: capt bullshot
author_rank: Super Contributor
timestamp: March 06, 2017, 12:56:12 pm
content: There are primary cells called "Lantern Battery"https://en.wikipedia.org/wiki/Lantern_batteryYou could combine two of the larger ones get your voltage/capacity ratingA 6V / 19Ah block is about 11 Euros here

author: bookaboo
author_rank: Frequent Contributor
timestamp: March 06, 2017, 01:11:37 pm
content: Highest non rechargable energy density I have found:http://uk.rs-online.com/web/p/products/6684550The 19Ah is calculated at nominal 4mA so you need to derate for 40mA, though putting these in parellel will help.Price around as RS may not be the cheapest.

author: BrianHG
author_rank: Super Contributor
timestamp: March 06, 2017, 01:53:13 pm
content: Take a look at good old fashioned D-Cell batteries.........Get the data sheets on Energizer or Duracell, go for a 18 cell bank...

author: mariush
author_rank: Super Contributor
timestamp: March 06, 2017, 02:56:05 pm
content: Yeah, you should look first at the power consumption of your circuit.Do you actually need 9v or you could go for lower voltage? If you have linear regulators that need 1-2v above the output voltage to work, you could probably change them for some very efficient linear regulators (there are some which need only 30-50mV above output voltage to work), which would allow you to lower then default voltage, so for example you could use a standard 7.2v lithium battery back, or maybe a 6v motorbike battery (lead acid)Alternatively, you could use high efficiency dc-dc switching regulators with low quiescent current, those will also give you 90%+ efficiency when converting the voltage to what your circuit needs.Is your product going to be somewhere in open space? Maybe you can add some solar cells that would slowly recharge internal batteries during the day?If you have no choice but to go with 12v, you should also look into lead-crystal, they're lighter and can survive going to very low voltagesex 2.3 Kg for this :http://uk.rs-online.com/web/p/lead-acid-rechargeable-batteries/8262723/but you'd probably need 2 of these if you don't optimize your power consumption

author: BrianHG
author_rank: Super Contributor
timestamp: March 06, 2017, 03:32:58 pm
content: You can always use that unbelievable batteroo with 6 D cells, that impossible 800% improvement will give you more than double what you need...

author: Zero999
author_rank: Super Contributor
timestamp: March 06, 2017, 07:27:32 pm
content: Quote from: sentry7 on March 06, 2017, 12:53:44 pmWhat do you mean by portable? How small does it need to be?You say 9V but what's the minimum/maximum acceptable voltage?What does the circuit do?

author: sentry7
author_rank: Regular Contributor
timestamp: March 06, 2017, 10:41:49 pm
content: Quote from: Hero999 on March 06, 2017, 07:27:32 pmIt's a mcu / led matrix display system. To drive the leds, the power has to be 9V at least. The dimensions of the matrix are relatively small (1.2"x1.2"), and I'm not planning on the enclosure to be much larger than that, so I needed the battery (or pack) to about the size of a 9V battery. Too bad an actual 9V battery won't work...not for long anyway.

author: Ian.M
author_rank: Super Contributor
timestamp: March 06, 2017, 11:52:45 pm
content: You are screwed - you want at least an order of magnitude greater energy density than any commercially available battery - seehttps://en.wikipedia.org/wiki/Energy_densityWhatever you do, its going to be big heavy and $expen$ive$, unless by some miracle you can reduce the average current by a factor of over 20.

author: frozenfrogz
author_rank: Frequent Contributor
timestamp: March 07, 2017, 12:19:44 am
content: May I ask what LEDs you are planning to drive in that matrix? 9V seems rather odd to me for any LED chip I played around with in the past – apart from high power COBs, but 40mA does not point to thatFrom the dimensions you mention, my guess would be something like the Adafruit 8x8 dot matrix, but these run on 2-3V depending on the LED color if I remember correctly.

author: sentry7
author_rank: Regular Contributor
timestamp: March 07, 2017, 12:30:36 am
content: Quote from: Ian.M on March 06, 2017, 11:52:45 pmI'm definitely working on it.Quote from: frozenfrogz on March 07, 2017, 12:19:44 amThat's right.Quote from: frozenfrogz on March 07, 2017, 12:19:44 amI've got the blue model, so more like 3-4V.

author: Ian.M
author_rank: Super Contributor
timestamp: March 07, 2017, 12:55:57 am
content: Are you using a linear regulator? If you can get everything after the regulator to run from 40mA @5V, you only need 600KJ with a 90% efficient buck converter, which you could *just* squeeze out of 8x Alkaline D Cells in series, for a nominal 12V dropping to 6.4V fully discharged.Rebuilding the matrix with discrete high efficiency LEDs so you can reduce the current might let you get that down to 4 D cells (2S2P) + a boost converter.

author: mikeselectricstuff
author_rank: Super Contributor
timestamp: March 07, 2017, 01:02:37 am
content: For maximum efficiency go for (cold) white LEDs, carefully select the current based on ambient light, and use an efficient drive method, dropping as little as possible over resistors, and an efficient buck regulator to get you close to the LED Vf.Also consider if the display needs to be on all the time- is there an input you could use ( light, movement etc.) to blank it when not needed.It does however sound like you are an order of magnitude or two away from where you want to be. E-ink or LCD is the only thing that's going to get you a display for a month from a standard 9V battery.

author: mikeselectricstuff
author_rank: Super Contributor
timestamp: March 07, 2017, 01:07:13 am
content: Quote from: sentry7 on March 07, 2017, 12:30:36 amThat type of display is not very optically efficient. discrete white LEDs will have good visibility at maybe10-25% of the current. A small LED concentrates the light in a small area so is visible at much less power.You will also want to put the MCU into a very low power state, only waking up to switch to the next row.

author: mariush
author_rank: Super Contributor
timestamp: March 07, 2017, 01:21:08 am
content: Do you HAVE to use LED matrix ?There are graphical LCD displays with transflexive or reflexive back that work with i2c or serial and are by default very high contrast, and the backlight runs on 3.3v or 3.6v or 5v and uses less current (configurable up to around 30-40mA)Your 8x8 matrix led display is 1.26" by 1.6" if I go by the pictures on adafruit, that's 41mm x 32mm .. here's something relatively the same size at 50mm x 27mm and runs at 3v (max 3.3v) :http://www.digikey.com/product-detail/en/newhaven-display-intl/NHD-C12864KGZ-FSW-GBW/NHD-C12864KGZ-FSW-GBW-ND/4073855If the size is not that critical, you have very sharp black on white 128x32 pixel displays with classic pins on it, easy to solder on base circuit board, and it also uses SPI requiring few pins on the mcu :http://www.digikey.com/product-detail/en/newhaven-display-intl/NHD-C12832A1Z-FSW-FBW-3V3/NHD-C12832A1Z-FSW-FBW-3V3-ND/2059236Or something like this 128x64 pixels , 54mm x 36mm , with pinshttp://www.digikey.com/product-detail/en/newhaven-display-intl/NHD-12864MZ-FSW-GBW-L/NHD-12864MZ-FSW-GBW-L-ND/1701268

author: sentry7
author_rank: Regular Contributor
timestamp: March 07, 2017, 02:54:40 am
content: Ok, so some ideas.It's true that the display need only be on when there is someone to see it, so I'll put in a motion sensor to turn off the matrix when there is no movement; I should be able to get a bunch more hours off of it since it will be off more often than on.I'm pretty much dead-set on an led matrix and the color, but the idea of using an LCD is attractive since it offers everything and more with equal or less current. I may even work on a TFT model in the near future.For those who were wondering, I am using a linear regulator; the LM1117-3v3. Looking at the datasheet, the input voltage has to be at least ~1.5V above the output voltage, so even if the leds looked alright at 5V, any battery solution would not work so well when the voltage starts to fall.Currently out of the 36-ish mA that I'm drawing, 11% of that is the mcu, and it's not even resting at all. The rest of the current is going to the matrix / transistor array.Worst case scenario, I can keep everything and just go with a plug pack. It's not very attractive, but if it's the only way to get everything I want without compromising...

author: mariush
author_rank: Super Contributor
timestamp: March 07, 2017, 05:54:14 am
content: You're reading the datasheet wrong.Seehttp://www.ti.com/lit/ds/symlink/lm1117.pdf- go to page 9 and look at Figure 1, dropout voltage (Vin - Vout) and you can see that at 25c, the dropout voltage is below 1.1v if the current is below 200mA ... and as it heats up, the dropout voltage decreases.Anyway, another thing... this LM1117 needs electrolytic capacitor on the output in order to be stable (0.1 ohm <= esr <= 1 ohm ) so if you use ceramic capacitors it may be unstable causing it to use more power than required.But I don't see why you'd have to use something like that when there's a huge amount of linear regulators that barely need any dropout voltage. For example, you could pick one optimized for low currents (let's say 100mA) and if you want to reduce the parts, with fixed output like you chose already.Here's a small selection that Digikey gives :http://www.digikey.com/short/324mppYou have there MIC5504-3.3YM5-TR at0.11$max 5.5v in, 3.3v out , 150mV dropout voltage at 300mA (less than 60mV drop at 100mA or less) :http://www.digikey.com/product-detail/en/microchip-technology/MIC5504-3.3YM5-TR/576-4764-1-ND/4864028You have AP2114H-3.3TRG1 at $0.35 which can do up to 1A with 0.45v drop, but less than 0.15v below 200mA (see datasheet, figure 30) :http://www.digikey.com/product-detail/en/diodes-incorporated/AP2114H-3.3TRG1/AP2114H-3.3TRG1DICT-ND/4505142Or MCP1700-3302E/TO if you want something through hole, 6v in , 3.3v out, max 250mA , 0.35v dropout voltage in worst case scenario at 250mA , about 0.1v at 100mA :http://www.digikey.com/product-detail/en/microchip-technology/MCP1700-3302E-TO/MCP1700-3302E-TO-ND/652680That would make it possible to use regular lithium batteries, like say 18650 batteries in parallel or something like that, or flat batterieslike this oneor some primary cells like this one:http://www.digikey.com/product-detail/en/tadiran-batteries/TL-5903-P/439-1015-ND/512517or this one slightly cheaper at 2.1 Ah (if you keep the current low) :http://www.digikey.com/product-detail/en/tadiran-batteries/TL-5104-P/439-1013-ND/512515One of those primary batteries will last about 400 hours at 5mA power consumption ... put two of those in parallel and if you can get your project to average below 5mA it would last a whole month.ps. Or you could use just 3 plain AAA or AA rechargeable batteries... that would give you 3 x 1.15v...1.35v = 3.45v .. 4.05v, which is perfectly fine with regulators like the ones I linked above. With AAA batteries you'd have maybe 1100mAh and with AA batteries you'd have up to around 2500-2800mAh or 2.8Ah ... if your project uses less than 10mA they may last a whole month.ps. or use 3 AA lithium batteries like Energizer Ultimate Lithium or Varta Professional Lithium .. these have about 3000 to 3300mAh of capacity in them .. for example one Energizer Ultimate Lithium lasted almost 16 hours at 200mA and the voltage was still above 1.2v ...http://batteryshowdown.com/graphs/200mA/Energizer-Ultimate-Lithium.pngso at 1/10 of that current and with 3 in series ... they should last a long time.

author: mikeselectricstuff
author_rank: Super Contributor
timestamp: March 07, 2017, 09:01:28 am
content: Higher battery voltage and a low power buck regulator is the way to go for maximum runtime.There is another approach for single-cell LiPo.You assume the battery voltage is 3.6v and use a resistive dropper. You then control brightness by adding blank periods to the multiplexing sequence, dimming it as much as possible to get it just bright enough.You then monitor the battery voltage, and increase the blank period at higher battery voltage to maintain constant brightness.This is simplest if your MCU will run from 3-5v, so you don't need a regulator. if it's a 3.3v MCU you need to seperate the MCU and display power rails.

author: mikeselectricstuff
author_rank: Super Contributor
timestamp: March 07, 2017, 09:04:46 am
content: Quote from: sentry7 on March 07, 2017, 02:54:40 amIf they do a white version it would be worth a try as it's possible you could save a lot of power for minimal effort, the eye is much more sensitive to white than blue.

author: sentry7
author_rank: Regular Contributor
timestamp: March 07, 2017, 12:41:38 pm
content: Quote from: mariush on March 07, 2017, 05:54:14 amPardon my ignorance...learning all the time. On page 5, output voltage, what do these conditions mean? (The bit about 4.75V <= Vin <= 10V)Quote from: mariush on March 07, 2017, 05:54:14 amIs this because of ripple? Why electrolytic? Is it because of the esr?

author: mikeselectricstuff
author_rank: Super Contributor
timestamp: March 07, 2017, 01:26:04 pm
content: Quote from: sentry7 on March 07, 2017, 12:41:38 pminput voltage betweem 4.75v and 10vQuotesome regulators can be unstable if the output cap ESR is too low or too high. Ceramics have very low ESR. Some regulators are designed to deal with this, they tend to mention this on the front page.

author: mariush
author_rank: Super Contributor
timestamp: March 07, 2017, 04:30:37 pm
content: Quote from: sentry7 on March 07, 2017, 12:41:38 pmI'm reading this as "The regulator works on a wide range of temperature, but as long as the temperature is between 0c and 125c, and the voltage is within 4.75v and 10v and the current is below 800mA, we guarantee the output voltage will be within 3.235v and 3.365v".It doesn't mean the regulator won't work below 0c (it's specified as low as -40c), it will work but they don't promise the voltage drop will be lower than 4.75v-3.3v at 800mA, at much less than 800mA it could reasonably low voltage drop.Also, the regulator works with up to 15v but with more than 10v, the regulator may not be able to output 800mA because of the amount of heat it produces [ (15v in - 3.3v out) x 0.8A = 9.36w , and the regulator has a junction to case thermal resistance of 16.6c per watt for the to-220 for example - see page 4 at the bottom - so the regulator will heat by more or less 16.6 c/w x 9.36w = 155 c over the ambient temperature, so a large heatsink would be required to keep internal temperature below 125c. (more or less because it's a tiny bit more complicated as you have to take in account junction about 0.2-0.5 c/w lost in thermal paste between chip and heatsink and other things but it's close enough)So that's why they say in the chart they guarantee that output voltage (at that wide current range) only up to 10v input voltage . With lower currents like 100mA or even 500mA, the regulator may work perfectly fine with up to 15v as long as it's kept below the maximum temperature, and may work perfectly fine with lower voltage drops allowing an input voltage less than 4.75vQuote from: mariush on March 07, 2017, 05:54:14 amIs this because of ripple? Why electrolytic? Is it because of the esr?The regulator expects a minimum ESR at the output, at least 0.1 ohm and below 1 ohm.. ceramic capacitors have super low esr, below 0.1 ohm. Some people just put a 0.2-0.5 ohm resistor in series with a ceramic capacitor to "fake" a higher esr capacitor, when they don't have room on the pcb for an electrolytic or polymer capacitor - the other alternatives would be tantalum capacitors but these are more expensive and have undesired properties like exploding or bursting in fire with nasty flames if the voltage exceeds their rating (which could happen if voltage regulator dies somehow) or being made from aconflict mineralAnyway, this LM1117 is old, popular and very cheap and everyone makes it, just like the old "jellybean" MC33063 / MC34063 ... but both are old and have such quirks you have to be careful about and their specifications aren't that great compared to modern stuff.You saw in the links I gave you that you can find more modern regulators like that 0.11 micrel regulator which can work perfectly fine with any ceramic capacitor on the output and has less voltage drop

author: mariush
author_rank: Super Contributor
timestamp: March 07, 2017, 05:17:44 pm
content: Quote from: mikeselectricstuff on March 07, 2017, 09:01:28 amI was thinking more of less of microcontrollers that are very sensitive to their maximum input voltage. For example, MSP430 is advertised as 3.3v, has an input voltage of up to 3.6v and absolute maximum input voltage of 4.1v - I wouldn't use this straight from a lithium battery that may have up to 4.2v, or straight from 3 alkaline batteries which may have up to 3x1.65v = 4.95v (even with rechargeable 3x1.35v =4.05v is very close)But yes, if the microcontroller is natively 3.3v or 3.6v but can support up to 5v (for example pic18f which may run at up to 6.5v but may only run at lower frequency like less than 32 Mhz if the voltage is below 3.6v) then it would make sense to connect directly to battery and do what you say.Though i'd have to think hard about how much complexity would that blanking tehnique add to the code. If that forces him to raise the microcontroller frequency and use more current this way, it's a loss.I'm also not sure about using switching dc-dc converters for this.With linear regulators, the efficiency is fairly predictable .. let's say you have 3.6v out and use 3 aa batteries or a lithium battery .. basically 3.7v.. 4.2v and the circuit uses 25mA ... with a linear regulator the efficiency is more or less (Vout / Vin * 100 because the excess is dissipated as heat .. so you have an efficiency of 88% .. 97% with most of the time above 90%.A dc-dc converter unless it's some fancy one will have losses in the inductor, in the diode, so the efficiency will probably be below 95% from the start... it will have some quiescent current (very small for a good one), the inductor will have to be carefully picked for the current range (let's say 10mA to 50mA output), the layout of the components around the switching regulator IC would matter a lot, all these are things a beginner wouldn't know how to address.Even a sync regulator likeTorex XC9263B75DMR-Gthat only needs an inductor and some ceramic resistors and capacitors would go just close to 90% efficiency with 12v in, 3.3v out.

author: Andreas
author_rank: Super Contributor
timestamp: March 07, 2017, 07:14:23 pm
content: Think big:https://www.weidezaun.info/voss-farming-alkaline-175ah-9v-weidezaunbatterie-gross.htmlShould be sufficient.And portable as they are listed with 3.1 kg.with best regardsAndreas

author: Ian.M
author_rank: Super Contributor
timestamp: March 07, 2017, 07:29:28 pm
content: Now that's a thought!'Weidezaunbatterie' is 'Electric fence battery'Googling:"electric fence battery" alkaline 9Vpulls up a lot of suppliers. Unfortunately not in the USA - it looks like the USA prefers 6V fence batteries.However , knowing such batteries are available could be very useful to anyone who has a one-off remote data-logging project they want to do on the cheap, with no budget for renewables, and with too tight a deadline to get the sleep mode power consumption low enough - it makes it viable to simply throw an Arduino, and RTC/data logging shield at the problem.

author: sentry7
author_rank: Regular Contributor
timestamp: March 08, 2017, 03:26:48 am
content: For all who are interested, here is the schematic of the design. Just the geometry for now--no component values. Will update tomorrow.

author: DBecker
author_rank: Frequent Contributor
timestamp: March 08, 2017, 03:58:11 am
content: Quote from: sentry7 on March 07, 2017, 12:41:38 pmThat part of the table is showing the output voltage accuracy. If you care about accuracy, you'll use that supply voltage range. But if actually care about accuracy, you probably will use a separate voltage reference.In your case you don't care how accurate the 3.3V supply is. Operate at the lowest input voltage that results in the MCU not crashing and the LEDs being acceptably bright.For your power efficiency requirement you shouldn't be looking at a linear regulator at all. Use a switching regulator, preferably a high efficiency type with synchronous rectification.

author: Ian.M
author_rank: Super Contributor
timestamp: March 08, 2017, 04:40:09 am
content: That schematic looks fairly disastrous. You've got a 4017 running from 9V driving emitter followers for the columns, for maybe 8V output swing, and the Propeller is driving the rows directly, which is over-voltaging the Propeller I/O pins when they are at logic '1'. Also the 555 is redundant, it would be better to use a level shifter to drive the 4017 from the Propeller.You are wasting power in the driver transistors and associated resistors, and throwing it away in the row resistors because the supply voltage is so much greater than the LED Vf.You should redesign it to run from 5V, use a 74HCT138 to provide both level shifting from 3.3V and a 1 of 8 active low drive to eight low Vgs threshold P channel MOSFETs to drive the columns. Three output lines from the Propeller would select the column, and a fourth would go to one of the enable pins to blank the whole lot for power saving. It would be a good idea to use eight very low Vgs threshold N channel MOSFETs to buffer the propeller row outputs, though as long as the LED Vf >>1.7V and you keep the current per LED well under 10mA you should be OK driving them direct.It isn't worth chasing down that last bit of energy wasted in the 3.3V LDO - just feed it from the 5V rail. Use a high efficiency switching regulator to provide the 5V rail.That will immediately save 45% of the power its using. Depending on the Vf of the LEDs at the current you intend to run them at, if you use a better LDO for the 3.3V rail it may be possible to drop the 5V rail to 4.5V and get the power saving up to 50%. If you can add further power-saveing with an ADC and a photocell to adjust for ambient lighting by varying the multiplex blanking period, and a low power PIR or other proximity sensor to blank it while noone is nearby, you can probably get it down to a battery pack the size of a cigarette packet.

author: mikeselectricstuff
author_rank: Super Contributor
timestamp: March 08, 2017, 08:46:38 am
content: Quote from: Ian.M on March 08, 2017, 04:40:09 amAgree.If you are really serious about getting a long battery life, you need to forget what you currently have and start from scratch.With the right LEDs, you'd be looking at driving a couple of mA per LED max, so you don't need any external drivers, just drive direct from the MCU pins.Use the cathodes as rows, as most MCUs will drive more current low than high.Connect the column anodes via series resistors.If you can't get enough current for the row cathode drive, or don't have enough pins, use an external NPIC6C596shift-register.Either:Run from a 3.6v Lipo with no regulator (Use a MCU that can run from 3-5v) ,and adjust the dead-time with supply voltage for constant brightnessOr :or a low-power buck regulator from a higher supply voltage, with the output set about 0.2v above the LED VfUse an MCU that can switch clock frequencies quickly and easily (this will be an on-chip RC oscillator as crystals take too long to start), and/or A low-power wake-up timer that can be set fast enough to use as the row scan timer. Nanowatt PIC8 devices would be ideal for this.For just updating the display, it shouldn't be hard to get the MCU'savaragedraw in the 20-100uA range

author: sentry7
author_rank: Regular Contributor
timestamp: March 08, 2017, 08:13:53 pm
content: Ok, new and improved? Let's hope so.

author: Ian.M
author_rank: Super Contributor
timestamp: March 08, 2017, 08:41:57 pm
content: No. As your LED Vf is around 3V you'd be FUBARed if you try to run the matrix directly at 3V Also, you *REALLY* need to get the LED matrix off the 3.3V LDO regulator output - that just wastes power. If you were using red or green low Vf LEDs I'd suggest a switching 3.3v regulator, but as you are using blue ones that need up to 4V Vf, I'd strongly recommend keeping the LDO and adding a switching pre-regulator for the LEDs and to reduce the power wasted by the LDO.A 74LS138 wont work properly from 3.3V. You need either 74HC138 for active low 1 of 8 output or 74HC238 for active high. Either can run from 3.3V. However their current sink or source capability isn't that great, so if you need much display brightness, you'll need to use the 74HCT138, or 74HCT238 which need 5V+/-0.5V, and have 3.3V logic compatible inputs, probably with external MOSFETs P channel high side for the '138 or N channel low side for the '238.Do you really need the full power of a Propeller? An 8 bit Microchip PIC or an Atmel ATmega would have significantly lower power consumption if you implement sleep mode correctly.

author: suicidaleggroll
author_rank: Super Contributor
timestamp: March 08, 2017, 08:48:38 pm
content: As soon as batteries and high efficiency requirements come into a project, linear regulators go OUT (apart from some very specific scenarios where you would still want one, yours is not one of those). You need to get rid of it. The efficiency of a linear regulator is Vout/Vin. With 3.3V out and 9V in, you have an efficiency of 37%, that means your battery will need to be roughly THREE TIMES larger than necessary because it's burning off two thirds of its energy in the linear regulator itself. That's abysmal. With a simple buck regulator you could get higher than 90% efficiency. Do you really want to make your project 3x larger and your battery 3x more expensive in order to save a dollar or two on your power regulator? You'll easily blow any savings on the extra battery capacity required, it makes no sense.

author: Ian.M
author_rank: Super Contributor
timestamp: March 08, 2017, 09:04:18 pm
content: +1, apart from the special case of a mixed logic voltage project with very low consumption on the 3.3V rail. If you are careful to avoid significant sourcing loads on the MCU's pins and if the average MCU current is only a small fraction of the load on the 5V rail, and the 5V rail is supplied by a high efficiency switching regulator, the cost, size and idle current of an extra switching 3.3V regulator may not be justifiable (5V-3.3V)*1mA is a hell of a lot more attractive than (9V-3.V)*40mA, and it may not be worth chasing that last milliwatt or two.

author: BrianHG
author_rank: Super Contributor
timestamp: March 09, 2017, 05:42:02 am
content: Quote from: Ian.M on March 08, 2017, 08:41:57 pmRunning an 8 bit pic at 5v with 16 free IOs makes this project an all in 1 chip solution, + some of the PICs have +12ma, - 24ma drive or others with +/-12ma. (Use a common cathode matrix here and get really good brightness with just series resistors on the anodes for the +12/-24 pics...)One Lipo battery pack with a regulated boost converter to 5v should power the device for a month.Careful when choosing the PIC, many newer lower power ones have +/-7ma. It's the older more power hungry 5v pics advertised to directly drive LEDs which have the +12/-24ma IOs. Don't worry about operating power consumption, they will still consume a fraction the power, going in and out of sleep or operating at low frequency, compared to the LEDs you are lighting...

author: mikeselectricstuff
author_rank: Super Contributor
timestamp: March 09, 2017, 11:53:30 am
content: Quote from: BrianHG on March 09, 2017, 05:42:02 amYou're wasting power running at 5V - you're just burning it in resistors and the boost reg. Run direct from a Lipo. If necessary measure the battery voltage and adjust gaps in your multiplex timing to maintain constant brightness over the typical 4.2-3.4v cell voltage.For a blue or white display, Vf is going to be pretty close to 3VPIC output drive current shouldn't be an issue at the sort of current you want to be driving the display at.

author: sentry7
author_rank: Regular Contributor
timestamp: March 09, 2017, 02:13:22 pm
content: Quote from: Ian.M on March 08, 2017, 08:41:57 pmI agree with you, however, the prototype is right here and the leds work. I'm not sure why they light up, but they do. Thedatasheetsays (typical) Vf=3.2V@20mA, but the actual current is less than that so they really shouldn't be lit. I'm measuring the voltage across them as 2.83V and the current is 2.7mA. If I read the datasheet correctly, this current gives less than 10% maximum luminosity, but I'm looking at them and they are really bright. Granted, they aren't being strobed, so the effective brightness will go down.Quote from: Ian.M on March 08, 2017, 08:41:57 pmI'm sorry, I'm not understanding what power is being wasted here. Can you give me an example of a switching regulator? Are these also called boost (or buck) converters?Quote from: Ian.M on March 08, 2017, 08:41:57 pmAlthough the schem says 74LS' the proto has a 74HC'.

author: sentry7
author_rank: Regular Contributor
timestamp: March 09, 2017, 02:30:14 pm
content: Are switching regulators more efficient than linear because they switch between being on and off versus linear which are essentially always on?

author: mikeselectricstuff
author_rank: Super Contributor
timestamp: March 09, 2017, 02:44:03 pm
content: Quote from: sentry7 on March 09, 2017, 02:30:14 pmLinear regs burn the difference between Vin and Vout as heat. Sitchers store energy in an inductor so they can translate high voltage, low current into low voltage at higher current.e.g. for 5V at 50mA out, and 9v in.A linear reg would draw 50mA at 9V, wasting 4*50=200mW as heat.A perfect switching reg would draw 50mA * 5/9 = 28mA.There will be some losses, and you need to choose a switcher that's designed for the current range you need for best efficiency, but efficiencies can be up to about 95%, so about 30mA in this case.

author: Ian.M
author_rank: Super Contributor
timestamp: March 09, 2017, 02:48:23 pm
content: I'll get back to your other questions later, but the switching regulator ones are fairly quick:Boost and buck converters are two different types (topologies) of switching regulator. There are others, some a lot more complex.Switching regulators in general cyclically store energy from their power feed in an inductor (or capacitor for some types) then transfer it to their load circuit. Think of them as transformers for DC. Neglecting switching losses etc, you get the same power out as you put in, just at a different voltage and current. This is totally different to a linear regulator that burns off the excess energy of its input to output voltage differential times the load current as heat.Edit: Mike's already come at this from a slightly different angle with mathsbut I'll let this stand anyway

author: bagnoz
author_rank: Newbie
timestamp: March 09, 2017, 03:05:23 pm
content: Quote from: Ian.M on March 08, 2017, 08:41:57 pmHaving never worked with PICs or ATmegas myself: What current consumption would one expect them to have?(For a correctly designed low-current application, I'd expect about 5 µA consumption for the Propeller.)

author: Ian.M
author_rank: Super Contributor
timestamp: March 09, 2017, 03:31:09 pm
content: I wasn't aware that Propellers could go that low.Depending on the PIC, you can get down in the hundreds of nA range. e.g. PIC16LF1507 can typically run at 8 KIPS for under 10uA and sleep at under 200nA, less at lower voltages.

author: mikeselectricstuff
author_rank: Super Contributor
timestamp: March 09, 2017, 03:49:09 pm
content: Quote from: bagnoz on March 09, 2017, 03:05:23 pmWhen sleeping, with a watchdog timer that can wake up in 1mS increments, well under 1uAe.g. for 16F182x :Running at 32KHz about 10uA at 3V, 500KHz about 150uA, 8MHz about 1mAFor just keeping the display updated, compared to the LED current, MCU current is going to be negligible if done properly.Using the internal RC oscillator ( about 2% accuate), wake-up from sleep only takes a few uS, so you can wake, do something and go back to sleep very quickly.

author: BrianHG
author_rank: Super Contributor
timestamp: March 09, 2017, 04:02:12 pm
content: Holy cow, the PIC16F15375, 40 pin dip, plenty of IOs has a whopping +/-50ma per pin on the IOs, though the maximum total dissipation at one time is limited to 350ma. This would cleanly drive your display on 2 free 8 bit ports without any dimming pattern.The current consumption is 32ua per Mhz while running. So, running at 8 Mhz, if you are sleeping for 99% the time, the current will be @1.8v = 2.6ua + 50na + 500na to keep the real time timer running, no deep sleep, power consumption would go down and additional 500na but its takes a us to wake up, not worth the savings. I need to check the specs, but if I remember, if you use the internal 500khz mode, you can sleep and wake up in a few clock cycles.Say around 10ua for 5v operation, 7ua for 4v, 5ua for 3.3v. Still, a fraction of just having a linear regulator with your processor which will eat 1ma even though the CPU will draw 5ua now that you might consider connecting a lipo battery directly to the PIC without any regulator.Now, this is not the lowest power pic, just a 1.50$ one with over 35 x 50ma IOs in a DIP package with 2.2v through 5v support, or, 1.8v through 3.6v for the LF version. (Don't use the LF, when charging a lipo battery, the voltage can get as high as 4.2v.)With such strong IOs and a wide supply range, just wire the battery pack directly to the PIC without any regulation & run the display. Sure, as the voltage power goes down, you can adjust the PWM width of the display to compensate. This pic has an internal Voltage reference so you can measure your battery voltage without any regulator...Note that you will basically be programming the wake-up timer's time to manipulate the length of the on-time for each row when refreshing the display to adjust brightness. I've done this with a PIC driven battery powered emergency light.Ok, for lower power, there are the newer 16 bit pics, but I need to look into that...

author: mikeselectricstuff
author_rank: Super Contributor
timestamp: March 09, 2017, 05:04:23 pm
content: Quote from: BrianHG on March 09, 2017, 04:02:12 pmif you any ( non- PLL) internal RC frequency you can wake in a few cycles. Just stay away from crystals and resonators as they take significant time to start.

author: BrianHG
author_rank: Super Contributor
timestamp: March 10, 2017, 05:04:56 am
content: Quote from: mikeselectricstuff on March 09, 2017, 05:04:23 pmThat's right, you can use a 32Khz XTAL to drive just the real-time counter, always on state, 500na draw only, and run the CPU in the internal RC mode so that it goes in and out of sleep within a few clock cycles. Feature of the XLP (extreme low power) pic I listed. This allows for a real time clock & a CPU which can go in and out of sleep extremely quick.

author: mikeselectricstuff
author_rank: Super Contributor
timestamp: March 10, 2017, 09:48:21 am
content: Quote from: BrianHG on March 10, 2017, 05:04:56 amUnless you need the accuracy you don't need the xtal, there's a low-frequency internal RC osc.