5V voltage regulator capacitor values

dvd

New Member
I want to drive a circuit with a 9V battery. This will power the picaxe, a shift register and some leds.

Anyway, I want to use a L78 voltage regulator to regulate the 9V to 5V. Looking at the picaxe manual, they recommend using 100nF + 100muF on both the input and the output side (4 capacitors in total).

So, my questions are
(1) why 4 capacitors and not 2 (one on each side)?
(2) How precise do I need to match the values of the capacitors? Could I use other values? I would like to order some capacitors in bulk to use with the L78 in other circuits. So, I'd rather buy some value(s) that is (are) good to use in other circuits.
 

hippy

Technical Support
Staff member
why 4 capacitors and not 2 (one on each side)?
That's usually a manufacturer's recommendation. The 100nF will mitigate against higher frequency noise, the 100uF mitigates against lower frequency noise and ripple. A bit like a warm coat protects against cold, a water-proof protects against the wet. Capacitors don't come in a form which do both so one needs one of each and, as the noise can occur on either side of the regulator, that's four in total.

As to whether one needs the full set is debatable, especially where there should be little noise or ripple with a battery supply, though if a particular regulator manufacturer insists or recommends them it's best to follow their advice.

The 100nF are pretty common as decoupling capacitors so if you're only buying one decoupling capacitor type; that's the one to get.

For the uF electrolytic capacitors that's a bit more complicated. It depends on what's feeding the regulator and what loads are powered by the regulator. One can always add more capacitance ( they are additive in parallel ) so buying 100uF in bulk would probably do, add additional as and if needed.

Some people don't use any capacitors at all, others use the full set. There is no absolute answer other than to do as recommended and adapt as appropriate.
 

westaust55

Moderator
Different voltage regulator datasheets often have differing suggestions on the capacitor requirements. By way of example:

1. An earlier datasheet (Fairchild are now part of On Semi):
https://www.fairchildsemi.com/datasheets/LM/LM7805.pdf
In terms of the ceramic 0.1uF cap on the output –
Co required if the regulator is an “appreciable” distance from the power supply filter.
And for the ceramic 0.33 uF cap on the input
Ci required to improve stability and transient response. I have read other datasheets in the past where the output side ceramic capacitor is indicated as being to prevent oscillation (= stability)

2. Later datasheet from On Semi:
http://www.onsemi.com/pub/Collateral/MC7800-D.PDF
Note this indicates on front sheet that “no external components required”

3. From ST Micro datasheet - see page 5:
http://www.st.com/content/ccc/resource/technical/document/datasheet/41/4f/b3/b0/12/d4/47/88/CD00000444.pdf/files/CD00000444.pdf/jcr:content/translations/en.CD00000444.pdf

4.from the National Semiconductor Datasheet (2004):
For the Input side ceramic capacitor - Required if the regulator is located far from the power supply filter.
For the output side ceramic capacitor - Although no output capacitor is needed for stability, it does help transient response. (If needed, use 0.1 μF, ceramic disc).


The larger electrolytic or tantalum capacitors are used for “bulk” energy to provide a reserve energy source to help prevent momentary voltage droop if there are high current surges from the connected load. Even power circuits supplied from small capacity batteries can sometimes benefit from an electro/tantalum type capacitor between the battery and regulator if there are brief high current demands.

For circuits with just a few IC's, I may sometimes use just a 10 uF Tantalum capacitor where as for more extensive circuits a 100 uF electrolytic and occasionally larger.
 
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eggdweather

Senior Member
The reason two capacitors are used is because large values constructed using electrolytic or tantalum materials tend to be highly inductive and as a result do-not respond well to transient events which by their nature have high frequency components. The large value capacitors therefore will not deliver stored energy quickly ( impedance = 2 x pi x F x L) so adding a small value such as 100nF with low impedance ( low inductive reactance ) solves the problem.

You can work out the values required for the large ( any ) capacitors from the equation Q = C x V or changing Q stored energy to I^2 x t becomes I^2 x t = C x V and C = (I^2.t)/V I changed to dot notation for 'x'. Then if you want to deliver 100mA for 10mS at 5v, a typical power supply dip, then the value would be 20uF.
 

eggdweather

Senior Member
Hint: unless it's a large capacity 9v battery it won't last long, maybe 24-72 hours, so your intended battery type is an important consideration as eclectic was asking, unless its for short term use.
 

westaust55

Moderator
Hint: unless it's a large capacity 9v battery it won't last long, maybe 24-72 hours, so your intended battery type is an important consideration as eclectic was asking, unless its for short term use.
As an extension of the need for further information:

There is mention by the OP of a shift register and driving LED's.

Lets assume in the absence of details:
74HC595 with 8 channels each driving an LED, 5 mA load by each LED and a nominal duty cycle of 50% of LED on time.

A conventional 9V "Energiser' battery with a capacity of around 600 mAh (when discharged at 25 mA rate) would last around 600 mAh / (8 LEDs x 50% x 5 mA) ==> 30 hrs


An Ultralife Lithium 9V with a capacity of 1200 mAh could achieve about twice that operating life but at sometime like 5 times the cost for the battery.
 

newplumber

Senior Member
Hi
would a buck converter save more power then L7805? as this cheap ebay one
http://www.ebay.com/itm/10X-Re-DC-DC-3A-Buck-Converter-Adjustable-Step-Down-Power-Supply-Module-LM2596S-/222245520202?epid=916442884&hash=item33bedd834a:g:UZsAAOSwknJX05VB

I'm guessing it would depend on applications and since dvd wants to use the L7805 maybe I should keep my garbage in the garbage can :)
but I was just curious if the buck converter would add more battery life then a L78.
As always its fun learning from the pros
BTW I am with westaust55 on using the 74hc595 .... I think every electronic thats invented should have a few on board whether it needs it or not
but I have been wrong before/after
your pay days never on friday only in the mail friend mark
 

newplumber

Senior Member
thanks stan74 another tool to add to my forget somewhere box
I know I don't understand the heat generated formula so its 1 amp output and
if a person was to run 12v dropped to 5v ... I soppuse without a heat sink you might be able to push 600 MA?
maybe its alot less ... but never seen one that has that low of a drop out cool to know
 

stan74

Senior Member
It will give 5V at 1A at 12V but will need a heat sink. Powering a pic from the reg at 12V will not need a heat sink as the pic only takes mA's even driving leds.
 

AllyCat

Senior Member
Hi,

Probably not, but (as often) we don't have enough information. The OP has only specified the load (current) as "some LEDs" and that he's using a "L78" regulator, so less than about 100 mA. The PICaxe core only needs around 600 uA (0.6 mA) at 4 MHz.

Therefore, perhaps the most important parameter is the "Quiescent" (no load) current of the regulator, because that may be the largest single drain in the circuit. It's not quoted in the ebay listing, but it is in the data sheets for the LM2596 and the LM2940T (post #9). The typical value for the LM2596 is 5 mA, but that is just for the internal circuit when NOT switching; it will probably be (much) larger with the output components connected (choke, diode, etc.) and switching at 150 kHz.

The LM2940T data is rather confusing, quoting the quiescent input current as 10 mA "at 5 mA load". IMHO 5 mA is not "no load", so it's not clear if the chip "internals" are taking 5 or 10 mA (but the "max" value is 20 mA anyway). Therefore, if the average "PICaxe circuit" current is less than, say, 10 mA, then neither regulator is really a "sensible" component to use. For comparison, the L78 drains a couple of mA, and there are more modern regulator chips with quiescent currents in the range of tens of microamps.

This also has a bearing on the OP's original question. It's not been confirmed, but several of us suspect that the "9 volt" battery is a PP3 type, which is probably NOT the best to use for (any) PICaxe applications. As mentioned in #7, it's only good for a few tens of mA at best, and may have an effective internal resistance of (many) tens of Ohms. A single lithium and/or rechargeable cell normally will have a resistance in tens of milliOhms. The writer of the application/data sheet has no idea what supply or load (resistance or current) will be used (with a potential range of perhaps thousands to one), so a reasonably "safe" combination of two capacitors is suggested. In practice, that may be more than needed, or perhaps inadequate if a PP3 is going to be thrashed beyond it capabilities. ;)

Cheers, Alan.
 

AllyCat

Senior Member
Hi,

Ah, yes. So we know even less about the OP battery/current ("some LEDs") requirements than I thought. :(

The ST data sheet shows that they use only 0.33 uF on input and 0.1 uF on output, but the Quiescent current may be up to 8 mA.

Still a matter of Horses for Courses. ;)

Cheers, Alan.
 

eggdweather

Senior Member
Using a buck converter would save power because efficiency is typically 97% for such devices, whereas a linear X05 type with a 9v input and supplying 100mA would be dissipating (9-5)x0.1 watts = 0.4w of wasted energy as heat so power out / power in = 0.5/0.9 or 55% then a battery should last about twice as long with a buck converter because the energy consumption is half that of a linear regulator.
 
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AllyCat

Senior Member
Hi,

Using a buck converter would save power because efficiency is typically 97% for such devices,
Hmm, very UNtypical I would have thought. Can you point us to some real measurements (at 5 volts output)?

The ebay listing in #8 claims a conversion efficiency of "92% (highest)" and the manufacturer's data sheet for the LM2596 chip which it uses, puts typical efficiencies nearer to 80%.


LM2596-efficiency.png

Figure 3 (page 7) shows the efficiency for 9 volts input, 5 volts output as 78%. That's at a rather high (and probably optimistic for the ebay module) 3 Amps output current. There, IR losses will be increasing, but at that current the Quiescent loss will be insignificant. At 100 mA / 5v output, the converted input current might be hopefully around 70 mA at 80% efficiency; add the 10 mA quiescent current and the input becomes 80 mA from 9v or 70% efficiency (cf 56% for linear).

Below about 50 mA average current drain, or with an alkaline battery a volt or two down (half-way through its expected life), and the switching regulator will probably waste more power than a (well-chosen) linear regulator. And then of course there's the switching ripple......

Cheers, Alan.
 

westaust55

Moderator
The OP (DVD) has, as at this moment, not logged back back in since posting the original question 10 days ago, so may have obtained the desired answer without logging into the forum.

The original post mentions "a shift register and some LED's".
Assuming, in the absence of firm information, a 75HC595 as the ubiquitous shift register then the absolute max current into Vcc or through the Gnd pin is a total of 70 mA (based on the Fairchild datasheet).

So I would envisiage the current draw to be of the order of 5 mA to 8 mA per LED and with 1 LED per shift register channel that comes to 40 to 64 mA plus quiescent current of the regulator and few uA for the PICAXE.
 

eggdweather

Senior Member
OK I may have been quoting the headline efficiencies of something like the Analogue Devices LTC series devices and which if correctly chosen can approach their quoted 97% efficiency, added to which, nearly all buck convertors have a huge input voltage range including in most cases the ability to boost to 5volts from 1.5 volts upwards to 15-20volts and overall efficiency is still going to be greater than a linear device with greater efficiency. I thought the suggestion to use a buck convertor was a good one to get maximum life from the battery, if we ever know what the final requirement was to be.
 

newplumber

Senior Member
Hi
Thanks for all the information I was on vacation (or in other words flushing visa money down a toilet) .... although I understand so little
so to get on the right track using a typical L7805 with no load with input at 12 volts ... will use 2 MA +/-
and a typical buck converter at input 12V with no load will use 5 MA +/- plus a ripple (I also understand it depends on what item but to keep it average)
I guess I will read all you quoted 3 more times so i become unconfused and its awesome all of you explain in more simple terms
I never use batterys in my projects I build because of my design efficiency is 10% - and its expensive to replace the batterys every 3 minutes :)
 

AllyCat

Senior Member
Hi,

a typical L7805 with no load with input at 12 volts ... will use 2 MA +/-
and a typical buck converter at input 12V with no load will use 5 MA +/- plus a ripple
Yes that's a good summary (except that strictly MA is Mega Amps and I think you mean milli). ;)

It happens that a few years ago I needed a 3 volt supply at a few amps and the exact buck converter you linked in #8 looked to be a nice simple solution. However, in practice it ran very hot and the chip/board is not really designed for easy heat-sinking. So I thought it was just a typical Chinese "fake" or at least "optimistic" specification and went for a different solution.

However, looking at the manufacturer's specification of the chip they only expect about 70% efficiency at 3 volts output. Even if the output diode is a Schottky type it will drop around 400 mV, so with a 3v output that's about 15% loss straight away. The output stage is actually an emitter follower (for good reason) which will drop around 1 volt, so more than 10% loss even from a reasonable input voltage like 9v. So actually that module is not too bad, although they have replaced the chip manufacturer's recommended Tantalum electrolytics with normal Aluminium electrolytics and replaced the 5A diode with a 3A (SS34), so it might be rather less efficient than the chip data sheet.

So I recently took the module out of my "sin bin" and actually tested it: At 9 volts, the Quiescent current is 6 mA, rising slightly at lower input voltages and dropping at higher voltages (but not to below 5 mA), so some of the losses are after (or in) the output stage. Probably worth using from a 12 volts input supply if your load is above about 10 mA continuous.

Then I tested for the "break even" point where the input current is the same as the output current (so an efficient non-switching regulator could be just as good). With an accurate 9 volts input, the break-even current measured as 20 mA, but that rises to 40 mA at 7.5 volts input (which might be expected from an Alkaline battery at ~50% discharge). Bear in mind that this break-even point is the average current drain over the whole time that the system is operating, not just the "peak" when lots of LEDS are alight or a transmitter (RF or IR) is operating. The regulator must of course be capable of these latter peak currents as well.

So, IMHO there won't be many PICaxe applications that would justify the use of a buck converter from a 9 volt battery, particularly a normal PP3 with its limited current capability. However, a boost converter from a single Lithium cell (3 or 3.7 volts) might be a good solution, provided that the Quiescent is sufficiently low. For example see this charge/discharge buck/boost battery controller which claims only 50uA in boost (5v output) mode. BUT note that it automatically shuts down after 30 seconds; follow their link to the "continue to output current" product and that appears to require 500 uA (still maybe a good solution though).

Cheers, Alan.
 

erco

Senior Member
@dvd: You're much better off with a 3x AAA or AA cell pack. Lasts much longer, no voltage regulator or caps to worry about. Just the usual decoupling caps near the Picaxe power pins for sport.
 

newplumber

Senior Member
Hi ally yes lol I mean milli amps ... its always awesome to read your explanations
I would have to try out a boost converter .. i bought so many china buck converters that never thought of the boost ones
then again I never worry about efficiency because i am just excited it works with out some plastic melting smell :)
but its good to know in case my electricity supplier feels i'm a unsafe customer.

premelec you ever wonder how they build that for 78 cents? its crazy ... I'm glad china can not do free installations for plumbing here :)
I am feeling rich so I buy 10 of them and try/burn them out ... maybe I can save my power bill from 220 to 219.99 a month (I blame it on kids leaving the lights on (joking))
 

AllyCat

Senior Member
Hi,

@erco: Agreed, but I assumed (probably wrongly) that the OP actually needed a regulated 5v supply (for some unknown reason). We don't know if he's still "watching" this thread or has lost interest, so cannot determine what the reason(s) might be.

@premelec: Yes those are nice "power banks". In UK, one of our "Pound store" chains has been selling them with a battery at the obvious price. The boxes are marked "1200 mA" but I've measured the cells as around 950 mAhr in my intelligent charger/tester. Of course that's less than an 18650 could hold, but the energy is still more than a good quality AA NiMH cell, which normally costs more than a Pound alone. However, there is said to be a more recent version which has a cell that is "heavy at one end" (i.e. the other end is empty), but I've been lucky enough not to find one of those to measure yet. ;)

They're worth the price just for the tiny buck/boost and battery protector circuit board which even has a data sheet (of sorts). Unlike some other desgns, it switches off the output when the cell's low threshold voltage is reached, not allowing it to over-discharge through the choke and diode (without boost). That means it's possible to cross-couple two of the little boxes so that one charges the other. Then when it runs out of juice, it switches off and the second recharges it until that runs out of power. Thus you can create an Ultra Low Frequency "Oscillator" (don't expect a constant frequency or for it to run for too long) which might be an "easy" way to estimate their conversion efficiency, using only a watch. ;)

However, they don't have an on/off switch; they are continually outputting 5 volts. So their Quiescent current drain is rather significant and they can't just be put in a pocket "until a rainy day" (well perhaps they could in the UK :) ). The first I measured as 700 uA (i.e. good for only a few weeks) and another two (from a different store) are 100 uA, so all were apparently "DOA" out of the box (but charged well). I don't know if it's a consequence or just a coincidence, but the first has its Red (charge) and Blue (discharge) LEDs central on the PCB to show through the opague white plastic of the lid, as intended. The other two (100uA) have their LEDs to one side, almost under the USB socket, not really shining anywhere, with their light just "escaping" through the hole for the USB sockets. I note that there is another (more expensive) version which has four (white?) LEDs in the same place that shine though a "window" in the side to indicate the state of charge of the cell.

Some YouTube videos have mentioned quiescent currents as high as 3 mA (and cells as low as 250 mAhr), so I've added a tiny slider switch ("cannibalised" from a Solar Garden Light from the same store) to disconnect the battery. It fits easily beside the USB socket (preferably glued in place) with its pins well-placed to connect to the PCB pad and battery wire. A slot can be cut in the plastic beside the socket with a drill to accommodate the "lever" of the switch. Of course the internal leakage current of the cells is unknown, but the switch may help to identify the wheat from the chaff.

Cheers, Alan.
 

premelec

Senior Member
When erco mentioned 'hi' quiescent current I pulled out several of these boxes that had been stored over a year - the cells were at 4v. It's evident the sense mechanism doesn't draw much current. Probably watching a pin for voltage change or pulse test but I haven't tracked it down... When I looked for static current from the battery I couldn't find any with the meter I had... thanks for the switch modification detail suggestions. . I have other battery banks that don't discharge in storage to any appreciable degree... I guess there are some units that draw 3 ma - assuming this isn't just a peak pulse current sometimes seen with digital meter [d'arsonval meter would smooth them out]. Too busy to dig into it just now - going by empirical - they work ;-0
 

newplumber

Senior Member
Hi Alan yes I think your right about the OP lost interest! lol thats funny ... you all give the OP all great information and no feed back
well at least I am some what learning and taking unreadable notes ...so for reals if you don't put a switch on your power bank ...it will slowly drain down?
another thing is awesome/horrible is that how can they say 1200 mA when you tested it to be 950mahr? (they must be using my calculator)
I have them ordered so I will be excited in 3 months - when they get here to try out.
 

hippy

Technical Support
Staff member
so for reals if you don't put a switch on your power bank ...it will slowly drain down?
Some will while some seem more resilient to this. It does depend on the circuitry within the power bank.

The main problem would seem to be where the battery bank is allowed to run the battery down to a voltage where it can no longer be recharged safely. Hopefully the circuitry would not allow such a recharging to be attempted but it would mean the battery is rendered unusable just through not using it.

Some rechargeable e-cigarettes suffer from that, purchased then 'dead', won't charge, unusable when taken out of the box some time after purchase.

I don't know if it's the same with power banks. Some of those I have hold their charge for many months ( larger ones with four LED's indicating capacity ). Others ( the 1200mA 'poundland' types ) don't but I can't tell if the LED indicates if they are charging or can't. Will report back later on that.
 

AllyCat

Senior Member
Hi,

@premelec: It's strange, because the first time I "tested" the first unit, it appeared to draw absolutely NO current. That seemed surprisng as it presumably must maintain 5 volts on the output terminal (to know when it needs to deliver some real power). However, the next time, it consumed 700uA continuous and I assumed that perhaps my first test had been upset by perhaps a too-high meter resistance. I guess one might expect the "boost" to run in short bursts (maybe with a very long off period) just to top up the output capacitor, so a storage 'scope is really needed to do the test properly.

@newplumber: Well, some of the cells are actually marked 1200 mAh so, by Chinese battery standards, 950 mA is really quite good. Yes, with our "Sale of Goods Act", I could return them for a refund, but IMHO they're worth more than a Pound even at 950 mAh. There are a few "reputable" Chinese sellers who advertise something like: "Rated 1200 mAh, actual 500 mA". The most extreme case I found was a NiMH AA cell marked "400 mAh" in a Solar Garden Light (i.e. hardly more than 10% of its "possible" capacity) and it actually measured 50 mAh !

Cheers, Alan.
 

premelec

Senior Member
Some years ago I bought a battery bank rated 3000 MaHr ... when I tested it it was less than 3000 MaHr at the 5 volts output but when i back calculated to 3.4 v lithium cell and likely up converter efficiency of around 85% it came close to 3000 MaHr - in short the current from the battery is higher than the 5v current to the load and apparently the rating was of the battery ;-0 One problem with inexpensive 'seems to work' electronics from China is low available specification of of the item... unless sometimes you can read Chinese...
 

newplumber

Senior Member
@ hippy well thats good to know so when I get my power banks and they won't work/charge at least I will have a little clue to why

@ Alan I guess everything I bought from china has always worked for me because of two reasons
1 : I have been lucky (fingers crossed)
2: I am not smart enough to know its not working
and btw your getting ripped off a pound in us dollars is like 1.29 so if you have 2 to 3 months you could pay .61 UK a piece since I paid 78 cents US
but over all i am with you for being worth it..of course its probably not apples to apples

@premelec yes that would be a big problem trying to read chinese or in other words for me trying to read any schematic/spec sheet english or not :)
I wish you could download any spec sheet in two types of tastes like... single page for the pros and 40 pages for the almost gifted people like me
 

AllyCat

Senior Member
Hi,

a pound in us dollars is like 1.29
Yes and it used to be around 1.60 until just over a year ago. :(

But my Power Banks come with a battery inside and I believe yours will not:

"Package Included:
1 x Battery Charger Case with Key Chain(18650 battery required to use the item, battery not include)"

A slightly misleading title mentioning "2600 mAh", perhaps rather like advertising "Champagne bottles" that are empty. Well perhaps you wanted to make a table lamp? :)

Cheers, Alan.
 

newplumber

Senior Member
Yes your right mine does not come with a battery so it was a good deal for you after all
and your right the title is misleading but its okay I have a few weeks to order 18650 batterys from a US seller
I might buy some 5800 mAh ones ... which in reality might be 4000 mAh
so a simple question since the battery is 3.7 volts
does a normal battery bank charger circuit charge with a buck converter on the input side
and then boost convert to the out put side to get 5.0 volts
or is it all 3.7 volts in/out ... OR this could be the first time some one asked a dumb question! :)
but its new to me so i figured i would ask

EDIT I found the answer in #24 of this post
so 5v in and 5 volt out I am guessing which is neat
 
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hippy

Technical Support
Staff member
so a simple question since the battery is 3.7 volts does a normal battery bank charger circuit charge with a buck converter on the input side and then boost convert to the out put side to get 5.0 volts or is it all 3.7 volts in/out ... OR this could be the first time some one asked a dumb question! :)
I am not sure how the circuitry is but all power banks I have, and those I have seen, take 5V in and put 5V out regardless of what batteries they contain.

Those which have USB output and USB charge sockets will be 5V. That's so they can be charged from a USB PSU or charger and sometimes even PC USB sockets and can power USB devices.

It's not really a dumb question because battery banks are rarely properly explained. Most people don't consider what's in them and just think of them as '5V batteries'.

As noted any mAh rating may commonly be what the batteries are capable of, not what the power bank itself is capable of, which may be lower. It's like 'music power' for amps and speakers which is sometimes used to advertise a figure which is higher than it could otherwise be stated. It's not 'false' but it's not necessarily meaningful or directly comparable with competing products. It appeals to 'the bigger number the better' approach which some consumers take to purchases which I imagine is why they do that.

It's worth noting that battery banks do not necessarily have to use 3.7V batteries.
 

AllyCat

Senior Member
Hi,

order 18650 batterys from a US seller
I might buy some 5800 mAh ones ... which in reality might be 4000 mAh
The problem with Chinese battery "specifications" is that you don't know if they worked on the principle of "Measure the capacity and double it, or treble it, or quadruple it....", or just invent a number. There are some Power Banks on ebay now claiming 300,000 mAh, that's about the energy in one or two car (auto) batteries, in the size of a cigarette packet! The sellers just work on the principle that most buyers won't know, or complain, and the few that do can have a refund (if they can afford to send the item back to China).

I would work on the basis that any cheap cells are (very) unlikely to be better than the best "branded" cells on the market, which is probably a little over 3000 mAh. Google found this review for me quite quickly. That test is some years old (the 2012 version looks even more comprehensive) but battery technology hasn't changed that fast. EDIT: Here's some more up-to-date information.

It's not something I've tried myself, but I understand that most laptop batteries contain 3 , 6 or 9 x 18650 cells. So try to find a "faulty" one and probably only one or two of the cells (or maybe the "protection" chip) will have failed. The others may well have more capacity than any new "unbranded" cell.

Cheers, Alan.
 
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premelec

Senior Member
Some more observations on step up 18650 USB battery bank: when very lightly loaded it first puts out 5.2 VDC then drops back to 3.3 VDC. When moderately loaded [not enough to keep it on] first 5.2 VDC then drops back to 30 MILLIVOLTS! Plugging in just short USB cord turns it on so small capacitance is sensed. I suspect the controller is doing an ADC read on the USB socket... It never continues to put out 5 volts with any load that won't hold it on.
 

newplumber

Senior Member
Thanks a lot for the information when i have them in hand I will be back reading this page a few times
like always ..great to read
lol 300,000 mAh I bet GM or FORD would love them
whats more funny is i'm the sucker who would try it out expecting it to start my truck :)
 

premelec

Senior Member
@newplumber... your well tuned up truck might start up with 300 amperes in 2 seconds [not N. Dakota _winter_...] - how many milliamere hours is that? (300 x 2)/(60x60) = 167 Miliamp hours... not a lot... what makes this possible in reality is a ulracapacitor, or supercapacitor of many farads across a battery - it can supply brief high current a small battery can't... many delivery trucks are installing supercaps though they are expensive - so many starts and stops in routine operation [let's face it they should be electric trucks but that's not yet the case...]. The supercap charges at relatively slow current from the battery it is in parallel with and then is ready for brief high current pulse.
 

newplumber

Senior Member
well premelec... I don't have a "well tuned truck yet" check engine light is on so its cool ..nice to see its checking the engine :) (joking) but its been on since I owned it
another thing is I tried to build a supercap after watching good/bad youtube learning videos ...of course I learned alot of useless ways
but all were very simple ... of course no numbers of capacitance was involved just seeing them holding a charge
 
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