Net Battery Capacity of 6 Volts into 3.3V regulator

[Veramacor]

Hello,  basic question but for some reason I can't get my head around it.

I have a circuit that is using a 3.3 Volt regulator.  The power source is a 6 volt battery pack with 4 C-size batteries (primary 1.5 volt each).

My circuit is measuring a 700 micro amp (.7 ma) current drain.  I am trying to figure out an approximate battery drain for my battery pack.

Lets say my voltage was 3 volts instead of 3.3 Volts.  C batteries from what I read hold 6000 milliamp hours of current capacity.  I've heard one should take 70% of that:

6000 * .7 = 4200 milliamp hours.   

If I were to use 2 C batteries with a 700 micro amp current drain:

6000 *.7 / .7 = 6000 hours or 250 days.


My battery pack is 4 C-sized batteries though.  I am not using 6 volts, but only 3.3 volts.  I am using a 'low quiescent current' regulator.  So how can i calculate the net current capacity of a 6 volt battery pack?  Are the extra (6-3.3V) = 2.7 volts just dissipated as heat?

Would I be better to just find a 3 Cell 4.5 battery pack instead of the 4 Cell pack? 

Thanks!

[hlavac]

If the regulator is linear, yes it will dissipate the extra voltage as heat, drawing the same current as you use on the 3.3V from the whole pack.
You would be better off using minimum number of cells to reach the voltage required by the regulator (taking into account minimum voltage of almost empty cell and minimum voltage drop on regulator)
You can add capacity by doubling each cell (put another in parallel).
No point in using more cells except maybe for minimum voltage when cells are almost dead.
With switched regulator it would be more efficient, more cells = more capacity / less current draw from them.

[Veramacor]

Thanks that's what I needed to hear.  Back to the drawing board, but in a good way.  4 Cells take up a lot of space in my design.  Maybe I might go with 2 x 3 AA in parallel.

tx again!

[digsys]

agree with hlavac, except that for the current you draw, NO DC-DC converter will have a low enough quiescent or efficiency to save energy.
Your best bet is to go for a ~say a 18650 Lithium cell. They have incredible storage energy (MUCH greater than a C cell) and stay pretty well
flat at 3.7V. You can even put 2-4 in parallel. If you go for a higher voltage, then ONLY a switched capacitor regulator "may" save you energy.
They have ultra-low quiescent.

[Veramacor]

Thanks for the suggestion!

I was thinking about using this awesome D size battery but balked at the price:

http://www.batteryjunction.com/xl-205f.html

I guess if I have :

19000 *.7 / .7 = 19,000 hours or 791 days! 

You have steered me back into this direction.  I may just say to people you can use a wall wart, but if you want remote capability, use THIS

Ok so $15 every two years isn't too shabby...

edit*** Any suggestion for reverse polarity protection diode.  I dont have much to drop at 3.6 V.  I think typical is .7 volt drop for a diode....

[digsys]

The reason for going to Lithium, either disposables or rechargeables is that you have a pretty well perfect voltage with ONE cell.
PLUS, the voltage will stay flat for most it's life PLUS a MUCH larger energy density than any other type of cell. 3.0AH +++

[digsys]

edit*** Any suggestion for reverse polarity protection diode.  I dont have much to drop at 3.6 V.  I think typical is .7 volt drop for a diode....

Don't waste power with a series diode. IF you want protection, use a say ~50mA polyfuse and a 4V3 1W Zener arrangement

[Veramacor]

Polyfuse and zener diode -  wire in series on the + of the battery? 

I might have to use like a 300 ma fuse as I will have a possible pulse current of 180 ma for about 2 seconds.   I ignored this pulse in my current capacity calc as this pulse may happen only 5 times a day for a total of 10 seconds.

I've seen a diode in parallel with  the battery + and - before acting like a fuse.  How would you wire a polyfuse and zener?  Would it be pretty low power consumption?

[Veramacor]

Whoops.  maybe the polyfuse is on the negative side of the battery with the zener in series?  so the fuse should really be a tiny value?

edit* answered my own question I think?

[digsys]

Example circuit -
You chose the polyfuse this way - A 100mA poly (or any "slow" fuse) will NEVER blow at 100mA.
It will likely take ~3-5secs+ to blow at 100% overload ie 200mA
So, in your case I'd look for app 150mA-200mA poly. NO MORE

edit: Yup, just saw your circuit

[Veramacor]

I'm good to go.  I feel smarter 

Mucho gracias!

[digsys]

I just noticed on your circuit that you marked your DC input 12-20VDC ??
The 4V3 Zener I suggested was for a 3.7VDC Input. IF you intend to use 12-20VDC In, then the Zener must be slightly higher ie 22-24VDC
Select a Zener app 2V+ higher than your MAX expected Input Voltage. Bonus is - it acts as reverse protection as well.

[mariush]

I've bookmarked ISL9111 for a project of mine :  http://www.intersil.com/content/dam/Intersil/documents/fn76/fn7602.pdf

20uA quiescent current, up to 95% efficiency at 5v output with a 3.6v input. 

There's another graph there showing 2.5v and 4v output :



Wouldn't it be more or about as much efficient to do 2 x 1.2v AA batteries -> 3.3v instead of converting 3.6-6v to 3.3v with linear regulators?


Just wondering... didn't do the math...

dc-dc boost converter -> supercap 5v - > fixed linear 3.3v

Would it be more efficient to combine this DC-DC boost  with a super capacitor, so that dc-dc converter pumps 5v @ 100-200mA at 95% efficiency until the super capacitor reaches 4.9-5v, then it's shut off until it starts again when the supercapacitor voltage goes down to about 3.5v?  A 3.3v ldo would work fine with as low as 3.5v input, right ? I'm not sure how the supercapacitor would discharge by itself while the linear regulator pulls from it...

This way, the dc-dc boost converter will run at 50-100mA, maybe more charging the supercap, then the low drop linear regulator would slowly drain the supercap but waste little power due to the small voltage drop

[Flávio V]

I do agreed with the super cap approach,it seems cheaper than using batteries and it will endure to a lifetime...

But for a longer non-feed some the super cap would need to be 350F 2.7V(2 of then for 5.4V max)which can be get quite cheap from digikey for 11$ each,but careful with the shorts on it because it can melt 1.5mm wire..

[mariush]

350F sounds like a huge value. How did you come to that value? (later edit: Oh, I just selected in digikey anything above 300F and that capacitor was the cheapest so I guess maybe that's how)

This boost regulator I had in mind is already capable of generating directly 3.3v but with a voltage input close to this value (2.4v or more) at a few mA output it has a high ripple of up to 100mV.

I was thinking that as the current draw of the device  is about 1mA, the boost regulator would be inefficient, at about a 65-70%. So I would use the super capacitor simply as a buffer of energy, to pump current into it at 100-200mA (the peak efficiency of such ISL9111) then turn it off until the super capacitor is going down to 3.5v or whereabouts.

There are LDOs with 10-100mA maximum current which have about 0.1v drop, maybe even less at 1 mA draw... for example MC78LC33NT ( http://www.farnell.com/datasheets/87792.pdf ) drops 30mV at 1mA.

The problem I have is... or better put the question...   Is the energy wasted by the linear regulator to convert the 3.5v - 5v from supercapacitor + the slow discharge of the super capacitor  LESS than the energy obtained by increasing the efficiency of the boost regulator to 95% ... or not?

Remember, when the supercapacitor is at 5v, the ldo converts to 3.3v wasting (5-3.3) x 2-3 mA .. but the super capacitor will decrease very fast down to about 3.4-3.5v so the losses are decreasing in a relatively linear way

I assume the the super capacitors would have to be low resistance kind, low esr, so not that battery cell types.. but I didn't have 350F in mind, 27-60F seemed like a more reasonable value to me...  a 22uF 2.7v  with 200mOhm ESR would be about 7$, 5$ at 100pcs order..... but I admit I didn't do the math... so I really don't know off the top of my head how many mA of current would 22 F equate to. 

Then it's a question of battery chemistry... would the battery prefer pulses of 100-200mA for a few seconds or minutes, or would it rather prefer long duration draw of a few mA?


I would be quite interested in getting some answers to these, if someone has more experience in this field.

[Veramacor]

digsys,

That circuit was just copied from another post on here as an example.  My voltage will be indeed 3.6V from a D-Cell Lithium.

Here are the first pass parts I order from Digikey:

F2109CT-ND           FUSE PTC RESET 24V .20A 1206
KDZTR3.9BCT-ND   DIODE ZENER 3.9V 1W SOD-123

Am I close?

[amyk]

Thanks for the suggestion!

I was thinking about using this awesome D size battery but balked at the price:

http://www.batteryjunction.com/xl-205f.html

I guess if I have :

19000 *.7 / .7 = 19,000 hours or 791 days! 

You have steered me back into this direction.  I may just say to people you can use a wall wart, but if you want remote capability, use THIS

That's LiSOCl2 which is a very specialised chemistry not intended for general use. From http://web.archive.org/web/20071212033213/http://www.rayovac.com/technical/wp_lithium.htm :

This 3.6 volt Lithium chemistry, originally designed to be a military cell, is a high energy, high current drain rate cell. Unlike most battery systems, thionyl chloride cells are hermetically sealed. Because of its high rate capability, the cell must be constructed with an internally fused design. If the battery is inadvertently charged or shorted, and the fuse mechanism fails, the cell could rupture and explode. Because of the relatively high volatility of this lithium chemistry, Underwriters Laboratory requires that only a trained technician replace the batteries in the applications where this chemistry has been designed. Environmentally, because of the highly toxic and corrosive thionyl chloride in the cell, this chemistry of lithium cell is not recognized as normal waste and requires special handling and disposal.

[Veramacor]

Bad news then again.  Ok thanks for the heads up looks like another redesign.  I may go back to maybe 3 C cells as heat is not a concert at 700 microamps.

Isn't design wonderful!

[digsys]

Bad news then again.  Ok thanks for the heads up looks like another redesign.  I may go back to maybe 3 C cells as heat is not a concert at 700 microamps.
Isn't design wonderful!

There are plenty of Lithium chemistries that are "user friendly". You just picked up a pretty fierce type :-)  Lithium cells are obviously used in millions of devices and sent by air regularly ie phones, laptops etc. As a designer, IF you send by air, it's your responsibility to make sure it's safe though. It is common to just supply the Lithium battery in it's own plastic carrier. There's heaps of cheap plastic holders on-line. Did you look at the 18650 Lithium cell I suggested, or CR123A .. These are extremely popular in high power torches and model cars / planes etc  Don't give up easy when designing, you'll get that hang of it.

[ptricks]

What is the circuit powering ? If it is a 3.3V micro , many of them run  fine on voltages down to 2.8V. If you don't take the lithium battery route then 2xAA batteries will work well too and lowers the parts count as well as being nearly 100% efficient.