What is the most power I can generate from a 1.2V battery

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I am using a tiny13, fastPWM 9,6mhz 50% duty cycle that is the highest current I can generate through a yellow led.

100uh inductor
2n7000 mosfet
BAT85 shottky diode
.22uf capacitor

I can drive one Yellow led at 10.05mA measuring the LED at that current I get 1.87V Battery measures 1.4V while circuit is running.

I have aome 1000uh and higher inductors but I get less power using the 1000 at any frequency I have tried so far.

1,4V goes into inductor -> thru mosfet, diode from that junction to capacitor cap to ground, I get 73V open circuit.

If I add a second inductor and fet I can get double the output and drive a white LED at 13.5mA.

How do I use a bigger inductor with just one mosfet to get up to 20mA? I have tried using two 100uh in series and parallel through the same mosfet but series I get the same and parallel I get 1/2 the power.

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Hi

Each battery in the world has a property: mAh;
mAh = (mili-ampers x hour) (Battery's capacity)

If u have a 1.2V battery with 1000mAh, it means if
u source 1000mA, after one hour, the batteries voltage
will drop from 1.2V to nearly 0 (maybe not zero, but forexample 0.4V).

So if u source for instance 5000mA, after 1000/5000 hourse (0.2Hour or 12 minutes) the battery will
deplete.

But on the other hand, 1.2Volt is not much voltage,
you might as well use a higher voltage battery, or
DC-to-DC converter, although they don't provide much
current.

Good Luck
Nasser

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Embedded Dreams
One day, knowledge will replace money.

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Have you looked at the datasheet for the battery? What does it say about peak and average draw, how long peak can be sustained, battery V drop at various loads, etc.?

Lee

You can put lipstick on a pig, but it is still a pig.

I've never met a pig I didn't like, as long as you have some salt and pepper.

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I think you are missing the point of my question, I guess I am not very skilled at asking a question.

I have built a DC/DC converter using PWM, a capacitor and an inductor. I know what mAH are and how to calculate the length of time a battery will last for a given ma draw.

Attached is a circuit that runs an atiny13 running a pwm signal to boost the battery voltage to drive a red LED at 14.5mA. I get about 7mA from each inductor, I could use an inverter and perhaps get more as the inductors would generate kickback alternately instead of the same time not sure about that.

What I was looking for is some discussion on frequency , inductor size, capacitor size, type, diodes type size etc to increase the output of the converter.

Note- After blowing up some LED's and actually blowing up a 10V zener diode, I now have the 10V zener in the circuit at startup (not on schematic I just whipped that up quickly to show the circuit) Should maby put a 5V zener but as long as the LED is in place at startup it is not necessary.

By the way, the reason I am using only 1.2V rechargeable batteries is that I am using the solar panels from three malibu solar garden lights to charge three 1.2V AA batteries in parallel. I tested the panels in series but they only have a fraction of the output in series and I am going to use one of the existing circuit boards from the garden light to act as the charge controller, It turns off the output and charges the batteries, during this time the tiny13 will be shut down. It may be possible to use the output that normally would go to the existing led thay have on that circuit to drive the tiny13, but I think I will use a FET and a small capacitor to turn on the juce to the atiny13.

So, if anyone has experience using small inductors to generate higher voltages or a link to inductors for dummies some law of physics that shows the maximum amperage attainable with a 1.2V rechargeable battery that would be great.

I have larger inductors but I don't understand the math involved in making them work. I just lucked out getting what I have because I bought a bunch of different size inductors to play with and just started sticking wires in a breadboard.

Lee, I don't think I need anything close to the maximum output of the battery, I have tested a 1ohm across the terminals and get about 1 AMP. The most I need is 100mA, I tested that using a 10ohm resistor started at 120mA and after 5 hours was still giving me 112mA, that's with three in parallel.

Nuno, That Daycounter is cool but I'm at a loss as to how to effectively use it. I put in the variables to simulate what I am getting but it shows a huge L 18115.412710007 uH compared to the two 100uH inductors I am using. Like I said I don't yet have a clue as to inductance and kickback math calculations yet to understand it. I also put in variables to get me to 100 mA and it has numbers but I can't make sense of them yet.

Empire11, I think with the above additional information you can see what I am after here.

Round II anyone?

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More math than what's on the wikipedia is "too much" for me :)
In fact, I guess in real life nobody does the math; some guys at the semiconductor manufacturer's R&D (DC-DC converters) did it, tested them and then deduced simplified formulas for the components, which you typically see in the datasheets.
I'm doing a CCD camera on my free spare times and I'm planning to do the DC-DC converters (a Sony CCD needs 3 different voltages!). Knowning the basic parameters (mainly from wikipedia, which has a superb page on DC-DC converters), I used a SPICE simulator to simulate the converters. In practice I get much lower current output of my boost converter (I'm doing 12V->15V) than I have simulated; I don't know yet what part(s) has a weak simulation model. I'm still working on it, but to have an idea you can take a look at the circuit and graphs at http://nsj.no.sapo.pt/astro/ccd_... .

Embedded Dreams
One day, knowledge will replace money.

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metron9 wrote:
What I was looking for is some discussion on frequency , inductor size, capacitor size, type, diodes type size etc to increase the output of the converter.

I think you want to use a Schottky diode and not a zener. This page has all the math you should need.

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Yes I am using a Schottky, Yes there is quite a bit of difference in a Schottky vs a standard signal diode. I will bet the ones I have are not the best ones either to use. The Zener is just a safety valve so the voltage can't get above 10V. When there is no load it passes about 1.8mA at 10V to ground. When there is a load the zener has nothing flowing through it. Because the load drops the voltage under 3V. I was swapping out different LED's and blowing them up because the capacitor was charging to 75V.

I emailed the daycounter folks to see if they can design what I need.

What would be nice is a 5V 50mA dc/dc converter that will simply take a 1.2V input.

I have my doubts this can be done and I have tried to get the person I am working on this for to consider a higher voltage solar panel. The spark fun panel 8V is perfect combined with a SLA 6V battery except it is just a bit large for the "look" of the product.

I thought about using some digital switches to switch the batteries from parallel to serial during the light output phase where the solar panel when energized would open two fets and close 2 fets using a hex inverter then I would have 3.6V to start with and that would be plenty to do what I need.

It boils down too, he has a product he wants more light output I really think it will take bigger solar panel and bigger battery or a better circuit designer than me. I can do the programming though, I am designing a virtual rotating light with no moving parts for his product.

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Perhaps a chip like LT1932 could solve your problem?

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The maximum power you can suck from the battery will be limited by the maximum current you can store in the inductor and the maximum frequency at which you can switch the inductor.

The 2N7000 is a lightweight -use the FET with the highest current rating that you can without exceeding the maximum output current spec for the ATtiny2313. Then, increase the frequency and decrease the inductance until increasting the frequency or decreasign the inductance doesn't bring an improvement.

BTW, be careful with those new high output white and blue LEDs, as they can permanently damage your eyes.

--
"Why am I so soft in the middle when the rest of my life is so hard?"
-Paul Simon

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First of all, the battery is not the problem. Normal batteries can deliver 1A or more (although they drain fast at this current), rechargable should be able to handle even several amps. 200mA or something is no problem at all.

As for your problem, here are my comments:

  • The voltage is determined by the duty cycle. A 50% cycle will double the input voltage in ideal situations, which is probably never but you should be able to come close with this duty cycle. Due to the load and several electrical losses the output voltage will differ from the math. Now normal boost regulators have a feedback line that compares the output voltage to the desired voltage, and adjusts the duty cycle up or down to compensate for the difference. You don't have such a feedback line which means you cannot regulate the line exactly. Since your load is probably fixed you might tweak the duty cycle manually so that the voltage is correct for your situation, so it may work this way. Still, you are creating a voltage source (that is what boost converters are) and LEDs don't want that. They operate at a certain current, not voltage. So you still have to use a resistor in combination with the LED to prevent it from failing. I suggest you first of all get the thing to work without a LED but with just a resistor a the load. See if you can get enough current and then replace it with a LED and the appropriate resistor.

  • Batteries can deliver a lot of current but not at high frequencies. Their resistance gets higher as the frequency gets higher (this happens with almost any voltage source), it just is too slow to deliver a swithed 9.6MHz current. You will need some capacitors with low ESR on the input, parallel with the battery. It will be able to deliver the fast pulsed current much better. Preferably use low ESR capacitors (special electrolytes or ceramic), and at least some ceramic capacitors in parallel as well.

  • 9.6MHz is a really high frequency for a DC-DC converter. Most basic converters start at around 50kHz up to a few MHz. Higher frequencies reduce the need for larger components but it does put a lot more strain on the components. FETs need to switch faster and more often, capacitors work worse at higher frequencies, inductors have to be able to handle the frequency and so on. First try things out at say 300kHz. This does probably mean however that your output capacitor (which was 0.22uF?) needs to be a lot larger. Try 47uF for a start, larger is almost never a problem. Low-ESR is preferred as well.

  • It has been mentioned already that the 2N7000 is not a powerful FET. Still, you only need 20mA so this should not be a problem. It does have an on-resistance (RDSon) of several ohms, which causes extra losses. For efficiency you'd better use a power mosfet. There may be a problem with the threshold voltage though. Vgs is min/typ/max 0.8V/2.1V/3V. You run your AVR at double the battery voltage, which is around 2.4V, close to the typical threshold voltage but even below the maximum. If you look at the transfers characteristics in the datasheet you will see that at 2.4V it will just barely conduct. Even at 4V on the gate it can only supply 300mA of its maximum 400mA. This FET was designed to be driven with a 5V signal. You will need a FET that conducts enough even at 2.4V on the gate, which is quite low.

  • What kind of inductor are you using? Do you have a type number? The inductor should be suited for the frequency that you operate, and handle the current that goes through it. Also, its DC resistance needs to be low for efficiency and maximum current.

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ezcomp wrote:
Perhaps a chip like LT1932 could solve your problem?

or LM2623?

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Quote:
First of all, the battery is not the problem.

It depends on the battery type.
AFAIK NiCd and Pb are not very critical. NiMh i don´t know, but LiIon don´t like very high current.

Best is to contact the battery manufacturer for reliable inforamtion.

Klaus
********************************
Look at: www.megausb.de (German)
********************************

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True, I assumed something like NiMH (which has no problems with high currents either).

Still there are many other flaws in the whole thing, and this isn't even about high currents (its just to drive a 20mA LED!).

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The LT1932 and the LM2623 I think would both work well. Then I try searching for the chip inductors and minimum quantitys (digikey was the only one that shows the LT1932 and no hit on the other one through findchips.com)

It looks like maybe only 100 units of the product my light would go into per year are being sold right now. The converter chips are all surface mount tiny chips with lots of other components required. I am going to see what kind of charging I can get with the batteries in series as madwizard points out the many flaws to start with.

Thank's madwizard, that information on inductors is just what I was looking for to experiment with. I appreciate the time you have taken to explain it.

Thanks to all for their input. I shall continue to experiment with various solutions.

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Quote:
The inductor should be suited for the frequency that you operate,

* and handle the current that goes through it *

This is very important. Since you are boosting the
voltage, remember the current through the inductor
must rise to a value *higher* than your intended
average load current. The inductor must be rated
to handle this "peak" current without saturating.
If/when the inductor saturates, it stops being an
inductor and turns into a "heater", causing your
circuit to draw more current without giving you any
more light output.

The current rise in the inductor when the switch
turns on depends on the voltage applied, the inductance,
and time. The current with one volt applied to
a 1 Henry inductor rises at 1 amp per second.
Apply this simple relationship to your circuit
and you can clearly see what is happening (or
trying to happen).

Also, the "boost" circuit is not really a voltage
source whose output is proportional to duty cycle.
The inductor will try to maintain a constant *current*
(whatever the inductor current rose to during
fet "on" time)through the led when the fet turns off.
The voltage will jump as high as it needs to (as you
have already observed). The higher the load voltage,
the shorter time the current pulse will last.

Tom Pappano
Tulsa, Oklahoma

Tom Pappano
Tulsa, Oklahoma

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tpappano wrote:

Also, the "boost" circuit is not really a voltage
source whose output is proportional to duty cycle.
The inductor will try to maintain a constant *current*
(whatever the inductor current rose to during
fet "on" time)through the led when the fet turns off.
The voltage will jump as high as it needs to (as you
have already observed). The higher the load voltage,
the shorter time the current pulse will last.

You are right, but in a normal case an output capacitor (much larger than 0.22uF) is used which stops the voltage from rising and falling rapidly. This way the voltage can be held pretty much constant with the feedback circuit. In essence you regulate the current with the switch and inductor that needs to flow into the capacitor and load so that the voltage stays the same. If the voltage on the output is held more or less the same you will get an output/input voltage ratio that is related to the duty cycle. If you don't hold the voltage on the output you will get whatever voltage is needed to conduct the current, like you said.

If you just want to regulate current then you should probably remove the output capacitor, otherwise it will use current and build up a voltage which would make it hard to regulate a constant current through the LED. But this would mean that during the on-state of the FET (when it is disconnected from the battery) the LED has no means of supply, so it will be pulsed. On the other hand, if you do have a capacitor it would not limit the current that got out of it during the on-state and probably burn up the LED.

You could use the LED directly on the inductor without a resistor if you get the duty cycle right so that the current at the end of the on-time is the current you want flowing through the LED.
I hope I calculate things correctly but here it goes: There is 1.2V on the input, you have an inductor of 100uH and want 20mA through the LED. If you don't have a capacitor so that it will lit in fast pulses, the 20mA is also the current that will flow through the inductor (unlike a normal boost converter where it is higher, but then you have a continuous supply which demands more current). You will get a lower average output current this way but you can increase the 20mA if it is pulsed. Current in an inductor rises as follows: V = L * dI/dt. 1.2 = 100e-6 * 0.020/dt. dt = 1.66uS. So you would have to switch on the FET for 1.66uS to get a 20mA current at the end of the cycle. When you switch the FET off this current will still flow, through the LED and slowly fall down to zero. The voltage will change to whatever it takes for the LED to conduct. You could try this with an on time of around the 1.66uS and then measure the current through the LED (which is pulsed now so you may need a scope for that). And then increase the on time if the current is too low.

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Very cool, when I get home I will try it, yes i have a 100mhz digital storage scope, i would be really lost without that thing.

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I bought the 100uH inductors from here:http://www.futurlec.com/Inductor...

I dont see any datasheets on them though.

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Metron9,

You might find this article interesting/scalable also.

http://www.elecdesign.com/Articl...

Ross McKenzie ValuSoft Melbourne Australia

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Thanks for the link, that is similar to the circuit used in the garden light.

Last night I ran down three NiCD batteries using a 10ohm resistor and a luxeon 1W led. I measured 61mA when they started and 8 hours later it was down to about 4mA so the batteries were pretty drained.

I connected the three solar panels from the garden lights in series with a schottky diode to the battery pack. I just got home, the batteries are fully charged and measure 4.17V open circuit. The same resistor and led measures 94mA of current. This is at a time of the year that is the worst for solar power. I really did not think these batteries had a chance at being fully recharged (probably overcharged) as it's 4 days to the winter solstice.

I am glad I have learned allot about inductors and dc/dc converters. This project has just taken a turn, I now plan to add a second set of three batteries to test charge rates. I was initially wrong on my measurements of the solar panels in series obviously.

My plan then from here is to use PWM directly from the processor to generate the required power for the led's

I will post a picture of the finished light, the customer wants 50 to start with. I may use a mega48 because I have them. I will have to monitor the voltage charge on the NiCad batteries and I know that is a little more complicated than checking the charge status of a lead acid battery. But I am sure I can figure it out. I know I plan to use a comparator across the diode to monitor when the solar power is higher than the battery for the on off control.

I just wish there was more time in a day so I could get more experimenting done.

Thanks again for all the help, I am sure I will be using inductors in some future project.

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Quote:
Last night I ran down three NiCD batteries using a 10ohm resistor and a luxeon 1W led. I measured 61mA when they started and 8 hours later it was down to about 4mA so the batteries were pretty drained.

going from 61mA down to 4mA at 10Ohms is only a voltage change of 570mV.
One has to look into datasheets if the batteries are really discharged.

Also mind the voltage drift vs. temperature.

Klaus
********************************
Look at: www.megausb.de (German)
********************************

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I have used LM2623, sorry about the bad picture:

http://www.national.com/ds/LM/LM...
I did exactly like it is put out in a first page.

http://www.national.com/pf/LM/LM...
...roll down and you'll find a few app.notes.

ELFA: http://www.elfa.se/en/

LM2623 nr. 73-068-49

Coil 4.7uH nr.58-817-27
http://www.elfa.se/pdf/58/058815...

Regards
heguli

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I have had great success lighting white leds using this connection:
http://www.emanator.demon.co.uk/...
Parts are cheap, but you need one for each led and winding those ferrite rings is timeconsuming. The connection accepts really low voltages. :)

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Perhaps I've misunderstood, but I think this is really about the design of switching supplies, not about battery capacity. Maybe the OP wants a brighter LED.

Switching regulators are about trading Voltage for Current or the opposite by storing energy in a magnetic field and then using it.

Some general principals:

- The higher the inductance the lower the switching frequency. You need to store more energy every cycle if the cycle rate (frequency) is lower. ( a larger bucket is needed to move more water if you use it less frequently).

- once energy is stored in the field, a higher voltage can be extracted by presenting a higher impedance load. similarly, higher current at a lower voltage can be extracted.

Question: would the OP be better off using an off-the shelf switching regulator instead of rolling his own?

HTH
DFR