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Using a Joule Thief as a bootstap

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Hero999

Banned
Just an idea but you could use a Joule Thief as a bootstap to allow an ordinary boost SMPS IC to work from a single AA cell draining it until it's dead.

This idea only uses the Joule Thief to get it going, once the output voltage is up to 5V the Joule Thief is shut down and the controller IC uses the output as a power supply.
 

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Sceadwian

Banned
Depends on the voltage the SMPS requires to start up and what it will actually run down to once it's started. Not sure if it'd be worth it. Sounds like an expensive tack on to get a marginally more amount of power out of a single AA. That's what Joule thiefs are after all, use 2 AA's and almost all your problems go away.
 

Hero999

Banned
The LM2585 here works down to 4V.

Yes you can buy some ICs that work down to much lower voltage and yes adding an extra AA solves the problem.

Expensive? I think you'll find an extra AA is more expensive than adding this circuit.
 

Sceadwian

Banned
With an SMPS an extra battery is a win win. Generally batteries are the problem of the consumer. As long as the efficiency is high you're adding free up time to the device, unless space is at a premium in which case you wouldn't be using AA's anyways you'd be using lithium cells which again solves the problem. Considering how much current is available once a battery gets that low too though I'm not sure how practical it would be. You'd need a micro power SMPS and effciency would be really low. I'm not sure what the internal resistnace of a AA is when it's nearly dead but given the voltages almost any resistance will really kill your ability to draw any useable current out of it. Joule thiefs are nice but once they battery gets that low you're not gonna get much out of it.
 

Quantised

Member
Hero999 - Your post started me thinking and so do you think its possible to build a joule thief that can extract enough juice from a dead car battery to start the engine? That would be useful bit of kit, some fairly large components maybe but is it possible?
Quantised
 

Hero999

Banned
That might be possible but I don't know how much current a dead car battery can provide and it certainly will kill the battery.
 

Nigel Goodwin

Super Moderator
Most Helpful Member
Hero999 - Your post started me thinking and so do you think its possible to build a joule thief that can extract enough juice from a dead car battery to start the engine? That would be useful bit of kit, some fairly large components maybe but is it possible?
Quantised
I wouldn't have thought so, you need a huge amount of energy to start a car, and a flat battery wouldn't contain enough. Bear in mind the reason the car won't start is because the battery can't provide enough current, so asking it to provide a great deal MORE current isn't going to work.
 

Quantised

Member
Maybe it is possible, the battery still has some amps left in it but the volts are down to say 8 or 10 on a 12v car battery, When you engage the starter the current draw is so high the volts drop too low to provide enough power but if the joule thief was able to up the volts it would be able get enough power - albeit for a very short time so better be sure the engine always starts on first kick. If you look at the internal resistance of a car battery it seems this is the show stopper but if the volts are raised the amps drawn will be shortened - is this how the new megajoule thief will work?
 

Hero999

Banned
If the voltage is raised, the current will also increase significantly so I don't think you'll gain anything.
 

Quantised

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If my starter takes 300amps at 12v (depressed to 10v due to internal battery resistance) then it supplies a theoretical 3KW but at 8v (6v depressed) it will only draw 180amps and so provide only 1.08KW which is not enough to spin the engine.

As most decent engines start within 2 seconds that means a battery with only 1 amp/hr left in it could provide 1800 amps for 2 seconds - if the new megajoule thief can keep the voltage up.

As you can only draw the amps if the voltage is high enough then it seems reasonable this will indeed work.
 

Hero999

Banned
The internal resistance might stop it from doing so.

3kW at 6V is 600A assuming 80% efficiency so the voltage drop will be more like 4V.

Also a tired car battery will have a higher impedance.

Maximum power transfer becomes a factor, if the voltage and internal resistance reach a point so that when the battery voltage is halved (by the current being drawn) the output power is under that required to start the engine, it won't work, whatever you do to it.

Your battery has an impedance of 6.6667mΩ.

Maximum power transfer will occur when the load impedance is equal to the battery's impedance, at this point the battery voltage will drop to half the open circuit voltage.

At what open circuit voltage will the battery be no longer capable of supplying the load with 3kW?

Calculate the voltage across a 6.6667mΩ load dissipating 3kW.

V = √(P×R) = √(3000×6.6667×10^-3) = 4.472V

This is the voltage across the load dissipating 3kW, the open circuit voltage will be twice this value so multiply by 2.

4.472×2 = 8.944V

So when the battery voltage drops below about 8.9V it will not start the car, adding a voltage booster won't do help, if anything to help, it'll make it worse. This is because: increasing the load on the battery will cause the terminal voltage to drop further causing the power dissipated by the load to be reduced.

If you do the calculations again using a DC-DC converter which is 80% efficient, the minimum start voltage increases to 9.8V.

So if your motor won't start when the voltage is 10V, the boost converter will only be any good until the open circuit voltage drops below 9.8V; is 200mV of extra starting capability really worth it?

Even if the converter is 100% efficient, the best you're going to get is 8.9V
 

Quantised

Member
We are on a big time shift here so its late now, I will need to study this in the morning.

The internal resistance of my batteries are dislayed directly at any voltage and load so I want to record this data in the morning. Its not a car but I used used that for simplicity and commonality as it is a case more likely to be seen widely.

I want to check if the specific gravity (SG) of the electrolyte influences the internal resistance so that as the battery is further discharged does it have a higher impedance. If it does then things looks bad for the megajoule.

I wonder if an SG of say 1.1 at low discharge versus 1.28 at fully charged influences the battery to provide the amperage drawn as you indicate may be the case. Hmmm.
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ronsimpson

Well-Known Member
Most Helpful Member
TPS61220 works down to 0.7 volts. There are many parts that work to 0.9 volts. Below 0.9 volts there is not much energy left in a battery.

I get pushed to make my products work below 0.7 volts but there is so little energy left that it only adds minutes to the battery life.
 

Sceadwian

Banned
Great for blinking or running a an LED though =) I used a fully charged Carbon C cell (accidentally bought carbon cells which are useless for most real devices) and hooked that up. The LED stayed lit for around 3 days, and when the voltage got low enough that it stopped oscilating it would start back up as the battery voltage recovered so it blinked for another 3-4 days, very slow and very dim though.
 
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Triode

Active Member
I think a preventative method would be more practical with the car battery. Because what you would be doing is hoping that the amount left after draining the battery by normals means could be boosted to start the engine, you would have to turn off this device after every use anyway. Something that cut off the battery when it was being drained too far without the engine running could be made, and I don't think it would be that difficult. It would need a feature that checked if the engine was running, because if you're alternator is out and your trying to make it to the repair station without having to stop, it would be a sucky time for the battery connection to be opened.
 

Sceadwian

Banned
Food for thought. Find out the capacitor value required to generate the starting current-time for a car. Super caps aren't cheap but they are available. A bank large enough to store the starting current of a car I don't think would be very large.
 

Quantised

Member
Also a tired car battery will have a higher impedance.

Maximum power transfer becomes a factor, if the voltage and internal resistance reach a point so that when the battery voltage is halved (by the current being drawn) the output power is under that required to start the engine, it won't work, whatever you do to it.

Your battery has an impedance of 6.6667mΩ.

Maximum power transfer will occur when the load impedance is equal to the battery's impedance, at this point the battery voltage will drop to half the open circuit voltage.

At what open circuit voltage will the battery be no longer capable of supplying the load with 3kW?

Calculate the voltage across a 6.6667mΩ load dissipating 3kW.

V = √(P×R) = √(3000×6.6667×10^-3) = 4.472V

This is the voltage across the load dissipating 3kW, the open circuit voltage will be twice this value so multiply by 2.

4.472×2 = 8.944V
Hero - I managed to measure the impedance at 90% discharge and it was about 40% worse than for a fully charged battery. One must assume if you go beyond the batteries stated capacity and into the real flat area it will get even worse. That would - as you say - be the demise of the megajoule.

But I would ask you explain for me why the open circuit voltage will be twice that of the loaded circuit? Why the exact factor of 2?

I would like to understand this as we use the amps currently being drawn corrected for the internal resistance to get an open circuit voltage, we then know how much power is left in the battery. As my main source of power is batteries it is quite important to get it right. Thanks in advance.

Maybe Sceadwian had an idea to steal the last remaining power to charge up the capactor bank and use that for starting - a nice solution but impractically expensive - still it was just a thought.

The alternative is what the real world use, often we see guys starting engines up to 115HP or even the 150HP with a rope on the flywheel. The Yamaha engines have a manual start option for flat batteries by removing the flywheel cover, but you do need to be quite a strong guy to pull it!!!
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Hero999

Banned
Hero - I managed to measure the impedance at 90% discharge and it was about 40% worse than for a fully charged battery. One must assume if you go beyond the batteries stated capacity and into the real flat area it will get even worse. That would - as you say - be the demise of the megajoule.

But I would ask you explain for me why the open circuit voltage will be twice that of the loaded circuit? Why the exact factor of 2?

I would like to understand this as we use the amps currently being drawn corrected for the internal resistance to get an open circuit voltage, we then know how much power is left in the battery. As my main source of power is batteries it is quite important to get it right. Thanks in advance.
Maximum power transfer theorem dictates that the maximum power is transferred to the load when its impedance is equal to the source.

At the point of maximum power transfer, the terminal voltage will be half the open circuit voltage due to the potential divider effect. If you connect two resistors of the same value in series, the voltage across each resistor will be half the voltage across them both.

If we know the power required by the starter motor and the impedance of the battery, we can work out what the minimum terminal voltage can output is required to output the required power.

A boost converter can vary its resistance to draw the power required by the load. At the minimum terminal voltage required to drive the motor, the boost converter's resistance will be equal to that off the battery. Once the terminal voltage becomes too low, the boost converter won't be able to draw enough power from the battery. Its resistance will drop below that of the battery and more power will be dissipated by the battery than the load.

Another thing that makes the situation worse is that the impedance of the battery is increasing further as it's discharged.
 

Quantised

Member
Thanks Hero - that all makes sense now and also answers why my 220AHr battery is better at starting than the 800AHr deep cycle unit, it has a much lower internal resistance - by a factor of nearly 10. Altough it is bigger now I see it cant deliver the amps at the higher voltage and so generate the power required, always wondered about that!

Thanks for you answer
Q
 
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