EEVblog #811 - How The Varta 15 Minute Battery Charger Works

[Don Hills]

... But Dave has discovered that the displayed voltage is lower than the measured voltage...we should investigate why this happen...?

Maybe it's measuring (and displaying) the voltage during the momentary "off" period?

Way back in the day of NiCD cells, I built a charger that charged at 1C for 55 seonds, then discharged at C/10 for 5 seconds. It measured the voltage at the end of each cycle and used the difference to determine charge state. I found that the difference between charge and discharge voltages was fairly constant until the cells reached full charge (confirmed by their temperature starting to rise), at which point the difference voltage as much as doubled.

[NiHaoMike]

Since resistive losses increase with the square of the current, I'm surprised the "4A" with 3 or 4 cells is implemented as a 50% duty cycle of 8A switched between the two banks. If they just did a constant 4A, it would theoretically cut the resistance losses in half.

[drussell]

Way back in the day of NiCD cells, I built a charger that charged at 1C for 55 seonds, then discharged at C/10 for 5 seconds. It measured the voltage at the end of each cycle and used the difference to determine charge state. I found that the difference between charge and discharge voltages was fairly constant until the cells reached full charge (confirmed by their temperature starting to rise), at which point the difference voltage as much as doubled.

That is a valid way of measuring and monitoring the true cell voltage during a charge cycle and should be quite accurate. 

I'd be willing to bet that your charger worked quite well on the vast majority of cells.

[coppice]

... But Dave has discovered that the displayed voltage is lower than the measured voltage...we should investigate why this happen...?

Maybe it's measuring (and displaying) the voltage during the momentary "off" period?

Way back in the day of NiCD cells, I built a charger that charged at 1C for 55 seonds, then discharged at C/10 for 5 seconds. It measured the voltage at the end of each cycle and used the difference to determine charge state. I found that the difference between charge and discharge voltages was fairly constant until the cells reached full charge (confirmed by their temperature starting to rise), at which point the difference voltage as much as doubled.

I think the 15 minute recharge NiCd battery packs and chargers Vivitar made for some of their flash guns in the 80s worked in this way. They used special batteries with heavier conductors, and, of course, they were charging at about 4 times your rate. They didn't need any cooling fans, and the batteries didn't get very warm.

[rr100]

You are the kind of people who like to cut hair in four !!! 

The internal resistance of the battery increases when the charging current is higher because the formation of bubbles of H² on the electrodes is greater and reduced the active area.

It isn't about cutting hairs, it is about not making unfounded assumptions. The internal resistance is not from a given resistor inside the battery, it is just an imaginary concept that help us to model how the battery behaves. It is nothing more and nothing less than the proportionality constant between the delta-v under charge and the charge current. Once you say "oh, but this constant isn't constant" there is no point in continuing with any argument that assumes this proportionality (including specifically the one about the loss being proportional powere-wise with the square of the current while the "charge" is proportional with the current).

[f4eru]

Hello,
The 15V gate drive of the mosfet is because with that arrangement they don't care that the 6V or so battery voltage is substracted from the gate, i.e. they don't need level translators.

[oldway]

It isn't about cutting hairs, it is about not making unfounded assumptions. The internal resistance is not from a given resistor inside the battery, it is just an imaginary concept that help us to model how the battery behaves. It is nothing more and nothing less than the proportionality constant between the delta-v under charge and the charge current. Once you say "oh, but this constant isn't constant" there is no point in continuing with any argument that assumes this proportionality (including specifically the one about the loss being proportional powere-wise with the square of the current while the "charge" is proportional with the current).

Your argument is nonsense. 
The internal resistance is not an imaginary concept.
It is real and due to internal connections of the battery and the ohmic resistance of the electrolyte.
This is not because the resistance can vary by a few percent that the fundamental principles of electricity are no longer aplicables.

[rr100]

If you are talking about the ohmic resistance of the electolyte itself then (even if this would be reasonably constant) it is incorrect to say the waste is proportional with the square of the current because this resistance isn't in series with the circuit, but in parallel. Actually, given the fact that the voltage doesn't increase that much (let's say at most 1.5 times) the "waste" on this "resistor" would increase just 2.25 times. That is even if you increase the current 100 times for example from 80 mA to 8000 mA.

The resistance of the internal (and not only) wires yes, that's correct - the wasted power does behave like you said, roughly.

In any case while the discussion is interesting I hope you agree that from the practical standpoint for people with access to "line power" it just doesn't matter. You'd be using much more electricity if you turn on the lights to take the batteries from the dead device and put them into the charger or even to RTFM :-)

[oldway]

If you are talking about the ohmic resistance of the electolyte itself then (even if this would be reasonably constant) it is incorrect to say the waste is proportional with the square of the current because this resistance isn't in series with the circuit, but in parallel.

The charging current passes through the electrolyte and his resistance is in series with the circuit and not in parallel.
NB: everybody knows that the internal resistance of a battery varies with the concentration of the electrolyte 

Actually, given the fact that the voltage doesn't increase that much (let's say at most 1.5 times) the "waste" on this "resistor" would increase just 2.25 times. That is even if you increase the current 100 times for example from 80 mA to 8000 mA.

Your reasoning is wrong.
Basic equation of a battery voltage with charging current I is:
U = E + RI
Joule losses/s (W) are RI²
E is far greater than RI.
For this reason, you can't base your reasoning on the increase of U to calculate the % increase of joule losses.

[f4eru]

Since resistive losses increase with the square of the current, I'm surprised the "4A" with 3 or 4 cells is implemented as a 50% duty cycle of 8A switched between the two banks. If they just did a constant 4A, it would theoretically cut the resistance losses in half.

Yes, but it would cost much more in the inductor of the DC/DC converter to allow for the higher dynamic in voltage
The additional loss per mosfet is in the order of 72mW, which is negligible, especially considering they have fan cooling

[VK3DRB]

NB: everybody knows that the internal resistance of a battery varies with the concentration of the electrolyte 

Everybody? Absolute rubbish. Most one year old babies would not even know what a battery is, let alone anything about the internal resistance varying with the concentration of the electrolyte. Even Justin Beiber who is "worth" $200 million might not know either. In fact, it would not surprise me that less than 1% of the world's population know.

The tone of some of the comments in this thread reminded me of when I walked into an electronic components store in 1979 to buy a replacement 6X5 vacuum tube (valve) for my Kikusui oscilloscope for $2.15. The old valve's filament had gone open circuit. Another customer, whom I had never met before, was standing next to me. He big noted himself and belittled me by raising his voice in public something like, "A 6X5 valve is a bridge rectifier valve! Why don't you just use a couple of diodes! You're hopeless. Ha Ha Ha. I guess we all had to start somewhere." Embarrassed as I was, I was tempted to give this know-all a black eye. Instead I tried to explain I did not have a schematic for the oscilloscope and it was not worth the time mucking around, but this bombastic clown just laughed in contempt.

Assuming someone else hasn't already "taken him out" in a drive-by shooting or laced one of his junior burgers with cyanide, it would be poetic to find he is has been long term unemployable due to his inability to get along with people. I went on to become an electronics engineer as a life long career, and whilst I don't know it all and am learning all the time, I certainly know a lot more than I did in 1979. I have come across similar dick heads to Mr. 6X5 in business, but fortunately they are rare. I fired one once.

By the way, analysing Mr. 6X5's argument of using the rectifier diodes...

- I would need a schematic to the oscilloscope or a valve datasheet to work out where the diodes go
- I would need two diodes that are rated for the PIV, and current drain for B+ line in the oscilloscope
- I would probably need some resistors as well
- Back then diodes were not that cheap, around 60 cents each
- Spending an afternoon for the sake of saving $1, hoping I got it right is not time put to good use. Instead with the 6X5 valve, I just plugged it in and the Kikusui came back to life.
- The new 6X5 would have outlasted the old oscilloscope anyway

So even if it happened today, I would have done the same thing in buying the 6X5 valve. The only thing different is I would have wiped that smile off Mr 6X5's face 

(I am not saying you guys are like Mr. 6X5, but I do come across them electronics and programming weblogs and on Youtube. I think it is called trolling.)

[Don Hills]

... That is a valid way of measuring and monitoring the true cell voltage during a charge cycle and should be quite accurate. 

I'd be willing to bet that your charger worked quite well on the vast majority of cells.

Yes, it was quite reliable. It was "microprocessor driven" - an XT class PC. I used pins on the parallel port to switch the charge/discharge relays, and a lamp / LDR connected to the joystick port to measure the voltage. The lamp conveniently applied the C/10 discharge load. It was all written in BASIC.

[oldway]

Everybody? Absolute rubbish. Most one year old babies would not even know what a battery is, let alone anything about the internal resistance varying with the concentration of the electrolyte. Even Justin Beiber who is "worth" $200 million might not know either. In fact, it would not surprise me that less than 1% of the world's population know.

You should understood that this mean "everybody on this forum" with the exception of those involved in the "beginners" section of this forum.
So far I know, one year babies, nor even Justin Bieber are not posting in this topic. 

When you have bought a Varta 15 minutes charger and you want to proove that charging batteries with high current rate does not causes higher losses, only because you don't like to read that, and for this, you use arguments totally absurd and against the fundamental laws of electricity, that's what I call to troll.

NB: speaking of trolling, what your post has to do with "How The Varta 15 minute Battery charger Works" ?

[rr100]

you want to proove that charging batteries with high current rate does not causes higher losses

It is you who came wanting to prove something and in fact provided just hot air and a verbose and strong worded answer to everybody doubting you.

As I said it is debatable. If you charge for example with a slow charger that has inside a regular transformer and you burn some watts for 16 hours you'll be about the same as going full tilt some tens of watts for 15 minutes. That is in practical terms.

If you want to prove your point the stage is yours. But please come either with experimental data or a proven model that can be applied here. No, I do not accept U = E + IR as a good enough model for a battery charging, and certainly not across orders of magnitude for current. If you have something better than "you are wrong" to back it up we can discuss it.

[oldway]

No, I do not accept U = E + IR as a good enough model for a battery charging

U = E - RI (discharge)
U = E + RI (charge)
Are basic equations for batteries, as U = RI is for Ohm's law.
I don't have to prove this !
If you don't accept them, whe don't have to loose our time discuting here.

[mikerj]

As I said it is debatable.

It absolutely is not debatable, this is very basic stuff.  If you increase the charging current, you increase I2R losses.

[rr100]

That is for the wires and anything resistive between the battery and power supply, yes - I agreed to that.

By the way many of the slow chargers have a linear regulator and the really slow ones I was referring to earlier have just a resistor as "regulator" - they just supply some voltage high enough that the voltage change of the battery and the variation of the internal resistance just won't have much effect. Frankly I find it totally ridiculous that we're fighting over whatever power loss might be over those tens of miliohms from the batteries and cables when nobody complained over 200% of the energy wasted on any dumb slow charger.

[IanB]

That is for the wires and anything resistive between the battery and power supply, yes - I agreed to that.

No, it is for any resistive losses inside the battery. As mikerj says, this is completely fundamental and not open to debate. When current flows through a device and there is power dissipated as heat, then the I²R law applies for the dissipated quantity, where R is a proportionality parameter. Of course R may vary with temperature, time and other factors, but to a first order approximation we may assume R is a constant.

[rr100]

What if the voltage doesn't increase proportionally with the current (like is the case here)?

[IanB]

What if the voltage doesn't increase proportionally with the current (like is the case here)?

But the voltage associated with resistance losses does increase proportionally with the current. What makes you think it doesn't?

[rr100]

As I said, I already agreed to the fact that the RESISTIVE loses, practically by definition, behave as we'd be expecting from a resistor.

BUT what I try to say is that those are just A PART of the loses when charging a battery. An eneloop has a specified 25 mOhm internal resistance [edit: AA "regular" ~2Ah ones]. If you charge at 0.2A that's a theoretical 5mV drop. To make things simple let's take the nominal voltage of the battery at 1V (we'll be exaggerating all relative losses by about 20%) so we're looking at 0.5% loss compared with the energy that goes into the "battery". Does this mean the charging is 99.5% efficient? Of course not, and we didn't even start to think about the charger. At 0.2A and 5 mV that's a 1 mW loss. Are we looking in the right place? 1mW is down in the noise compared to what a charger eats just by sitting in the wall.

YES, if we want to take THIS 0.5% loss and say that we waste 20x more if charging at 4A and 40x at 8A yea, sure, that's correct. But a slow charger (or for that matter ANY charger lest in the wall) can easily and plausibly waste 3000x.

Back to this one for example eats 80mA (@15V) just sitting there. That's 1.2W, you'd need theoretically more than 10xAA eneloops to feed it ever day if you just leave it sitting there. And that is at the DC output, we need to add the wall-wart too but I can't measure that power with any reasonable degree of accuracy.

[oldway]

You are an incredible bad faith.

To justify your mistakes regarding batteries losses during charging, you now mix it with losses of the batteries charger.

But again your explanations are wrong.

We can now make low power switching power supplies (eg with STRA6069H SANKEN) that are as efficient (efficiency, no-load current) as SMPS 20 times more powerful (for the same output voltage).

Your reasoning is wrong again.

The 0.4A charger losses will therefore be 20x lower than the 8A charger but last 20 times longer, which means that there is no difference in energy loss regarding battery chargers.

In fact, there is a small difference in favor of the 0.4A charger as the charger 8A will have to provide more energy to compensate the additional losses from the battery, which will cause some additional losses in the 8A charger.

[rr100]

I do not try to justify anything, I'm just trying to clarify that your oversimplification ("huge waste of energy") based on the fact that you lose 20x or 40x more over the RESISTIVE LOSES IN THE BATTERY is an exaggeration. Yea, sure if you're using a 1/10C charger for a 2Ah battery you're pissing away only 1mW "resistively", sure. And energy-wise you'll be using 40x more with a 4C charger. But it won't be 40 times the TOTAL losses, it will be 40x those tiny resistive loses. There are more loses in the battery itself (no, you don't get a charging efficiency of 99.5% with NiMh, even at 1/10C), losses in the charger, etc. You need to be looking at the whole package.

[mikerj]

YES, if we want to take THIS 0.5% loss and say that we waste 20x more if charging at 4A and 40x at 8A yea, sure, that's correct. But a slow charger (or for that matter ANY charger lest in the wall) can easily and plausibly waste 3000x.

Only if it's left switched on continuously after the battery is charged, which is completely irrelevant to the charging efficiency of the battery.  If you apply the same conditions to a slow and fast charger (i.e. charge battery to the same level) the slow charger is always the more efficient one. 

You might want to read this PowerStream article:

The coulometric charging efficiency of nickel metal hydride batteries is typically 66%, meaning that you must put 150 amp hours into the battery for every 100 amp hours you get out. The faster you charge the worse this gets.

[rr100]

Only if it's left switched on continuously after the battery is charged

DURING the charge, of course. Do you think it is implausible for a charger to lose 3W (that is 3000x the 1mW waste) without doing anything? This one eats 1.2W without taking into account the wall-wart.

If you apply the same conditions to a slow and fast charger (i.e. charge battery to the same level) the slow charger is always the more efficient one. 

First of all there are no "same conditions" in practical terms between a 4C charger and a (let's say) 0.1C charger. They'll differ wildly and those differences would matter much more than the internal resistance of the battery. But even if we have two chargers as close as we can build them, let's say that this Varta charger has a 0.1C mode. It'll waste only (see previous approximation) 0.5% energy compared to the battery capacity when charging at 0.1C while it'll waste 40x more at 4C. That's 20%. FINE. Now please estimate how much it'll waste by eating even one extra watt for the rest of the 10-16 hours extra it takes to charge at 0.1C! How much is that, x00%+ ?