Hi
I am charging a single cell LIPO using a wireless charger and a regulator module. Here's a diagram.



Everything has to be permanently connected because it's going to be encapsulated. The problem is that when the LIPO is not charging (i.e. the receiver coil is not near the transmitter coil), everything to the left of the battery on this diagram drains about 15uA - which is not much but it's significant given that the battery has such a small capacity. Any ideas of how to reduce this? I feel it ought to be possible with a diode so that the current can only flow from left to right, so to speak, from the regulator to the battery rather than in reverse, but when I put a schottky diode in the circuit virtually no current flowed in either direction.

 

15uA is too much drain !!!??? ... A typical capacity for a small Li-po is 2.5 Ahr ......... The drain from your circuit in one week is 2.52mAhr

So in 10 weeks you have only lost 1% capacity .... the listed self discharge for these cells is 1% in 3 weeks which is amazingly low , if they have a safety circuit the drain is 1% in 1 week...

Older cells have higher self discharge , the main factor is age of cell , your circuit has an insignificant effect

 

Chip datasheet Me4056...

https://www.google.co.uk/url?sa=t&s...FjABegQIBxAB&usg=AOvVaw1WkGWhvKYd6p48DZwTxYht

 

Thanks
The batteries I am using are 30mAH not 2.5AH
The UUT in sleep mode draws about 5uA - i.e. the battery will drain in about 6000 hours which is good (not taking into account self-discharge)
Adding in everything else brings it down to 1500 hours which is not so good. The application requires the battery to be charged as infrequently as possible. Appreciate the point about self-discharge but I still want to minimise the draw.
The datasheet says 2uA when the supply is removed but that's not my measurement. It's possible that the main draw is coming from the receiver module (I can't tell what chip this is - it came with the coil) but I haven't measured this yet.
Why is it not possible to use a diode to allow current to flow into the battery from the charger (and receiver), and out of the battery into the UUT, but not out of the battery into the charger?

 

wireless chargers are always less efficient than those with wired connections. Thus no matter what you will suffer inefficiency. For this particular application, if you are able to put a magneticly operated reed switch in to disconnect the charger from the battery when the device is not near the charger and magnet, that could be one way to solve your problem. It would be simple and not waste any power.

 

15 µA does seem far too high. The datasheets from two 4046 device makers both say 2 µA maximum if the input supply is not present, but that may be at 25 °C and might increase at higher temperature, depending on whether it "leakage current" in semiconductors or there is a resistive path. For your cell and average load, even 2 µA is a nasty overhead. I have seen claims there are "counterfeit" 4046 devices around. Perhaps they behave badly.

Are you using the 4046 device in a circuit of your own design or a commercially made module of some sort? If you are using your own circuit, can you post a schematic?

It is probably worthwhile to try to determine where the 15 µA is going - just "disappearing" (to ground, presumably) in the 4046 circuit or somehow making it out of the input of the 4046 circuit.

It is probably possible to reduce the current to well under a microamp by means of a carefully chosen MOSFET switch. The problem with most "power" MOSFETs is that the OFF state leakage current isn't very well specified. An analog switch, such as the TS5A3166 (haven't looked carefully at the data; you have to watch for issues like "leakage" current when there is no power to the switch) might be better suited, though you might actually have to keep it powered with your cell (still an improvement on 2 µA). If you can tolerate a rather bulky part, you can probably find a opto coupler with a FET output. In that case the OFF resistance should be high and there would be no supply current from your cell. The input supply would probably have to deliver at least a milliamp or two to turn the switch on.

Why the conventional diode should completely block charging is not clear to me. Normally the 4046 starts charging at power-up by essentially testing the battery to see if it will come up to a threshold voltage by using a very small current. I would expect that the diode wouldn't interfere with that.

 

Once again, if any of the excess drain is coming from the power receiver section, a reed switch for isolation will work very well with less loss than any other device. The extra burden is needing a magnet. Perhaps that is an OK trade-off.

 

ebp - Thanks. It's a commercially made module, not my own. There are pictures and a description of it here:
http://budgetlightforum.com/node/57414
But they are available all over ebay. It looks to me very similar to the schematic shown in the datasheet.
I have changed the resistor on pin 3 to 80k to bring the charging current down to 15mA (0.5C for this cell) according to the formula in the data sheet
There is no current drain from the "receiver" unit attached to the coil, so the problem is all in this charging module

(There's no room for a reed switch unfortunately, and also they are not robust enough for this operation since the unit will get some pretty rough handling.)

Can you help me with the diode question: are you saying that my idea of a diode between the 4056 module and the cell OUGHT to work and you can't see why it doesn't; or that you can't see why I should imagine that it ever WOULD work? If the idea is essentially sound, can you suggest a diode that I might try? I thought a Schottky one - I used 1N5819 - would be best because of the lower voltage drop but as I say no significant current passes when it is forward biased.

With the TS5A3166 are you suggesting something like this below? So it is only switched on when there is a signal from the receiver unit. If so, I don't see why it can't also get its power from the receiver unit rather than the cell. In any case, as far as I can see from the datasheet, the leakage currents when off appear to be in the nA range




Many thanks again

 

I guess:
A relay will also do the job (SSD would be with lower losess), but no matter what you use it will still have losses. If we consider your device will be 50% on and 50% off, its the losses when on that you should also consider (depends on how strong your wireless signall is). An easy solution is a second battery. As long as there is the coil has lower potential it will keep draining the current from the battery. If you find a way to stabilize the voltage before the battery to a little higher one, it will not drain anymore. This is electrical diffusion.

A bipolar transistor and the battery in the emitter will work, but it will still have losses. An impulse MOS FET control with mOhms of resistance is your best option. It can be a PMOS one.

 

No, there's no room for a second battery or a relay. Anyway, the whole device is sealed (that's the reason for the wireless charging) so I'm not sure how you'd charge the second battery.
Unless I'm missing something, there seems to be some misunderstanding (in the thread). I know that wireless transmission is inefficient. That doesn't matter to me at all. So long as the battery can charge up in a reasonable time I don't care at all about those losses. It's the losses that occur while the battery is not charging that are key - which is almost all the time. Charging should be a pretty rare event.
The scenario is something like this: the device is on for maybe an hour a week. The current draw while it's on is maybe 0.7mA
The rest of the time it's in sleep mode, and the draw is around 5uA. So an extra 15uA is significant and undesirable.
The MOSFET solution is worth looking into if the TS5A3166 idea doesn't work. Thanks

 

Sorry - I'm saying I can't see any reason why the diode shouldn't work. There is nothing I can see in the datasheet that suggests the circuit won't at least try to start with no battery connected. The blocking diode would make it appear that there was no battery in terms of there essentially being no voltage on the BAT pin, but as soon as the chip enters the trickle charge phase, the voltage would be pulled up. The drawback to the diode is that it will reduced the final voltage on the battery by a few hundred millivolts. That is actually beneficial from the point of view of the number of charge-discharge cycles the battery will deliver and actually will yield greater total lifetime capacity (in terms of total deliverable coulombs) - but shorter run time on each charge.

Is there any chance you weren't getting any current into the battery just because it was near full charge and with the diode in series the charger chip would interpret that as truly fully charged and end charging?

Caution with Schottky diodes - relative to PN junction types, the reverse leakage current is usually considerably higher.

There was something I saw in the datasheet for the analog switch that gave me the notion that the leakage current would be higher in the unpowered state, which is why I suggested it might need to be powered from the battery. I really only skimmed the datasheet, so I'm not at all sure of that. It was also the first switch in that general class that I looked at at all. There may be others that are better. In these sorts of applications it really is necessary to study the datasheet carefully to be sure there isn't some detail that could bite you when you use it something a little different from the usual application for the part. Again, the reason I suggested an analog switch is primarily because they are well specified for OFF state current, whereas small general-purpose MOSFETs tend to have spec's along the lines of "it'll be less than X, but we don't make any promises as to actual value." Picking the right part can be quite time consuming because of the need to look at all the little details for each part you consider.

Be careful with the analog switch not to exceed the voltage rating on any of the pins. This is where powering it and driving the control input from the power input to the charger board may be an issue. When I was doing a quick search for possible parts at Digi-Key, there were some parts with a name that suggested they could handle voltage outside of the supply rail voltage, but I didn't look at them.

You probably could use an ordinary NPN bipolar transistor between the battery negative and the "ground" of the charger board - collector to the battery, emitter to ground, base with perhaps 20k or so to emitter (to be sure the transistor is OFF when input is off) and one to the raw voltage into the charger board, sufficient for about 0.5 mA of base current (with 15 mA of collector current). This would operate the transistor in saturation when the input supply was present, costing only a few tens of millivolts. The OFF leakage current would be very low. An ordinary 2N3904 or similar could be used. The drawback is the base current and the small collector-emitter voltage drop, but the part is extremely cheap and the collector leakage current is spec'd at 50 nA maximum at 25 °C. You could find even lower leakage BJTs - types intended as amplifiers rather than switches tend to be a bit less leaky.

People do have a tendency to see things in terms of what is familiar to them and go off on tangents. I've done it myself. From my point of view, the only detail missing in your original post was the cell capacity, and I rather assumed that if you were concerned with 15 µA you had a small cell and could do arithmetic.

I know someone who had to do a charger permanently attached to a very small lithium cell. I don't know what charger chip he used, but I'll ask next time I talk to him. It may be a few days. A different charger chip might fix the entire problem.

 

Yes you could well be right - the battery was probably very near fully charged. With the charger module alone, current tends to carry on trickling in for a very long time at a very low value and it the module hardly ever gives up and admits the cell is full. I've got used to that but as you say, with a diode in the circuit it might behave differently. I'll try a non-Schottky diode too - as you say, not charging to full capacity would be a bonus so the voltage drop not such a big issue.

The analog switch you mention is cheap and I'll have it by Monday so it's worth a go. If possible I would prefer not to power it from the battery in case the battery goes totally flat and then I am stuck: no way to turn the switch on but no way to charge the battery if the switch is off.

Transistor suggestions also noted. Thanks.

 

OK, so I swapped the Schottky for a IN4001 and used a less charged battery. Only about 5mA flowed as opposed to the 15mA with no diode but the good news is that when not charging the leakage current was <1uA. The battery charged slower, but it reached an acceptable voltage so this is a solution, albeit not altogether optimal.

 

The analog switch must be powered from the charger so that when the charger is off it is off as well. At least that is how it looks to me.

 

Hmm, well the analog switch didn't work - the drainage was much bigger when the charger was off, whether or not the power supply to the chip came from the battery or the unregulated supply. But to be honest I hadn't appreciated how tiny a package it came, and soldering those 5 pins pushed my eyesight and soldering skills to be absolute limit, and it's not really a viable option. The diode method works, but I think the voltage drop confuses the charger and it slows down charging a lot. So I want to try the NPN transistor idea - but I have to admit I don't understand how you (ebp) are suggesting the wiring should be. Any chance of a simple diagram?

 

Sorry, I forgot about you. This reply is just to kick this thread "up the list." I need to consider some things.

I'm baffled why the analog switch wouldn't work. Did you have it arranged so that you could be sure the control signal was in the appropriate state when the switch should have been off? The control signal is a CMOS input with extremely high impedance, so it cannot be left open. Typically it would be pulled either HIGH or LOW with a resistor and actively the other way.

 

Thanks. I didn't have a resistor on the input so yes it would have been floating. I guess it would have floated high and so was never off. I should have tied it low. (I connected as per my diagram a few posts up.) I can go back to it. But the packaging is so small that I'm going to struggle even to make a PCB to accommodate it though of course I could one made. That's why I wanted to explore the transistor option. I do appreciate your help.

 

Sorry, I forgot about you. This reply is just to kick this thread "up the list." I need to consider some things.

I'm baffled why the analog switch wouldn't work. Did you have it arranged so that you could be sure the control signal was in the appropriate state when the switch should have been off? The control signal is a CMOS input with extremely high impedance, so it cannot be left open. Typically it would be pulled either HIGH or LOW with a resistor and actively the other way.

You will have losses if you pull it low as it will flow directly to ground.

 

Hmm, well the analog switch didn't work - the drainage was much bigger when the charger was off, whether or not the power supply to the chip came from the battery or the unregulated supply. But to be honest I hadn't appreciated how tiny a package it came, and soldering those 5 pins pushed my eyesight and soldering skills to be absolute limit, and it's not really a viable option. The diode method works, but I think the voltage drop confuses the charger and it slows down charging a lot. So I want to try the NPN transistor idea - but I have to admit I don't understand how you (ebp) are suggesting the wiring should be. Any chance of a simple diagram?

This can be do to the voltage not being enough to charge the battery properly. It might even damage your battery and you need it to work for a long time (i think it was 2 years) ?

 

You will have losses if you pull it low as it will flow directly to ground.

I'm talking about pulling the switch input to low using a high value resistor eg 10M just to stop it floating. As you would with a mosfet. Won't any losses be negligible? And in any case if the switch was powered from the "raw" voltage input as i'd hope to do (rather than the cell) they won't matter?