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?

Yes, the loss would be negligible. You do need to be careful not to pull the control input to excessively high voltage from the input source, but you can probably manage that with a small zener if necessary. As long as the input is driven through another high resistance, you could likely just rely on the input protection diode built into the part, BUT this can pull the supply pin of the switch up and will put the current into "something" which would be the cell if the cell were used to power the switch.

Given your construction issues, I think perhaps the BJT solution is the best compromise.

I should have mentioned the package size. The problem with all the "good" parts these days is that they get put in ever-diminishing surface mount packages. I hope you got the SOT-23 and not the SC-70 part.

I'll try to hand-draw the circuit for the BJT. I don't have any schematic tools on the computer I use for AAC.


[EDIT]: I need to rethink this. I was forgetting about the fact that BJTs will still behave as transistors, albeit with low gain, in reverse. This means that the resistor I had between the base and emitter actually partially turns the transistor ON because the battery allows it to work in reverse and what would be the base-emitter resistor becomes in effect a base-collector resistor for the reverse transistor I think this is soluble with a diode outside of the charging path, but I need to consider it very carefully.

e.g. 2N3904 transistor (very common and very inexpensive)

• collector to the negative terminal of the cell
• emitter to the negative output terminal of the charger board
• resistor of about 1.5k ohms per volt between base and input power to the charger board (e.g. if the input voltage to the charger were 8 V, you would use about 8 x 1.5k = 12k; if it were 12 V you'd use about 18k) - you want something around 0.7 mA into the base if you are using 15 mA of charging current to get the transistor reasonably well into saturation.

Switching the "low side" like this won't be acceptable if you need to have a normal "ground" connection from the cell all the way back to the input source, but it sounds as if that isn't necessary.

 

Got it (I think). No, the cell doesn't need to be ground connected back to the input source.
Thanks

 

What voltage do you get at the output of the wireless charger module when the transmitter side is off but the charger and the cell are connected? Assuming it goes to some low voltage, does it drop quite quickly (e.g. a few seconds or less)?

What voltage do you get from the wireless module when it is ON and your full charging current is going into your cell?

What is the maximum operating temperature you expect in your charger and cell assembly?

I'm struggling with trying to find a simple way to use bipolar transistors without running into grief from the "backwards transistor" problem.

I'm still at a loss for why your charger board is allowing so much current from the cell when the input source is off. It just doesn't make sense in view of the spec's.

Have a look at some of the hits from the following search:
https://www.mouser.com/Optoelectron...OSFET-Output-Optocouplers/_/N-awss6?P=1yzxnag
These are opto MOS switches and at least some have two opto MOSFETs "back to back" which makes them good bidirectional switches (block current regradless of which side is positive). You would have to be able to supply a few milliamps from you wireless charger to turn them on, which may be prohibitive.
I glanced at the data for the TLP176AM which might be acceptable - not too large, not too small, reasonable ON resistance. The maximum OFF leakage is 1 µA, but typical is orders of magnitude lower. There may be better parts both from Toshiba & others.

 

What voltage do you get at the output of the wireless charger module when the transmitter side is off but the charger and the cell are connected? Assuming it goes to some low voltage, does it drop quite quickly (e.g. a few seconds or less)?
About 2.2V. Yes it drops apparently instantaneously

What voltage do you get from the wireless module when it is ON and your full charging current is going into your cell?
About 4.2V. You didn't ask, but when the cell and charger are not connected, it's about 5V

What is the maximum operating temperature you expect in your charger and cell assembly?
Hard to say. I would hope not to have it above about 30-40 degrees.

I went back to the analog switch, and used a high value resistor to tie the input down. Unconnected to anything, it works perfectly. When on, the resistance is about 1 ohm. When off, is about 200kohm.
When connected to the regulator, it looks as though it should work. With COM connected to the regulator output, there's about 4.2V on the NO pin when switched on, and 0V when switched off. But if I connect the NO to the cell, the current flowing to the cell is about right when the switch is on (15mA), but as soon as the switch is off, I get about -7mA - i.e. in the wrong direction. It doesn't make any difference whether the switch is powered from the cell or the raw output from the wireless module.

Next thing to try is a transistor switch.

 

Try adding a resistor right across the input to the charger board. I'd start with something in the 50 k range. See if it has any influence on the leakage current from the cell without the analog switch and when the wireless charger is off.

I'm slightly suspicious that there is a weird sort of self-sustaining "leakage" thing going on - you get a small amount of leakage from the cell which keeps the voltage at the input to charger high enough that what should be turning off fully to stop the leakage isn't. The intent of this experiment is to see if "loading" the leakage actually reduces the current - which to me is completely counter-intuitive, but maybe ... One of the things that loads the cell is the voltage divider chain for the feedback. There is a FET switch that should turn off to disconnect it, but if it is partially enhanced, that could account for the excessive leakage current you are seeing.

 

OK - I configured the 2N3904 as you suggested, with a 6k resistor into the base. The transistor seems to reach saturation OK and allow the cell to charge, but when the charger is off, the leakage is about 50uA

Onto the photorelay. I'm not really familiar with these but is it just analogous to a mechanical relay, with the LED pins equivalent to the coil and the two other pins like contacts (except there is a polarity for both)? I assume a resistor also needed for the LED?

 

As I mentioned, I couldn't come up with a way to use a bipolar transistor without the problem that I initially overlooked with the transistor acting in reverse. There are ways to do the job, but it requires a minimum of 2 transistors (only one in the current path) and some passives, so it's a bit messy without a PCB.

The opto MOSFET device does have the drawback of requiring a few milliamps of input power (yes, through a resistor, just as for any LED), but the total isolation of the input fixes any possibility of problems with sneak paths. Some of the surface mount packages should be big enough to be managable without being to bulky.

I keep forgetting to mention: There are tiny "adpater" PCBs available for many popular surface mount packages. They are intended, I think, primarily to allow plugging the parts into solderless breadboards, but might be OK for what you might need. I've never bought any, I just know they exist. I'm sure you can find them on ebay and probably Amazon. If they are of interest but you can't find them, I'd suggest starting a new thread requesting help finding them and you'll probably get a ton of recommendations.

 

Try adding a resistor right across the input to the charger board. I'd start with something in the 50 k range. See if it has any influence on the leakage current from the cell without the analog switch and when the wireless charger is off.

I'm slightly suspicious that there is a weird sort of self-sustaining "leakage" thing going on - you get a small amount of leakage from the cell which keeps the voltage at the input to charger high enough that what should be turning off fully to stop the leakage isn't. The intent of this experiment is to see if "loading" the leakage actually reduces the current - which to me is completely counter-intuitive, but maybe ... One of the things that loads the cell is the voltage divider chain for the feedback. There is a FET switch that should turn off to disconnect it, but if it is partially enhanced, that could account for the excessive leakage current you are seeing.

I tried a 42k resistor - the leakage current shot up to about 50ua!
Pinning my hopes on an opto-isolator now, or if not, the original diode idea.

 

The opto-isolator works - had to get the LED current down to about 2mA otherwise the output voltage from the charger dropped too much. Looks like problem solved ...