Hi all! First post here, please excuse the lengthy read and thanks in advance for any help you can offer.

For background, I'm a hobby-level tech nerd, mostly with audio equipment though... tends to be a bit modular, haven't designed many circuits. Super handy with a soldering iron, quick learner, I can build something from schematic with the components supplied, but figuring out the correct components requires an understanding that's beyond me so far... I like building guitars, the circuits are pretty simple.

Thing: TronicalTune Plus, which is a device mounted to the head of the guitar that automatically tunes the strings by way of tiny motors in the tuners connected to a piezo transducer and a little electric brain [mainboard]

Brief: the mainboard is normally powered by a small lipo (1s, 4.2v) battery in a proprietary plastic casing that prevents reverse polarity connection because of its shape. These batteries are no longer manufactured and the company appears to be selling off their old stock, which are likely too old to be useful. They are also selling old mainboards PCBs, so... snatching those up while I can!
In any case, the batteries barely last, and the only thing that makes sense as a replacement is...

Idea: 18650s. I've mounted a holder for one on the front of my guitar's headstock and it looks siiiiick. They hold plenty of power and the 18650 works perfectly when connected straight to the mainboard, the discharge curve seems to suit the thing.

Problem: I didn't polarity protect the mainboard, so it fried when the battery was inserted backwards (not by me)
as outlined above, parts from this manufacturer are becoming scarce.
Finally received a replacement mainboard after nearly a year and don't want to fry this one too...

Proposed solution: reverse polarity protection circuit something like this one... https://circuitdigest.com/electronic-circuits/reverse-polarity-protection-circuit-diagram. I'd solder it up so nice, give it a little perspex viewing window in the guitar itself, maybe even set it in resin.
Ideally, the circuit would simply pass the battery's voltage correctly regardless of which polarity the battery is inserted in. If it just protects the circuit and stays off when inserted one way; cool. If it works both ways around, ideal use case!

Where you come in: the theory/maths etc concerned with choosing the handful of components with the properties necessary to build the circuit. And perhaps sketching the circuit
Goes right the way over my head! I don't know if what I'm asking is 60secs thought for someone with the right understanding, or if I'm actually asking for something that would be a fair investment of time/effort - happy to compensate you for your time if you can help me find the right components.

•Relevant resources•
Jaycar Electronics - https://www.jaycar.com.au - these are the larger chain electronics stores available to me locally (Australia).
Mouser and Element14 are totally viable for me as well, I've been trying to sort this out for a couple years so a little shipping time won't hurt.

Thanks for any help you can offer and have a great day!
Jack Shepherd

 

Hi Jack
Welcome to AAC.

The circuit you propose is massive overkill plus it'll be tricky to get it to work down to 3v.

The second circuit on that page with just a single diode will do what you want. Yes you'll lose a little voltage over the diode but your board is only drawing a few tens of mA. What's the capacity of the 18650 and how long a run time do you get fully charged?

A suitable diode from Jaycar is the 1N5822. Connect in the + battery lead, stripe end to your board.

An alternative approach it to use a protection board like this one from ebay.

This will charge the battery from a USB port or charger. While this doesn't itself provide reverse protection it removes the need to take the battery out. Be sure to get the type that has separate B+/B- and out+/out- connections as shown. This can also be used with a 3.7/4.2v li-ion pouch battery (though looking at the scarcity of them on eBay Australia maybe that's not an option for you?).

 

Hi Jack
Welcome to AAC.

The circuit you propose... ...Australia maybe that's not an option for you?).

Hello Irving,

Thanks so much for taking the time to help me with this.

Overkill is usually the go for me, I'm happy to route pockets in the guitar for the circuits and show them from behind perspex - I want the best possible solution, even if it means butchering the instrument a little.

The ideal use case would allow me to insert the 18650 in either direction and successfully power the circuit, but the solution you describe with a single Schottky would be fine too; I'll see if I can make that work... previous attempts with single diodes have led to the tuners malfunctioning, not receiving enough current etc.

As an update which may explain this, I have just measured the initial current when all 6 strings are being tuned, and it appears to be as high as 2 amps momentarily; around 800mA when tuning single strings. I am using a Samsung 30Q -Samsung 30Q 18650 3000mAh 15A Batteryhttps://www.imrbatteries.com › samsung-30q-18650-3000... - I basically want to charge the cell once a fortnight or once a month, carrying a charged spare in the guitar case. I am a gigging musician and I need the solution to be absolutely idiot-proof for both safety and the continued functionality of my instrument. As much as I love to mess with soldering irons, not so much while gigging.

Whatever I end up using as a solution, the battery has to be able to be removed and inserted at will - the physical shape of the 18650 makes it perfect for use as a "steel" - a slide to play "lap steel" style guitar.

The 30Q works perfectly well without polarity protection, but as described, leaving the circuit unprotected is asking for trouble and the circuits are hard to come by...

Any idea how to choose the correct components to build a proper polarity reversal circuit?

 

You could use 4 schottky diodes in a bridge arrangement

 

You could use 4 schottky diodes in a bridge arrangement

In cases where the voltage drop and parts count are not issues, I am a big fan of bridge rectifiers as polarity protection. There is something really cool about how they operate.

 

As an update which may explain this, I have just measured the initial current when all 6 strings are being tuned, and it appears to be as high as 2 amps momentarily; around 800mA when tuning single strings. I am using a Samsung 30Q -Samsung 30Q 18650 3000mAh 15A Batteryhttps://www.imrbatteries.com › samsung-30q-18650-3000... - I basically want to charge the cell once a fortnight or once a month, carrying a charged spare in the guitar case

How long does it take to tune up?

 

I

How long does it take to tune up?

the device is usually on for anything from 10 to 30 seconds. Max current measured so far with multimeter (I don’t have a scope) was 2.2amps.
I’m on my way to jaycar right now for the part you described, would that work for a 2A draw?

More info; device has a number of different modes - some utilise all 6 motors simultaneously, and some make small adjustments to on string at a time. Sometimes it would be on for minutes at a time while setting up, adjusting tunings, or even just completely slackening all the strings when the instrument needs a fresh set. Basically whatever the solution is I want to engineer it to deal with every use case - trying to give the thing a large power reserve and make it equal-or-better in terms of functionality - the computer seems a little sensitive and my attempts to give it the correct voltage by butchering existing circuits, adding buck converters etc have all resulted in some sort of glitchy malfunction

 

The diode I spec'd is OK for 2A. It'll get warm over 30secs but not unduly. Try to keep the longest lead length you can, at least >10mm, keep the body 5mm away from the board to allow as much free air circulation as possible and solder to the largest area of copper you can get...

 

As much as I’m tempted to wire that up, the idea that this thing will have to get any kind of hot in normal operation isn’t thrilling, it’s mounted to a timber guitar… guess I’m gonna have to figure out how to build the overkill circuit…?

i tried the single resistor from jaycar and it appears to work with a minuscule forward voltage drop, but has reverse voltage of 0.6v - is that normal? And could it damage the circuit? Haven’t been game to test it in reverse with the circuit in there as they’re quite expensive

 

warm is not hot... it'll dissipate 0.7W so will reach about 56degC above ambient internally. Lead temperature will be 51degC... just warm... timber won't even notice it...


What do you mean by single resistor? did you mean diode? Forward drop will be 0.3 - 0.4v, I don't understand what you mean by reverse voltage - how were you measuring it?

To test without circuit, find a flashlight bulb... one way it lights, the other it won't!

 

warm is not hot... it'll dissipate 0.7W so will reach about 56degC above ambient internally. Lead temperature will be 51degC... just warm... timber won't even notice it...


What do you mean by single resistor? did you mean diode? Forward drop will be 0.3 - 0.4v, I don't understand what you mean by reverse voltage - how were you measuring it?

To test without circuit, find a flashlight bulb... one way it lights, the other it won't!

On the way home from a hundredth jaycar visit now with 4 of the suggested diodes to try that circuit you’ve suggested I must’ve been doing something wrong but I’ll get into it properly tonight and be able to give more useful information. Very much appreciate the attention you’ve already give.

 

The diode I spec'd is OK for 2A. It'll get warm over 30secs but not unduly. Try to keep the longest lead length you can, at least >10mm, keep the body 5mm away from the board to allow as much free air circulation as possible and solder to the largest area of copper you can get...

View attachment 241972

Could I confirm that [bat-] only connects to the blocking/right end of D4? I.e. the path crossover visually to the left of “D4” in the diagram is just the paths crossing over visually, not connecting electrically?

I don’t read a lot of circuits but I’m assuming the four green dots are the only actual solder joints intended

 

Correct. The easiest way to do this is:

Lay out 2 diodes horizontally in a row, bar ends pointing towards each other, middle wires overlapping .
below, lay out 2 more in a row, bar ends pointing away from each other, middle wires overlapping.

Excuse rubbish pic, Im doing this on phone.

The middle top junction is + to board, the middle bottom is - to board.

The end junctions go to battery terminals on holder. Keep wires as long as possible and avoid shorting the middle junctions together!

 

warm is not hot... it'll dissipate 0.7W so will reach about 56degC above ambient internally. Lead temperature will be 51degC... just warm... timber won't even notice it...


What do you mean by single resistor? did you mean diode? Forward drop will be 0.3 - 0.4v, I don't understand what you mean by reverse voltage - how were you measuring it?

To test without circuit, find a flashlight bulb... one way it lights, the other it won't!

Okay, so!

yes I did mean diode in that previous statement.

The cell sits fully charged at 4.1v. Running cell directly to the tuner circuit, the drop during operation is less than 0.1v

Trying the method you outlined with the four diodes results in a voltage drop down to about 3.

Correct. The easiest way to do this is:

Lay out 2 diodes horizontally in a row, bar ends pointing towards each other, middle wires overlapping .
below, lay out 2 more in a row, bar ends pointing away from each other, middle wires overlapping.

Excuse rubbish pic, Im doing this on phone.

The middle top junction is + to board, the middle bottom is - to board.

The end junctions go to battery terminals on holder. Keep wires as long as possible and avoid shorting the middle junctions together!
View attachment 242051

alright!
Okay, so!

yes I did mean diode in that previous statement.

The cell sits fully charged at 4.1v. Running cell directly to the tuner circuit, the drop during operation is less than 0.1v, tuners respond lightning fast and all is well.

Trying the method you outlined with the four diodes between battery + and tuner + results in a voltage drop down to about 3.45v; the tuner circuit boots up but glitches out as soon as it needs to turn a tuner motor

Using a single diode in series between battery + and tuner circuit + resulted in a drop down to 3.8v, sometimes 3.6v, and while the tuner circuit functions, it works slowly and can only tune one string at once - not ideal

So voltage drop with any of these simple diode solutions is currently the biggest issue, as the circuit seems to dislike anything too low in voltage.

The reverse voltage I was referring to was with the single diode in series - measuring the battery directly with only the diode in series, I got -0.6v with the cell inserted backwards which makes me wonder what would happen to the tuner circuit if it saw that -0.6v… the voltage drop with the single diode and the cell inserted correctly makes that solution unviable anyway, so I’m sorta back at square one!

 

Hmmm, that circuit is very sensitive to battery volts - it must be the high current those motors need to start moving, assuming they're not binding in any way.

OK, lets revisit the original circuit. As it stands that won't work... let me have a think on it... if you really need >3.7v and room for a 3+Amp initial current draw that's going to be pretty tricky. Remember that a Li-Po battery is nominally 3.7v, not 4.1v which is its 100% charged open circuit voltage. It will rapidly drop to 3.7v as soon as it comes off 100% SoC and drops to ~3.4v at 50% SoC and 3v at 20% SoC typical.

Back shortly...

 

Right, I've analysed the original circuit - not only would it not work at 3v, it never worked at 12v either. The reason is simple - all MOSFETs have an intrinsic body diode from source to drain, so when that battery was reversed, yes the MOSFET was off... but the current now flowed through the body diode. Either the original designer didn't know that or just ignored it...

Basically you always need a series diode...

 

Right, there are only two answers I can see to this (other than figure a way to stop the battery being inserted incorrectly) and you're not going to like either I suspect...

Option 1 is raise the voltage using 2 batteries and then regulate down to 4v. You'll get longer run time as well, and you could use smaller cells.

Option 2 is use a boost converter to stabilise the output at 4v for an input between say 3.7v down to 3v. Again you'll get longer run time as you'll be better utilising the cell charge... I wouldn't recommend using a smaller cell.

 

Right, there are only two answers I can see to this (other than figure a way to stop the battery being inserted incorrectly) and you're not going to like either I suspect...

Option 1 is raise the voltage using 2 batteries and then regulate down to 4v. You'll get longer run time as well, and you could use smaller cells.

Option 2 is use a boost converter to stabilise the output at 4v for an input between say 3.7v down to 3v. Again you'll get longer run time as you'll be better utilising the cell charge... I wouldn't recommend using a smaller cell.

Boost converter sounds like an opportunity to screw something electronic to my guitar - I'm all for it!

I did find this last night though, any use you reckon? https://www.pololu.com/product/2810

 

OK, I'll look into that & get back to you.

The switch module has the same issue, one body diode drop, which oddly he doesn't mention though its obvious from the schematic.

 

Quick update - there's nothing on eBay comes close due to the high starting current of the motors. I've found a candidate chip that will do the job however I'm starting to think the two-cell option with a buck converter is a safer, easier option, and you might find something off the shelf... your thoughts?