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Minky

Motor braking with relays and/or FETs

Although it seems all I do is try and sponge Nerf info at the moment I only just came across this..
https://youtu.be/0UVpDUibDEE
...that's based off..
http://nerfscience.blogspot.co.uk...the-works-guide-elite-rayven.html
...has anyone tried implementing motor braking through a relay and if so what was the results?

Is it,  as Derek suggests,  not worth it for 180s due to those motors having quicker run down times anyway?

Could this be used in a Mosfet loom?

Apologies,  lots of questions.

MERGED:

Am I right in thinking that if you wanted to keep amps away from your hand in a full auto blaster such as a rapidstrike or Hyperfire would you have to run two mosfets?
SSGT

Re: Engine Braking

EDIT

I've merged your threads since the FET question actually leads on quite well from the relay question - both involve methods of braking a motor.



Minky wrote:
Has anyone tried implementing motor braking through a relay and if so what was the results?


Not personally, although hopefully I can explain a few things.

I personally don't really see the point (once you've fired you've given away your position anyway) although others may have use cases where running braking on the flywheels, allowing them to come to a complete stop more rapidly, is useful. Be aware though that it is very hard on the motors and switchgear - you're effectively reversing the current flowing through the motors from at least no-load current forwards, to close to peak stall current in reverse, in a fraction of a second (and a motor is effectively an inductor - it'll try to resist rapid changes in current). This is a necessary evil with pusher circuits if you want snappy cycle control but is IMO unnecessary wear on flywheel motors which may be revved and braked multiple times before you actually decide to fire a shot.

If you do want to do this then you actually only need the relay if you want keep the functionality of the magazine and jam door interlocks so that the blaster won't function when a mag isn't fitted and/or the jam door is open and will come to a rapid stop if you remove/open them. Braking only when letting go of the trigger can be achieved with the standard SPDT microswitch that you'd probably be fitting anyway.

Minky wrote:
Is it, as Derek suggests, not worth it for 180s due to those motors having quicker run down times anyway?


If anything it'd be the opposite since a 180 motor has a longer rotor and armature (which means more mass in a similar distribution from the centre) which means a greater moment of inertia i.e. more resistance to changes in rotational speed (the difference is probably negligible in practice though). Under braking, however, a 180 will generally stop quicker since they tend to produce more torque.

Minky wrote:
Could this be used in a Mosfet loom?


It could but that sort of defeats the point since you'd then have a relay switching a FET (or FETs) switching the motors. At that point you might as well just make the switch(es) switch the FET (or FETs), or let the relay switch the motors.

Minky wrote:
Am I right in thinking that if you wanted to keep amps away from your hand in a full auto blaster such as a rapidstrike or Hyperfire would you have to run two mosfets?


You'd actually need three (assuming you aren't braking the flywheel motors - if you are you'd need four). The pusher motor (or any motor that needs to be braked) requires a half-bridge circuit. To be honest it's unnecessary and overly complicated to run the pusher motor on FETs unless you're using a microcontroller anyway for some other reason. It can be done without a microcontroller but you need to know what you're doing. This is an example circuit I drew up for someone who asked this a while ago but to be honest I'd sooner recommend using a relay or just using a pair of 10A microswitches for the trigger and cycle control switches. Extra complexity should be avoided if possible IMO and, with FETs on a circuit like this that doesn't use a microcontroller to individually switch them "on"/"off", you always end up with some overlap as one FET turns "off" and the other FET turns "on". It shouldn't be an issue (other than possibly having the FETs getting a little warm since both will be partially conductive for a very brief period) but IMO it's not the best solution. Using a relay, or just using the microswitches themselves, is a far better solution since the driving circuit is always broken before the braking circuit is closed thereby reducing the chance of a dangerous short-circuit failure mode.
Minky

First I just want to say I really appreciate your replies,  truly.  They stretch my knowledge of circuits and that's no bad thing at all. On reading this I actually went off and read BBC Bitesize Physics revision pages on the topic then some other pages of definitions and descriptions of Relays. Turns out I didn't know what they were and now I realise how some of that series of questions were really daft. Thanks for your patience in replying.
OldNoob

No such thing as daft questions- they are how we all learn. I once had to ask Boff how to remove the flywheels on a Stryfe..
Boff

Agreed. No daft questions in this thread. Smile You did a bit of research before hand and asked a question that has added to the body of knowledge on the forum. SSGT's explanation is an asset because he's articulated something I couldn't as usual!

To add to the thread from field experience, unevenly ever used braking once and its really not worth it. Yes you avoid run down but wear and tear was obvious after just a weekend event. Smile
Minky

Actually I may have a follow up daft question... I have not done any research soo..

When power is cut to an electric motor don't they act as generators whilst the inertia lasts?  I thought that the reason Derek thought the bigger motors would run down quicker was because they were trying to push more electricity or is that the aforementioned 'daft'?

EDIT: I don't know what Derek thought. I'm not psychic. There is not such a thing as being psychic.  When Derek said bigger motors span down quicker I jumped to the above conclusion.  All me, no Derek.
SSGT

Not at all, it's actually a very good question. As above, you don't learn if you don't ask.

You are right, an electric motor can act as a generator. In fact, since they are effectively the same thing, electrical engineers don't always distinguish and simply refer to both as "electrical machines" (an electrical machine may be designed/optimised to be used as a motor or a generator but it is still capable of performing as either). The key thing to remember is that a motor is actually generating a voltage whenever it is spinning and the magnitude of this voltage is relative to the rate of change of magnetic flux passing through the circuit - if you have a wire and move a magnet alongside it you generate a voltage or "electromotive force"/"EMF", the faster you move the magnet relative to the wire the greater the EMF that's generated (this is referred to as Faraday's Law). The interesting part is that the voltage generated by a spinning motor opposes the voltage applied to it by the power source because the generated EMF acts to oppose the change that produced it (this is referred to as Lenz's Law). This is why a motor draws less current at full speed than it does at stall - it's generating an EMF (referred to as "back EMF" as it's acting against the power supply) that effectively reduces the voltage across the motor - less voltage across the motor equals less current through it as per Ohm's Law (V=I*R or I=V/R).

So why doesn't this cause a more powerful motor to coast to a stop more quickly than a less powerful motor? It's actually quite simple - whilst a spinning motor does generate a voltage it cannot induce a current since you haven't completed the circuit. It's like having a battery that isn't connected to anything - there is a voltage across it's terminals but it can't use that voltage to do anything unless you put something between it's terminals. When we brake a motor by shorting it's terminals, we complete the circuit and so a current can flow - since the voltage generated opposes the voltage that was applied to the motor by the battery this current flows through the motor backwards relative to the current that flowed through the motor when it was being powered by the battery. The motor is effectively powering itself with the electricity it's generating but because the current is flowing in reverse, the torque it generates is also in reverse meaning the torque it generates acts to slow itself down - this is why a more powerful motor, that can produce more torque, will slow itself down more quickly under braking than a less powerful motor.

Hopefully this image, and the explanation below, makes things a little clearer. The graphs show voltage across the motor, current through the motor and speed of the motor over time (I was going to add a torque graph aswell but the shape would be pretty much the same as the current graph other than reaching zero at full speed). The vertical black line across all graphs shows the instant at which the switches were flipped. The top scenario is a coasting motor with no braking, the bottom is a motor that is being braked by having it's terminals shorted.

Looking at the voltage graph under no braking, at the instant of switching, you can see that the voltage drops from 9V (being applied to the motor by the battery) to just under 8V - this is the back EMF being generated by the motor at that instant (you can't see this under braking due to the ground pulling the voltage down just know that it is still being generated). Notice that the current through the motor under no braking just stops (since the motor is also effectively an inductor it will try and keep this forward current flowing, albeit momentarily - if you didn't have that diode in parallel with the motor this would cause arcing across the switch contacts as you opened the switch). It was flowing from the power supply, through the motor and down to ground but now the circuit is open it can't flow even though the motor is generating that back EMF. Because there's no current flowing through the motor, the motor isn't generating any torque so it slows down steadily based solely on the moment of inertia of the rotating mass inside it. When you short the terminals the current now has a path to flow. The current graph in the braking scenario shows that the current flows in the reverse direction hence why the torque generated acts to slow the motor down rather than keep it spinning.

Back to braking flywheels, one idea that might be worth considering is an "emergency brake" button that only brakes the flywheels when pressed. That way you can brake the flywheels if you do need to stop them quickly for some reason, without adding significant wear on the motors every time you release the acceleration trigger and without removing the ability to fire trailing shots as the flywheels spin down. Momentary SPDT push switches are hard to come by so you might need a momentary SPST push switch that switches a relay. Alternatively you could use a SPDT (rocker/toggle/slide) switch as a safety "catch" that when switched to "safe" both prevents you from accidentally revving/firing and allows you to stop the motors quickly if you need to - if you switch to "safe" when the motors aren't already spinning it won't increase wear at all. I probably wouldn't recommend using a braking safety catch in a full-auto though as, if you brake the whole circuit before letting go of the trigger, the pusher may end up partially extended and may (if it's a live-centre setup) try to fire off a shot, potentially into static flywheels, when you switch to "fire"/release the brake. An e-brake button on just the flywheels will work as long as you don't pull the trigger while the flywheels aren't spinning (although that should go without saying) - if you make it so you need your dominant-hand/trigger-finger to activate it that will stop you from doing that by accident.
tbr

And the above is why I really like it here at Britnerf.

I personally have no interest in the above topic, but the fact that the reply above exists, is well written by someone who clearly knows what they are talking about and cares enough to write a proper, coherent answer to the question make me happy inside Smile

Now back to the intelligent conversation.
Minky

Fantastic explanation and I'm kicking myself for not making the logical jump to the fact that the circuit isn't closed!

Although I can see that there's some good reasons to not use electric breaking I'm thinking that perhaps I can harness the spinning down motors energy in a cosmetic way whilst perhaps inducing some braking. My thought is to use two sets of LED looms,  one from motor feed to negative and the other on an alternative circuit from a relay feeding the motors. The thought being that you pull the rev trigger,  relay connects to positive and one loom goes from dim to bright as the motors come up from 0 to no load,  release trigger,  relay connects to 2nd loom and energy generated by motors passes through 2nd loom whose lights go from bright to dim to off.

... And the motors would experience Braking due to the resistance of the 2nd LED loom?  (See?  This is still on topic!)

I think explaining it I've just realised I've put the LEDs in the 2nd loom (bottom left) the wrong way around however if you were to bear that in mind could you have a look and see whether the attached circuit diagram would work?

Daft circuit diagram http://iob.imgur.com/uYGg/ReeK66uWxy

EDIT: Sorry I realise I didn't properly address your last point about the emergency brake. In honesty I was thinking of putting in motor breaking for showing off/the challenge and for 'stealth'.  I realise I'm living in hope with that one one when I will have 180's wailing directly out of my motor cover however I'm comfortable living there Smile

EDIT 2: @TBR I know exactly how you feel.  I read everything new put up here and I am trying to read through the tome of info BritNerf is. There's a lot I'm interested in and a bit I'm not,  but I read it all trying to sponge it all up because sometimes the answers are so well written and informative of their own subject it'd be a waste not to!  I hope it's all rubbing off..  And not pushing out anything important out of the memory banks.. When did I get married again?

EDIT 3: I instantly understood the graphs incidentally..  Maybe it is rubbing off?
Boff

If you wanted to utilise the back EMF of the motors for LEDs then just wire them in the same way you'd wire your normal motor diodes. The back EMF will flow through nicely and provide some current that will produce a fade off effect as the motor spins down. I've done it with blasters very recently in fact. Smile

Ignore the above, sleepy Boff shouldn't be allowed to post. See SSGT's post further down for accurate information.
Minky

Thanks Boff

So they'd run in parallel to the motors like the motor diode?
Would they be fitted in the opposite direction to the motor diode?
Boff

Wire them in parallel to the motor diode facing the same way. Then they'll only come on when the back EMF is being drained out of the motors. If you're clever, you can run a wire between the anode of the LED and the jam door switch to give you LEDs that light up in both circumstances: jam door open and motor spin down.
Minky

Erm... Nope.  No more follow questions, just a lot of thanks for everyone who contributed.  Thanks guys!
SSGT

Minky wrote:
Fantastic explanation and I'm kicking myself for not making the logical jump to the fact that the circuit isn't closed!

Although I can see that there's some good reasons to not use electric breaking I'm thinking that perhaps I can harness the spinning down motors energy in a cosmetic way whilst perhaps inducing some braking.


No worries. Again, no stupid questions here - if you never ask you'll never know. Most of us have made stupid mistakes, or missed something that seemed obvious later, at some point.

If you want the LEDs to light when you pull the acceleration trigger, and dim as the motors slowly spin down (and the back EMF generated reduces), then you'd want to wire the LEDs in parallel with the motors and oriented forwards so that they do turn on both when powered by the battery and the motor as per this diagram (note that you should still use a reverse-biased "flyback" diode to allow a path for the current to freewheel momentarily when you open the switch):



This is effectively how the LEDs in the top-right of your diagram are wired - when the motors are powered they'll turn on and they'll slowly dim when you cut power to the motors and the motors slowly spin down. I'm not quite sure what's going on with the relay in the diagram though - as far as I can tell the motors run until you close the switch and the LEDs in the bottom-left will be on whenever the motors as long as you aren't holding the rev trigger - they'll turn off when you press it and turn off again when you let go.

If you wire the LEDs in reverse (i.e. as you would wire the flyback diode) you will only get a brief flash as you release the acceleration trigger - how long that lasts will depend on the inductance of the motors (i.e. how resistant the motor coils are to changes in current). It may be so brief that you don't notice it. Like I mentioned before, when you release the trigger, the motors (as inductors) try to resist the change in current - they try to keep it flowing forwards as long as they can. Without the flyback diode there is no path for the current to continue to flow when you open the switch, so the motors will try instead to pull the current through the opening switch contacts for as long as the contacts are close enough together for it to make the jump - this is "arcing". In this case the LEDs take the place of the flyback diode however you may still experience some arcing across the contacts (the LEDs will only want around 20mA each so the motors will either try and force more through them, or try and draw the rest across the switch contacts - if the inductance is pretty low they may not do too much of either but still, neither of those is a good thing even if it they only happen a little bit).

Note that neither LED configuration will do much in terms of actually braking the motors. In the latter the LED or diodes will block the current flow under braking (just as the flyback diode they replace does) and in the former current will flow but it won't be enough to generate enough torque to slow the motors down significantly - the resistor that's being used to protect the LED will also the limit the braking current (it will slow down more quickly than without LEDs but, as we're talking tens of milliamps of current as opposed to tens of Amps, it probably won't be too noticeable... unless you use a lot of LEDs in parallel). For braking, as with normal running, you want as little resistance in the circuit as possible.

A flyback diode blocks the current flow under braking, as demonstrated below (left image), because the current flows in the reverse direction compared to under normal running:


(Click image for larger version - graphs show motor speed over time)

A diode in the forward direction will brake the motors (as per centre image) but it won't do it as well as a simple wire (right image) and will short the battery under normal use so you'd have to switch it in anyway. Note that a diode has a forward voltage drop of approximately 0.7V and so you effectively lose 0.7V of the generated back EMF across it (when the generated back EMF drops below 0.7V the diode no longer conducts at all).
Minky

Thanks SSGT.  For the purpose of calculating the resistor for the LEDs do you think the max EMF would be a volt off the amount the batteries would have been providing,  rapidly decreasing?
SSGT

You'll want to spec the resistors for the supply voltage anyway (since they will switch on from the moment you send power to the motors and be exposed to full pack voltage) but to answer the question (based on stall and no-load currents at 11.1V) the back-EMF generated at full speed at 11.1V should be around 10.9V for Hellcats and 10.7V for Rhinos. At full speed motors do almost generate the voltage you apply to them (hence why no-load currents are often so low, especially compared to stall when no back-EMF is generated).

Like I say though, I'd spec for pack voltage and maybe add a little for peak charge to be on the safe side (i.e. use between 12-12.6V minus the LED forward voltage for your resistor calcs).
Minky

Ah right,  I see where my mangling of language and circuit diagrams has confused things.  

The effect I had imagined was one set going dim to bright on trigger pull and a separate set soaking up the back EMF dimming from bright to off. (So imagine Green LED loom lighting up with trigger pull then and switching off on release,  when a red set light up and dim out).

That was what I hoped I'd drew in the circuit diagram and why I put in the relay to close a circuit on the 2nd set as power was cut from the battery to the motors but now I realise that there's nothing there that would extinguish the first LED loom. So perhaps I should have put in a DPDT relay and put the 1st loom through it so that Batt to Motors and 1st LED loom is completed on trigger pull and only motors to 2nd LED loom is completed on release?

Im sorry if im making this a lot more complicated than it has to be, I just imagine it could make for a very cool effect.

EDIT: I wrote 'Double throw'  and meant DPDT.
SSGT

Minky wrote:
The effect I had imagined was one set going dim to bright on trigger pull and a separate set soaking up the back EMF dimming from bright to off. (So imagine Green LED loom lighting up with trigger pull then and switching off on release,  when a red set light up and dim out).


If I'm honest I'm not entirely sure what you want is possible, at least not when relying purely on the back EMF generated by the motors. The LEDs in parallel with the motors will light up instantly when you press the rev trigger and apply power to the motors (the voltage/brigthness won't slowly "ramp up" since the LEDs won't "see" the back EMF, they'll simply see the voltage supplied by the pack) although they will slowly fade out when you release the rev trigger and the motors spin down. Likewise the other LEDs will turn on instantly when you apply power to them and, since there aren't any motors in that circuit to provide power as they spin down, will switch off as soon as power is cut.

If you want the LEDs to slowly light up on acceleration trigger pull then you'd need a capacitor and a resistor to create an RC circuit. To control the LED properly, and allow you to choose the resistance and capacitance to vary the time constant and therefore the time taken to power up/down the LED, you'd also need a transistor (either a BJT or a FET - you can make use of the fact that either of these can act as variable resistors instead of simple on/off switches).

If you only want one set of LEDs to both slowly light up when you press the switch, and slowly power down when you release the switch, then this circuit is probably about as simple as you can get away with:



When you close the switch, the capacitor charges up but the rate at which it charges is limited by the resistor. As the voltage across the capacitor rises the transistor (I've used an N-channel FET here) begins conducting. Transistors don't actually operate as simple on/off switches - if you only apply a small voltage, or current in the case of BJTs, to the gate/collector pin they don't "switch on" fully (you can actually vary the degree to which they conduct by varying the voltage/current applied). This allows the LED to slowly light up. When you open the switch the capacitor discharges (in this case through the motor although later on it will have it's own path to ground - again, negligible current so it won't do much to keep the motor spinning) and the transistor, and therefore LED, slowly turns off (again, the resistor limits the rate of capacitor discharge).

If you want anything more complicated you'll probably need a relay. The following circuit effectively does the same as above but with a SPDT (single-pole double-throw) relay (circuit shown purely to make the transition between designs a little easier - I'll try to avoid changing too many things at once):



There is a minor quirk with this circuit, and all of the circuits with the relay in parallel with the motors, but I'll get to that (and how to fix it if you don't like what it does) later.

If you want the LED to turn on instantly when you press the switch, but slowly power down when you release it, then move the resistor to the other side of the relay on the NC side to ground (this circuit effectively achieves the same thing, other than the additional quirk and the ability to vary the time period, as simply having the LEDs in parallel to the motor as shown in my previous post):



If you want the LED to slowly light up when you press the switch, but instantly turn off when you release it, then move the resistor to the NO side connecting to the positive side of the power supply:



If you want it to do both, but you want to independently vary the on/off period, then merge the above two circuits and use a different value resistor for the power up/power down branch (in this case the resistance of the resistor that charges the capacitor is greater than that of the resistor that discharges the capacitor so the LED will take longer to turn on than turn off):



If you do still want two sets of LEDs, with one set slowly lighting up when you press the switch and turn off instantly after you release the switch, and another set of LEDs that switch on instantly and then slowly power down after you release the switch then you'll probably need something with a DPDT relay that's more akin to this:



As I said before though, in all of the above cases using a relay in parallel with the motors, the LEDs won't turn off/start dimming precisely when you release the switch as the back-EMF of the motor will keep the relay energised for a short time (shouldn't be more than a second if you're using a 12V relay with an 11.1V supply). This might be desirable but if you don't want this then you need to allow the motor to also be switched by the relay (Yenka doesn't have 3PDT relays so I used a bit of paint magic to make this one - you may struggle to find one that meets your needs though):



The same applies to previous versions - you should be able to satisfy yourself that the circuit linked here effectively achieves the same thing as the first circuit at the top of this post.

Again, if you want to vary rise/fall times you need to replace the single resistor with two resistors on the opposite side of the relay:



In this case the LED on the left will slowly light up when you press the switch. It will turn off, and the LED on the right will turn on, as soon as you release the switch. As the capacitor discharges the LED on the right will slowly fade off. The LED on the left will also take longer to turn on than the LED on the right will take to turn off owing to the different resistances.

Something to bear in mind, however, is that a 12V relay will not work in a 2S system. If you still plan on using both 2S and 3S you'll need to use a 6V relay. Also bear in mind that if you do use the relay to switch the motors you need to find one capable of handling the inrush current of the motors you are using. If you can't you will need to default to placing the relay coil in parallel with the motors - if you are using a 6V relay (to allow both 2S and 3S use) this may result in a fairly significant delay (at least on 3S) between releasing the switch and the LEDs turning off/starting to dim as the back-EMF generated by the motor will be able to keep the the relay coil energised for longer.
Minky

Sorry I feel like I'd taking up far too much of your time with this.  If you want to call it a day and just say,  'Go and get an education then come back and talk to me!' that would be absolutely fine.  

On the 'dim to bright' on revving,  I watched Hawki007 say that he had deliberately chosen a second power supply for 'Ethereal Fortress' because when the motors had to work harder the LEDs dimmed so I thought that once the rev trigger was depressed the LEDs would turn on but be momentarily dimmer than normal.  

Would you look at the below?



I can't yet draw these simply but the idea is to wire lighting loom one so it activates on rev in the way you and Boff described, but then have a second identical loom that the circuit switches over to when the the trigger is released so the back EMF lights those rather than the the first one.  Would it work?

I like the idea of doing it properly with Capacitors..  I think it'd look great but the above was all I was thinking about was the above.

EDIT: I tried to use the Img thing but it didn't work.

EDIT 2: Please ignore the values on the Resistors and even the Battery I just threw them in.

EDIT

Fixed the image code for you. URL must end in an image file extension e.g. .png/.jpg/.bmp etc.

In this case I've also added an "l" (lowercase "L") before the file extension to force the image into 640 pixel wide "large thumbnail" (only works with imgur) as per forum rules.

Full instructions in "How to add images to forum posts" thread.

SSGT

SSGT

Minky wrote:
Sorry I feel like I'd taking up far too much of your time with this.  If you want to call it a day and just say,  'Go and get an education then come back and talk to me!' that would be absolutely fine.


Honestly, I don't mind too much. If it's posted on here it becomes a resource that people (myself included) can look back to and reference later. Besides, just thinking of a way to explain something can often help you cement your own understanding.

Minky wrote:
Would you look at the below?


OK, yeah, I see what you're getting at. You are relying on the power supply voltage sagging under load though, which isn't ideal. You should be speccing the power supply to be able to handle the transient current load of the motors and most LiPos tend to have a pretty low internal resistance which should make the sag negligible. That said, Hawki007 used a 2S 1000mAh 65C-130C graphene LiPo in that build - if he experienced a significant enough voltage sag to cause a momentary dim in LED light output then you may well get the effect you're after but it'll depend on the motors and pack you're using and even then it would only be for a fraction of a second. To be honest, it's a new one on me - I've never had anyone ask if they can use the voltage sag - as I say, normally you try and reduce it as much as possible. Internal resistance also changes with battery wear, load and temperature so it's not something you can necessarily predictably work with.

The pack Hawki used lists an internal resistance of "as low as" 1.2mΩ - he used a 2S pack and XP180s but if you use a pair of Hellcats, and a 3S pack with an equivalent internal resistance, then, taken on face value, that should correspond to a voltage drop of up to only 0.053V at the moment you press the rev trigger (i.e. the voltage should only sag to 11.047V). If that was meant to be 1.2mΩ per cell that would still only be 3.6mΩ for a 3S which still equates to only 0.16V sag (down to 10.94V). If it were a standard LiPo, with an IR of anywhere up to 5mΩ per cell (at least when new), that would be 0.62V sag for a 3S (i.e. down to 10.48V). I wouldn't have thought even that would result in a noticeable dimming of the LEDs but then there may have been some other significant drain in that build somewhere, there may be something wrong with his pack (the specs may not be entirely truthful for instance) or, as mentioned, the IR may be significantly greater than that at that kind of load. A pretty dead standard LiPo may have an IR of 25mΩ per cell - in that case a 3S pack would sag by 2.55V (down to 8.55V) - that sort of drop would be more noticeable but, again, only for a fraction of a second. I think it may be more noticeable in Hawki's build since I believe his LEDs are switched on independently of the motors - a momentary dip will likely be more noticeable if the LEDs were already on than it would if the dip happens as the LEDs turn on.

It's probably worth trying though. Maybe wire an LED in parallel between a pair of pack connectors and place it between the motor block and the rest of the loom and see what happens when you run the motors? That way you don't have to commit to building the whole thing first. If you do get the effect you're after then you're diagram looks mostly good except (assuming the relay is at rest and pin 3 and 5 are NC contacts) you'll need to swap the connection from pin 3 of the relay to pin 4 otherwise the motors will spin when you aren't pushing the rev trigger - this will make D2 the LED (or LEDs) that fade(s) out with the motor rather than D3.

EDIT: You will also need a diode across the relay coil - like the motors, it's also an inductor and will resist changes in current so you need to give the current somewhere to freewheel until it dissipates.
Minky

Arrgh.  Drawing these,  what can seem,  very simple circuit diagrams is like trying to learn another language with it's own alphabet and very specific grammar..  And that's before I've learnt the underlying maths!  I had assumed (been using that word Far too much) that the dashed lines that come out of the relay coil indicated it was on and so what the switch positions would be on activation but to draw that correctly I should of had the switch prior closed! Sigh.  Still, im glad that,  even though it must seem very basic to you it would have a chance of flying..  Whether it would be worth it remains to be seen. Im glad this is here now though because I really like the idea of fade up,  fade out lights and once I learn a bit more I can use your post above as reference to attempt one properly.

Thank you so much for the time you've given on such detailed answers. I'm going to give everything we've talked about a go,  probably not all in the build I'm doing now,  I should probably make sure I can make a basic circuit first and ensure my soldering is up to spec but this all really interests me.  Thanks again to everyone who contributed.
SSGT

The problem is that it also varies a fair bit depending on when/by whom you were taught and what programs you use to draw things (continuing the language analogy, it's a little like different dialects - even the symbol for resistors and lamps can vary depending on who you ask). In general it's a safe bet to assume everything in a diagram is at rest. The dotted lines from the relay just show which switches the coil activates and that the different switches all switch together at the same time - most of the time, though, a diagram of a relay at rest would be switched the other way so that when it energises the coil "pulls" the contacts over and when it de-energises they spring back (or it'll be like like this one and you have no choice but to assume how it's drawn is where it sits at rest).

One thing I forgot to mention - in your diagram, you would also need to put a diode across the relay just as you would with the motors. The relay is a coil and therefore an inductor and so, just like the motors, it will try and resist the change in current when you open the switch. If it can't arc that current across the switch (which is bad in of itself) you'll likely get a nasty reverse voltage spike that could well damage something else.

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