Reply With QuoteHi, Many thanks, I have not really done a video for this stuff before but if a lot of interest is shown I might try and put something together to show how easy it is. cheers
Exactly - so when the motor wants to pull more current, you dont want the resistor droping more volts and resucing the power further.Originally Posted by DorrisMancer
Hi, Many thanks, I have not really done a video for this stuff before but if a lot of interest is shown I might try and put something together to show how easy it is. cheers
I believe that will actually work the other way around because the motor's internal resistance (impedance) is practically zero in rest. The impedance increases when the motor speeds up because of the magnetic flux created. Thus in theory a serial resistor would make it harder for the motor to start up but have a very reduced effect on the final speed of the motor. I think if you place the resistor in parallel this would not hinder the motor starting but it will reduce it's maximum speed as the current will favour the resistor over the increasing motor impedance more and more. One should be aware though that the resistor might become quite hot, depending on the value required to achieve the desired behaviour, and it will probably consume batteries like crazy.
Just a thought: instead of regular diodes you could also use LEDs (light emitting diodes). By using two different colours you can make it visible whether the valve is opening or closing.
Sorry, the resistor suggestion just doesn't make sense. Also, the forward voltage of an LED is significantly more than a silicon diode so what sounds like an attractive idea would not work. The OP's suggestion of a couple of back-back 1N4001s is the way to reduce the voltage across the motor ... but whether that would allow the motor to generate enough torque to operate the valve, especially with an ageing battery or after a summer lay-off, is an open question. The "correct" way to reduce the noise is to improve the mechanics or acoustic damping, but this is rather more difficult.
Would fitting these diodes result in an increase in battery life by any chance?
Unfortunately not, the motor runs slower so also needs to run longer so it will balance out. Not sure at what speed the motor has maximum efficiency but even if it runs slighlty more efficiently it wont make a noticable difference. cheers
Indeed. Don't know what I was thinking. LEDs do in fact have a voltage drop of around 2V (depending on the colour) so that would leave very little to spin the motor.
@DIY2: so did you try simply replacing the batteries with rechargeable ones? That will also give you the 0.6V drop. Of course in that case this lower voltage would also apply to the electronics and an additional problem is that they don't gradually deteriorate but sort of instantly die, so when (if even) the controller shows you the low battery warning it will already have gone offline.
Hi, no I didnt try that but switching the batteries would work too. I didnt want that route for the reason you have stated plus they dont last the same and I have 11 Rads so dont want the hassle. Also as one of the previous posters has mentioned, by reducing the voltage hence drive power, there is a risk that on a stickier valve operation then you run out of power. They are also correct that if your right on the limit then as the battery voltage drops further near the end of its life, again you might run out of power, although that would be unlucky.
With that in mind I wanted to try the current solution and see how it runs - I can always pick different diodes or qty with different voltage drops to tweek it later - but for now seems fine. If I had the time I would drive the motor directly with batteries out, and vary it to characterise where its performance drops off. Im snowed under at the moment so have not been scientific about it so far.
If someone else has the time - perhaps they could give it a go?
Wish Honeywell had just picked a better motor/gearbox combination.
Cheers
I would say battery life would be reduced as it takes (roughly) the same total amount of energy (Joules) to operate the valve whether you do it quickly or slowly, and the voltage drop of the diodes represents power dissipation as heat. Power spent heating up the diodes is power not contributing to operating the valve.
How much effect it would have on battery life depends on how much of the total battery consumption is actually turning the motor (vs long term wireless comms, etc) and what proportion of the supply voltage that would normally power the motor is being consumed by the 0.7v drop of the diode...
That's quite simple: it's V*I. The main question here is: how do the curves for `I` relate to each other when you feed the motor with 3V versus 2.3V?
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