RMweb Gold Dagworth Posted November 20, 2012 RMweb Gold Share Posted November 20, 2012 The "soaking up backfeed" is not going to work either, basically what happens is that the back emf generated by the motor if allowed to try to charge the capacitor would act as a dirty great rheostatic brake on the loco. Try shorting the brushes of a motor and try to turn it and see how stiff it becomes. (BR used this technique very effectively on real electric locos - and class 50s before refurbishment - except the power was fed to resistances instead of into a capacitor). That the cap is already charged from the track power means that this doesn't happen as the Back emf never has a chance to try to charge the cap. It has NO effect on "motor friction" at all, let's not try to kid anyone. The ONLY part that makes any difference is the flywheel! The cap being only 1000uf at 16v is going to provide power to the motor for fractions of milliseconds and will make no discernable difference to the performance of the loco. Andi Link to post Share on other sites More sharing options...
RMweb Premium melmerby Posted November 21, 2012 RMweb Premium Share Posted November 21, 2012 It's obvious you don't know what you're on about and you are destroying what could be an interesting subject for those who would like to learn. From what you have said above it's pretty clear that you have not tried this system out and that you know nothing about motors and capacitors. What is behind your motivation i don't know. If anyone else would like to know more please feel free to send me a PM. I have been following this thread with some interest (and amusement!) What exactly is rectifying an already rectified current going to do? If you know your electronics you should know it will do precisely - nothing! You cannot rectify DC. Just how much energy is stored in a capacitor and how much energy is stored in a flywheel? Until you can answer that and show why what you are saying is correct (or otherwise) You should be careful of slagging off others. Keith Link to post Share on other sites More sharing options...
RMweb Gold Dagworth Posted November 21, 2012 RMweb Gold Share Posted November 21, 2012 Apologies to our later readers that the thread no longer makes sense. As you can guess from the quoted items we have been subject to a troll who, when all his ideas were questioned, simply deleted all his posts... Andi Link to post Share on other sites More sharing options...
renovater 1 Posted November 21, 2012 Share Posted November 21, 2012 I have deleted my previous posts because instead of having an informative discussion this is turning into a battle and that is not what i am looking for. No matter what is said when a system works, it works. When talking about "back emf" obviously the capacitor is not going to do much when there is incoming current, but will soak up the "emf" if there is a break in current or when the incomming current is switched off ! As for "motor friction" or "motor spike" it's well known that the right capacitor will relieve this greatly freeing up the motor and the flywheel. Concerning the phrase "if the incomming current is AC or DC rectify it" this is to ensure that the terminals for the first cap will always be the correct polarity and won't risk being reversed. Lastly, it dosn't cost much to buy these caps and try out for yourselves, then you will see how well it works. As i said before if anyone should need any more info please send pm. Link to post Share on other sites More sharing options...
RMweb Gold Dagworth Posted November 21, 2012 RMweb Gold Share Posted November 21, 2012 I have deleted my previous posts because instead of having an informative discussion this is turning into a battle and that is not what i am looking for. No matter what is said when a system works, it works. When talking about "back emf" obviously the capacitor is not going to do much when there is incoming current, but will soak up the "emf" if there is a break in current or when the incomming current is switched off ! As for "motor friction" or "motor spike" it's well known that the right capacitor will relieve this greatly freeing up the motor and the flywheel. Lastly, it dosn't cost much to buy these caps and try out for yourselves, then you will see how well it works. As i said before if anyone should need any more info please send pm. " As for "motor friction" or "motor spike" it's well known that the right capacitor will relieve this" It's quite surprising then that googling "motor friction capacitor" and "motor spike capacitor" bring up precisely ###### all results of any relevance then, the only thing even vaguely relevant is about suppression of arcing at the brushes which is more to do with the tiny little suppression caps that the manufacturers already fit to our models and that most DCC users cut off... You deleted your posts because: Your arguments had been proven to be hopeless. You are a troll... Andi Link to post Share on other sites More sharing options...
RMweb Gold Dagworth Posted November 21, 2012 RMweb Gold Share Posted November 21, 2012 This is the circuit proposed by "Dagworth" to get you over bad spots, in fact considered to be an "electric booster" in case of loss of incomming current ! Spot the fault yet ? Before giving others lessons and accusations of trolling which only serve to disrupt an interesting discussion for others it's best to know your stuff first ! Uploaded with ImageShack.us You mean the one the I pointed out in the other topic??? http://www.rmweb.co...._50#entry450541 or have you missed that one too and failed to read the description on the circuit and how it works? Otherwise please do tell me why it won't work as no-one else could a year ago. Andi Link to post Share on other sites More sharing options...
RMweb Gold Dagworth Posted November 21, 2012 RMweb Gold Share Posted November 21, 2012 Enough said, have fun !! Meaning what precisely? That I'd already answered your point over a year ago and that there is nothing wrong with my circuit? You are proving my point that you are a troll, not that I really need to prove it when you did such a good job your self here http://www.rmweb.co.uk/community/index.php?/topic/64466-is-there-a-future-for-oo/&do=findComment&comment=849206 Andi Link to post Share on other sites More sharing options...
RMweb Gold Dagworth Posted November 21, 2012 RMweb Gold Share Posted November 21, 2012 I do not troll, i may ask questions which may upset certain people, that is not my intention, having said that, i consider them to be justified. Now how about you explaining why your circuit is going to work ? I've already done that at length in the other thread, care to tell my why it wont? Andi Link to post Share on other sites More sharing options...
RMweb Gold Dagworth Posted November 21, 2012 RMweb Gold Share Posted November 21, 2012 Let's not spoil it for others, the "new" version ! Uploaded with ImageShack.us yes, and what is wrong with it please? Edit: Add circuit description from the other topic "The power first hits a relay coil of a latching relay which will determine polarity and therefore direction required. The main power is then rectified and charges a small capacitor. The relay contacts are arranged as a reversor so that the power hits the motor in the correct direction. In normal operation the track voltage will be higher than that allowed through the transistor from the battery so the diode above the transistor will be reverse biased and no power will flow from the battery. Cut track power and the battery/transistor is now at a higher voltage and current will flow through the motor from the battery. As the charge in the capacitor drains through the transistor the motor will slow to a stop until track power is regained and the cap recharged. The diode in the main flow line at the top is to prevent the battery from charging the cap." Andi Link to post Share on other sites More sharing options...
renovater 1 Posted November 21, 2012 Share Posted November 21, 2012 yes, and what is wrong with it please? Andi Firstly i ask you to apologise for falsly accusing me of trolling and that we can continue to discuss the subject without it turning into a battle ! Link to post Share on other sites More sharing options...
RMweb Gold Dagworth Posted November 21, 2012 RMweb Gold Share Posted November 21, 2012 Firstly i ask you to apologise for falsly accusing me of trolling and that we can continue to discuss the subject without it turning into a battle ! You prove yourself and I'll withdraw my statement. You went on the attack against me with your statement (since deleted but quoted above) "It's obvious you don't know what you're on about and you are destroying what could be an interesting subject for those who would like to learn. From what you have said above it's pretty clear that you have not tried this system out and that you know nothing about motors and capacitors. What is behind your motivation i don't know." I have very adequately proved that I DO know what I'm talking about. Find and prove a problem in my circuit that has not already been discounted in the other topic and I'll quite happily admit I was wrong. I won't go deleting my posts to hide my mistakes and will openly admit when I get it wrong as the other thread proves. Andi Link to post Share on other sites More sharing options...
RMweb Gold Dagworth Posted November 21, 2012 RMweb Gold Share Posted November 21, 2012 Well, i didn't get very far into looking into your circuit because a major problem hit me straight away, what tells the coil to change direction of the relay ? One would need two coils with two seperate circuits, one to induce the contacts in one way and the other to induce the contacts in the opposite direction, this would simply require diodes and an additional circuit for the second coil. But then again nothing is going to work at all, ever, because the moment current is cut off even momentarily, the whole lot falls to bits as the circuit is supplied and run by the one supply !! Now after this i didn't even get around to working out the resistors and the rest of it because it's got more chance of flying than having any effect whatsoever on the motor even if the current mananged to get that far in the first place !!! Sorry. Read the circuit description... The latching relay was dealt with in the other topic, it is such that current flowing one way will cause the contacts to change to one position, current flowing the other will cause them to change to the other position. Loss of current will leave them in the last place they were switched to. I put a link to such a device in the other thread. Loss of track power through dirt (the whole circuit is in the loco) allows the transistor fed from the capacitor to come into conduction permitting power from the battery to reach the motor and keep the loco moving. As the cap discharges through the transistor the loco will slow to a halt unless track power is regained in which case the motor will run from track power and the cap recharge ready for the next dirty patch. You prove by your statement that you didn't read what I wrote in the circuit description, or simply don't understand schoolboy physics. Maybe you are a troll who doesn't read the thing he is replying to, or perhaps you haven't done that part of your GCSEs yet? Andi Link to post Share on other sites More sharing options...
Dicky W Posted November 21, 2012 Share Posted November 21, 2012 Having been notified by more than one person, I'm watching this with interest as Andy is 'out of office' today. While not understanding much of the technical side of what has gone on, I do think the emotional side is becoming more to the fore and this is slowly turning into a slanging match. If this carries on I'll carry out my first 'official' function and lock this topic..... Link to post Share on other sites More sharing options...
RMweb Gold Dagworth Posted November 21, 2012 RMweb Gold Share Posted November 21, 2012 , Fine, if there is a latching relay that can sense the direction of current based as drawn in your diagram then that's news to me, it would have to have at least two inbuilt coils with a positive (live) sensor, would need an additional wire to earth or be fixed to an earthed chassis and be extremely expensive. In anycase i'm out. (edit: this reply to another now deleted post...) http://uk.rs-online....relays/7181948/ £2.27 is extremely expensive? I put a link to this item in the other thread. It has a single coil, it doesn't need any kind of connection to earth or the chassis (?) As I've stated all along there is nothing wrong with my circuit. Possibly it may need another diode in series with the resistor that charges the cap to prevent the voltage in the cap feeding back through the motor rather than it biasing the transistor but the principle is sound. Andi Link to post Share on other sites More sharing options...
RMweb Gold Dagworth Posted November 21, 2012 RMweb Gold Share Posted November 21, 2012 OK, I've had a look unless i'm blind i can't for the life of me see the two coil relay ? I can see the one coil version with reset and in DPDT. I am interested to seeing this, i need to see the internal circuit which changes the relay by the polarity of the coil supply or at least an explanation. Thanks It is the one coil version that I'm talking about, have a look at http://docs-europe.e...66b80f95686.pdf which is the data sheet for the range I linked to, it shows a bistable version as snipped here. (Bistable: Stable in both positions even when coil current removed) I use similar relays a lot. They have an internal magnet that sticks them in the last position asked for until the current is reversed. Hope that helps to clear things up. Andi Link to post Share on other sites More sharing options...
RMweb Gold RedgateModels Posted November 21, 2012 RMweb Gold Share Posted November 21, 2012 wouldn't mind having a go at this circuit if someone could suggest some suitable values for the components ... Link to post Share on other sites More sharing options...
RMweb Gold Dagworth Posted November 21, 2012 RMweb Gold Share Posted November 21, 2012 wouldn't mind having a go at this circuit if someone could suggest some suitable values for the components ... I want to have a go at it myself to see if it can actually be made to work. When I do I'll post a video link and updated circuit. Andi Link to post Share on other sites More sharing options...
Kal Posted November 21, 2012 Share Posted November 21, 2012 Hi Be interested too see the end result Regards Kal Link to post Share on other sites More sharing options...
RMweb Gold Dagworth Posted November 21, 2012 RMweb Gold Share Posted November 21, 2012 Ignoring the relay part of the circuit... Track power on: The track power passes through the rectifier. it spilts two ways, some goes to charge the capacitor (via the resistor R1 and the additional diode D3 that we discussed earlier) The resistor is there to make sure that the cap charges less than instantaneously, if it was not there then the cap would cause the wheels to spark after loss of contact as it recharged. The rest goes straight to the motor and the loco moves. The battery does nothing as the transistor at this point is switched off. This is because the voltage at the top of the transistor above D2 is greater than that at the 'base', (the terminal nearest the capacitor). The reason for the voltage difference is that each diode will cause a 0.7 volt drop. So if the track is at 12 volts then we can expect to see 10.6 volts after the rectifier (two diode drops), and 9.9 after D1 and thus at the motor. In ordewr for the transistor to turn on the current would have to flow through D3, R1, R2 and D2 as well as the transistor. There is a much easier path through just D1 so it won't bother. (Electricity will always take the path of least resistance) However, it will charge the capacitor up to - with 12 volts on the track - 9.9 volts (Rectifier drop and D3 drop) Loss of track power: NOW, the voltage at the capacitor is higher then that at D2, it can only get out by going through R2, the transistor, D2, the motor and back to the other side of the capacitor. As it flows through the transistor it switches it on allowing the battery voltage to also flow through the transistor, D2, motor and back to the other side of the battery. Our loco keeps moving, in effect an electronic fly wheel. R2 ensures that the full current available cannot hit the transistor in one hit and means that the transistor will stay on for a few seconds. As the cap discharges through the transistor etc. its voltage gets less, the transistor gradually turns off and the loco slows to a gentle stop. if the power is restored then the cap recharges, the transistor switches off (higher voltage from the track again) and the track power runs the motor. All the cap is doing is turning on the transistor when needed, it is not actually providing motor power, that comes from the battery. The fact that the cap is charged from the track power means that it can only ever get to track voltage minus the 0.7v diode drops, so if you have the controller on slow rather than full speed then the cap will only charge to a low voltage and on loss of power will only allow a low voltage from the battery to get to the motor. The system will compensate for whatever speed you ask it to drive at. I hope that makes sense... Andi Link to post Share on other sites More sharing options...
Grovenor Posted November 21, 2012 Share Posted November 21, 2012 As was mentioned in the other topic, the main difficulty here is going to be the sensitivity of the relay. To be practical the relay needs to change state below the starting voltage of the motor or you will get jackrabbit starts. Choosing a low voltage relay will make survivability of the relay difficult when full volts is applied to speed the train up. It looks as though you will need to limit the maximum voltage applied to the relay somehow or add series diodes to increase the start voltage of the motor,which will then limit its top speed. Regards Keith Link to post Share on other sites More sharing options...
RMweb Gold Dagworth Posted November 21, 2012 RMweb Gold Share Posted November 21, 2012 The datasheet that I linked to earlier shows that these relays will switch at 3.25 volts, well below anything that the motor is going to see, especially as the coil is before the rectifier/D1. Effectively for the relay to see 3.25volts the motor will be getting 1.15v..... I want to see what happens if I put 12v across a 5volt relay coil. Andi Link to post Share on other sites More sharing options...
RMweb Gold Dagworth Posted November 21, 2012 RMweb Gold Share Posted November 21, 2012 Could you possibly give me the value of the cap and two resistors either side, that would help me to understand better and also it would be interesting to know what kind of battery you would be using, rechargeable maybe ? Thanks I haven't actually built it yet, but I reckon on around 1k for R1 and R2, D1, D2 and D3 can be any rectifier diode, the transistor would need to be a reasonable current darlington device and the cap probably somewhere around 1000uf at 16v, maybe less, that would be subject to experimentation on discharge times. if you need the loco to stop more quickly then lower values, the higher value the longer the loco will run for. For the battery I would use a nnormal square 9v type, maybe two in parallel if one couldn't provide enough current. Nothing is huge and the whole lot should be able to fit in any Lima type loco, you might have more of a job to fit it into a modern loco but modern locos don't need it as they have flywheels anyway. Andi Link to post Share on other sites More sharing options...
RMweb Gold Dagworth Posted November 21, 2012 RMweb Gold Share Posted November 21, 2012 I see you intend to use the same value cap as me ! Is the battery the rechargable type or not and how do you regulate the battery, you didn't say ? Thanks Just an ordinary 9v battery, it doesn't need any regulation as it is controlled by the transistor. When the cap is discharged the transistor switches off, no current can flow from the battery and the motor stops. Andi Link to post Share on other sites More sharing options...
renovater 1 Posted November 21, 2012 Share Posted November 21, 2012 Just an ordinary 9v battery, it doesn't need any regulation as it is controlled by the transistor. When the cap is discharged the transistor switches off, no current can flow from the battery and the motor stops. Andi So, in otherwords if we take it that the circuit works, each time there is a loss of current the battery automatically feeds the motor with 9V regardless of the speed beforehand solely regulated by the output of the cap and it's resistor, what is the transistor type ? interesting all this, thanks Link to post Share on other sites More sharing options...
RMweb Gold Dagworth Posted November 21, 2012 RMweb Gold Share Posted November 21, 2012 So, in otherwords if we take it that the circuit works, each time there is a loss of current the battery automatically feeds the motor with 9V regardless of the speed beforehand ? No, again the transistor takes care of this. Because the transistor is controlled by the capacitor and the cap can only be charged to the voltage on the track it will only switch on as much as the cap will let it. Andi Link to post Share on other sites More sharing options...
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