renovater 1 Posted November 21, 2012 Share Posted November 21, 2012 The problem as i see it is that the charge from the cap is on dimunition whilst the charge from the battery is constant. The cap will not be controlling the amount of current coming from the battery but simply the amount of time it does so ? Am i wrong ? 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 transistor should control the voltage coming from the battery. As the cap drains and its voltage reduces then so should the voltage that the transistor allows through from the battery. I'm just building a test circuit now, although I've just been caught out... The Lima 47 that I grabbed from out of the cupboard as the test loco turns out to be one that I had DCC fitted and then forgotten having done so, got to go and find another 47 now... Andi Link to post Share on other sites More sharing options...
renovater 1 Posted November 21, 2012 Share Posted November 21, 2012 The thing is how to control the cap which in turn controls the transistor, if the loco had been running for a while then the cap would be full so letting max current from the battery which might not of been needed, in otherwords the cap charges up with time so won't be discharging in relation to the speed of the loco at any given moment, the amount of charge held by the cap at any given time is unknown rendering the battery discharge an unknown factor !! You really have to build it, i'm sure you will find faults as i did with mine, at the end of the day, simple is best. Link to post Share on other sites More sharing options...
Suzie Posted November 21, 2012 Share Posted November 21, 2012 The transistor is in the correct configuration that the voltage on the emitter should follow the base (and therefore the voltage on the capacitor). I suspect that there could be a problem of non-linear discharge of the capacitor though via a transistor configured this way and a simple discharge resistor across the capacitor might be required (with no discharge resistor a low voltage on the capacitor will see low current from the battery in to the motor and therefore low discharge current, loco will appear to slow quickly following loss of power and then crawl on like forever as the capacitor stops discharging). Link to post Share on other sites More sharing options...
renovater 1 Posted November 21, 2012 Share Posted November 21, 2012 The transistor is in the correct configuration that the voltage on the emitter should follow the base (and therefore the voltage on the capacitor). I suspect that there could be a problem of non-linear discharge of the capacitor though via a transistor configured this way and a simple discharge resistor across the capacitor might be required (with no discharge resistor a low voltage on the capacitor will see low current from the battery in to the motor and therefore low discharge current, loco will appear to slow quickly following loss of power and then crawl on like forever as the capacitor stops discharging). ..And what would happen if it was the reverse, too much current from the cap ? With this system there has to be a fixed constant relation between the cap charge, the battery charge and the loco speed, not counting current for the resistors and the relay !! 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 the proof of the pudding is in the eating, it works after a fashion... With R1 at 10 ohms and R2 at 220 and the a 500uf cap (I don't have a 1000 to hand) and using a TIP41C as the transistor the circuit does seem to work as expected. I only have some rather flat 9volt batteries so I've got three hooked up in series but switching the Duette rapidly off results in the motor (lima 87 power bogie) taking about three seconds to stop. Lower settings on the Duette give lower speed both on and off power. 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 The thing is how to control the cap which in turn controls the transistor, if the loco had been running for a while then the cap would be full so letting max current from the battery which might not of been needed, in otherwords the cap charges up with time so won't be discharging in relation to the speed of the loco at any given moment, the amount of charge held by the cap at any given time is unknown rendering the battery discharge an unknown factor !! You really have to build it, i'm sure you will find faults as i did with mine, at the end of the day, simple is best. The cap charges up to controller voltage and no more, (minus the diode drops as discussed earlier) if the controller is at half voltage then the cap can only charge to half voltage and no more. It is all about what voltage the cap is at when power is lost. Andi Link to post Share on other sites More sharing options...
renovater 1 Posted November 21, 2012 Share Posted November 21, 2012 Does the length of cap discharge change if so under what circumstances ? 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 cap, at the values I have at the moment (and I got one wrong, R2 is 2200 ohms) takes about three seconds to discharge from full speed, less from lower speeds Andi (in the pub now :-) ) Link to post Share on other sites More sharing options...
renovater 1 Posted November 22, 2012 Share Posted November 22, 2012 All very interesting, what i would like to know is, when the locomotive is running normally with the controller what is to stop the current from the controller running straight pass the cap and opening up the transistor allowing the battery to discharge, if you tell me it's because of the second resistor the same thing would apply to the cap ? In that case is there a need for a switch NO between the cap and the transistor which closes in case of loss of power from the controller and if so how would the switch work ? All good stuff this ! Link to post Share on other sites More sharing options...
RMweb Gold Dagworth Posted November 22, 2012 RMweb Gold Share Posted November 22, 2012 All very interesting, what i would like to know is, when the locomotive is running normally with the controller what is to stop the current from the controller running straight pass the cap and opening up the transistor allowing the battery to discharge, if you tell me it's because of the second resistor the same thing would apply to the cap ? In that case is there a need for a switch NO between the cap and the transistor which closes in case of loss of power from the controller and if so how would the switch work ? All good stuff this ! I did already cover this in my earlier description of how the circuit worked. Track power on: 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 order 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) Andi Link to post Share on other sites More sharing options...
renovater 1 Posted November 22, 2012 Share Posted November 22, 2012 I'm surprised to see that the controller current will get as far as the cap considering that the route via D1 is the "path of least resistance" ! As you said the locomotive runs on for three seconds, you still have a long way to go by comparison to the system i use. For example without the motor connected to the transmission i am at one minute twenty seconds over run with cut off at 12V using a faulhaber 2235 and a 6X36 flywheel and i know i could do better if i can balance the flywheel exactly. This i know is completely excessive but it does show how well the system works. Link to post Share on other sites More sharing options...
RMweb Gold Dagworth Posted November 22, 2012 RMweb Gold Share Posted November 22, 2012 I'm surprised to see that the controller current will get as far as the cap considering that the route via D1 is the "path of least resistance" ! As you said the locomotive runs on for three seconds, you still have a long way to go by comparison to the system i use. For example without the motor connected to the transmission i am at one minute twenty seconds over run with cut off at 12V using a faulhaber 2235 and a 6X36 flywheel and i know i could do better if i can balance the flywheel exactly. This i know is completely excessive but it does show how well the system works. Ah but remember that the path via D1 is also via the motor. Any path through the transistor also has to go via the motor. The path to charge the capacitor is not via the motor, and a discharged capacitor is virtually a short circuit, it is so hungry to be charged. Follow the current path from the rectifier positive to negative. It has two means of getting there. Via D1 and the motor Via D3, a 10ohm resistor (R1) and the cap (as I say, the discharged cap is virtually a short circuit and as a current path it can be ignored) This short circuit effect of a discharged cap is crucial to various comments that I have already made in the thread, it is the reason why we need R1 in the first place, otherwise the wheels would spark on regaining contact after a dirty patch - leading to dirtier track - and also it is the reason why it will have a braking effect if put in parallel with a DC motor. (the tiny suppression caps have such a low capacity that they charge fully in incredibly short times and can be ignored in this respect) As for the flywheel, well this is an old Lima power bogie that hasn't run for ten years, with a gear train and driven wheels. Hardly a fair comparison and this topic is about electronic means of getting over dirty patches not mechanical ones. I will be taking my design to club on Monday so I can get some video of it in action on the club test track as a working loco chassis. I expect it to be able to adequately run over a piece of masking tape covering 2ft of one rail. Can you do likewise? Andi Link to post Share on other sites More sharing options...
renovater 1 Posted November 22, 2012 Share Posted November 22, 2012 The thing to take into account is that without my little caps i only managed half the time !! I would like to see a video of your test. Link to post Share on other sites More sharing options...
RMweb Gold Dagworth Posted November 22, 2012 RMweb Gold Share Posted November 22, 2012 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 The thing to take into account is that without my little caps i only managed half the time !! I would like to see a video of your test. Andi Link to post Share on other sites More sharing options...
RMweb Gold Dagworth Posted November 24, 2012 RMweb Gold Share Posted November 24, 2012 Just wondering if my design would be eligible for inclusion in the 2012 innovation challenge? Andi Link to post Share on other sites More sharing options...
Andy Y Posted November 24, 2012 Share Posted November 24, 2012 It would be too late to do it now I'm afraid. Link to post Share on other sites More sharing options...
jdp298 Posted November 24, 2012 Share Posted November 24, 2012 I have a couple of questions about how it works. The self-setting relay is brilliant, that solves a stack of problems in its own right. It also means you don't really need a battery to solve the momentary short problem. If you've got space for a 9v battery, you've got space for the Maplin 1F capcitors. The modification is then to have the +ve terminal of the cap attached to both the base and collector, but with an additional resistor to the base. As your stands, with the Darlington comment, you really ought to be getting greater gain through the cap and the transistor. I'm guessing your actual build involves using the transistor in the (mostly) linear region, where the varying voltage of the cap switches the battery more gradually. And a final pedantic point which may have confused those who haven't realy studied your diagram; I think the battery symbol is the wrong way round. You do need the +ve terminal of the battery connected to the collector don't you? Link to post Share on other sites More sharing options...
RMweb Gold Dagworth Posted November 24, 2012 RMweb Gold Share Posted November 24, 2012 Yep, fair comment about the battery, I tend to think that - as the small terminal on an ordinary battery is positive - the small line on the symbol should be the positive! At the moment the version of it that I have just has the TIP41C, when I get a chance to rebuild it neater and with the 1000uf cap on Monday before club I want to add a BC108 into the circuit to make it a darlington pair. Andi Link to post Share on other sites More sharing options...
renovater 1 Posted November 25, 2012 Share Posted November 25, 2012 I have a couple of questions about how it works. The self-setting relay is brilliant, that solves a stack of problems in its own right. It also means you don't really need a battery to solve the momentary short problem. If you've got space for a 9v battery, you've got space for the Maplin 1F capcitors. The modification is then to have the +ve terminal of the cap attached to both the base and collector, but with an additional resistor to the base. As your stands, with the Darlington comment, you really ought to be getting greater gain through the cap and the transistor. I'm guessing your actual build involves using the transistor in the (mostly) linear region, where the varying voltage of the cap switches the battery more gradually. And a final pedantic point which may have confused those who haven't realy studied your diagram; I think the battery symbol is the wrong way round. You do need the +ve terminal of the battery connected to the collector don't you? That is why the this subject is just going around in circles ! All you need to do is to use an on board Bi-polar cap in parallel with the motor terminals in connection with a decent sized flywheel. The additional off board cap is to smooth the system out even more. Theres no need for complicated circuits, simple is best. Link to post Share on other sites More sharing options...
jdp298 Posted November 25, 2012 Share Posted November 25, 2012 I'd not use a Darlington pair if I were you. It's operating in about the right analogue way at the moment. If you get the D-pair sorted out, then the gain of them with even a small cap voltage will turn on the battery fully, pretty much regardless of prevailing controller conditions. In other words, it'll become a bit to much of a switch rather than a controller, which is what you have now. The on-board bi-polar cap is still my preferred option, but because models go both ways [sic] then at some point the bi-polar will get all unhappy and let out the magic smoke because it's been reverse-biased. You need at least a pair of bi-polars of sufficient capacity to store enough energy to move a few mm. I covered the numbers, Joules, current, voltage etc earlier in the thread. This is a bit more sophisticated and it's ultimately a personal choice about what you do. I have similar discussions about making circuits for bicycle dynamo headlights. Mine are simple but can't switch LEDs in and out. Others are complex, use rechargeable batteries and MOSFET H-bridges. Link to post Share on other sites More sharing options...
renovater 1 Posted November 25, 2012 Share Posted November 25, 2012 I'd not use a Darlington pair if I were you. It's operating in about the right analogue way at the moment. If you get the D-pair sorted out, then the gain of them with even a small cap voltage will turn on the battery fully, pretty much regardless of prevailing controller conditions. In other words, it'll become a bit to much of a switch rather than a controller, which is what you have now. The on-board bi-polar cap is still my preferred option, but because models go both ways [sic] then at some point the bi-polar will get all unhappy and let out the magic smoke because it's been reverse-biased. You need at least a pair of bi-polars of sufficient capacity to store enough energy to move a few mm. I covered the numbers, Joules, current, voltage etc earlier in the thread. This is a bit more sophisticated and it's ultimately a personal choice about what you do. I have similar discussions about making circuits for bicycle dynamo headlights. Mine are simple but can't switch LEDs in and out. Others are complex, use rechargeable batteries and MOSFET H-bridges. Wasn't there an electric loco that blew up at Rugby somewhere because the driver decided to put it into reverse before it had stopped ? I have no problems in any shape or form with the on board cap i use, just excellent running, acceleration and slow down. These days with production models having an all ready fitted flywheel if there is room for a 10X20mm cap then put it in, the off board cap can come later if you so wish, with just the one cap you should see a difference straight away. Link to post Share on other sites More sharing options...
Suzie Posted November 25, 2012 Share Posted November 25, 2012 The transistor circuit is self regulating. If the transistor tries to turn on more, the emitter voltage will rise - consequently reducing the base-emitter voltage and therefore the base-emitter current turning off the transistor. It is a traditional bipolar transistor circuit which includes inherent negative feedback. Adding a darlington arrangement will do two things. 1. Increase the base-emitter voltage thereby reducing the voltage to the motor when power is lost 2. Reduce the required base current from the capacitor allowing the use of a (significantly) smaller capacitor. I think just finding the highest gain transistor will be the way to go. The TIP41C is a bit of a whopper, 3A and has I think a minimum gain of 15, I am sure that something better can be found (like perhaps a ZTX650 hFE of 100 I think, and 2A Ic. Might get a bit warm though). I think additional simplification could occur with the removal of the emitter diode (transistor will always be turned off when voltage coming from the track is higher than the capacitor voltage), and also the base resistor can go since there is current limiting in the capacitor charge resistor anyway, and all the while there is a good charge in the battery the emitter will be kept up to a high enough voltage that there will be no base current. This will additionally prevent the capacitor charging to a higher voltage than the battery will supply making the slow-down following loss of track power a bit more graceful. Link to post Share on other sites More sharing options...
RMweb Gold Dagworth Posted November 26, 2012 RMweb Gold Share Posted November 26, 2012 OK, once again we get shouting that my circuit is more than needed and you only need some bipolar caps. Well come on then one of you advocating that route, build it and show us that it works. Andi Link to post Share on other sites More sharing options...
RMweb Gold Dagworth Posted November 26, 2012 RMweb Gold Share Posted November 26, 2012 The transistor circuit is self regulating. If the transistor tries to turn on more, the emitter voltage will rise - consequently reducing the base-emitter voltage and therefore the base-emitter current turning off the transistor. It is a traditional bipolar transistor circuit which includes inherent negative feedback. Adding a darlington arrangement will do two things. 1. Increase the base-emitter voltage thereby reducing the voltage to the motor when power is lost 2. Reduce the required base current from the capacitor allowing the use of a (significantly) smaller capacitor. I think just finding the highest gain transistor will be the way to go. The TIP41C is a bit of a whopper, 3A and has I think a minimum gain of 15, I am sure that something better can be found (like perhaps a ZTX650 hFE of 100 I think, and 2A Ic. Might get a bit warm though). I think additional simplification could occur with the removal of the emitter diode (transistor will always be turned off when voltage coming from the track is higher than the capacitor voltage), and also the base resistor can go since there is current limiting in the capacitor charge resistor anyway, and all the while there is a good charge in the battery the emitter will be kept up to a high enough voltage that there will be no base current. This will additionally prevent the capacitor charging to a higher voltage than the battery will supply making the slow-down following loss of track power a bit more graceful. Thanks Suzie, I'm going to build mark2 of it in the morning, I'll try different options and see what happens. Andi Link to post Share on other sites More sharing options...
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