Model railway controller

Discussion in 'The Projects Forum' started by JohnnyD, Aug 29, 2006.

  1. JohnnyD

    Thread Starter Well-Known Member

    Aug 29, 2006
    79
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    Hi,

    I'm new to this forum so I thought I'd better say hello before getting down to details... so hi!

    I have no qualifications in electronics but I've been tinkering with the subject for a long time, since I got one of those 1001 toy project boards when i was about 10.

    The main reason for posting is this....

    My dad has this really old model railway controller wich isn't made anymore. I took it apart and looked at the circuit which seemed very simple so I thought I'd have a go at making one myself. I translated the circuit board into a schematic and have now built a working prototype on breadboard. It's just, because of my lack of knowlege in certain areas, I'm not quite sure why some of the components are connected the way they are. So I was wondering if someone might be able to explain it to me.

    Anyway, here is the schematic

    I understand what each bit of the circuit does, apart from one thing. I'm just not sure why there is a 10uf capacitor (C3) connected between the emitter of Q2 and the base of Q1. It doesn't seem to make any difference to the controller if I leave it out.

    If anyone could shed some light on this I'd be very grateful.

    Thanks a lot,

    John.
     
  2. Gadget

    Distinguished Member

    Jan 10, 2006
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    The circuit is a basic series passed voltage divider. The controller pot divides the voltage, and the transistors boost the current from that divider.

    I think you may find C3 is for "Inertia", so that the train rolls on for a while after the controller is turned to "0"..(C3 holds the voltage up for a while). S1 and R7 speed up the discharge of C3 so forming a Brake circuit.
     
  3. JohnnyD

    Thread Starter Well-Known Member

    Aug 29, 2006
    79
    0
    Hi,

    Thanks for the reply. So basically I can do without C3 really because S1 turns the 'simulator' on and off. With the simulator off I want the locos to stop dead if i turn the voltage right down. With the simulator on I can control the rate of decelaration anyway so I don't see C3 having much benefit.

    Another thing I was pondering is the location of F1. The schematic isn't entirely correct because at the moment F1 is located straight after the +ve side of the bridge rectifier. I was unsure whether the circuit breaker would operate the same with AC as it does for DC. If it will work with AC then I'll put it where it is in the schematic.

    Thanks again,

    John.
     
  4. Gadget

    Distinguished Member

    Jan 10, 2006
    613
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    Either will work, but I would put it on the AC side of the rectifier to protect in case the rectifier goes shorted.
    Does the simulator still work without C3..???? Odd.
     
  5. JohnnyD

    Thread Starter Well-Known Member

    Aug 29, 2006
    79
    0
    OK brilliant, I'll put the circuit breaker on the AC side then, thanks.

    As far as I can tell, it is the charging/discharging of C1 that makes the simulator work. S1 either bypasses R4, D3, C1 and R5 hence turning the simulator off, or with S1 connected to R6 allows C1 to be charged through R4 and discharged through R5, R6 and R7. I think D3 means that C1 has to discharge through the transistors, not through the voltage divider.

    This is all just what I've worked out with my limited knowlege of electronics but if I am right, then I have no idea what C3 does!
     
  6. n9352527

    AAC Fanatic!

    Oct 14, 2005
    1,198
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    I think that C3 is there to provide the extra juice during acceleration and extra braking force during deceleration without any jerkiness (smooth fast acceleration and deceleration). It bootstraps the base drive of Q1 both ways from the output.
     
  7. Spoggles

    Well-Known Member

    Dec 2, 2005
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    Is there something missing? I do not see how M1 "sees" the negative side of the power supply.

    What is the function of the 2pdt switch going to no where?

    Spoggles
     
  8. JohnnyD

    Thread Starter Well-Known Member

    Aug 29, 2006
    79
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    M1 is an ammeter. the program i used to draw the schematic wouldn't let me swap the + and -. they should be the other way round.

    The 2PDT switch is a polarity reversing switch to make the model trains go in both directions.

    I've heard of the term bootstrap but i'm not quite sure what it means. Does it mean that when you start accelerating, C3 has to charge first before the base of Q1. and then if you decelerate really quickly, C3 will discharge through Q1, making the deceleration slightly smoother?
     
  9. n9352527

    AAC Fanatic!

    Oct 14, 2005
    1,198
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    The voltage at the emitter of Q2 follows the voltage at the base of Q1, minus the Q1 VBE, R8 drop and Q2 VBE. When the voltage at the base of Q1 increases, the voltage at Q2 emitter also increases by roughly the same amount a moment later. This increase then actually push the voltage at the base of Q1 up even further, because it takes a finite time to charge/discharge C3. This is a controlled positive feedback loop of the rate of change of the output (derivative of the output), so that when you increase the speed, the actual initial increase in output voltage is faster than the setting/charging of C1 would allowed. Except that the speed equilibrium (setting point) follows the setting from the R2 pot and is not affected by C3.

    Reducing the speed results in similar effect, except the voltages change the opposite ways.

    It's like when you are driving a car, you can step on the accelerator pedal hard when you want to accelerate and ease up gradually as you approach the required speed. The result is you get to the required speed faster.
     
  10. JohnnyD

    Thread Starter Well-Known Member

    Aug 29, 2006
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    Thanks for explaining in such detail.

    Is this what they call closed-loop feedback? Where, if the load on the motor increased, the voltage would try and increase to maintain a constant speed?
     
  11. n9352527

    AAC Fanatic!

    Oct 14, 2005
    1,198
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    It's not actually a closed-loop feedback in the sense that you mentioned above. In this circuit, the voltage across the motor during constant speed is dictated by the pot. The output voltage is fairly constant with varying motor load as long as the circuit (especially Q1 and Q2) does not run out of gain for the output current (i.e. the output current can be mantained by Q1 and Q2). So, there is no closed-loop feedback to maintain motor speed according to load.

    The feedback across C3 is the _derivative_ of the acceleration/deceleration (or increase/decrease of output voltage more correctly) which would result in faster acceleration/deceleration as per the example in my previous post.
     
  12. JohnnyD

    Thread Starter Well-Known Member

    Aug 29, 2006
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    Ahhh right I see. Thanks a lot for taking the time to explain.

    I'd really like to add a closed-loop feedback element to this controller but I've got no idea how. I've seen other schematics with feedback but they are pretty different, using a PNP darlington power transistor with a smaller NPN connected between the collector and base. I've drawn one from memory so it may not be right but here is the feedback part of the circuit

    I'm not sure how it works but i'd love to have feedback on my controller.

    Anyway, i'm just ranting now i should really get on and actually make one rather than tweaking and adjusting everything for weeks!
     
  13. Montgue

    New Member

    Nov 16, 2011
    8
    1
    Hi John, i have been following the thread very closely, as i have been looking for a controller for a while. I need something simple as this is. i do have a question though, R7 is it a trim pot or a linear control and could it be changed to a horizontal trim pot. I take it that R2 is the main control. I am a brand new member and have only just found your listing. Hope you don't mind. Thanks Montgue
     
  14. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    Well, I wish you luck in getting a reply from these fellows.

    JohnnyD's been dormant since May of 2009.
    n9352527 hasn't been heard from since May of 2010.
    This thread went dormant over 5 years ago.
     
    Montgue likes this.
  15. JohnnyD

    Thread Starter Well-Known Member

    Aug 29, 2006
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    Weirdly enough, I just checked an old email account and spotted this!

    Hi Monague. To answer your question, R7 is another standard pot. It is designed to be the 'brake / momentum' control. It adjusts the rate at which C3 discharges and hence how quickly the motor will get up to speed / lose speed when you change the throttle (R2).

    Incidentally, S1 should be labelled 'Simulator on/off'. When it is off, the throttle (R2) directly controls the motor speed.

    With the simulator on, you can turn the brake (R7) all the way on and then set the throttle to a particular speed. The motor won't start moving until you gradually release the brake. If you suddenly release the brake, the motor will speed up gradually until it reaches the point where the throttle is set. As you approach the station, you can kill the throttle with the brake fully off and the motor will gradually coast into the station. You can then just ride the brake as you get into the station to come to a realistic stop.

    It is a great circuit and the momentum/braking is pretty realistic. Be aware though that with the simulator on and the brake full off, the loco will coast for a very long distance if it's going at a fair speed and you kill the throttle.

    I have built and tested this circuit now and it works. The only things wrong in the schematic are - the transformer shouldn't be 15v secondary, more like 12v. And the ammeter (M1) is shown wired the wrong way round (although I left it out of mine due to cost).

    Good luck with it!
     
  16. Montgue

    New Member

    Nov 16, 2011
    8
    1
    Ok, thanks john for coming back. I now understand the reason for R7. I will make a board for it and see how it works. Take care.
     
  17. John P

    Senior Member

    Oct 14, 2008
    1,603
    213
    There is another circuit out there that does provide a feedback function, which is useful for moving trains smoothly at low speed. In discussions a few years back, the designer admitted that he couldn't explain how it works! But it definitely does have the feature that if you stall out a moving train, the voltage on the rails will increase. It's called the Cooler Crawler:

    http://users.rcn.com/weyand/tractronics/articles/ccartcl/ccartcl.htm
     
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