Model train controller

[Crackle]

I have been reading the article in the link which Al posted.

I think I understand what the author is talking about. There are 3 basic circuits;
1. Voltage control using a Darlington pair, with current sense and limiting (TR3), back EMF diode protection and fuse.
2. 555 ic giving variable PWM modulation of the voltage to give accurate and realistic starting and slow speed control.
3. A circuit to automatically control the PWM to give predetermined acceleration and breaking of the train.
I have added the 3 separate circuits together in the attached diagram.

Please can you confirm my diagram is a correct interpretation of the notes in the article.
The author does not fully explain the operation of S2 and S3 in the section 3 part of the circuit. Can you help here.
link to article; https://www.instructables.com/id/Simple-...M-inertia/

combined circuit diagram.
   


Thanks
Mike

[Refugee]

Are those not braking and emergency braking?
They look like they are in the inertia simulator.

[Crackle]

I think you may be right, I initially thought they were start and stop buttons but they do both appear to be wired to discharge the capacitor.

thanks
Mike

[Crackle]

I constructed a test prototype of the circuit in Al's link.
I dont have the 555 ic's as yet so I am feeding the control voltage straight into the Darlington circuit.
But I was not impressed with the load control and voltage stability.
I am looking for around 1 Amp supply at 12 volt. But with various test resistors I am getting the following results.
Initial voltage on smoothing cap 17.8v
full voltage on output 13.2v no load.
47R load 240mA @ 11.3v
22R load 450mA @ 10v
12R load 750mA @ 9v
4.3R load 1300mA @ 5.5v at this load the protection circuit has cut in.

So no real voltage stability.

I am now thinking of using a mosfet transistor for the final control, I believe these have a far better loss and lower resistance when in conducting state than the 2N3055 transistor used in my prototype.

This is the basic circuit I am now considering.
   

But I am confused by the 2 diodes in the variable resistor circuit. What are these for.

But I would like to add the inertia function of the first speed control circuit, and overload protection as well.
The variable resistor would be used to set the final max speed.
I am thinking to use a 7815 regulator for the initial supply for the motor and mosfet circuit, then feed that to the 7805 to get the supply for the 555 PWM controller.
If I was going to use a R C combination to build up the 555 control voltage slowly where would I connect this into the variable resistor circuit.

Thanks
Mike

[Refugee]

I found that a UC3842 SMPSU chip can be run at much lower frequencies and works for all sorts of pulse driven things and are also more widely sold.
When I was young we had a model train that had been converted from clock work.
A vehicle lamp was connected in series with a resistive controller. The cold lamp passed just enough current to get the train started and then progress was smooth.

[radioman]

But I am confused by the 2 diodes in the variable resistor circuit. What are these for.

They allow different charge/discharge paths for C5 and therefore alter the mark/space ratio of the output waveform giving the PWM.

Andy

[Terry]

Mike, it might be a good idea to repeat the test but this time, check the voltages both before and after the Darlington pair. Also the voltage across R4/R5 - if you are using the same 1R3 value (0R65 total), it shouldn't reach 700mV (the voltage at which Tr3 starts to conduct) until the current reaches ~1A.

There is no reason for the circuit not to work, so there must be a fault somewhere.

There should be a constant 1.4V drop between the base of Tr1 and the emitter of Tr2 when the Darlington pair are conducting.

When the protection circuit cuts in, it should short circuit the input to TR1, reducing the output to nil, so your 1300mA @ 5.5v figure is doubly wrong!

Maximum voltage for the 555 timer is 18V, so your 7815 regulator will be fine but not the 7805! That will reduce the 555 output to ~5V which, with the 1.4V drop in the Darlington pair will reduce the maximum output to 3.6V peak!

As for those diodes, D3 prevents R5 from shunting the output of VR1 with Sw1 in the normal position but I can't see any logical reason for D1 and D2.

[Crackle]

Mike, it might be a good idea to repeat the test but this time, check the voltages both before and after the Darlington pair. Also the voltage across R4/R5 - if you are using the same 1R3 value (0R65 total),  it shouldn't reach 700mV (the voltage at which Tr3 starts to conduct) until the current reaches ~1A.

Hi Terry,
I am using a .47 ohm resistor for the current sense R in the first circuit.

There is no reason for the circuit not to work, so there must be a fault somewhere.
When the protection circuit cuts in, it should short circuit the input to TR1, reducing the output to nil, so your 1300mA @ 5.5v figure is doubly wrong!

Not sure what is happening there then, when I said the protection circuit had cut in, I was going by the fact the LED was alight.

Maximum voltage for the 555 timer is 18V, so your 7815 regulator will be fine but not the 7805! That will reduce the 555 output to ~5V which, with the 1.4V drop in the Darlington pair will reduce the maximum output to 3.6V peak!

I was only intending to use the 7805 regulator for the mosfet circuit, as I believe the gate voltage needs to be less than the 15v supply. (as per second circuit.)

As for those diodes, D3 prevents R5 from shunting the output of VR1 with Sw1 in the normal position but I can't see any logical reason for D1 and D2.

Terry, I was confused by the 2 diodes, D3 & D5 in the second circuit (with mosfet) around R3, the 50k pot .

I will check the voltages in the first (darlington) circuit again and before the darlington as well.
Thanks
Mike

[Crackle]

Hi Terry
I have measured the voltages again, values in (brackets) are taken before the darlington pair.
Initial voltage on smoothing cap 17.6v
full voltage on output 13v no load.
47R load 240mA @ 11.2v (16.76)
22R load 450mA @ 9.9v (16.2)
12R load 750mA @ 9v (15.8)
4.3R load 1300mA @ 5.5v (14.75) at this load the protection circuit has cut in, voltage across the current sense resistor was 560mV.
thanks
Mike

[Terry]

Mike, my initial thought was that Tr2 was dropping an excessive voltage as the current increases but now I'm not so sure.

Look at the difference between the two voltages and calculate the effective resistance of Tr2.

At 240mA it is 23R16, at 450 mA it is 14R, and at 750mA it is 9R06, so it is trying to compensate for increasing current by reducing its effective resistance, as it should

However, your PSU isn't stabilised so, as it sags, the control voltage will also sag. You haven't said how you are providing this voltage but, if it is from the 10k pot, the upper portion of the track will be supplying the base - emitter current of the two transistors, so this will further drop the control voltage.

I would suggest that you repeat the test but also measure the voltage at Tr1's base and see how that varies.

Your Tr3 base - emitter voltage suggests that it is limiting at 1A2 whereas 700mA would indicate a limit of 1A5.

However there must be a sort of yo-yo effect because when Tr3 turns on because of an overload, Tr2 and Tr3 are turned off, so there is no output and thus no overload so it must turn back on again! Not something I've ever investigated but it might be interesting to put a 'scope on the output!

As for the diodes in the other circuit, depending on the state of the discharge pin either one or other of the diodes will conduct and determine which bit of R3 is in circuit to vary the mark -space ratio - at least that's my interpretation!