Thanks Bob for the technical information. Let me knock some of the dust off and impose upon you some more technical questions. Going to Tanner's in Dallas they may not have the exact cap value, thus can I sub for a higher/lower value? See where T will follow C and am I correct in thinking a higher T value is better? More questions, I have some CM 327 lamps with holders, can these be used? CM 327 lamps specification are 28v .04a, do not think the MSCP value is used here. Can we not use the power value here, 1.7 to 1.1, is a .6 loss going to cause problems? If I was to throw two turnouts at one time, what can of current draw at 18v dc can be expected?
For enquiring minds, here's the derivation:
The differential equation to be solved is
dv/dt = i / C
where v is the capacitor voltage, i is the current that the lamp delivers to the capacitor, and C is the capacitance. Assume that the solution and its derivative have the forms
v = Vsup - b * (-t)^a
dv/dt = b * a * (-t)^(a-1)
where Vsup is the positive supply voltage, t is time relative to full charge, and a and b are constants whose values are to be determined. The lamp current is proportional to the kth power of lamp voltage, usually .55:
i = [(Vsup-v) / V0]^k * I0
where V0 and I0 are the specified operating point of the lamp. Plugging the expressions for i and v into the differential equation gives
b * a * (-t)^(a-1) = b^k * (-t)^(a*k) / V0^k * I0 / C
In order for the equation to be true for all t <= 0, the exponents of -t must be equal:
a - 1 = a * k
a = 1 / (1-k)
Then, cancelling the -t terms,
b * a = b^k * I0 / V0^k / C
b = [I0 / V0^k / C * (1-k)]^[1/(1-k)]
These constants are not functions of Vsup.
Now the time to recharge T can be calculated from the original assumed solution:
0 = Vsup - b * T^a
T = Vsup^(1-k) * C * V0^k / I0 / (1-k)
If Vsup is set equal the specified lamp voltage V0, the recharge time is
T = C * V0 / I0 / (1-k)
Bob Nelson
Understand your reason for the lamp. Will be looking for some caps and lamps when I go to Dallas, also looking forward to this winter project. Thanks again. cah
You can substitute a 100-ohm 5-watt resistor for the number-53 lamp; but I don't recommend it.
The lamp recharges the capacitor significantly faster than the resistor. The resistor or the lamp provides about the same recharging current at first; but, as the capacitor voltage approaches the supply voltage the resistor's current drops in proportion to the voltage difference (Ohm's law). As the capacitor gets closer to the supply voltage, the charging current and therefore the charging rate gets pretty slow; so the resistor-capacitor circuit takes a relatively long time to reach full charge. The lamp, on the other hand, reduces its resistance as the voltage difference drops. The charging current still drops, but much less than it would with the resistor; so the capacitor recharges fully very quickly.
You could of course use a much lower resistance, to shorten the resistor's charging time. The problem with that is that the initial charging current would be rather high. And your power supply would have to supply it continuously while any part of the train is on the turnout.
Why not use a lamp? If you don't want to see it, you can easily cover it up.
Thanks for the replies, now I understand where I was looking at the setup wrong. Can you substitute a diode or resistor for the #53 lamp? If so what would the resistor value be in ohms and watts? If you substitute will the cap still stay at 4700 µf 35v?
Many thanks to Larry, Bruce and Bob.
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When you disassemble the 1122 switches to extract the power wire for the coils, I recommend you restore the switches. I have posted a procedure for doing that at http://www.modeltrainforum.com/showthread.php?t=5513&highlight=1122+switches
I think Bob Nelson also has a procedure on this forum.
I have never found an 1122 switch that did not need restoration. Age and the fact that the plastic cold flows cause the crimp connections to become high resistance. This can damage the plastic that the switch is made from.
I think that the confusion here comes from the fact that, in that earlier topic, he wanted to include a latching relay along with the turnout. The capacitive-discharge circuit can be used with either a positive or a negative supply; and I described it to him first with a positive supply. However, when the latching relay was introduced, it turned out to be polarized and in such a way that it needed a negative supply. So we switched horses in midstream.
I'll describe it for you anew assuming a positive supply: Connect the negative terminal of the DC supply to the layout common, that is, the outside rails (other than the control rails). Connect the negative terminal of the capacitor also to the layout common. Connect the lamp between the positive terminal of the DC supply and the positive terminal of the capacitor. Connect the positive terminal of the capacitor to the coil wires--the ones that you have to extract from the insides of the turnout.
I would locate the capacitor close to the turnout, with its negative connection to the track nearby. That keeps the current impulse that throws the turnout within a small area and eleminates any concern that wire impedance will affect the operation. The lamp and the power supply can go anywhere; and the power supply's negative terminal can be connected to the track or the transformer. (I suspect that the transformer will be more convenient.)
If for some reason you (or anyone else trying it) want to use a negative supply, just swap the words "positive" and "negative" throughout the description above.
Thanks Larry for the reply. Still need to add the capacitor discharge parts. From what I read on Bob N. posting I think when we add the cap and lamp, the + and - from the dc supply are reverse. See his posting: http://cs.trains.com/TRCCS/forums/p/173684/1915557.aspx#1915557 With the outside rails being ground can the capacitor just go to the ground terminal post? Thought I read somewhere if you stop on the turnout you could burn up the coils, thus the cap and lamp solves this problem. Correct me if I am going up the wrong tree here. Many thanks cah
The conversion to external power will work with the 1122E turnouts, and everything will work normally. You need to disconnect the power wire for the coils from the center rails of the turnout, and hook it to the positive of the power supply. Connect the negative to the outside rails of your layout, and your done.
Larry
cah - Welcome to Trains.com!
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I have the 1122E RC switch turnouts with the 1122-100 controllers. Would like to place them on a separate dc supply with the capacitor discharge circuit. Reading Bob N. inputs, I think\hope I can do it. Some questions, 1. Will this work with 1122E switches? 2. Can I still use the 1122-100 controllers? 3. Is there a wiring diagram available showing this conversion; if available how can I get it? 3. Correct in saying the lantern, light, and anti-derailing will still work correctly? Thanks for any and all reply's.
Just a FYI, could not enter a post message in this forum with IE9 32 or 64bit, had to use Firefox.
cah
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