I have read that you can add a capacitor to a model loco to keep it going over a dead spot on the track - not DCC - just DC. Is this accurate? On one forum someone has been experimenting with 2000mF 12V DC polarised capacitors. However what happens if the power is reversed - I assume a short circuit - can anyone help with advice?
This is easy for digitally controlled railroads, where you can connect the capacitors to the DC part of the circuit behind the rectifier.
I've done it for LGB garden railroads with nominal 24V (in reality up to 28V), using ten or eleven 3 Farad supercaps in series, which gives about 0.3F = 300mF. The supercaps need to have low internal resistance.
With that the train can continue to run for about a meter, which IMHO is a bit much (but there are no suitable smaller capacitors, so I can't easily change it). If control is lost near the end of the track this can result in a crash.
A more detailed description of the circuit: The circuit uses a diode and a resistor to limit current when charging, but not when discharging. There is also a choke in series with the capacitor, so that the fast digital data signals are not filtered out. Resistors across each capacitor even out the voltage and discharge it slowly when not in use.
Connecting a regular electrolytic capacitor in reverse results in that capacitor blowing up (or just creating a short circuit if there is not enough power for it to blow up).
However, there are non-polarized electrolytic capacitors. if you really need to, you can make a non-polarized capacitor by taking two capacitors of equal capacity and connecting their positive leads together (while connecting the negative leads to the circuit). The resulting capacity will be half of that of a single capacitor and the voltage rating will be the same as a single capacitor.
As @Pentium100 says - connecting two std polarised capacitors "back to back" is a reasonably good way of making a non polarised capacitor of half the value.
Note that 1 mF = 1 milli-Farad = 1000 uF = 1000 microFarad.
Was your forumite using a 2,000 uF cap or a 2,000,000 uF one ? Probably the former.
As a guide A capacitor will droop APPROXIMATELY 1 volt in time t where
- t = C/I
It will droop V volts in V times as long.
C = capacitance in FARADS and I = current in amps.
On a 12 Volt system I ~= W/12
where W = motor watts.
So a 1000 uF (not mF) cap will droop 1 V in 1 mS at 1A or in 10 mS at 100 mA or in 100 mS at 10 mA or in 1 second at 1 mA
In a model railway system you can probably run over a dead spot if V >= 6V (at a guess) so
- t ~~= 6C/I [s, F, A]
- C = t x I /6 [F, s, A]
If time is in mS and capacitance in uF then
- t (ms) ~~= 6 x C (uF) / (1000 x I) = C/(166 x I)
So if you want a 300 mA motor to hold up for 100 mS at 6V droop
t = C/166I
So C = 166 x t x I
= 166 x 100ms x 0.3A = 5000 uF