Supercapacitor backup?

Background: as part of a project, I want to have a camera transmitting video from a model railroad car. 18v AC power can be obtained from the railroad tracks. However, contact with the track can be interrupted due to intermittent contact of the wheels, so I'd like a way to provide power for short (say 1 second) interruptions in power.

The video module I plan to use is described here. The module is said to accept 7.4 to 24 V as input, drawing 200 mA @ 12V.

Power can be supplied by an adjustable AC-DC converter like this one, which can supply about 3A.

I was thinking I could use a supercapacitor and a boost module (like the Pololu designs) to supply the power. I calculate that I need about 2.4 joules, which seems within the range of a supercapacitor.

Assuming this is feasible, the question is what supercapacitor? Is it better to use a relatively high voltage (9v) capacitor because it stores more energy, or a higher capacity 5v capacitor? And how much ESR can I tolerate? To make things simpler, let's say I want a capacitor costing no more than \$10? I'm not sure how to find the right balance of capacity, voltage, ESR, and price.

A less high-tech approach could be to use a 9V battery backup. In that case, I would need some circuit that would cut the battery out when power was interrupted for a long time, since I wouldn't want the battery to be powering the module when the car is removed from the track.

Is it better to use a relatively high voltage (9v) capacitor because it stores more energy, or a higher capacity 5v capacitor?

I can't answer everything, but just one comment - A higher voltage is better because a capacitor discharge curve is the opposite of an ideal battery. It discharges rapidly from the maximum, then the voltage levels-off the more it discharges. So the higher the voltage the longer you can get usable energy out of it.

It would be much easier and less expensive to use a 9V battery as backup. Use a diode OR switch, as in this:

If the supply voltage is higher than the backup battery voltage, no current will be drawn from the backup battery.

Would there be a way to cut the battery out of the circuit if power was removed for a prolonged time?

The battery should be supplying power only when power from the track is interrupted momentarily. I would not want to have to retrieve the car from the track and turn it off manually every time I turned the track power off.

lefstin:
Is it better to use a relatively high voltage (9v) capacitor because it stores more energy, or a higher capacity 5v capacitor?

You want to maximize the working range of voltage in the capacitor. Assuming this is set by the working range of your video module (7.4V to 24V), you want a capacitor and charging system that goes to 24V full charge. This allows the use of (24 - 7.4)/24 = 69% of the total capacity of the capacitor.

MrMark:
You want to maximize the working range of voltage in the capacitor. Assuming this is set by the working range of your video module (7.4V to 24V), you want a capacitor and charging system that goes to 24V full charge. This allows the use of (24 - 7.4)/24 = 69% of the total capacity of the capacitor.

Except that even a 20V supercapacitor costs at least \$13 and has a capacity of only 4.7mf.

As stated in post #1, I thought of using a boost converter to achieve 9v or 12v. I'm trying to find the optimal combination of voltage, capacity, and ESR for the supercapacitor, but don't know how to figure that out.

The stored energy in a capacitor is 1/2CV2, but the effective voltage range of the device limits the useful range to 1/2C(V2max - V2min) where Vmax and Vmin are the maximum and minimum working voltages of the voltage regulator or boost converter following the capacitor as per post #4.

If you add a boost converter then there is an efficiency multiplier (typically around 80%) for that component.

OK, but what about the ESR? The higher-voltage supercaps I see (~ 11v) have ESR from 6-20 ohms.... does that imply a pretty hefty drop at 200 mA or more?

So a micro reed relay such as HE3300 uses about 35ma to engage.

You could use a reasonably small capacitor in the feed line to that relay so that when power is lost the capacitor keeps the relay closed.

Have your battery circuit pass through the relay closed contact into the OR circuit above and you will have auto shutdown when the capacitor can no longer hold the relay closed.

Also millions of 555 circuits that could keep that relay powered for a timed period powered off the battery.

Thanks Slumpert. I was thinking about a relay solution like that if I went the battery route.

Is there some clever way to do it with a P-channel MOSFET instead of a relay?

No tricks, but be sure to use a pull down resistor or it may not shut down for long, long time..

I love supercaps. I buy Vishay 1F, 5.5V caps. They make a perfect battery change/power-failure failsafe device, and I use them anywhere that could be an issue. I think I paid a buck fifty each for 40 of them from ... mouser, newegg, element14? Or one of the US based suppliers:

does that imply a pretty hefty drop at 200 mA or more?

Yes, and you can even use Ohm's law to calculate it.

Hint: 0.2 A*20 Ohms = 4 V.

I'm going to experiment with one of these in conjunction with a boost module.

They're rated 9V, 1F, and have a stated ESR of 350 mOhm. Reasonably priced at \$6.50.