The one you posted (TJ) and # 12 modified is fine with a cap like this.
http://www.mouser.com/ds/2/420/al-kyb-e-140701-514759.pdf

 

If 5 seconds is enough, 10,000 uf @ 10 V
Here's a page of them starting at $1.42
You can search for the size and shape you want.

http://www.mouser.com/Passive-Compo...tors/_/N-75hqt?P=1z0si9qZ1z0x6es&Ns=Pricing|0

 

Like ships in the night.

 

If you care about the voltage balance then you can in parallel with a resistor for each capacitor, or you have some other better ideas?

I care about voltage balance because I don't want capacitors exploding on him. I know, silly me. Passive balancing with resistors should be fine for this application.

 

The one you posted (TJ) and # 12 modified is fine with a cap like this.
http://www.mouser.com/ds/2/420/al-kyb-e-140701-514759.pdf

Thanks for the clarification and the circuit diagram.

If 5 seconds is enough, 10,000 uf @ 10 V
Here's a page of them starting at $1.42
You can search for the size and shape you want.

http://www.mouser.com/Passive-Compo...tors/_/N-75hqt?P=1z0si9qZ1z0x6es&Ns=Pricing|0

Thanks for the link. I hadn't heard of Mouser before but it looks like they have some good deals.

Regrettably, it seems like my idea is unrealistic. Any 8200μF capacitor (it seems) necessarily has a diameter greater than ½-inch, which is too large to fit on my hovercraft. So I think I'll have to give up on this idea for now.

 

the cap value could be reduced down below 1000 uF if a transistor is added to the circuit

 

the cap value could be reduced down below 1000 uF if a transistor is added to the circuit

Where exactly would it need to be placed in the circuit? Like is shown in the section "Basic capacitance multiplier circuit" at http://www.radio-electronics.com/info/circuits/transistor/capacitance-multiplier-circuit.php?

 

yes, drawing 2, except your R1 is a diode.

 

yes, drawing 2, except your R1 is a diode.

Sorry if this seems like a basic question, but how would I know what size capacitor I need if I use a transistor?

 

Your discharge path used to be a resistor and an LED. Now R2 is your discharge path for the capacitor, and you get to choose both components! Put your original capacitor in and experiment. Try 62k.

 

What I had in mind was this:

Bias resistor should be <= 10k; Not 100k

 

What I had in mind was this:

Bias resistor should be <= 10k; Not 100k

Thanks for the very helpful circuit diagram!
I'm curious why there are two resistors in it instead of just one?

 

Yes, having the LED fade when I disconnect the battery is exactly what I want.

No battery, no power.

 

I'm curious why there are two resistors in it rather than just one?

One is used to bias the transistor, limiting cap discharges current, and the other one is used to limit LED current

 

@ronv: Do you mean that the LED won't fade when I disconnect the battery if I use a transistor?

@iimagine: Thanks.

 

Alas, no. It will just go out. The power must be stored to power it after the battery is removed.

 

Uhmm... If you completely disconnect the battery, nothing will lit. My circuit will work if 'disconnection' by mean of a switch and battery remains intact

 

I'm confused. I know a transistor won't make the LED fade out by itself, but isn't it OK to use it in combination with a capacitor? If I include a transistor in my circuit, I shouldn't need to use as large a capacitor then, right?

 

The problem is when you disconnect the battery there is no power to light the LED. So you need to store some power in something like a battery or capacitor. You can't get more power out than you put in.
If you could use a red or green LED you could maybe power it with a small coin cell. Something like this or smaller. The blue LEDs needs over 3 volts to turn on so they wouldn't be bright with a 3 volt battery.
http://www.batteryjunction.com/renata-cr2325-cu.html?gclid=CKDNlqem5cICFUdefgodGbQA3w
Problem is they are kind of heavy for their size.

 

the cap value could be reduced down below 1000 uF if a transistor is added to the circuit

But how would that help out given that the energy from the LED is coming from the capacitor in question?

Now, if you add a coin cell this this might work to fade out the LED without having to physically switch off the battery quickly to avoid draining it too much. Which brings things back to the circuit I suggested to make it so that the power comes from the external battery until it is disconnected and then from the coin cell until this simple timer circuit kills it off.