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I want to know the numbers needed to build a capacitor bank to sustain a load of 36 V, 1500 W up to 5 - 10 minutes at time before needing recharge. I have four 10,000 μF, 40 V caps connected in parallel to a 12 - 36 V booster with a 36 V, 1500 W load connect to the other end of the cap bank.

enter image description here

Ok, thank you all for your input, let me add some more details. The Project is being powered by a 12 V car battery, the battery is powering a 12V DC motor turning a Self Exciting 12 V, 96 A alternator, the alternator is powering a 12 V to 36 V booster at 30 A and in turn charging the CAP bank and powering the 36 V, 1500 W load, while also keeping the battery charged, at IDLE the load is drawing 100 W.

OK, with the comments so far , i need to give more details again.

The battery is only powering the 12 V motor, cooling fnas and power monitors.

The alternator is supplying power to a separate set of power rails for the 12 V to 36 V booster from the booster to the cap bank and then to the 36 V 1500 W load, my thought was to have the cap bank as a buffer between the alternator and actual load device. A seperate 12 V mini charger will also take power from the alt and send it back to the battery to help keep it charged.

The motor driving the alt is 400 mA, 12 v, 3500 rpm with no load, the cooling fans are 120 mm computer fans so the draw from them is also minor, like a car the battery is basically only for starting the system while the alt is providing all the main power and returning some to the battery, NOT perpetual motion.

Does all this sound feasible and is there anything else I might need, like high-wattage balancing resistors and such?

What else do I need to know?

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    \$\begingroup\$ The amount of time 40000µF can power that load would be expressed in milliseconds rather then minutes... \$\endgroup\$ Apr 24 at 14:51
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    \$\begingroup\$ You need around 275 farads... i.e. around 27500 of those 4 capacitors that you have. (capacitor charged to 40V, discharged to 12V in 5 minutes, supplying about 1500W). Calculator here mustcalculate.com/electronics/capacitorchargeanddischarge.php \$\endgroup\$
    – Indraneel
    Apr 24 at 15:01
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    \$\begingroup\$ You're asking how many toothpicks it'll take to build a shed. Even if they're the finest toothpicks known to man, they're still the wrong tool for the job. You need batteries for this, not capacitors \$\endgroup\$
    – brhans
    Apr 24 at 15:08
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    \$\begingroup\$ Are you saying that you have a battery running a motor-alternator set which is running a 1500 W load and charging the battery that is powering the circuit? That's a perpetual motion machine and they don't exist. "... is there anything else I might need?" Maybe an exemption from the laws of physics. \$\endgroup\$
    – Transistor
    Apr 24 at 17:37
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    \$\begingroup\$ If the motor draws 400 mA at 12 V, that's 4.8 Watts - you can't get more than 4.8 Watts out of the alternator that it is driving. You are making a very complex battery discharger. \$\endgroup\$ Apr 24 at 20:16
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Does all this sound feasible and is there anything else I might need like high watt balancing resistors and such?

No.

What else do I need to know?

Perpetual motion machines don't work, no matter how many magnets you use.

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You're looking for an energy storage of 1500 W for 600 s so that's \$ E = 1500 \times 600 = 900,000 \ \text J = 900 \ \mathrm {kJ} \$.

You've got capacity for \$ E = \frac 1 2 C V^2 = \frac 1 2 \times 40000 \mu \times 36^2 = 26 \ \text J \$. And this assumes that you are able to extract all the energy from the caps - which you won't.

You're rather short of capacitors.


enter image description here

Figure 1. The perpetual motion part of your concept.

... like a car the battery is basically only for starting the system while the alt is providing all the main power and returning some to the battery, NOT perpetual motion.

No, in a car the internal combustion engine drives the alternator. When the engine revs are high enough the alternator output will be 14 V or so and will power the car's electrical circuits as well as recharging the battery. In your scheme the battery is providing the power to drive the alternator which powers the charger which feeds back into the battery.

If you were to try this and get 5 A out of your charger at 14 V:

  • You would need \$ 12 \times 5 = 60 \ \text W \$ from the alternator +, say, 10 W for losses in the charger so that's 70 W.
  • If your alternator is 80% efficient you will need \$ \frac {70}{0.80} = 87.5 \ \text W \$ mechanical input to the alternator.
  • If the DC motor is 80% efficient you would need \$ \frac {87.5}{0.80} = 109 \ \text W \$ into your motor.
  • The result is that you would be taking 109 W out to put 60 W back in.

Can you see why this doesn't work? It's the same reason you can't run an electric car or bike using a dynamo to power the motor.


OK, so with a battery having 650 CCA, and if the number are where you calculate them at, about how long could i run this configuration before needing outside power to recharge the battery.

CCA (cold-cranking amps) doesn't tell you how much energy the battery supplies. It just tells you the maximum current it can supply (but not for how long).

You need to work out the capacity of the battery. For this you need the voltage (12 V) and the ampere-hour (Ah) capacity. From these you can calculate the energy stored as \$ E = V \times Ah \$. The result is watt-hours (Wh).

Then figure out how much power (W or watts) your load draws. \$ P = VI \$. Divide the result by the efficiency of the motor (say 0.8) and again by the efficiency of the alternator (say 0.8 again) to calculate the power drawn from the battery.

The run time is then given by \$ \frac {battery\ capacity}{power\ required} \$.

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  • \$\begingroup\$ ok, so with a battery having 650 CCA, and if the number are where you calculate them at, about how long could i run this configuration before needing outside power to recharge the battery. Im a mechanic, not a electrical engineer, so the help is appreciated. \$\endgroup\$
    – frank
    Apr 24 at 20:31
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    \$\begingroup\$ See the update. \$\endgroup\$
    – Transistor
    Apr 24 at 21:31
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You need some basic math skills. 1500 W for 600 seconds is 900 kJ. A 40 V, 10 mF capacitor holds a maximum of 0.5 × C × V2 = 8 J. Even if your power conversion were 100% efficient all the way down to zero charge, you'd need about 120,000 capacitors.

If you're charging the capacitors to 36 V and discharging them to 12 V, you're only getting 5.76 J out of each one. If your boost converter is 80% efficient over that range, then you'll need a total of about 200,000 capacitors.

This is why "supercapacitors" and "ultracapacitors" exist. I have a couple of boxes of 3000 F, 2.7 V ultracapacitors. These store 10 kJ each, so it would still take roughly 100 of them to hold the energy you're asking for.

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You wanted up to 10Ah @ 36V with some losses or sag or an equivalent of 900,000 J for 10 min. One car battery has about 50 Ah @ 12V or 600 Wh. That is what you need.

This is almost kinetic energy of a car going 100 MPH.

A cap can store \$ E= \frac 1 2 CV^2 \$ [joules] initially but must cutoff and retain energy at your Vmin tolerance.

Thus is useful for small energy demand but is not cost-effective for large energy demand.

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