I plan on powering a 240VAC mobile candy floss making machine from a 12VDC leisure battery using an inverter... the problem is my understanding in this area is rudimentary.

Candy Floss (cotton candy) is made by heating sugar and spinning it. The machines have a heating element to get the device up to temperature - usually around 1200W. Once the device is at temperature they can be dialed down to a maintenance voltage to keep the machine hot.

The machines I have seen have a voltage dial to control the temperature. You'd start them at 240V and then dial down to, say 90V when it is hot.

Here's my reckoning - I'd appreciate it if someone can correct me if I'm mistaken...

Current drawn by the candy floss machine at full power...

I=5 Amps

Let's say it takes 60 second to get up to temperature...

As = A x t
   = 5 * 60
   = 300 Amp seconds
   = 0.83 Amp hours

Finding the load of the candy floss machine...

R = V/I
R = 240 / 5
R = 48 Ohms

When the machine is running on 90VAC...

I = V/R
I = 90 / 48
I = 1.9 Amps

If we're running the machine for 6 hours at this current...

Ah = A * h
   = 1.9 * 6
   = 11.43 Ah

Allowing for the inefficiency of an inverter, say 75%...

Ah = 4 * 11.43 / 3
   = 15.2 Ah

Assuming a maximum 50% battery discharge...

Ah = 15.2 * 2
   = 30.4 Ah

So I only need to find a leisure battery that can handle 30+Ah.

Am I on the right track here? Something tells me I've missed something important :)


  • 1
    \$\begingroup\$ You're right, you missed something important. 1A at 240V = 20A at 12V so you're looking for 600Ah (and then some, inverters aren't that efficient, and lead acid batteries don't like being fully discharged) so you probably want at least 1200Ah at 12V. (Or a generator). (Transistor's correct that there are savings to be made in the 90V stage that I didn't account for here, so this is too pessimistic) \$\endgroup\$ – Brian Drummond May 15 '18 at 18:55

You've forgotten that the battery is at 12 V.

Let's work it out in watts instead and then see what current we need at 12 V.

  • Warm up: 1200 W for one minute. We can really ignore this except for the piak current calculation.
  • Running: Since \$ P \propto V^2 \$ we can say that running power = \$ \frac {90^2}{240^2}1200 = 168 \ \text W \$. From \$ P = VI \$ we can calculate that \$ I = \frac {P}{V} = \frac {168}{12} = 14 \ \text A \$.
  • 14 A for 6 hours = 84 Ah at 240 V.
  • Input current will be \$ \frac {output}{efficiency} \$ so at, say, 80% inverter efficiency you will require 17.5 A in and 105 Ah capacity.

Now let's look at the peak demand again:

  • 1200 W at 12 V will require 100 A for a minute. When we factor in the inverter efficiency again we get 125 A peak current from the 12 V battery.

You probably require 2 x 125 Ah batteries. In that case go for a 24 V inverter and connect the batteries in series. The current and cable sizes will be lower.

  • \$\begingroup\$ Thanks, Transistor. I think I pretty much follow your logic here, but I'm not getting why you jump to 2 batteries. It looks like the warm up cycle doesn't really drain the battery too much compared to the lower voltage, extended time cycle. Are you doubling it so that the batteries aren't drained by more than 50%? \$\endgroup\$ – Doug May 16 '18 at 17:12
  • \$\begingroup\$ Correct. Go for a good safety factor. \$\endgroup\$ – Transistor May 16 '18 at 17:14
  • \$\begingroup\$ Nice one. After a little more research, and speaking to the manufacturers of these things, I'll be able to plug more accurate figures into these equations and see what pops out the other end :) Thanks. \$\endgroup\$ – Doug May 16 '18 at 18:51

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