The Calvin cycle converts fixed carbon into a reduced three-carbon product while regenerating the $CO_2$ acceptor RuBP.

After RuBisCO fixes three $CO_2$ molecules, six molecules of 3-phosphoglycerate (3-PGA) are present. The cycle then has two remaining jobs:

1. reduce 3-PGA to the carbohydrate-level intermediate glyceraldehyde-3-phosphate (G3P);
2. regenerate RuBP so new $CO_2$ can be fixed.

Reduction uses ATP and NADPH

The six 3-PGA molecules are first phosphorylated using six ATP and then reduced using six NADPH, producing six G3P molecules.

At the level of carrier bookkeeping,

$$6\ \mathrm{ATP}\rightarrow6\ \mathrm{ADP}$$

and

$$6\ \mathrm{NADPH}\rightarrow6\ \mathrm{NADP^+}.$$

NADPH supplies reducing equivalents; ATP supplies thermodynamic driving force and phosphoryl-transfer chemistry.

Only one G3P is net output per three CO2

Six G3P molecules contain

$$6\times3=18\ \text{carbon atoms}.$$

One G3P, containing three carbons, can leave the cycle as net fixed-carbon output. The remaining five G3P molecules contain

$$5\times3=15\ \text{carbons},$$

exactly enough carbon to regenerate three five-carbon RuBP molecules:

$$15=3\times5.$$

RuBP regeneration requires three additional ATP.

Therefore, producing one net G3P while regenerating RuBP requires the fixation of

$$3\ CO_2,$$

consumption of

$$9\ ATP,$$

and oxidation of

$$6\ NADPH\rightarrow6\ NADP^+.$$

These are pathway resource counts, not a complete balanced molecular equation for every proton, phosphate and water molecule involved.

Why it is a cycle

The carbon flow is

3 RuBP + 3 CO2
      ↓ fixation
6 3-PGA
      ↓ ATP + NADPH
6 G3P
  ↙       ↘
1 net      5 rearranged
output      ↓ + ATP
           3 RuBP regenerated

The Calvin cycle therefore does not turn every newly formed intermediate directly into sugar. Most fixed carbon temporarily remains in the cycle so the $CO_2$ acceptor can be rebuilt.