Voltage (V) is how hard a battery pushes; capacity (mAh) is how much charge it holds. Run time ≈ capacity ÷ current, and plan on using about 80% of it. A battery for a robot has to meet every criterion: voltage in range, enough run time, light enough to carry.
Your robot drives beautifully for four minutes, then crawls. A 9 V battery, more volts than the AA pack: shouldn't that be better?
Voltage (V): how hard it pushes. The motor driver needs it within a range.
Capacity (mAh, milliamp-hours): how much charge it holds. 2000 mAh could give 2000 mA for an hour, or 500 mA for 4 hours.
Run time (hours) ≈ capacity (mAh) ÷ current (mA).
Predict: Read the problem and decide what you think before you try anything. A wrong guess is fine: it shows you what to look for.
Try: Try it on the board and compare the result with your prediction.
Investigate: Work out why it happens, one step at a time.
Modify: Change the problem on purpose and see what happens.
Practise: A few on your own, to check you have it.
Make: Make your own problem using what you have learned. Expect a few tries; that is normal.
Make your prediction first. Help opens once you've locked it in.
PredictRead the problem and decide what you think before you try anything. A wrong guess is fine: it shows you what to look for.
The 9 V battery
TryTry it on the board and compare the result with your prediction.
Try: compare them
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The PP3 is fitted. Note its run time, then try 4 × AA.
Use the board. This ticks when you get there.
InvestigateWork out why it happens, one step at a time.
Worked example: run time
Worked exampleBefore each step, say what you think comes next.
2000 mAh÷350 mA≈5.7 hCapacity ÷ current gives hours.
5.7×60≈343 minIn minutes.
2000×0.8=1600 mAhPlan on about 80%: voltage sags near empty, and rechargeables are damaged by running flat.
1600÷350×60≈274 minA safer estimate.
Now the LiPo.
A 1000 mAh LiPo powers a robot drawing 900 mA. Using only 80% of it, how many minutes does it last? (nearest whole minute)
Sam's robot is no faster with the new pack. Click the line where Sam's reasoning first goes wrong.
ModifyChange the problem on purpose and see what happens.
Vary: a hungrier robot
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The robot now draws 900 mA and must run for an hour. Switch on 'use only 80%'. Does 4 × AA still pass?
Use the board. This ticks when you get there.
Try 4 × AA NiMH rechargeables. Why do they fail?
Use the board. This ticks when you get there.
PractiseA few on your own, to check you have it.
Run times
A 2200 mAh pack at 1100 mA, using all of it. How many hours?
You need 90 minutes at 600 mA, using 80% of the battery. What capacity, at least? (mAh)
MakeMake your own problem using what you have learned. Expect a few tries; that is normal.
A light robot
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Need a hint?
Check each criterion for each battery: voltage, run time (at 80%), weight.
Another hint
AA packs are too heavy for 80 g.
Another hint
The 1000 mAh 2S LiPo: 7.4 V, 53 minutes, 60 g.
A small robot: 900 mA, at least 45 minutes using 80%, 6–9 V, at most 80 g. Find the battery that meets every criterion.
Use the board. This ticks when you get there.
For teachers
Students compare a 9 V PP3 with 4 × AA for a robot, work out run time from capacity and draw with an 80% margin, find why NiMH cells fail a voltage range, and choose a battery against voltage, run time and weight, which leads to a LiPo. About 40 minutes.
I can work out a battery's run time from its capacity and the current.
I can choose a battery against a brief's criteria.
I can explain the difference between voltage and capacity.
Before class
Figures are typical, not exact: alkaline AA capacity falls a lot at high current, and PP3s are poor at motor currents (often well under 500 mAh). Check the packs you buy. The board's run time is capacity × usable % ÷ draw.
LiPos appear here as the battery that fits; the next lesson is required before anyone uses one.
Running the lesson
Predict: ask what the 9 in 9 V tells you, and what it doesn't.
Make: ask students to say which criterion rules out each battery.
The hard part (on purpose)
Students rank batteries by one number. Keep all three criteria on the board for every candidate.
When students get stuck
A student divides current by capacity. Ask: Which is bigger, and which way gives hours? Fix: Capacity ÷ current.
Stretch and support
Stretch: Two batteries in series: what happens to the voltage, and to the capacity? In parallel?
Support: A table with a row per battery and a column per criterion.
the 9 V battery: Committing to which battery lasts longer tests the new move: run time comes from capacity and current, not voltage.
Try: compare them: Switching between the two batteries on the board settles the prediction.
Worked example: run time: Working run time from capacity and draw, then planning on 80%, models the move. Sam's 'more mAh is more powerful' is planted after the method.
Vary: a hungrier robot: A bigger current draw, a longer run and the 80% rule change which batteries pass, one criterion at a time.
run times: Run time both ways.
a light robot: A brief where weight rules out the AA pack and run time rules out the PP3 leads to the LiPo, and to why it needs the next lesson.
Now you can
Work out a battery's run time from its capacity and the current.
Choose a battery against a brief's criteria.
Explain the difference between voltage and capacity.