Electronic – Comparing rated, stall, and inrush current in DC motor

Some motor basics.

A motor's stall current is determined by the resistance of the coil. (quoted at the nominal operating voltage). If the shaft is clamped so it cannot rotate then the 'motor' will act like any other resistor and follow Ohm's law - so yes, the stall current will increase with increasing applied voltage.

Once the rotor starts to rotate it will induce a back emf so the voltage 'seen' across the rotor will be reduced. The net current through the motor will be reduced.

Increasing the applied voltage will increase the speed (how much depends on the motor). With increasing speed comes increasing loss (friction, windage etc.) so there will be a corresponding increase in current.

Applying a load (torque) to the output shaft slows the rotation speed and reduces the back emf. This in turn will increase the motor current. Applying too much torque may reduce the shaft speed to 0 and this is once again back to a stall.

At 5V it draws about 150mA when running

That's a power of 0.75 watts.

The basic operation is to drive a small air pump for a very brief period (about 200ms)

That's an energy of 0.75 watts x 0.2 seconds = 150 mJ.

Using a capacitor charged at 5V means it must hold possibly ten times the energy needed to be supplied to avoid the terminal voltage dropping too low when the motor is connected.

Let's say the capacitor needs to hold 1.5 joules at 5 volts. Energy held by a capacitor is: -

E = \$\dfrac{CV^2}{2}\$ so rearranging and inserting the numbers we get: -

Capacitance = 120,000 uF.

You are probably about 100 times out in your expectations. So maybe you could use a 120,000uF cap? How long would it take to charge that cap to 5V from dead with 100mA.

Q = CV and rearranging

\$\dfrac{dQ}{dt} = C \dfrac{dv}{dt}\$ = current = 100mA

Time to reach 5 volts = 0.12 farads x 5 volts divided by 0.1 amps = 6 seconds.

That sounds OK because this is just to get the big cap charged up to 5 volts - you'll only be taking 0.15 joules in 0.2 seconds and I guess, if you did the math it will take something like 0.5 seconds to recharge the cap to the full 5V before you need to activate the motor again but this may still be too long?

EDIT - can you use the 100mA from your power source AND the power that the capacitor can deliver to get this working as you hope? It's going to be a trade-off - you can't steal power to the motor without extending capacitor charge time. It's also very easy to over-estimate the ability of the motor given the low currents assumed.