Electronic – External power supply for a DSLR flash

Fundamental to this type of diode protection is a large capacitor that can soak up the energy from the motors, via the diodes when all transistors or switched off. Going deeper, you can also use a high power zener diode across the rails and, going deeper still you can switch a resistor load directly across the power supply lines to dissipate the excess energy for the duration expected.

You can also protect your supply with a diode so that excessive voltages don't force current into the power supply.

The method you suggest will still attempt to raise the supply voltage so this might still cause problems on a sensitive power supply.

If you can provide a link to the motor it may be possible to estimate energy stored but, in most cases it will depend on the mechanical load free wheeling the armature and causing it to act like a generator.

2 amps for 2 seconds at 6V is 24 joules of energy and if this was provided by a capacitor the capacitance should be at least 5 times that required to meet the energy equation.

Energy in a capacitor is \$\dfrac{C\cdot V^2}{2}\$ therefore \$\dfrac{2 \cdot 24}{6^2} = C\$

C is therefore 1.333 farads but realistically it needs to be 5 times that so it doesn't become totally depleted during the time when the energy is needed in the 2 second burst. More like 7 F is needed. Make sure it's a low ESR type!!!

A problem occurs when the the requirement to drive the heavy load occurs too often. You said in your question that there might be a 10 second gap between "usages" and I infer only 200mA is available for restoring charge in the capacitor. 200mA at 6V for ten seconds is an energy availability of 12 joules and the only way around this is to have the capacitor as large as possible so that even if two "big demand" periods occur over a ten second period there is still enough residual change in the capacitor to service both requirements. Maybe go for a 14 F cap.

Can you get these?