Let me start by saying that @JImDearden's answer is the correct one. You don't need a regulator for this motor. The answer I'm going to give assumes that you, for some reason, still need a regulator. While this doesn't exactly directly apply to this question, it brings up some things to consider that are worth knowing.
Don't Use A Linear Regulator:
Going from 7.4V to 6V will require a voltage drop of 1.4 volts. At six amps, that works out to be 8.4 watts. This means that your linear regulator will be outputting 8.4 watts of heat. That might not sound like a lot, but when it is concentrated into a small area it is enough to break things. You would certainly need a reasonable heat sink on this, just to get rid of the heat. But more importantly, that 8.4 watts of power will be wasted and since you are running on batteries this is not a good thing.
You Could Use A Switching Regulator, But:
A switching regulator can easily handle the 6 amp output that you require. However, you would want to carefully design it for maximum efficiency. If you don't take care in designing it, it would be about 80% efficient. With 36 watts of output, an 80% efficient switching regulator would waste 7.2 watts-- not significantly better than the linear regulator. With care, a switching regulator can be 90-95% efficient, and only waste 1.8 to 3.6 watts.
Switching regulators are also complex, and is beyond most hobbyist-level EE's. But buying a switcher module is an easy way to do it, without the difficult part of designing a proper PCB.
But PWM or Current Limiting is a good alternative:
Motors rarely care a lot about the voltage, it is total power (and thus, heat) that they care about. (There are some important caveats that I'll cover later.) It is possible, and common, to run motors at a higher voltage than what they are rated for. Two ways to achieve this is by using PWM or current limiting be within the rated power.
Let's say that you run the motor at roughly double the rated voltage. In that event, you can run the motor with a 50% duty cycle. Total power is essentially the same as if you ran it at the rated voltage but with a 100% duty cycle.
Alternatively, you can somewhat ignore voltage but run the motor at the rated current. Many motor driver chips can automatically measure and limit the current to the motor for just this purpose.
Caveats: Of course things are rarely that easy, and there are many factors to consider. I've just given a super quick overview. Here are some things to be aware of... Brushless DC motors (like muffin fans) have IC's in them that often can't handle voltages that are too high. Brushed-DC motors might wear out sooner due to increased arcing on the brushes at higher voltages. A voltage that is very high could cause the insulation to break down, so don't run a 6v motor off of 100 volts. Motors are highly inductive, which could help or hurt you when PWMing. And motor control is a complex subject, and you can make it as simple or as difficult as you want. But the closer to the edge that you push your motors, the more you have to pay attention to the details.
This is a hard question to answer without knowing a few more things about your components.
- If you run the motors at 3.6V, or close to that, how much current do they require? You could measure this using a multimeter in series with your motor and an about 3.6V power supply.
Since they run at a pretty low voltage, I'd guess the motors require quite a bit of current, or they're relatively low power machines. That's probably why the 9V batteries don't work. Regular 9V alkaline batteries can't provide much current.
- Is the red light blinking on your shield an overcurrent indication? Trying to run a 3.6 v motor at 18v is probably going to be bad for the motor in the long run (or not so long), and it could certainly result in a lot of current going through the motor driver and causing it to shut off.
One thing you could try is using your 18v battery, but running the motor at a very low duty cycle setting. Are you using the shield based on the MC33932? If so, it supports duty cycle control. Theoretically, you shouldn't exceed 20% duty cycle with an 18v battery (3.6/18=0.2 or 20%).
Best Answer
The eBay link to the motor did not yield any datasheet. So I will assume for the purpose of this answer that the motors are similar to the Precision Microdrives 106-002 motor, which has nearly identical dimensions and electrical specifications.
The datasheet indicates a starting current of 180 mA at 3 Volts, and a no-load current of 17 mA at 3 Volts. Extrapolate that to the 22 mA current rating from the eBay listing, to a 233 mA starting current for the motors in the question.
If the 4 motors are likely to be started or stalled simultaneously, it helps to design for this maximum current:
233 x 4 = 932 mA = ~ 1 A. For normal operation, this value becomes22 x 4 = 88 mA = ~ 100 mA.For a 5 Volt supply, we should thus allow for 1.5333 Amperes current, or at least 1.5 Amperes if we need to cut corners.
The second one: Just use a MOSFET (or 4 of them, one for each if you plan to control them separately) to switch the low-side of the 4 motors.
Well, you do need to get the supply voltage down to within the motor's specified voltage range somehow. I might even get parsimonious and use 4 diodes such as 1n4007 in series to do the voltage reduction: End result, 4.6 Volts, so the motors will live a bit longer. After that, I'd drive all 4 motors from this rail, with the capacitors and diodes in place of course.
simulate this circuit – Schematic created using CircuitLab
Option 1: Single regulator for all 4 motors:
simulate this circuit
22 * 4 = 88 mA = ~ 100 mA. For a linear regulator (e.g. 7805) the normal running dissipation would be around(7.4 - 5) x 0.1 = 0.24 Watts, which isn't much for a TO220 regulator package.Option 2: Separate regulators for each motor:
Recommendation: Note that this is a personal view... I would go with a single regulator and add the diodes + capacitors as close as possible to each motor