I need to use a SN754410 motor driver with a 4.5V microchip and 12V motor all running on batteries.

I want to power the microchip with 3x1.5v AAA batteries. The motor requires 12V and I want to use a larger 12V battery for this.

I'm reluctant to use a voltage regulator because testing the voltage drop from 12V to 4.5V I seem to use excessive current converting the voltage when the microchip is doing nothing.

Here's where I'm confused - do I need to connect the two ground (-ve) terminals of the 2 sets of batteries together?

Picture attached from the SN754410 data sheet.

image of circuit http://jimsrobot.com/wp-content/uploads/2013/2-battery.jpg

  • \$\begingroup\$ Short answer: Yes. I'll let someone else elaborate why. \$\endgroup\$ – AndrejaKo Feb 10 '13 at 23:40

Russell has covered the ground connection. I would just like to mention the option of a switching regulator in a case like this.

Instead of using the 3 x 1.5V batteries for your 4.5V supply, you could consider using a buck converter. This type of switching regulator is more efficient (typically >80%) than a simple linear regulator (which will be less than 50% efficient with such a drop).

Here is an example circuit using a linear regulator (LT1085) and a switching regulator (LT1933)
Both have the same input voltage (12V) and are set to ~5V output and 50mA current draw:

Linear vs Switching

Here is the simulation. The linear regulator traces are in the top graph, and the switching regulators in the bottom. They show the power from the supply, the power consumed by the load, and the efficiency calculated from these (Load_Power/Input_Power). We can see the linear regulator is only around 36% efficient, compared to the 82% efficiency of the buck converter (open in a new tab for larger view):


  • \$\begingroup\$ Thanks Oli, I might have my first go at a switching regulator. It will be my first use of an inductor as well... Quick question, what will draw less current, dropping to 5V or 3V for the microchip? It can run from 2V8 to 5V5 \$\endgroup\$ – headingwest Feb 11 '13 at 6:30

You MUST connect the negative terminals of the two supplies together.

A regulator should give Iout ~+ Iin to a close approximation in MOST cases but you do lose the extra voltage so a separate supply makes sense IF I_low_voltage ~= I_motor. If the low voltage average supply current is much less than average motor current then a egu;lator makes sense.

Note that if you use 3 x alkaline cells then Vbattery new ~= 3 x 1.55V = 4.65V and Vbattery_end_of_life ~= 3 x 1V = 3V. Alkaline cells vary from about 1.55V down to about 1.0V.

NimH and NiCd cells vary from about 1.3V (briefly) down to about 1.0V.

If your low voltage supply needs 5V and you want to use all of the battery capacity then you will need at least 5 x Alkaline or 5 x NimH or NiCd cells. You may then need a low voltage regulator.

In many cases a regulator from the 12V supply will be a sensible option.
I_regulator_in ~= I_5V.
Power_12V_in ~= 12/5 x Power 5V or about 2.4:1. usually this is acceptable.


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