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I have an Arduino and a bunch of components, among which, a dc motor (actually an aquarium pump) and a heater. These are connected via a relay module, so that the Arduino can switch on the motor or the heating element by tripping the relay. The circuit is laid out as follows:

enter image description here

As you can see, a battery pack supplies 6v to the Vin pin of the Arduino and to all the components.

However, the Arduino became unreliable as soon as the motor was switched on. I understand this has something to do with noise the motor introduces to the circuit. Figuring I could isolate both the motor and the heater completely from the rest of the circuit, I added a second battery pack and connected the heater and the motor to it via the relays:

enter image description here

This works! However, now I need two battery packs (and essentially double the batteries) to make this circuit work. How can I make this work using only a single battery pack that powers both the Arduino and all the components, but without the motor introducing "noise" (or rather, whatever the specific problem in that setup is)?

Edit: a diagram showing what I think is the situation where turning on the motor causes the Arduino to behave unexpectedly, followed by a diagram of the situation in which this doesn't happen but the motor requires a second, isolated power source:

enter image description here enter image description here

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  • \$\begingroup\$ Can you please draw a circuit diagram instead of a mess of wires on a prototype board? Also, what you need is probably better decoupling. Use your oscilloscope to plot Vcc over time as the motor starts. \$\endgroup\$ – winny Jun 10 '18 at 18:32
  • \$\begingroup\$ Consider battery ESR, Motor DCR , a full MOSFET bridge RdsOn and choose a 6V motor rated for 1% of battery current limit since your start current can reach 10x rated current. \$\endgroup\$ – Sunnyskyguy EE75 Jun 10 '18 at 18:36
  • \$\begingroup\$ I could draw a diagram, but I might make a mistake in doing so, since it won't match my circuit exactly. That would only muddy the question and we might not find out that I made a mistake with the circuit until after much discussion, and then we still will not have found the answer. At least with these Fritzing images, I know it's exactly the configuration of my circuit, because I can compare them one on one. But, I'll make two diagrams as well. \$\endgroup\$ – Bas Jun 10 '18 at 18:59
  • \$\begingroup\$ Question updated to include diagrams. \$\endgroup\$ – Bas Jun 10 '18 at 19:29
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    \$\begingroup\$ Pay attention to @TonyStewartolderthandirt comment above - most likely your problem not in motor noise but in huge current draw by motors and heater (never heard of a heater running on batteries 8O ) that brings your voltage down and resets Arduino. If this is supposed to be any kind of permanent installation you need good power supply in the first place \$\endgroup\$ – Maple Jun 10 '18 at 20:06
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The Arduino Nano V 3.0 has an LT1117 voltage regulator at VIN. The dropout is somewhere between ~1.1-1.5V. According to the Arduino tech spec. the lowest VIN should be 7 Volts, so you´re pushing the margins. What probably happens is that your unspecified motor pulls more current than the internal resistance of your circuit can handle, thereby dropping the voltage to the Arduino even more. You might get away with just adding one or two cells to the first setup.

If you´re really using one of those batteryholders in the Fritzing picture, note that they can have quite high resistance.

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  • \$\begingroup\$ I guess that's probably it, but where are you getting this information about the lowest VIN being 7 volts? The documentation specifies that VIN can be 6-20V: "The Arduino Nano can be powered via the Mini-B USB connection, 6-20V unregulated external power supply (pin 30), or 5V regulated external power supply (pin 27)." store.arduino.cc/arduino-nano \$\endgroup\$ – Bas Jun 10 '18 at 21:01
  • \$\begingroup\$ The tech spec tab says 7-12V, the documentation tab says 6-20V. The regulator has a max rating of 20V so that seems right, but 6V would place it in the drop out region. With a low impedance power source it's possible to stay close to the drop out voltage. But you don't have that. \$\endgroup\$ – Dejvid_no1 Jun 10 '18 at 22:40

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