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I'm trying to better understand Ohm's law as it relates to general microelectronics projects by taking a simple project I'm working on now.

I have a 9V battery powering a small tower pro sg90 servo. The specs on the motor is that it takes 5Volts, and has a running current draw of about 220 milliamps (+-50). So if I use ohm's law to calculate the resistance needed if I'm supplying 9V, that'd be 9V=.220*x or 40 ohms. That seems really small to add a resistor on. The smallest I can find are about 110ohm resistors.

Should I add a resistor before the motor? And if so, how do I calculate what size I need?

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    \$\begingroup\$ You need a voltage regulator. When you measure your servo working you will notice it draws a varying amount of current. Plug that back into Ohm's law and check how much the voltage will vary with the resistor "solution". \$\endgroup\$ – Wesley Lee Oct 15 '17 at 12:18
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    \$\begingroup\$ Using a resistor to step down the voltage will make the circuit \$\frac{(9-5)^2}{9^2}=19.7\%\$ efficient. This means that roughly 80% of the energy will be wasted in the form of heat. That's nasty. Surely you got a 5 V DC outlet? Or you can just split open a USB-cable and connect that to the servo and put the USB-cable into a regular smartphone USB charger. \$\endgroup\$ – Harry Svensson Oct 15 '17 at 12:19
  • \$\begingroup\$ Plus motors will generally require more current when they have a higher load. A resistor does not work in this application. A voltage regulator is a lot better. Even a linear one. \$\endgroup\$ – vidarlo Oct 15 '17 at 12:21
  • \$\begingroup\$ @vidarlo The sg90 servo I got next to me has a stall current of 0.75 A. \$\endgroup\$ – Harry Svensson Oct 15 '17 at 12:23
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    \$\begingroup\$ You probably won't be happy if you run that servo from a 9V battery. The battery doesn't have much capacity. Drawing 220mA from it will run it down very quickly. Probably better to use 4 AA cells in series. That gets you 6V (which the servo can handle) and much more capacity (longer running time.) \$\endgroup\$ – JRE Oct 15 '17 at 12:35
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Assuming you should do this, your math is a little off.

\$R=V_{Drop}/I_{Min} = 4/(.220-0.05) = 23.5\Omega\$

Power \$P = I_{Max}^2R = (.220+0.05)^2*23.5 = 1.7W\$

Be aware though...

Start current will be much higher than that, with the above R in place closer to \$9/23.5 = .380A\$. That means a stalled motor will cause ALMOST 3.5W of heat to be dissipated in the resistor. So you need to up the wattage to closer to 4W.

And yes those resistors are quite small. You may need to put several in parallel to obtain the value and power rating you need. Ultimately, using a resistor for this purpose is not the best way to go.

The biggest issue you will have with it is your startup current will be limited to the value I mentioned. That means the acceleration will be severely limited and in some cases, the motor may not even start.

As such you really ought to be powering the motor with something more active. A simple 1A linear voltage regulator would work well, provided you have enough heat-sinking to keep it cool. Or you could use a simple voltage follower circuit like the one below.

schematic

simulate this circuit – Schematic created using CircuitLab

Ultimately though a buck-boost regulator would get you the most efficiency and power from your battery.

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  • \$\begingroup\$ Thanks Trevor. I didn't think about stall current. I looked it up for this motor and it's rated at 650 ±80mA. I think what I'm confused about is, if I put the stall current into Ohm's law, it actually makes the resistance in the equation lower. But, if the motor is stuck, I feel like you would need even more resistance, so you aren't burning out the motor on the servo. I would love an explanation of the calculation of power and how that factors in with resistors. But maybe that's another question on the exchange... \$\endgroup\$ – mheavers Oct 15 '17 at 13:30
  • \$\begingroup\$ @mheavers assuming the motor will handle being stalled indefinitely you will not need to worry about it overheating unless you enclose it in such a way that heat can not escape. See my edit for better solutions to deliver the voltages you need. \$\endgroup\$ – Trevor_G Oct 15 '17 at 13:59
  • \$\begingroup\$ @mheavers re power.. read this excellent page on ohms law and power electronics-tutorials.ws/dccircuits/dcp_2.html \$\endgroup\$ – Trevor_G Oct 15 '17 at 14:12
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    \$\begingroup\$ There's been a major error on my part, I used the \$P=\frac{V^2}{R}\$ to calculate \$P_{efficiency}=\frac{P_{out}}{P_{in}}\$. Where the R canceled out. But they don't cancel out. And after that I made the other error of using 9V - 5V to calculate the drop voltage thinking that it was equal to the waste power. Which it's not. - With a whiteboard in front of me I can think more clearly and come to the conclusion that \$P_{eff}=\frac{V_{out}×I}{V_{in}×I}\$ and see that the currents are the same through the whole circuit. And knowing that \$V_{out}\$=5V and \$V_{in}\$=9V makes \$P_{eff}\$=5/9=55% \$\endgroup\$ – Harry Svensson Oct 16 '17 at 3:27
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    \$\begingroup\$ @HarrySvensson - I definitely have taken your advice into consideration and ordered a 3 & 4xAA battery holder with switch. But Trevor's solution was indeed what I asked about and was very helpful in my understanding of Ohm's law, power, and my project setup in general. Thanks to both of you! \$\endgroup\$ – mheavers Oct 16 '17 at 13:46
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220mA and dropping about 4V means 0.880W of power dissipation.

That's a lot for a single resistor.

Instead look for a voltage regulator.

Though if you don't mind plugging it into a wall grab a generic USB power supply. They supply a regulated 5V DC.

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  • \$\begingroup\$ Thanks - how did you determine the .880w of power dissipation? \$\endgroup\$ – mheavers Oct 15 '17 at 13:16
  • \$\begingroup\$ Multiply the amps with the voltage dropped over the resistor. \$\endgroup\$ – ratchet freak Oct 15 '17 at 13:23
  • \$\begingroup\$ You don't specify what sort of voltage regulator you're recommending. A linear type will dissipate the same amount of power as a resistor... \$\endgroup\$ – brhans Oct 15 '17 at 13:24
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To perform calculations you can simply go with this calculator: enter image description here Link is: Ohm's law calculator

But your case is not that simple, you need to understand a little bit of voltage divider as well and then you can go for it. The second choice is obviously a voltage regulator which will provide a continuous supply of mA. Actually, a voltage divider has a fault the with the battery life the current going to your servo will vary as well. It will cross your servo limit one day as well: 220 +- 50. So you should opt for a regulator.

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