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I am a beginner, so I may have made a horrible mistake in my diagram, but I am trying to learn.

According as I have figured it, I am having an issue wherein I need a resistor rated for a minimum of about 1.07 Watts; this is well, however, I do not have any resistors rated that highly heat-wise. I figured that I could just use a group of five 1/4W resistors, similarly to how a pair of two different resistors adds to a higher resistance.

enter image description here

*That does say 20kΩ and 29.6kΩ in the schem.; I have no idea why I opted to not write the capital omega in those two places.

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    \$\begingroup\$ Be careful of the voltage limit of a resistor. It may be 200 volts for a large SMD resistor and 400 volts for a 1/4 watt thru-hole. This determines whether multiple resistor are best in series or parallel (to dissipate the same wattage). \$\endgroup\$
    – user105652
    Aug 29 '18 at 0:58
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Why yes, you can certainly combine multiple resistors to spread out the power dissipation.

I see you've planned to have 177.5V across and 6mA through R1.

You can parallel 5 resistors that will take 1.2mA each.

177.5V / 1.2mA gives 147916 ohms, probably 150k ohms will be close enough. Then put 5 of them in parallel. The overall resistance will work out to 30k ohms (150k / 5).

Note that 1/4W resistors, even though they're rated for 1/4W, can still get rather hot when dissipating 1/4W, enough to burn you. Make sure to leave room in between for airflow (don't just bunch them all up together).

Also note that you probably won't be able to turn the anode off with this circuit. And if you do manage to turn it off, your Arduino will be fried.

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  • \$\begingroup\$ What do you mean it would fry an Arduino, if I might ask? What can I do differently (other than NPN transistors?) I did figure I might lose it, but I did not think I would. \$\endgroup\$ Aug 29 '18 at 1:22
  • \$\begingroup\$ @MatthewT.Scarbrough Well you're connecting 180V to one of the D pins. As long as the pin is outputting LOW the voltage will be dropped across the resistor. If it outputs HIGH, I'm not sure what will happen - one possibility is that it outputs 5V, and absorbs current, so the anode is still on - another possibility is that it doesn't absorb current, and the voltage goes up to 180V. \$\endgroup\$
    – user253751
    Aug 29 '18 at 1:22
  • \$\begingroup\$ Hmm... that is true... but does not a resistor resist voltage as well as limit amperage? Again, I am new, so forgive me if it is a foolish question. \$\endgroup\$ Aug 29 '18 at 1:24
  • \$\begingroup\$ @MatthewT.Scarbrough The resistor makes V=IR. If I is 0 then V is 0 (across the resistor) which means the V on both sides are the same. If I isn't 0 then the anode is still on. \$\endgroup\$
    – user253751
    Aug 29 '18 at 1:27
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    \$\begingroup\$ Remember to make sure the NPN transistor is suitable for 180 volts. \$\endgroup\$
    – user253751
    Aug 29 '18 at 3:37
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Yes, you can make a high power resistor from multiple low power ones.

The simplest way to do this is to put them all either in series or all in parallel. With N resistors in series, make each 1/N the total desired resistance. Similarly, with N resistors in parallel, make each N times the desired resistance.

You aren't limited to only series or only parallel, but then computing the final resistance is a little more complex. If you don't make all the resistors equal and all in series or parallel, the resistors may not be sharing the power equally. That's OK, but must be taken into account to guarantee that no individual resistor has its power rating exceeded.

A simple trick is to put two of the desired resistances in parallel. That makes half the desired resistance. Now put two of those parallel combinations in series. That doubles the result, so you're back to the original resistance. Here is a diagram of this:

If R1 thru R4 are the same, then the resulting composite resistor is also that same value, but with 4x the power capability.

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  • \$\begingroup\$ For the readers -- to add just a little bit to the "trick" that Olin shows above regarding the 2x2 resistor network, this works not just for 2x2, but also for any NxN network of the same-valued resistors. So a 3x3, 4x4, 5x5, etc. network of 1-ohm resistors would still be 1-ohm, but if having 1/2 watt resistors, would handle 3x3=>4.5W, 4x4=>8W, 8x8=>32W. And each square array of 1-ohm resistors is still 1-ohm. So an array of 16x16 (256) 0.5W 1-ohm resistors will dissipate 256 x 0.5W = 128W and will only drop 1 ohm. Also, for this, through-hole resistors are often the cheapest solution. \$\endgroup\$ Nov 27 '20 at 20:33

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