Ohm’s Law – Resistors for Fan Speed - Page 1
Forum section: electronics / beginners

author: the offset voltage
author_rank: Regular Contributor
timestamp: July 23, 2025, 01:50:03 pm
content: Hi,I am working through the book Principles of Electric Circuits by Floyd.Now I'm on section Ohm's Law. I need your help with some support please.I am working on this application where I have to design a simple switch speed controller for a ventilator where I have to use a 12V ventilator, a rotary switch and I have to calculate the values of the resistors (I have to use 3 resistors to have 3 selectable speeds for the ventilator).The specs are:Vventilator = 12VIventilator = 80mANow, to calculate the resistors value, I think like this:R1 = V/I = 12/60mA = 200 Ohms. P = I²R =720mWR2 = V/I = 12/40mA = 300 Ohms. P = I²R = 480mWR3 = V/I = 12/20mA = 600 Ohms. P = I²R = 240mWBefore I worder the components, can someone say if my approach is ok? Thank you

author: tunk
author_rank: Super Contributor
timestamp: July 23, 2025, 02:41:52 pm
content: When the resistor is in series with the fan, the voltage drop over theresistor isn't 12V. If you have the fan datasheet, then it might justhave a voltage vs. current table/figure. If not, you have to experimentwith either different resistors, or if you have an adjustable PSU, usedifferent voltages and record the current.

author: TimFox
author_rank: Super Contributor
timestamp: July 23, 2025, 02:43:55 pm
content: The voltage across the motor terminals is a function of the motor speed (rpm).Do you have a data sheet for the ventilator that gives the motor voltage and current for the air flow?

author: the offset voltage
author_rank: Regular Contributor
timestamp: July 23, 2025, 02:47:21 pm
content: Thank you,The fan that i have it's from a old equipment, and i don't have the datasheet for it. I have a power supply, but i don't know what should i look for when testing. Thank you

author: ledtester
author_rank: Super Contributor
timestamp: July 23, 2025, 02:47:29 pm
content: Controlling the speed of a DC fan is usually done with with PWM control which involves turning the fan on and off in quick pulses.Another way DC fans are controlled is by altering the voltage supplied to the fan but the relationship between supplied voltage and fan speed is highly dependent on the fan itself. The usable voltage range for a 12V fan may be in the range of 7-12V -- lower than that and it may not even start.Here is a video which demonstrates the effect of putting a resistor in series with a fan:DRAFT Fan Speed with dropping resistors -- Daniel Sullivanhttps://youtu.be/MIkGLtPQYU0Note that even a very small resistance ( < 1 ohm) can have a dramatic effect on the fan's speed.If you provide more information about the fan perhaps we can suggest the best way to control it.

author: the offset voltage
author_rank: Regular Contributor
timestamp: July 23, 2025, 04:21:25 pm
content: Thank you, the only info that i have for the Fan are the ones writhed on it:Vventilator = 12VIventilator = 80mAI saw some tutorials about controlling Fan's with PWM, but the application circuit part of Ohm's law suggest to use resistors (calculate the R and P Value) and a rotary switch. And i want to go through all the materials from this book (i learn by myself).

author: wraper
author_rank: Unknown
timestamp: July 23, 2025, 04:30:18 pm
content: You cannot even really calculate it other than for rough estimate (unless you characterize particular fan used in particular conditions). Fan won't act like a resistor. Reduction of voltage will not result in equal reduction of current.

author: themadhippy
author_rank: Super Contributor
timestamp: July 23, 2025, 04:34:17 pm
content: Personally i dont think a motor is a good way to grasp ohms law as theirs to many other things that could effect the results.If its only to grasp ohm law use a 150 ohm 1 watt resistor instead of the motor ,or a 12v 1 watt incandescent lamp if you want to see the results as well as just measuring them.

author: wraper
author_rank: Unknown
timestamp: July 23, 2025, 04:41:22 pm
content: Quote from: ledtester on July 23, 2025, 02:47:29 pmDepends on what type of DC fan. If it's brushless like the vast majority of DC fans, then hell no, their power must not be PWMed.

author: the offset voltage
author_rank: Regular Contributor
timestamp: July 23, 2025, 05:01:37 pm
content: Thank you for all your suggestions. Should i skip this application project in this case? (i will also reach PWM and other exotic topics in my future study)

author: tunk
author_rank: Super Contributor
timestamp: July 23, 2025, 05:17:58 pm
content: You could get an 1-100 Ohm resistor assortment and test it.You will learn Ohm's law and get an idea about how electric (fan) motors work.

author: Analog Kid
author_rank: Super Contributor
timestamp: July 23, 2025, 05:23:21 pm
content: Wellll, since you have the motor, and you have the curiosity to explore this very important aspect of electricity/electronics, why not go ahead and experiment? It's not like you're going to destroy anything, or at least anything valuable.You didn't say you had a power supply; obviously you'll need one. Shouldn't be hard to get your hands on a low-current 12-volt supply, like a wall wart.And do you have any test equipment? It would help to have at least 1 meter; a DMM (digital multimeter) would do nicely, and can be acquired for pretty cheap. That way you can measure voltages and currents.Try it out; use the resistor values you calculated, connect it up and see what voltages/currents you get. (Keep in mind that you'll have to connect the ammeter inserieswith the load to measure current; you may need to set up ashunt resistorto do this, since DMMs typically only measure to a maximum of ~200mA).And of course observe what happens to the motor speed as you do this. And then report your findings back to us.And have fun!

author: magic
author_rank: Super Contributor
timestamp: July 23, 2025, 05:33:35 pm
content: If the fan draws 80mA then you can use 4V/0.080A = 50Ω resistor to reduce its supply voltage by 4V from 12V to 8V. This will give a meaningful speed reduction and the fan will likely still be able to start.The calculation is not entirely accurate, because current will decrease at reduced supply voltage. But I would start with 47Ω (a common value) and go from there.

author: Analog Kid
author_rank: Super Contributor
timestamp: July 23, 2025, 05:52:05 pm
content: By your own calculations, you'll need to use (at least) 1 watt resistors.2 watt ones wouldn't hurt.You might want to monitor how hot they get (see if they're too hot to touch).

author: RJSV
author_rank: Super Contributor
timestamp: July 25, 2025, 05:26:05 am
content: I noticed when you kind of switched your topic focus, to involve power formula. So, which is it ? My GUESS is you desired certain power levels as a goal, replacing the speed approximate values. That's valid, as an approach, but I'm not sure if you are expressing this, because you haven't gone through the equations yet.Like other folks have noted, you could use a voltage drop in a set of guesses. I would start way up near 12 volts, and not go much below 9 volts, so there you can experiment with whichever resistor will cause a 2 volt drop, and observe motor response.I would be cautious below about 8 volts, so as not to stall the motor (causes overheat).For initially, try even a 1 ohm resistor in series, reading the voltage across it. At 40 mA you would obtain a 40 mV reading that gives you a glimpse of initial conditions. That tiny voltage drop won't affect the motor much, really, so in that example you could assume that the motor basically draws 40 mA, at 12 V nominal.To start there and drop an arbitrary 2 Volts, you would try R = 2/0.040 which is 50 ohms needed. You could then calculate the other two speed levels approximately needed, guessing drop 4 volts or 5, etc.Notice I didn't even TOUCH any formula for POWER. Clearly not needed.

author: Analog Kid
author_rank: Super Contributor
timestamp: July 25, 2025, 05:29:31 am
content: Quote from: RJSV on July 25, 2025, 05:26:05 amWell, ackshooly it is; a 1/8 watt resistor might object to being subjected to excessive power by burning to a crisp.But it's easy enough to calculate:P = V * A.

author: RJSV
author_rank: Super Contributor
timestamp: July 25, 2025, 05:42:40 am
content: Nice, accurate response, missing the point though.The question itself helps mislead beginners into thinking they need to consider power, TO OBTAIN AN ANSWER,...not merely to check as an afterthought. Yes, 80 milliwatts is, safely, below 125 mW.I used V = I X R and POWER can be checked after calculation is made, to avoid your 'safety warnjng'. Thank you

author: Analog Kid
author_rank: Super Contributor
timestamp: July 25, 2025, 06:06:20 am
content: Quote from: RJSV on July 25, 2025, 05:42:40 amWell, why not do both? It's easy enough, and then you know exactly what resistor you need:X ohms, Y watts. Then go order it online. Easy peasy.

author: RJSV
author_rank: Super Contributor
timestamp: July 25, 2025, 06:27:40 am
content: You still missing my point. Simple Google 'Ohms Law' relates NOTHING on the power formula. Thanks to your detail-oriented scrutiny, the second step, of checking estimated power, has been described. BUT, a beginner, like the OP here, has the incorrect assumption that we use power in the initial evaluation. Not true. We use Ohms Law, V = IR for initial calculation.Many folks with LED stuff don't even bother to check, the thermal overheat angle, in their day to day activities.I tweak, (just a little) on this point, because I've often observed that beginning students can get mixed little n the jargon, and think that almost ANY technical sounding word or formula is appropriate.You don't use a hammer thingy, to turn a screw, (in carpentry), and you don't use POWER as an appropriate parameter, here.I know you enjoy goading people. Notice I didn't revert to your flawed argument(s).

author: Analog Kid
author_rank: Super Contributor
timestamp: July 25, 2025, 06:43:06 am
content: Quote from: RJSV on July 25, 2025, 06:27:40 amMy arguments are not flawed at all here.Hopefully not to belabor this point too much, but:Yes, Ohm's Law gives us a resistance value when we know the voltage and current, as you correctly stated:R = E / ISimple enough, right? Shouldn't strain anyone's brain who's at least taken high-school algebra.But since we already know voltage and current, why not go ahead and calculate power? It's justP = V x A(OK, I was a little pedantic with the Ohm's Law variables: I prefer using "E" for voltage and "I" for current, but same difference between E & V and I and A.)It's so simple, there's no reason not to do it at the same time: just one simple multiplication.I appreciate your trying not to confuse beginners, I really do. But I don't think this is really going to trip them up.Just keep the two formulas separate.

author: themadhippy
author_rank: Super Contributor
timestamp: July 25, 2025, 10:17:15 am
content: Quoteya must use a diffrent google to me,second hit

author: iMo
author_rank: Super Contributor
timestamp: July 25, 2025, 10:29:48 am
content: Take an 1k trimmer (or similar), wire it as the rheostat (in series with the motor), look at the resistor with an oscilloscope - you will certainly see some periodic changes due to rotation of the motor, based on that calculate the RPM, and simply adjust your trimmer for the required RPM. Measure the trimmer's value for the required RPM and replace with fixed resistor(s).

author: MrAl
author_rank: Super Contributor
timestamp: July 25, 2025, 10:30:45 am
content: Quote from: the offset voltage on July 23, 2025, 01:50:03 pmHi,These relationships can be slightly nonlinear but let's assume a general case where it is not that bad.The power demand is related to the cube of the RPM.The airflow is related to the RPM, approximately.What this means is that if you use power as a goal and work it in a linear way, you will not get the air flow you want unless you want to change it by an exponential factor.This also means that if you want the air flow to increase by a constant factor K, you have to increase the power by a factor of K^3.For example, if you want to double the air flow you have to increase the power by 2^3=8.To triple it, the power has to increase by 3^3=27.If the fan takes 10 watts, to increase the air flow by two times that means 80 watts.This means that your table of power levels may not equate to what you want to happen regarding the air flow. You would have to increase the power in a different way if you want to step the air flow rate in equal increments. Equal increments would mean something like 1.00 to 1.20 to 1.40 to 1.60, etc., where each change in air flow is 0.20 units.Note you may still be able to use Ohm's Law, but pay more attention to the power levels.Since the above is only linear to a certain point and may even be a bit confusing if it was perfect, it's good to try different values and see if you get the air flow results you want.Also, sometimes air flow is not the goal sometimes pressure is and that would be related to the square of the RPM.

author: iMo
author_rank: Super Contributor
timestamp: July 25, 2025, 11:06:53 am
content: Quote from: iMo on July 25, 2025, 10:29:48 amThis is what I see on a 220ohm resistor in series with a small CPU fan (12V).. By turning the blades with finger you may get how many magnet poles are there per revolution, imho.. You may even measure the frequency on the resistor with your DMM if none oscilloscope is handy..

author: WattsThat
author_rank: Frequent Contributor
timestamp: July 25, 2025, 01:38:20 pm
content: Ok, let’s go back to your original post. As stated, the fan with 12v applied will need 80ma. So you calculate the resistance of the fan. 12 / 0.08 = 150 ohmsIt looks like you decided on three different ranges of 60, 40 and 20 ma. What you missed was the resistance of the fan itself, now known to be 150 ohms. So if you want to apply 12 volts and have 60 ma flow, you get: 12/0.06=200 which we now have to subtract the 150 giving us 50 ohms since you’ll be placing the 50 ohms in series with the fan. The remaining two values would be (12/0.04)-150=150 and (12/0.02)-150=450.For power dissipation, your formula is correct, you just need to use the resistor actual value, not the total resistance.Now, finally, the gotcha in all of this. As some described in other terms, a dc motor does not behave as a pure resistance. The speed is not is not going to be linear with voltage applied due to a number of factors. What will you get? No idea really. It would depend on the motor. Here are a couple of known things:The speed of a dc permanent magnet motor is proportional to the applied voltage.The current in a dc motor is proportional to the torque required by the load on the shaft.The torque required by a fan is proportional to the speed squared.Given the above rules, you can deduce that the speed will go down linearly with applied voltage at the motor but the current will go down at the square root of that applied voltage. This means your calculated values are not going to give you linear speed changes with each different calculated value.All in all, as mentioned, a fan is lousy example device for this exercise. But, the same rules still apply with regard to Ohms law if look past the motor issues.

author: BeBuLamar
author_rank: Super Contributor
timestamp: July 25, 2025, 01:59:12 pm
content: If you have a variable power supply then simply connect it to the motor and vary the voltage. Adjust the voltage for the speeds you like then measure the the current. That way you would know what the equivalent resistance for the motor at those speeds.

author: eugene
author_rank: Frequent Contributor
timestamp: July 25, 2025, 02:03:28 pm
content: Quote from: the offset voltage on July 23, 2025, 01:50:03 pmThe right way to apply Ohm's Law is to first calculate the resistance of the fan motor:R = V/I = 12V / 0.080A = 150 OhmsNow you use an additional resistor in series with the fan to make what amounts to a voltage divider with the fan as the bottom resistor. From here you can find the voltage across the fan, which is what you want to vary to control the speed (which is a reasonable way to control fan speed if the motor is brushed DC.) You could start with a desired voltage across the fan and calculate what added resistor you need...But this entire analysis falls apart because the fan is not "Ohmic." What that means is that it doesn't obey Ohm's Law: the current through it is not going to be directly proportional to the voltage across it under all reasonable condiions. The calculation of resistance above is valid for only one combination of operating conditions, so all of your resistor divider equations will be wrong.The only thing you can do is buy a bunch of resistors - or a large enough rheostat - and test to see what works. In the end, the only thing that you'll learn about Ohm's Law is that it doesn't apply to everything.

author: TimFox
author_rank: Super Contributor
timestamp: July 25, 2025, 03:19:38 pm
content: DC motors have a voltage (“back emf”) proportional to rpm, added to the copper loss (current times winding resistance), where the current is proportional to the torque (useful plus bearing friction).

author: Analog Kid
author_rank: Super Contributor
timestamp: July 25, 2025, 09:30:25 pm
content: Quote from: TimFox on July 25, 2025, 03:19:38 pmThis is one rare instance where I'm going to say Tim, what you write is no doubt true, but it's total information overload for the poor O.P. here, and therefore not helpful.Consider your audience.

author: RJSV
author_rank: Super Contributor
timestamp: July 25, 2025, 10:16:28 pm
content: A.k. I've been attempting to hold my tongue here, and wondering when some other (member) gets enough annoyed. What's the term used ? TROLL ? maybe. You skirt the rules on common courtesy, while another member wonders, Are you being 'deliberately obtuse' ?Analog Kid, you maybe, should wait like more that 3 minutes, before popping back in with the snarky and rude comments. That's why I've speculated that maybe you are an AI bot designed to annoy. TRY WAITJNG 15 minutes, so we can have our regular FORUM first...THEN publish your snide ridicule.I wonder what benefit a person in Asia gets, reading your 'detail fact checking' that you do so well. (And don't forget to remind /reprimand myself, when you feel things go 'Off Topic'.)Exactly the right (wrong) posts, to get me disgusted enough to quit posting my stuff. Only one person, or stupid bot.We don't need the hear your need to mangle the math terms and symbols.

author: TimFox
author_rank: Super Contributor
timestamp: July 25, 2025, 10:30:10 pm
content: Quote from: Analog Kid on July 25, 2025, 09:30:25 pmI totally disagree. The OP needs to know about basic motor operation to analyze his application.Ohm’s Law does not suffice for every electrical application.My opinion about advice to novices is that it is important to mention the “terms of art” so they know what to look up to learn about their questions.

author: Analog Kid
author_rank: Super Contributor
timestamp: July 25, 2025, 10:44:48 pm
content: Quote from: RJSV on July 25, 2025, 10:16:28 pmLook, Rick, why don't you go somewhere and concentrate on fixing your word-saladitis instead? You'd be doing all of us a great favor.Meanwhile, don't like what I write? Call 1-800-SO-SUE-ME.

author: fourfathom
author_rank: Super Contributor
timestamp: July 26, 2025, 02:17:18 am
content: Calculations are great, but sometimes you just want to try something and see what happens. For these situations I keep on hand a few power potentiometers (also useful for load-testing power supplies and amplifiers.) Here's an example:https://www.amazon.com/Electronics-Salon-Wirewound-Potentiometer-Rheostat-Resistor/dp/B012S844ASThese are often poorly constructed, but still can be very handy.

author: radiolistener
author_rank: Super Contributor
timestamp: July 26, 2025, 02:46:06 am
content: Quote from: fourfathom on July 26, 2025, 02:17:18 amI have a USSR-made 50 Watt ППБ-50Г 6.8 Ω resistor, which is very useful for testing. It can withstand even a 100–150 Watt load for 5–10 seconds, but then it heats up significantly and needs to be cooled down. The construction is quite robust and performs well. However, while experimenting with low-resistance loads and a high-current source (about 50 Amps), I accidentally burned out one the jumper connecting the terminal to the resistor's conductor.

author: 5U4GB
author_rank: Super Contributor
timestamp: July 26, 2025, 09:17:05 am
content: Quote from: tunk on July 23, 2025, 05:17:58 pmAnother reason for doing that is that at lowish voltages some fans won't start/run, or will spin if manually started but won't self-start. There's also a threshold region where they may run in one environment but not another, and the only way to determine that is empirically. Google should turn up various discussions of this in the fan-undervolting-to-silence-them community.

author: tggzzz
author_rank: Super Contributor
timestamp: July 26, 2025, 09:26:35 am
content: Be very aware that Ohm's law is only valid for components that are ohmic, i.e. where the voltage across a component is proportional to the current through it (and vice versa).Most components are not ohmic.Some components even have anegativeresistance.Examples of non-ohmic copmponents: incandescent light bulbs, diodes, motors, transistors, capacitors, inductors, batteries/cells etc etc etc.

author: iMo
author_rank: Super Contributor
timestamp: July 26, 2025, 10:34:19 am
content: I've gone through the Book, section Ohm's law, and I have not found that application with "ventilator and its speed control" (there are many editions of that Book, however). From what I see there the task is simply calculate the resistors "assuming the fan is of resistive nature, and the fan's rpms are roughly proportional to the voltage at the motor, or something like that". Everything in that section deals with the simplistic Ohm's law, so none specialties there - as we broadly discuss the nuances here..

author: tggzzz
author_rank: Super Contributor
timestamp: July 26, 2025, 10:40:02 am
content: Quote from: iMo on July 26, 2025, 10:34:19 amSo the equivalent of a "light inextensible string" or "spherical cow"?They could have picked better examples to illustrate and encourage the use of Ohm's law and network/nodal analysis.

author: BeBuLamar
author_rank: Super Contributor
timestamp: July 26, 2025, 01:17:22 pm
content: Quote from: tggzzz on July 26, 2025, 10:40:02 amWell I have seen books (and training courses) used incandescent light bulb to teach Ohm law. So when the students measure the resistance of the light bulbs and then power it up and measure the current it doesn't match the formula as the filament resistance increases many fold when heated up. And the when they series the bulb it doesn't work either.They should avoid to use these examples if they want to keep things simple.

author: MrAl
author_rank: Super Contributor
timestamp: July 26, 2025, 02:23:18 pm
content: Quote from: Analog Kid on July 25, 2025, 10:44:48 pmHi there,Why so nasty with the attitude aimed at someone who volunteers information that could be relevant? Is it really that bad? I don't see how.It's really an ongoing debate whether or not a beginner gets overloaded or appreciates the extra information, assuming it is extra information which is kind of hard to judge sometimes anyway.

author: BeBuLamar
author_rank: Super Contributor
timestamp: July 26, 2025, 04:01:15 pm
content: Quote from: the offset voltage on July 23, 2025, 01:50:03 pmIf we assume that the motor is a simple resistor that draws 80mA at 12V then its resistance is 150Ohms.So the valuke or R1 would be 50, R2 is 150 and R3 is 450.

author: KrudyZ
author_rank: Frequent Contributor
timestamp: July 26, 2025, 07:15:19 pm
content: QuoteA motor does not behave like this. As Tim said above, its impedance is dominated by the back EMF.In the spirit of using the real world as a simulator, I just took a 48V rated brushed DC motor.The winding resistance measured 6.8 Ohm. If the motor behaved like a DC resistor then Ohm's law would result in a current of about 7 A at 48 V.When I actually hooked up 48 V to it the current I got was 90 mA.When I lower the voltage to 30, 20 , 10 V the current stays very nearly the same, while the speed reduces.What's going on here?90mA * 6.8 Ohm is about 600 mV. That is your resistive loss in the winding and generates heat.The remaining voltage 47.4 V is countered by the motor spinning and generating back EMF, the same voltage you would get if you would drive the motor at the same RPM and use it as a generator. A brushed DC motor is switching (commutating) its internal magnetic field as the motor rotates via the brushes.This reverses the direction of current in the windings which present an inductance. The impedance of an inductor is proportionate to the switching frequency. The faster you switch (higher RPM) the higher the impedance. Since this is a complex impedance dominated by inductance the majority of the energy is preserved and therefore does not get converted into heat. So even with a very low friction and otherwise lossless motor there will be a maximum RPM that you can reach. This is when the motor generate zero excess torque.As you add friction the motor will be asked to perform more mechanical work. This energy of course has to come from the power supply, so as the motor slows down, it generates less back EMF and the phase angle between the internal (commutated) voltage over the winding vs the current through it reduces from near 90 degrees to lower and lower numbers. This corresponds to a change in the magnetic vector from near axial to more tangential which is the part that generates torque.Remember that power is the dot product of force and velocity.Why does the current stay basically constant over supply voltage with no mechanical load?This relationship means that power goes up linearly with voltage.Some of the losses in the motor go up with velocity, but not linearly, thereby requiring more torque.However, back EMF is strictly linear with velocity.The solution is that the no load motor speed doesn't actually double with voltage, so not all of it is countered by back EMF.The excess voltage is used to generate the required additional torque.The reason the current draw is so insensitive to the supply voltage is because back EMF is the dominating controlling element.p.s. if you stalled the motor that I tested it would indeed draw 7 A, while converting 336 W into heat.

author: TimFox
author_rank: Super Contributor
timestamp: July 26, 2025, 08:13:04 pm
content: There are motors designed to operate under stalled conditions, where the back emf is zero and the current is given by your 7 A example. The torque fighting the “brick wall” load that holds the rotor stationary is proportional to that “stall current”, and the motor construction must survive the high power dissipation due to Ohmic loss.When choosing motors for servo operation, I concentrated on the velocity constant (RPM/V), the torque constant (N-m/A), and the resistance, then checking for power rating. Final design also needed inductance, inertia, and bearing (friction) torque.Also, the back emf is proportional to the magnetic field in the motor, which is usually fixed. When I was working on MRI systems, one of my colleagues built a non-ferrous motor, using the main field (either 0.5 or 1.5 T) and copper winding, so that it did not perturb the homogeneous field when powered off. He was surprised that it ran slower at the higher field value, but had more available torque, until I explained the dependence on field. This motor was used to tune a vacuum variable capacitor in the RF circuit. His final design used a piezoelectric motor instead.

author: Analog Kid
author_rank: Super Contributor
timestamp: July 26, 2025, 08:26:49 pm
content: Quote from: MrAl on July 26, 2025, 02:23:18 pmNo. The person I was responding to there was not offering any information, relevant or not; they were just dinging me because they didn't like my attitude. Hence my (intentionally) rude reply.QuoteYes it is, isn't it?I still contend that the best way to deal with dispensing such information is to consider the audience, especially if it's a n00b (no harm, no foul: I'm a n00b on some subjects myself), and not dump the entire encyclopedia of facts on them at once.Some things are hard enough to understand without piling on extra layers of true, but only remotely relevant information. Go easy here.It's the whole thing of starting with a first approximation of a situation, which is not totally accurate but explains the thing (Ohm's Law in this case) well enough to start. Then move to a second approximation, revealing more interesting nuances, and so on. Don't just throw the entire textbook at them at once and expect them to absorb it.I know if it were me that I'd be somewhat overwhelmed, and therefore somewhat frustated.

author: TimFox
author_rank: Super Contributor
timestamp: July 26, 2025, 08:32:42 pm
content: My point was that the most basic description of the motor involves back emf and torque, not Ohm’s Law that is far from predicting the operation. The novice is better to start from that simple description and then add copper loss and other features.

author: Analog Kid
author_rank: Super Contributor
timestamp: July 26, 2025, 08:50:47 pm
content: Why not take the far simpler approach (that's already been suggested here more than once) of simplyputting a resistor in series with the motor and seeing how that affects its operation?Totally empirical exercise.That way they can find out how much the voltage/current can be restricted before the motor ceases to start on its own, for one thing, and about how fast it runs.Then they can delve into the more nuanced aspects of how resistance affects motor operation.But first things first. Do you disagree with that approach?

author: TimFox
author_rank: Super Contributor
timestamp: July 26, 2025, 09:00:59 pm
content: Since he doesn’t have a useful data sheet for the unit, that is probably the only thing that will work.I really don’t understand why his reference gave such bad advice, since the Ohmic scaling won’t work.One basic unit in my home lab is a power resistor decade box (Clarostat 240C) that is out of production.I bought it at a reasonable price through eBay, but they seem to be quite expensive now.I stand by my basic explanation of DC motor operation.

author: iMo
author_rank: Super Contributor
timestamp: July 26, 2025, 09:10:48 pm
content: IMHO, before we start to generate many ideas how to solve a problem, we have to ask the OP some additional questions, especially if he/she/it is a beginner studying that Book (have a look in the book, the Ohm's law chapter is absolutely basic stuff). So, let us wait on the OP, perhaps he/she/it may clarify..

author: RJSV
author_rank: Super Contributor
timestamp: July 26, 2025, 09:37:55 pm
content: Maybe a glance at OP posting frequency, (2 times here) can help as a part of our interpretation, suggesting, in effect, that we wait for more guidance, from Original post on what the intended approach is. Several replys have expressed enough 'guessing' about the OP's intentions that it seems to be a 'Passive or nearly dormant' involvement here (maybe rightfully confused, but never expressed that, either.BUT "When discussion next delves into PWM...".I just don't think, in total, that we are capable, of demonstrating or teaching at that level of subtle complexity, (other than the good replies we see here, in between the quarral-fest embarrassments).

author: the offset voltage
author_rank: Regular Contributor
timestamp: July 27, 2025, 08:00:51 am
content: Thank you all,I went on an i did some tests on my breadboard using two resistors, and i came out with this results:R1= 47Ohm in series with 12V power supply and the fan:Ifan = 60mAVfan = 8.9VVR1 = 2.8VR2 = 100Ohm in series with 12V power supply and the fan:Ifan = 50mAVfan = 6.9VVR2 = 4.8VThis resistor are low power and gets hot on my breadboard. I want now to order this two resistors (i will go with 2W as a member suggested ) because i want to learn now to design also the PCB and i saw that i need the datasheet to know the precise dimensions of the terminal and pitch (i do now a tutorial on how to design PCB from Robert Feranec Youtube channel).Regarding ordering the components, i have a question, other than the value and power, should i take into account others parameters?On Mouser when i have a look there are many other parameters, and i don't know what else to take into account.Thank you

author: RJSV
author_rank: Super Contributor
timestamp: July 27, 2025, 08:17:30 am
content: Might be surface mount varieties ?My experience a 2 Watt size is something like 5/8 inch spacing, assuming both leads have 90 degree bend. The smaller power rated stuff could get small; maybe at 5/16 inch, for the similar type of install having both leads bent 90 degrees.BTW, use caution as motors stall below about 50 or 60 % (percent), maybe runs at 7.9 V but without a load.

author: tunk
author_rank: Super Contributor
timestamp: July 27, 2025, 10:10:52 am
content: 2.8V * 60mA = 0.168W4.8V * 50mA = 0.24WSo 1/4W resistors will work in these two cases.If you have 1/8W resistors, you could use two in parallel.

author: the offset voltage
author_rank: Regular Contributor
timestamp: July 27, 2025, 10:34:19 am
content: Great.What other parameters should i take into accent when ordering the resistors from mouser?other than the value and power (this is my first mouser order).Thank you,

author: BeBuLamar
author_rank: Super Contributor
timestamp: July 27, 2025, 11:44:50 am
content: Quote from: KrudyZ on July 26, 2025, 07:15:19 pmI know that and that is why I said "Assume" which I believe the book did that. Because otherwise it's not a question about Ohm law and there isn't enough information to solve the problem. As I have pointed out I have seen many books and training courses use light bulbs as example for ohm law and when the students did the experiment it doesn't work.

author: BeBuLamar
author_rank: Super Contributor
timestamp: July 27, 2025, 11:47:16 am
content: Quote from: iMo on July 26, 2025, 09:10:48 pmI don't think the OP has the motor. It's just a make up problem from the book and is a bad one.

author: MrAl
author_rank: Super Contributor
timestamp: July 27, 2025, 01:42:21 pm
content: Quote from: Analog Kid on July 26, 2025, 08:26:49 pmHello again,Well of course I have to agree in part, but you have to realize that you can't use an extreme exaggeration to prove your point. In fact, it really depends on the nature of the problem and the information given. For example, if someone asks for a 4.6k resistor and we offer a suggestion of 4.7k, that's extra information but it can't be too much more can it.As I have said time and time again, I have met both types, those that take it as too much and those that appreciate the information. Maybe we should ask them first ha ha ? That could be the best procedure,maybe

author: the offset voltage
author_rank: Regular Contributor
timestamp: July 27, 2025, 03:39:32 pm
content: Thank you all,As i mentioned, I went on an i did some tests on my breadboard using two resistors, the fan works, and i came out with this results:R1= 47Ohm in series with 12V power supply and the fan:Ifan = 60mAVfan = 8.9VVR1 = 2.8VR2 = 100Ohm in series with 12V power supply and the fan:Ifan = 50mAVfan = 6.9VVR2 = 4.8VThis resistors on my breadboard are low power and gets hot after some minutes (i have 1/4W i guess). I want now to order this two resistors (i will go with 2 W) because i want to learn now to design also the PCB and i saw that i need the datasheet to know the precise dimensions of the terminal and pitch (i do now a tutorial on how to design PCB from Robert Feranec Youtube channel).Regarding ordering the components, i have a question, other than the value and power, should i take into account others parameters?This will be my first mouser order.Thank you

author: fourfathom
author_rank: Super Contributor
timestamp: July 27, 2025, 04:17:43 pm
content: At 60 mA that 100 Ohm resistor is dissipating 1/4W, and a quarter-Watt resistor *will* get quite hot. In general you want to keep the actual dissipation to less than half the rated power, so a 1/2W resistor is OK, but 1W should have plenty of margin. Do be careful about current drawn by a stalled fan motor -- this depend on the type of motor though. The stall current (if any) will be limited by your series resistor, so I suggest you use that 100 Ohm resistor and physically stop the fan rotation, and then measure the fan voltage. This will let you calculate the stall current so you can choose your resistor power rating.As for resistor types you should probably look at inexpensive 5% or 10% tolerance parts -- this isn't a critical application.

author: the offset voltage
author_rank: Regular Contributor
timestamp: July 27, 2025, 04:28:19 pm
content: Hello, thank youCan i use SMD resistors or should i go with through hole resistors?Thank you,

author: fourfathom
author_rank: Super Contributor
timestamp: July 27, 2025, 04:42:47 pm
content: Quote from: the offset voltage on July 27, 2025, 04:28:19 pmYes, you can use a 2W surface mount resistor in the "2512" size, something like this:https://www.digikey.com/en/products/detail/bourns-inc/CRM2512-JW-101ELF/2563945It's still going to get warm (Watts are Watts), but that's OK. While the 2512 size is easy enough to handle, unless you're familiar with surface mount a through-hole resistor might be easier to experiment withBut I do suggest using surface-mount where practical. That's the way the world is going (has gone) and you might as well get comfortable with it. I seldom breadboard with through-hole parts anymore.

author: the offset voltage
author_rank: Regular Contributor
timestamp: July 27, 2025, 04:53:33 pm
content: Great,I will select and order the components that i need, in order to start the PCB design.Thank you

author: Analog Kid
author_rank: Super Contributor
timestamp: July 27, 2025, 07:10:59 pm
content: Quote from: fourfathom on July 27, 2025, 04:42:47 pmI certainly don't mean to start an argument here, but I have a different view:Since you're essentially a hobbyist, as I am, I find through-hole parts much easier to deal with.You can stick them into solderless breadboards for testing.For construction of such simple circuits, I just use regular old perfboard without any copper, with standard 0.1" hole spacing (easily available from any electronics supplier), stick the parts in and make connections on the back side with wire. Either just use wire to make outside connections, or for a more "modular" board, put in header pins for connections.Easy peasy.But do whatever you feel comfortable with here.

author: TimFox
author_rank: Super Contributor
timestamp: July 27, 2025, 07:14:20 pm
content: When resistors might get hot, I also prefer THT units.Back in the day, Heathkit recommended cutting a bit of cardboard from the carton to temporarily space power resistors away from the PCB surface during soldering, to be removed before use.

author: tggzzz
author_rank: Super Contributor
timestamp: July 27, 2025, 07:48:37 pm
content: Quote from: Analog Kid on July 27, 2025, 07:10:59 pmFor simple circuits (and complex ones if you know what you are doing), delete the cardboard and just solder components together in mid air.For other techniques, start withhttps://entertaininghacks.wordpress.com/2020/07/22/prototyping-circuits-easy-cheap-fast-reliable-techniques/and note "rat's nest".

author: RJSV
author_rank: Super Contributor
timestamp: July 27, 2025, 08:09:37 pm
content: OP, I glanced at your location, there, and wondering:There in your local town, (in Austria) what's the (electronics hobby) retail situation ? For obtaining parts and technical books, those in SF Bay area would, traditionally, go into a store, like ZAKS ELECTRONICS to buy and browse. Of course there was, also, Radio Shack at one time.Also, in Europe there, do hobbyist folks cross borders to shop, ok n German towns nearby, etc. ?

author: Analog Kid
author_rank: Super Contributor
timestamp: July 27, 2025, 08:18:14 pm
content: Quote from: RJSV on July 27, 2025, 08:09:37 pmRick, just curious: where is Zak's Electronics? Are they still in business?I've written here many times how incredible I find it that, for the most part, the SF Bay Area hasalmost no electronics stores!Really pretty close to zero.When I left there (2019), there were, to my knowledge, onlyJameco in Belmont right off 101 (still in business)San Mateo ElectronicsOne place up in Marin County, forget the nameWeird Stuff down in Sunnyvale/Santa Clara, but they went tits-up shortly afterAnd that was it.Incredible, dontcha think, for the place that's home to Silicon Valley?My favorite store of all, Al Lasher's Electronics in Bezerkeley, went out of business around that same time. Gone.

author: bson
author_rank: Unknown
timestamp: July 29, 2025, 06:27:17 pm
content: A fan is a motor, and as such it converts electric power to mechanical work. Its impedance depends on its mechanical resistance: the higher the mechanical resistance, the less is the impedance and hence the higher the current. In other words, it needs to work harder and the impedance drops. A resistor isn't going to substantially change this, all that will happen is it dissipates power as heat, and the motor needs to drop its impedance further to pull sufficient power to overcome the mechanical resistance. The resistor will set a minimum impedance and limit the max current, so could save the motor if mechanically blocked from moving. But it does so at the expense of power wasted as heat, and there are better ways to accomplish this, for example with a power or current limiting circuit.

author: TimFox
author_rank: Super Contributor
timestamp: July 29, 2025, 06:49:13 pm
content: Quote from: bson on July 29, 2025, 06:27:17 pmA better way to analyze a motor is vaguely transformer-like.The electrical input is measured in voltage and current (product is power).The mechanical output is measured in speed (radians/sec) and torque (N-m), where the product is also power, and less than the input power, with efficiency < 1.The motor has two important parameters: speed constant in (rad/s)/V and torque constant in (N-m)/A. (Sometimes quoted in reciprocals thereof.)Also, two important loss parameters: winding resistance inand bearing friction torque in N-m (some dependence on speed).For steady speed, the total torque (load + friction) requires a current that flows through the resistance, dissipating power for loss, and reduces the voltage that determines the speed. Thence the efficiency.For acceleration, the winding inductance and rotor inertia need to be included in the calculations.This is not well-represented by Ohmic resistance or impedance values.It’s not common, but the two speed and torque constants in a DC motor can be changed by varying the static magnetic field. Increasing the field gives less speed at constant voltage and more torque at constant current (analogous to changing the turns ratio in a transformer).

author: BeBuLamar
author_rank: Super Contributor
timestamp: July 29, 2025, 07:18:48 pm
content: Quote from: the offset voltage on July 27, 2025, 04:53:33 pm80mA at 12V is very little power and almost any variable power supply would run that motor well. So I would suggest connect the motor to the power supply and also connect whatever load you intend to use the motor for. Run the motor at 12V notice the current draw. Then reduce the voltage until the motor slows to the speed you want and notice the voltage and current. Do that for the other speeds. With that information now you know the resistance values and wattage rating needed for the resistors.

author: Analog Kid
author_rank: Super Contributor
timestamp: July 29, 2025, 10:46:59 pm
content: Quote from: TimFox on July 29, 2025, 06:49:13 pmIt occurred to me; can't we treat a motor as a transducer? Like a speaker, microphone, etc.?Something that converts electrical energy to mechanical energy (or v/v).

author: TimFox
author_rank: Super Contributor
timestamp: July 29, 2025, 11:11:28 pm
content: Yes, a transducer typically converts electrical variables on one end (usually voltage and current) to mechanical variables (e.g. torque and speed, or pressure and displacement), or vice-versa. The DC motor parameters in my post are a good example of linear operations of a transducer.

author: MrAl
author_rank: Super Contributor
timestamp: July 30, 2025, 10:07:29 am
content: Hello again,Modeling a motor is not that hard to do. In some control theory, the frequency domain model is:w(s)=Va*Km/((s*L+R)*(s*J+F)+Km*Kb)wherew(s) is the speed,Va is the armature voltage,Km is the motor constant,Kb is the back EMF constant,L is the armature inductance,R is the armature resistance,J is the angular inertia,F is the friction.This allows us to use this in some control circuit to evaluate the performance.

author: the offset voltage
author_rank: Regular Contributor
timestamp: August 04, 2025, 03:50:13 pm
content: Hi, i am almost done with the PCB, but i have a question. Should i connect a diode in parallel with the fan for protection?I saw in other schematics, that they have a diode in parallel with the motor, and i don't know if should i add one, i see many circuits uses 1n4007.Thank you,