Voltage Divider - Doesn't power fan - Page 1
Forum section: electronics / beginners

author: arkufahl
author_rank: Newbie
timestamp: December 29, 2022, 02:06:56 am
content: I setup the circuit as shown in the attached image... a 9V source with two 10k resistors in series. I measure the voltage between wires A and B, and it shows 4.5V as I expected. However, when I connect a 5V DC fan to wires A and B, the fan doesn't operate. I know the fan will work with 4.5V because I tested it with a plain 4.5V circuit (with only the voltage source of 4.5V and the fan... nothing else).The only thing I noticed, is that measuring the voltage between A and B without the fan resulted in a reading of 4.5V, and measuring the same points with the fan resulted in a reading of 1.10V. But if that is the reason for the fan not working, I'll need it explained because my knowledge is limited.Thank you,Andrew

author: Vincent
author_rank: Regular Contributor
timestamp: December 29, 2022, 02:18:25 am
content: Don't forget currents!10K ohm on 9V leads to quite low currents, and certainly not enough to power that fan! In fact, a 10K ohm resistor connected to a 9V supply draws 0.9mA.If you want to stick to the voltage divider approach, you'll need to use much lower value resistors (and higher power). But if you can find a 7805 voltage regulator, that'll work much better, with less wasted power.What happens right now is that your fan is a comparatively much lower resistance component than the lower half of your voltage divider. And you're connecting that fan in parallel to it. The fans pulls the voltage at the divider's centre down to 1.10V as a result. So the fan only "sees" 1.10V, no wonder it doesn't work!I can create a little schematic to clarify things even more if needed.

author: arkufahl
author_rank: Newbie
timestamp: December 29, 2022, 02:27:27 am
content: Well, I'm happy to know that it's not supposed to work the way I have it setup. When I started down this path, I had an initial hunch it was not enough amperage... as the fan states 5V 0.09A... and at some point I attempted to measure the amps and could see it might be an issue.I'm just starting out, so maybe I skipped ahead a little further than I should have. I was merely hoping to power-up a component or two while learning, without supplying them with too much voltage/amps.A schematic would be much appreciated. And I thank you for the reply.

author: MikeK
author_rank: Super Contributor
timestamp: December 29, 2022, 02:34:03 am
content: A voltage divider is a poor way to power the fan. The fan's resistance changes the bottom leg of the voltage divider. What if the fan's resistance is 100 ohms...What then is the voltage at A? A voltage divider that is less affected by the fan's resistance will, of course, be wasting more current.

author: BillyO
author_rank: Super Contributor
timestamp: December 29, 2022, 02:34:27 am
content: A single 51 ohm .5 watt resistor should do the job for you.

author: BeBuLamar
author_rank: Super Contributor
timestamp: December 29, 2022, 02:46:11 am
content: voltage divider would only work for things that consume almost no power.

author: arkufahl
author_rank: Newbie
timestamp: December 29, 2022, 03:00:51 am
content: Quote from: BeBuLamar on December 29, 2022, 02:46:11 amThank you for the reply.Would you have a somewhat simple "for example" where using a voltage divider would make sense? If its not a simple answer, that's fine, just figured I would ask.When I was reading up on voltage dividers, it seemed like it would come in handy if I was making smaller projects that required little voltage, but wanted to keep it simple and use a 9V battery. So, I guess I need to understand the concept better.

author: BeBuLamar
author_rank: Super Contributor
timestamp: December 29, 2022, 03:09:55 am
content: One thing the voltage dividers are often use is for volume control in a pre amp level which is basically a voltage amplification with very low current. The current is typical less than 1mA.Another thing is to feed a variable voltage reference which is typical 0-10V and the current is about 1mA or so. I have some mismatch and the input draw something like 10mA and the source couldn't do it and voltage drop badly.

author: Vincent
author_rank: Regular Contributor
timestamp: December 29, 2022, 03:13:58 am
content: (This schematic is so half-assed it's cringeworthy)So here's the thing:We've got a voltage divider with equal resistor values and with 9V applied to it, which by itself creates 4.5V at its centre tap, as you yourself measured. Then you connect a 5V fan drawing 90mA @ 5V between the aforementioned centre tap and ground. Using Ohm's law we can easily calculate that your fan acts roughly like a 55.555 ohm resistor (5V / 0.09A = 55.555 ohm). That resistance is connected in parallel to the lower 10K resistor. Your new setup is now kind of like a 10K resistor in series with a 55.2480 ohm resistor!Or in numbers: 1/(1/10 000 + 1/55.555) = 55.2480 ohmNow the full analysis of this circuit is incredibly more complex than this, lots of nuances and details are intentionally ignored, but my goal is to keep things easy to understand as much as possible.Hopefully this sheds some light on the issue!

author: BeBuLamar
author_rank: Super Contributor
timestamp: December 29, 2022, 03:18:33 am
content: And in this case you would do better using a single 50 ohms resistor.

author: BillyO
author_rank: Super Contributor
timestamp: December 29, 2022, 03:30:18 am
content: Quote from: Vincent on December 29, 2022, 03:13:58 amMore like 5mV - not 1.1V

author: Vincent
author_rank: Regular Contributor
timestamp: December 29, 2022, 03:59:27 am
content: Quote from: BillyO on December 29, 2022, 03:30:18 amThat 1.1V was the measurement from OP. The hypothetical 55.555 ohm would indeed pull the voltage much lower.Assuming OP's fan is brushless DC, it probably doesn't load the circuit quite enough to pull the divider all the way down to the theoretical value, but still low enough not to start. But as I previously mentioned, I want to keep things within OP's knowledge level as much as possible.

author: Zero999
author_rank: Super Contributor
timestamp: December 29, 2022, 12:00:44 pm
content: The output of a potential divider looks like a voltage source, in series with a resistor, equal to the value of both of the resistors in the divider connected in parallel.Here's an example, showing a potential divider using different value resistors and a 5V source.R1 = 5k6R2 = 2k4VIN= 5VVOUT= R2*VIN*/(R1+R2) = 5*2.4k/(5.6k+2.4k) = 5*2.4k/8k = 1.5VROUT= R1||R2 = (R1+R2)/(R1*R2) = (5.6k+2.4k)/(5.4k*2.4k) = 13.44k/8 = 1k68In your case you have 9V in and two 10k resistors, giving an equivalent circuit of 4.5V and 5k.

author: arkufahl
author_rank: Newbie
timestamp: December 29, 2022, 03:43:18 pm
content: For the most part I understand most of what everyone had explained, thank you all for taking the time.One thing that confuses me though, is where my fan and the second resistor are parallel... is there not a way to take that 5V path NOT parallel? Otherwise, to me, it seems goofy to say "I need to knock this 9V down to 5V that automatically get a resistor". I would think most people would want an un-resisted 5V which they could then resistor it or not resistor it. One of the schematics shown as an example seemed to imply that might be possible, but I'm still learning, so maybe I'm reading it incorrectly.

author: BeBuLamar
author_rank: Super Contributor
timestamp: December 29, 2022, 04:16:57 pm
content: While the fan doesn't behave like a resistor but as other have calculate an approximation that the fan is equivalent to a 50 ohms or so resistor. So to drop the voltage down from 9V you just need to have resistor 1 of 50 ohms and the motor is the resitor 2.

author: Zero999
author_rank: Super Contributor
timestamp: December 29, 2022, 06:30:26 pm
content: Quote from: arkufahl on December 29, 2022, 03:43:18 pmThat's true.QuoteIt's not possible to do that with passive components. Use a voltage regulator such as the LM7805, to give a stable 5V power supply from the 9V. A switched mode regulator is another option and is more efficient, but a bit more complicated. Fortunately there is no shortage of switched mode regulator modules available.

author: BillyO
author_rank: Super Contributor
timestamp: December 29, 2022, 07:05:10 pm
content: The fan draws 90mA @ ~5V. If you use just a 50 ohm resistor Ohms law tells you the voltage drop across the resistor will be 50*0.09 = 4.5V so that you original 9V will be dropped to 5.4V which should be enough to drive the fan.A word of warning though. That resistor will need to be at least .5 Watt. If all you have is .25 W resistors you could get a 1W 50 ohm resistor by putting four 200 ohm resistors in parallel.Alternatively, as suggested, you could use an LM7805 regulator to reduce the 9V to 5V.You could also build your own regulator with a resistor a 5.6V Zener diode and a transistor. (see attached)You could also look for a switch mode buck regulator on eBay. They are both cheap and efficient. All the other methods mentioned above will dissipate about .5W as heat.

author: wasedadoc
author_rank: Super Contributor
timestamp: December 29, 2022, 07:17:39 pm
content: Quote from: BillyO on December 29, 2022, 07:05:10 pmNo! 1000 times 0.09 mA.The OP is confused enough without such bloomers.

author: BillyO
author_rank: Super Contributor
timestamp: December 29, 2022, 07:28:11 pm
content: Quote from: wasedadoc on December 29, 2022, 07:17:39 pmOppps, corrected that!! I have a lot of small "m" lying around. One just got in there by accident!

author: Gyro
author_rank: Super Contributor
timestamp: December 29, 2022, 09:00:53 pm
content: We definitely need a bold sticky in this section, saying that 'voltage dividers can't be used topoweranything'. It comes up so often.

author: arkufahl
author_rank: Newbie
timestamp: January 01, 2023, 07:55:27 am
content: Not to beat a dead horse, but I re-visited this after studying a bit. I ended up changing the first resistor to 56 ohms, and the second resistor to 470 ohms. This ended up giving me about 4.4V at the split for the fan. Which, unlike my original circuit, powered the fan. Measuring the amps, it was a little low (0.06A), but I'm not complaining about that. Based on my calculations, I believe the second resistor should theoretically have been about 550 ohms... which I don't possess, so just opted for the 470 ohm resistor.I can see the value of the other solutions described (I did in fact give the 7805 a shot), but it was good to see this work with the original setup (with more suitable resistance, of course).Feedback is appreciated.

author: Zero999
author_rank: Super Contributor
timestamp: January 01, 2023, 09:02:20 am
content: Quote from: arkufahl on January 01, 2023, 07:55:27 amThe second resistor isn't needed, just use a 68R resistor. The fan will act as the second resistor.

author: BeBuLamar
author_rank: Super Contributor
timestamp: January 01, 2023, 02:43:39 pm
content: Quote from: arkufahl on January 01, 2023, 07:55:27 amWhy do you need the 550 ohms resistor in parallel with the fan? It only waste power and reduce the voltage to the fan a little bit. If you want the voltage lower than what you get with a single 56 ohms resistor then simply use a higher value resistor. In this case you get slightly less voltage than ideal so remove the 470 ohms resistor is the thing to do.

author: BillyO
author_rank: Super Contributor
timestamp: January 01, 2023, 03:38:01 pm
content: Quote from: Zero999 on January 01, 2023, 09:02:20 amToo much resistor. @ 90mA there will be a 6.12V drop across that resistor. A 47R is the correct size for a common value. But it needs to be 1/2 watt or better. Four 200R 1/4 watt in parallel would work too. But your right. The 2nd resistor is just wasting power and lowering teh voltage a bit more than is required.

author: Pfriemler
author_rank: Regular Contributor
timestamp: January 01, 2023, 03:50:28 pm
content: wasting power ... you will waste nearly the same energy necessary for the fan in the resistor.If the overall question is: "how to use a 5V fan on a 9V source", why not consider a cheap usb output stepdown plug for cars?For short periods of use, a resistor will do of course. jm2c

author: arkufahl
author_rank: Newbie
timestamp: January 01, 2023, 04:23:49 pm
content: The original question was actually "how come the voltage divider doesn't power the fan?" I know I can NOT use a voltage divider and make this work, but that doesn't help me learn/understand voltage dividers. So decoupling the voltage divider isn't really what I'm after. While I don't at the moment understand the "waste" involved (but I do believe there is), the goal, again, was not necessarily building the ideal circuit to power the fan, but how to make a voltage divider do it.There was mention that a voltage divider can't power anything, and that seemed odd to me, and it appears that it can, so I'm a little confused now about that statement. Don't misunderstand me, its good for me to hear what is frowned upon, but in learning I like to know both what can be done, and why something shouldn't be done. Knowing this is an option, as inefficient or impractical as it may be, may come in handy some day... who knows 🤔 Now, if it was dangerous... different story. But still, an understanding of why it's dangerous is good information/knowledge to have.

author: Zero999
author_rank: Super Contributor
timestamp: January 01, 2023, 04:55:26 pm
content: Quote from: arkufahl on January 01, 2023, 04:23:49 pmIt's still a voltage divider, even if only one series resistor is used, because the fan is acting like the second resistor.

author: Vincent
author_rank: Regular Contributor
timestamp: January 01, 2023, 05:03:46 pm
content: Then let's word things differently:That 2X10K divider has a relatively high resistance and "allows" relatively little current to flow. The fan on the other hand "resists" current flow much less. But in this configuration it has to work with whatever current the upper 10K resistor allows to flow, i.e. not enough, and voltage across the fan drops as a result.In fact the lower 10K resistor hardly makes any difference, hence why it's been suggested to use a single 51/68 ohm resistor in series instead. Both said resistor and your fan have roughly the same opposition to current flow, so as with two water pipe restrictions of roughly the same diameter, the pressures (voltages) will roughly balance.Hope this helps.

author: arkufahl
author_rank: Newbie
timestamp: January 01, 2023, 05:14:38 pm
content: You are technically correct, but again, that goes back to a simple circuit, and not the typical example you see when voltage dividers are discussed, in terms of teaching beginners. I apologize for not specifically stating that I want a voltage divider that has to use 2 resistors to power a fan. I get that the fan has resistance, but it doesn't count as a resistor... I'm a beginner... a resistor is a resistor, a fan is a device that has resistance. The circuit must use 2 resistors and a fan. If a resistor is taken out of the solution, then its off-topic.

author: arkufahl
author_rank: Newbie
timestamp: January 01, 2023, 05:16:10 pm
content: Quote from: Vincent on January 01, 2023, 05:03:46 pmYes, that does help. Thank you.

author: Zero999
author_rank: Super Contributor
timestamp: January 01, 2023, 05:18:00 pm
content: Quote from: arkufahl on January 01, 2023, 05:14:38 pmWhy do you want two resistors? It's a bad idea to use more components, than necessary, especially if it makes it worse, not better, as in this case.The fan is behaving as a resistor, so taking the second resistor out is the best solution.

author: Gyro
author_rank: Super Contributor
timestamp: January 01, 2023, 05:42:47 pm
content: Quote from: arkufahl on January 01, 2023, 05:14:38 pmIt is inapropriate to power a fan from a resistive voltage divider, it is not a satisfactory engineering solution. It is wasteful of power and doesn't provide the fan with the required low impedance supply - the fan takes current in pulses as the coils are switched, worst case, it will malfunction. I'm not sure if that makes the premise of the thread OT?As I mentioned previously, resistive voltage dividers are not used for powering things. You use them for signal purposes, not where any sigificant current is drawn.

author: mariush
author_rank: Super Contributor
timestamp: January 01, 2023, 07:14:00 pm
content: A fan will be quite flexible when it comes to voltage ... a 5v fan will run at 4.5v or 5.5v maybe even higher voltage. 12v fans are often specified to run at 12v +/- 10% because of ATX fan specification (for computers) which says fans should tolerate up to 13.2vAs others explained, a voltage divider is pretty much only useful to measure the voltage, because you don't need a lot of current for that.Fan has its own resistance so when you connect it to the output of the voltage divider, that resistance is in parallel with one of the resistors in the voltage divider and will change the output voltage.In addition to it, fans have a nominal (standard) power consumption, but when they start spinning the motor consumes much more power for those few seconds until the motor is up to speed. So the voltage of the voltage divider will be even more affected during this period.If you want to limit the voltage in a rough/basic way probably the easiest would be to have a bunch of diodes in series, as each diode will cause a voltage drop to occur.For example, a basic 1n400x (where x=1..7, bigger number just means higher voltage but even 1n4001 can do 30v) diode will have a voltage drop of around 0.7v at 100mA or more, and probably as low as 0.5v at 1mA-ish.So, if you have a 12v input and you want 5v output, you need to drop 7v which can be safely done with around 8-10 diodes in series. Diodes are pennies and you can alternate orientation to reduce the amount of space used by the diodes.You only need to be careful about how much power is dissipated across each diode but that's easy to measure P = V x I = ~0.7v x 0.1A in your case, or around 0.07w which is way below the maximum a diode could tolerate.But, keep in mind that at low currents this voltage drop can be much smaller - this doesn't matter for powering a fan but may matter for other things like microcontrollers or other things, because a fan will start with high current draw and then current draw will go down but will still be reasonably high.Another method that would work here (and you use less components) would be to use a 5.1v or 5.6v zener diode to limit the voltage the fan "sees" close to 5v - it will work because the fan doesn't consume a lot of power :The safest method would be to use a fixed 5v output linear regulator, or an adjustable regulator.

author: Zero999
author_rank: Super Contributor
timestamp: January 01, 2023, 09:32:47 pm
content: If you must use two resistors, to answer a homework question, then add an emitter follower between them and the fan.R1 & R2 drop the voltage down to 6V, but the base current from the transistor drops it by a few hundred mV and the base-emitter drops another 0.7V, leaving about 5V for the fan.

author: BeBuLamar
author_rank: Super Contributor
timestamp: January 01, 2023, 10:31:41 pm
content: Quote from: arkufahl on January 01, 2023, 04:23:49 pmVoltage dividers are not used for anything with power. Their use is primarily to provide a varying voltage but supply almost no current. The simplest way to reduce voltage to a load is using a series resistor and not a divider.You said you want a voltage divider where the resistance of the fan has little effect. To do that basically the design is 90% or more current will flow thru the divider and 10% or less current flowing thru the fan. So to do that the resistor in parallel with the fan would be about 5 or 6 ohms and the resistore #1 is also about that. So you have a current flow approximate 1A while you only need about 0.1 A to run the fan. And the resistors would have to have quite high power rating.

author: Pfriemler
author_rank: Regular Contributor
timestamp: January 02, 2023, 09:15:37 am
content: Not knowing your initial intention, I was about to show that the solution to a problem should be searched more globally. This is clear now.The lesson to be learned from this to here remains simply: A voltage divider is not to be used to power anything. Even an overdimensioned divider will be "detune" by any load (and a fan is not an ideal resistor, even more a current sink).As shown in the transistor circuit, it may be used to generate a "regulater rule" to the transistor (spoken simply, although it's not correct here, such circuits were and are common by now, not only for beginners, but nowadays it is more useful to use a complete ic for that). But the load dependent basis current will detune the divider ratio in the same way, reduced by the factor of the current amplification of the transistor.