Voltage Divider - Doesn't power fan

[arkufahl]

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

[Vincent]

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. 

[arkufahl]

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.

[MikeK]

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.

[BillyO]

A single 51 ohm .5 watt resistor should do the job for you.

[BeBuLamar]

voltage divider would only work for things that consume almost no power.

[arkufahl]

voltage divider would only work for things that consume almost no power.

Thank 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.

[BeBuLamar]

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.

[Vincent]

(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 ohm

Now 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!

[BeBuLamar]

And in this case you would do better using a single 50 ohms resistor.

[BillyO]

(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 ohm

Now 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!

More like 5mV - not 1.1V

[Vincent]

More like 5mV - not 1.1V

That 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.

[Zero999]

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 = 5k6
R2 = 2k4
V_IN = 5V

V_OUT = R2*V_IN*/(R1+R2) = 5*2.4k/(5.6k+2.4k) = 5*2.4k/8k = 1.5V

R_OUT = R1||R2 = (R1+R2)/(R1*R2) = (5.6k+2.4k)/(5.4k*2.4k) = 13.44k/8 = 1k68

In your case you have 9V in and two 10k resistors, giving an equivalent circuit of 4.5V and 5k.

[arkufahl]

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.

[BeBuLamar]

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.

[Zero999]

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".

That's true.

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.

It'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.

[BillyO]

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.

[wasedadoc]

The fan draws 0.09 ma @ ~5V.

No! 1000 times 0.09 mA.

The OP is confused enough without such bloomers. 

[BillyO]

The fan draws 0.09 ma @ ~5V.

No! 1000 times 0.09 mA.

The OP is confused enough without such bloomers. 

Oppps, corrected that!!  I have a lot of small "m" lying around.  One just got in there by accident!

[Gyro]

We definitely need a bold sticky in this section, saying that 'voltage dividers can't be used to power anything'. It comes up so often.

[arkufahl]

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.

[Zero999]

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.

The second resistor isn't needed, just use a 68R resistor. The fan will act as the second resistor.

[BeBuLamar]

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.

Why 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.

[BillyO]

The second resistor isn't needed, just use a 68R resistor.

Too 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.

[Pfriemler]

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