# voltage divider

Discussion in 'General Electronics Chat' started by fupersly, Dec 4, 2008.

1. ### fupersly Thread Starter New Member

Dec 4, 2008
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Hello, not a homework question for me, but I figure this is a good enough forum to ask.

I'm a bit confused by a voltage divider circuit I was working on, perhaps someone can shed some light on the subject for me?

I have a 5 volt source (through a 7805) and I want to provide power to a pager motor that is rated at 95ma @ 3 volts.

I breadboarded a voltage divider using two 1k resistors (i have a ton of 1k resistors). From my calculation the 2 1k resistors will cut the voltage in half.

I measured the voltage drop across each resistor at about 2.45 volts, which is fine as the voltage from the regulator is around 4.9 volts or there abouts.

Here's the rub. I hooked the motor up incorrectly at first, in series with the resistors (rather than in parallel with the second resistor). The motor didn't spin at all. And this is the crux; I cannot figure out why!

I know i hooked it up wrong, but I cannot figure out why the motor doesn't move at all, unless I'm not giving it enough juice to start moving? I hooked up a spare LED in series and it lit up, not strongly but not so dim as I had trouble seeing it.

The other question I have about voltage dividers, is: How do I know what the optimal resistor value is for my circuit? Why not use a 10k resistor instead of a 1k resistor? I know about using the 10% rule for accounting for a load, but I just read that last night (after I put everything away) so perhaps my 1k resistors are the wrong choice here.

Cheers and Thanks!

Shane

2. ### Audioguru New Member

Dec 20, 2007
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896
Two series resistors make a voltage divider but cannot supply much current.
Your motor probably needs a lot of current.

4.9V through two 1k resistors and in series with the low resistance motor gave the motor almost no voltage and a current of only 2.45mA. If the LED is a 2V red one then it got 2V and 1.45mA which would make it visible but dim.

3. ### mik3 Senior Member

Feb 4, 2008
4,846
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The motor needs 95mA to operate at normal speed (at start up it requires more current). With 5V and 2K resistor (two 1K) in series (ignoring the motor) the maximum current is 2.5mA. This is not enough even to run the motor at normal speed, thus it can't start.

4. ### mik3 Senior Member

Feb 4, 2008
4,846
63
You don't really need a voltage divider here.Just a series resistor with the motor will do the job.

Its value equals:

R=(5-3)/95mA=21 ohms, its power rating has to be 1/2 watts or greater.

5. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
Diodes would probably work better, say 3 1N4001 in series. Each diode drops around .7 volts each. Since the diodes don't care what the current is (within limits of course) you'll be left with 3V.

6. ### mik3 Senior Member

Feb 4, 2008
4,846
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Bill's idea would be better if you require high starting torque. If you don't require high starting torque I think a resistor would be better because it limits the starting current and the power supply is not over stressed.

7. ### fupersly Thread Starter New Member

Dec 4, 2008
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R=(5-3)/95mA=21 ohms, its power rating has to be 1/2 watts or greater.

Ok I understand now why my motor was not running when connected in series with 2k resistors in total. Thanks I appreciate it!

Back to the question of the voltage divider/dropping-voltage. In Mik3 example, I am (correct if wrong please) calculating what resistor I would need to drop 2 volt source to supply a 95 ma load?

For power rating of resistor: Given P=V * I => 2 * 95ma => .57 W or a half a watt source?

Thanks!
Shane

8. ### leftyretro Active Member

Nov 25, 2008
394
2
To calculate the needed resistor in ohms, R= 2/.095 = 21 ohms (use nearest standard value avalible).

To calculate power dissaption rating in watts, use did it correctly:

"For power rating of resistor: Given P=V * I => 2 * 95ma => .57 W or a half a watt source?"

In practice you should select the next higher or better standard wattage rating, in this case 1 watt or even 2 watts.

9. ### fupersly Thread Starter New Member

Dec 4, 2008
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Okay - I think I have everything now for my particular circuit, as simple as it is. One thing remains, when dealing with low voltages like I am and small motors and ICs when would you ever bother with a voltage divider? Under what conditions would that be more suitable rather than just adding a resister in series with the load?

Shane

10. ### fupersly Thread Starter New Member

Dec 4, 2008
7
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Hmm I think I did my calculation wrong:

Is the power rating for the resistor calculated as follows:

P = V * I => 2 * .095 = .19 watt
or
P = V * I = 5 * .095 .475 watt

I would think that I would be using the .19 watt solution given I'm dropping 2 volts across the resistor?

Sorry, still find this confusing!

11. ### Jassper Active Member

Sep 24, 2008
84
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I use This when working with voltage dividers, makes "Headache caused by math" much easier.

12. ### MikeD_72 Active Member

Nov 11, 2008
46
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The power dissipated across the resistor is the voltage across the resistor times the current through the resistor. In other words, your first calculation.

13. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
The first equation is correct, since we are talking about the wattage across the resistor, and therefore the voltage across it only.

The second equation is the wattage the entire circuit is using.

14. ### Audioguru New Member

Dec 20, 2007
9,411
896
2V times 95mA is 0.19W.
But when the motor starts and when it is stalled its current is much higher and the resistor will heat up to more power dissipation quickly.

15. ### Metalfan1185 Active Member

Sep 12, 2008
146
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You may also consider using a Variable resistor on the Ground pin of the 7805 to vary the output Voltage.

The 7805 uses the Ground pin as a reference for the regulation of the 5V, by using a Variable resistor n series with this pin, you make the 7805 regulator "think" that the voltage is closer to + than it really is, so it adjusts the output to 0V - 5V depending on the setting of the resistor.

can tell ya what value though, maybe a 1K adjustble?

or scratch the 7805 and use a LM337, which is the same idea, but is designed to be adjustable

16. ### trrobbie New Member

Dec 11, 2008
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Your on the right track! 2 devided by .095= 21ohms. This is to find the value of the resistor you need. 2x.095= .57 is used to give you the power rating of the device.(that is how much heat it can dissapate to the air before dying due to gettting too hot) Hope this helps!

17. ### SgtWookie Expert

Jul 17, 2007
22,183
1,729
That is the power dissipation in just the resistor.
That is the total power dissipation of both the motor and the resistor.

Note that the "rule of thumb" for selecting a resistor's wattage is to double the result you receive, then select a resistor that has a wattage rating equal to or higher than the doubled result. For example:
Since 2V * 0.095A = 0.19W, then 2 * 0.19W = 0.38W, you will need a resistor rated for 1/2 Watt.