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Voltage Divider circuits

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juststarting

New Member
I need to drop 5 volts to 2 volts. After lots of reading I come across voltage divider circuits,
It seems the perfect solution,
I have worked out that I need a 5k resister on feed and a 3.3k resister to earth, which should give me 1.9 volts between the resisters.
Is this right ? what about power consumption ? what wattage resisters do I need ?
I am using a 5 volt 2.5 amp supply, and running a motor from a kids toy,
"second thought" could I use a variable resister to adjust the output volts ?
Is there anything I need to beware of ?
A bit of expert advice would be much appreciated.

Thanks for the advice on LEDs you all, now converted to Piranha, (perfic)

Baz

What could possibly go wrong ?
 
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ronsimpson

Well-Known Member
Most Helpful Member
Your math is good.

A voltage divider is effected by the load. Here the load (your electric motor) is only 33 ohms.
3300 ohms and 33 ohms in parallel = 32 ohms. So Rbottom is now 32 ohms and the voltage is very small.
upload_2016-8-19_6-45-52.png
Watts = Voltage X Current
Voltage = 1.9, Current = 1.9 /3300, (W=1.9 X 1.9/3300)
Watts = Voltage X Voltage/resistance.
Watts burned in Rbottom (with out load resistor):
1.9 X 1.9/3300=0.001W
 

cowboybob

Well-Known Member
Most Helpful Member
Baz,
A voltage divider (VD) circuit is generally used to provide a low power voltage source, i.e., a reference voltage or some similar low current usage circuit. It is rarely used to provide the power needed to perform work (like operate a motor).

What is your goal for the output of the VD you have used as an example?
 

juststarting

New Member
Thought a VD was too easy.
To cut a long story shortish, I am playing with hydroponics,
Thanks to all the help I received, I have now got the LED lighting good, (all growing well)
however ! I am using a fish tank pump at mains voltage (240v)
As my set up is in a cupboard, and left for days on end, I am not over the moon with this set up.
After sorting through the "bits box" I have found a 4 ohm resistance motor at 3 volt (changed from original spec)
I have a 5 volt 2.5 amp power supply,
I need to run the motor (pump)

HELP ! ! ! !

Baz
 

JonSea

Well-Known Member
Most Helpful Member
A voltage divider won't work, as explained above.

What could work is a series resistor, if the motor current is roughly constant. We need to drop 3 volts across the resistor to provide 2 volts across the motor. By Ohm's Law,

V = IR

rearranging,

R = V / I = 5 / I

Measure the motor current, and you can calculate the needed resistance.

You need to consider the power dissipated by the resistor, which may be considerable.

Pd = I^2 × R

Select a resistor rated to handle more power than this, and don't be alarmed if it gets warm while the motor is running.

SmartSelectImage_2016-08-19-08-02-14.png
 

juststarting

New Member
Maths not my strongest subject, (Must have been sick that day at school),
but will try to work it all out,
Yes ! ! ! I could buy a proper pump, instead of making one, but that wouldn't be so much fun,
(this could be trial and error)

WHAT COULD POSSIBLY GO WRONG ?

Baz
 

Tony Stewart

Well-Known Member
Most Helpful Member
it is always best to compute power dissipation and to avoid heat rise, match the power source to the load rather than dump 50% or more in series.
 

juststarting

New Member
"You could let out the "Magic smoke"!
Know all about that, Smelt it, felt it, worn the T shirt.
Seriously,
I have found the motor (no load) draws 0.03 amp, the resistance of the motor is 4 ohm,
so working on "Ohms law"
V=IR the voltage should be 0.12 volt, mmmn this can't be right, cos It wouldn't run on 5 volt,
sorry to be a pain, please excuse my ignorance,
but I realy need some help on this,
From Tony's comment, perhaps I should "buy"(dirty word) a pump,

Baz
 
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KeepItSimpleStupid

Well-Known Member
Most Helpful Member
Ohms law doesn't work on motors. There's this other quantity called inductance that effectively raises the effective resistance seen by the power supply, The winding resistance is also a "tiny bit" dependent on where the brushes are landing (shaft position). You typically have to be able to supply a bit more current than the motor is rated for to start it.

In a DC motor the voltage is proportional to RPM when unloaded. meaning, if you drive the shaft with another motor the voltage produced will be proportional to RPM. Torque is proportional to current.

V=Vw-I*Rw where w is winding is often used to control speed. The V is the voltage applied to the motor terminals, I is the current through the motor and R is the resistance you measured.

The R that you measured is likely wrong because of technique and the low value. very low resistances usually require measuring the voltage across a known "length/resistance" and the current through the length. Resistance "shunts" which are used to measure high currents often drop say 50 mV at 100 Amps. There are BIG bolts to connect the 0-100 A circuit, but small critically placed screws to measure the voltage across the very low known resistance.

So, ohms law doesn't work very well on circuits that contain inductance. In AC circuits, the load has to be resistive and the AC waveform needs the RMS value. Cheap meters cheat and assume the input is a sine wave.

When measuring current with a multi-meter, there is usually a voltage drop associated with the range and the amount of current. < 0.6 V is common.

Most of this probably went way over the top of your head,
 

Les Jones

Well-Known Member
Most Helpful Member
Hi Baz,
You can't treat a motor as a simple resistor. I am assuming you motor is a permanent magnet brush motor. The fact that the motor is rotating generated a voltage. For a theoretically perfect motor with no load on the shaft it would rotate at a speed where the voltage generated would be equal to the applied voltage and in this state it would not be drawing any current. In a real motor it would never reach this state due to friction and other losses. The current drawn by the motor is proportional to the torque it is producing overcoming friction and driving a load. If the resistance of the motor is 4 ohms then when it was stalled with a supply voltage of 2 volts it would draw 0.5 amps. When the motor is rotating the EMF produced by the rotation is subtracted from the supply voltage and the difference is developed across the motor's internal resistance. This is one reason why a real motor slows down when a load is applied to the shaft. This is a simplified explanation but it should help you to understand what is happening.

Les.
 

juststarting

New Member
Yes ! !
Most of it was over my head, k,i,s,s, but did get the general jist of what you were saying,
Thanks Les this did help explain.
Seem to have come a long way since my original question,
due to your help I have changed the motor, (took several things to bits) this one feels stronger.
My conclusion is, that my motor running on 0.03 amp at rest, will draw more when under load, as I have a supply of 2.5 amp I don't think this will be to much of a problem.
Due to your comments I realized the size of the impeller will make a difference, this I can cope with having a lathe (made from spin dryer motor,
seems I am better at engineering than electrics)
So it seems I am on my way to a solution,
thank you all again, it has been an insight to how complicated this electrikery can be,

Baz

Magic smoke says you got it wrong.
 

Tony Stewart

Well-Known Member
Most Helpful Member
THe starting current is V+/DCR of motor then with no load the back EMF generated almost matches and opposes the forward voltage into the motor so the difference from friction is the no-load current.

The steady state current at full rated load will be about 20% of the starting current.
 
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Couldn't you just build a simple rheostat and gradually lower the resistance until the motor is running properly? Then unhook the power and measure the resistance of only the rheostat and use that as your resistor ohm rating for the permanent setup? You could also measure the current going into the motor at that point too and voltage drop across the motor and rheostat. Seems like getting all critical numbers shouldn't be too hard.
 

cowboybob

Well-Known Member
Most Helpful Member
Yes, you could.

But for all the reasons stated above, using a voltage divider for this application is not a good idea.
 

Tony Stewart

Well-Known Member
Most Helpful Member
A 2V adjustable LDO is the preferred solution

What we know.

0.03 amp, the resistance of the motor is 4 ohm = DCR
But no load running current is 2V?/0.03A = 67Ω
Motor rating 2V ?
Start surge = V+/DCR e.g. 2V/4Ω = 0.5 A
Ratio of start to no load running current =0.5 / 0.03 = 16.7

Typical full load current is 12% to 20% of start surge current or 8:1 to 5:1 ratio for DC motors (depending on design)
Thus motor current expect when loaded to rated power is 1/5 start surge of 0.5A or 0.1A

In order to have good speed control, a voltage source must have an impedance much lower than the load, usually 1% of the load.
Using a a large rheostat was used in old railroad train toys and were like variacs with AC then rectified to DC.

These days it is more effective to use adjustable regulators which like Power Amps have very low output impedance.
.
 
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