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Need simple voltage drop help...

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Scratch

New Member
Here's a simple one... I think...
I need help with a project I'm doing. I know it needs resistors, but I don't know what value, and where to put them.

I have a 12VDC battery, and I want to control a 4VDC motor. The battery says 12VDC 4AH, The motor only says 4VDC. I don't know what the current is.

Any help...?
 

ChrisP

Member
What is the make and model of the motor? The best answers to your question really depend upon knowing the design current draw of the motor in question.

The source voltage could be droppped via a series resistor, but there will be considerable wasted energy that way. For example, suppose that the motor draws 1.5A. In that case, a 5.1 ohm resistor in series with the motor would effectively reduce the source by dropping approximately 8 volts, leaving roughly 4 volts for the motor. However, the dropping resistor would then have to dissipate slightly less than 11-1/2 watts.

Consider the following circuit:



We can calculate the value for R1 as follows:

R1 = (voltage to be dropped) / (circuit current)

If the motor is rated 1.5A at 4VDC, then the equation becomes:

R1 = 8 / 1.5 = 5.333

If we use a 5.1 ohm resistor (the closest standard value) for R1, we end up with 7.65V being dropped across R1 (1.5A x 5.1 ohms), leaving 4.35V (from a perfect 12.0VDC source) for the motor. This would give us 11.475 watts to be dissipated by R1 (7.65V x 1.5A).
 

Russlk

New Member
You don't want to use a resistor because the motor draws more current when you load it which will reduce the motor voltage and torque. You could use an 8 volt zener but the power loss will be twice as much as the power used. If efficiency is important, a 12 volt to 4 volt switching regulator could be used.
 

Exo

Active Member
You could use a 5V zener to generate a stable 5V source and buffer this with a transistor. The 0.8 - 1V voltage drop over the transistor makes it about 4 volts for the motor
 

Styx

Active Member
You could use a 5V zener to generate a stable 5V source and buffer this with a transistor. The 0.8 - 1V voltage drop over the transistor makes it about 4 volts for the motor
Since a battery is being used and there is a likelyhood that as much life out of it will probably be needed a 5V zener will disipate alot of power.

The one thing that has stuck with me from Uni abt electrical machines is

"volts equals speed, Current equal torque"

since it is a 4V DC motor the chances are the winding insulation can handle alot more. Can you run the 4V motor straight off the 12V but use a current shunt resistor in parallel. Unloaded ok the machine will go ~3x it designed speed and probably ruin it. But if you know what load it is on you should be able to over-shunt it so it doesnt have enough torque to reach its max speed so it must reduce speed to maintain that torque

ok that resistor will also loose power like the zener but if you are after a resistor solution rather than a silicon solution (ie a voltage reg) then this should work.

make sense anyone?
 

Russlk

New Member
Styx: You are correct, "volts=speed, current = torque, but I don't understand putting a resistor in parallel with the motor. If you draw the schematic, you will see that it won't work.
 

Nigel Goodwin

Super Moderator
Most Helpful Member
All the previously mentioned ideas have the big drawback that they are wasting 2/3 of the power (as heat) and only providing 1/3 to the motor.

You could make a simple PWM driver (using a 555 and a driver transistor) to give 66% off and 34% on for a vastly more efficient design - but no where to make your toast :lol:
 

Styx

Active Member
Russlk said:
Styx: You are correct, "volts=speed, current = torque, but I don't understand putting a resistor in parallel with the motor. If you draw the schematic, you will see that it won't work.
lol. I know. But from what Scratch sayed it sounded like he wanted a resistor solution and a shunt was the closest way of doing it otherwise impedance probs with votage dividers.

A 5V regulator would be teh way I would do it (5v on a 4v rated motor only 25% extra speed). PWM if I wanted more control.

The shunt was a throw causion to the wind, that is why i put

"make sense anyone?" at the end
 

ChrisP

Member
Yeah -- I agree that a resistive method may well be the least desirable method here -- as I hinted when I stated that there would be a lot of wasted energy. I simply posted what I did re: a series ressitor to help the OP understand that knowing the motor design current would make it easier to devise a fix, as well as to show the OP how it figures in the calculation if he were to use a series resistor.

In reality, depending upon the motor size and current needs, a simple series resistor may well work just fine, though. This is a method commonly used with many miniature "hobby" type motors in limited-life low-cost consumer electronics these days.
 

bogdanfirst

New Member
the problem here is efficency, the circuit is powered from a battery.
if you would power the motor from a 12V adapter or so, then you wouldn't mind wasting 2/3 of the power. but powering the motor from a battery means that you want efficency. i think that a simple switching circuit that generaters 5V(more available) or 4V, can be used. if the motor doesn't draw much current, than this type of suply would not be very expensive.
personally, i would use a PWM. a simple 555 circuit can do the trick. but i don't think that you will need to use the circuit for 33% ON time and 66% off time. just like said before, current=torque. even if you apply the same power to the motor, you apply 3 times more current. this means that you should have more torque than you have when you power the motor from 4V, but matba another speed. also, what are you useing the motor to drive?
what is the motor from? maybe we can tell the aproximate current. also it would be good if you could power the motor from a stabilized 4V suply and mesure the current when the motor goes with load and when it is maintained powered and held still(not allowed to turn). i think that this is called the short-circuit current of the motor. this way you will know what to use as final transistor in the circuit.
if efficency is not important, you should use a 5V regulator and a diode in series with the motor. this is much better than using a resistor for the voltage drop.
the idea with the shunt doesn't work. all that the shunt will do is to add more load to the motor. the motor is still powered from 12V and is still draws all the current it needs.
 

Scratch

New Member
The motor is a volume control for a stereo, when you hit volume up on the remote control, the volume knob physically turns, increasing the volume. I want to control that from other parts of the house.
The 12 volts I have is actually from a regulated power supply, I probably should have said that but I was hoping it would be simpler, so wasting power is not a problem.
I thought this would be a simple problem. Who knew I would get this many responses!! :eek:
 

seeker

New Member
Hi scratch,
since it's only going to be momentary use you should be fine with a 5 volt regulator like an LM7805. If you want to be really safe you can put a 1N4001 diode inline with the output of the regulator as someone suggested to get the voltage to about 4.3-4.4 volts.

(Don't want to step on anyones toes who have have replied earlier)

Your last post just made everything so clear :D

Good luck with it! Tony
 

Styx

Active Member
The motor is a volume control for a stereo, when you hit volume up on the remote control, the volume knob physically turns, increasing the volume. I want to control that from other parts of the house.
The 12 volts I have is actually from a regulated power supply, I probably should have said that but I was hoping it would be simpler, so wasting power is not a problem.
I thought this would be a simple problem. Who knew I would get this many responses!!

Well on this note Im gonna have to say use a PWM full bridge inverter.

It is a DC machine and thus the direction of rotation is defined by the polarity of the voltage applied. A straight resistor (if it worked), zener, reg would only provide positive volts thus only one direction of rotation.

Use two P-N type inverter legs to make a full bridge thus making easy control. You can get 15V logic (I think it is the 4000 series) so no need to get a 12v-5V reg to power logic. all you need then is some logic to say what gate on teh inverter to fire to PWM the right polarity onto the DC machine. This way bi-directonal and you improve efficiency


I have drawn the cct that would be needed but I havn't got anywhere to host it. It is very simple and you dont even need a 555 timer. I am using an LM311 comparator as a PWM generator and by running the limited logic (one inverter and two AMD gates) also off 12V

2x N-type MOSFET - these will have anti-parallel diode within
2x P-type MOSFET - these will have anti-parallel diode within
1x LM311
1x Inverter package
1x AND package
resistors and caps as required.
possible PNP,NPN as push-pull on the output of the AND gates that feed teh FETS if they cannot source enough.

Should be enough source/sink since only low amp FETS. You should be able to set the switching freq pretty high (high 10s of kHz) so the ripple is low and have a duty of 33% for an average of 4V. These are all set with the resistors/cap around teh LM311

Also since you say a 4V motor, 4V will probably have quite a high RPM for its size and also for your application. By changing the comparator feedback threshholds you can drop the duty => drop the machine volts thus drop the speed.

If anyone can host I have a bmp of the cct
 
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