Reduce Voltage to Motor

Madman

Madman

Registered
Country flag
There must be a way to reduce the DC voltage to a small motor. The scenario I have is pictured below. Basically, it uses one of those R/C motor reversers to operate two functions on a crane i am working on. For clarity, I have shown only one of the exiting circuits, the one that powers and reverses the hook motor.

Power goes into the R/C motor reverser and exits +/- or -/+ depending on which button is pressed on the transmitter. 14.8 volts in and 14.8 volts out. I would like to reduce the power going out. I experimented with one of those tiny voltage reducers, wired into the exiting wiring. The problem with those is that they are polarity sensitive, so no matter command the motor reverser is given, the polarity does not change to the motor after exiting the voyage reducer. A resistor didn't prove successful either.

IMG_5008.jpg
 
try using a few diodes in series. If you have two sets, one each way then you can keep your reversing. Or just put them in the feed before the reverser if that is not voltage sensitive.
Each diode will drop the voltage by about 0.7V
 
Last edited:
Greg Elmassian said:
easy, cheap, polarity insensitive, no heat

greg
Click to expand...

Generally agreed, but with respect, I beg to differ on one point. Diodes can get quite warm. With 1 amp flowing and 0.7V dropped 0.7 Watts will need to be dispersed by each diode. With multiple diodes this may be significant but probably of no consequence with a small low power motor of the type mentioned in the first post.
 
Last edited:
For the length of time it will be running (hook motor), I doubt you will even feel a difference in the temperature of the dioides..
However, if you do this in the motor feed to a G scale loco, a different matter. - They will get warm enough (over time) to come loose if you use 'sticky pads'..
Luckily, I found this our whilst running as an open chassis.. :rolleyes:
 
Madman said:
There must be a way to reduce the DC voltage to a small motor.
Power goes into the R/C motor reverser and exits +/- or -/+ depending on which button is pressed on the transmitter. 14.8 volts in and 14.8 volts out. I would like to reduce the power going out. I experimented with one of those tiny voltage reducers, wired into the existing wiring. The problem with those is that they are polarity sensitive, so no matter command the motor reverser is given, the polarity does not change to the motor after exiting the voltage reducer. A resistor didn't prove successful either.
Click to expand...

DC to DC Step Down.png
 
Last edited by a moderator:
Madman said:
Looks promising Greg. So this gets wired into the exiting leads, the right side in my photo. If I read the description correctly, I can still reverse the polarity to the motor ?
Click to expand...
I am thinking that you could use two of the "voltage reducers" if you hook a pair of diodes (opposing polarities) to each pole of your reverser switch and then feed the reducers with the required polarity. Just to play it safe, I would use the same diode arrangement at both motor terminals. I believe this could work with either IC voltage regulators or "buck converters" (which I am having a lot of fun with). Could be more precise than a diode chain.
 
The circuit shown works for any polarity and can be "inserted" in either motor lead. If you trace through the circuit with positive on either lead you will see there is a path through (you can trace a path with the arrows all pointing in the same direction)

It's more compact and less work to wire up than two separate strings of diodes, and cheaper.

I would not use a buck/boost inverter, because they have a minimum voltage before they start working, AND they have a regulated output, what you do not what. You would have no voltage to the motor until it turned on, and then you would have a fixed voltage to the motor after that.

In addition, since most systems put out PWM to a motor, feeding PWM to the buck/boost unit would most likely result in erratic operation.

What is asked for is a voltage reducer, not a power supply.

Greg
 
Madman said:
There must be a way to reduce the DC voltage to a small motor. The scenario I have is pictured below. Basically, it uses one of those R/C motor reversers to operate two functions on a crane i am working on. For clarity, I have shown only one of the exiting circuits, the one that powers and reverses the hook motor.

Power goes into the R/C motor reverser and exits +/- or -/+ depending on which button is pressed on the transmitter. 14.8 volts in and 14.8 volts out. I would like to reduce the power going out. I experimented with one of those tiny voltage reducers, wired into the exiting wiring. The problem with those is that they are polarity sensitive, so no matter command the motor reverser is given, the polarity does not change to the motor after exiting the voyage reducer. A resistor didn't prove successful either.

View attachment 222104
Click to expand...

Is there a possibility that the input voltage can be reduced?

Maybe via something as simple as a primitive voltage divider the current draw from infrequent use would allow that.

Or is it battery powered? The 14.8V suggests a 4S Lipo if that is the case then try a 3S, 11.1V.

Lower V in is going to give a lower V out unless there is a voltage amplifier in the box.

What is the input voltage spec on the box (reverser) eg 10V - 16V?
 
Last edited:
GAP said:
Is there a possibility that the input voltage can be reduced?

Maybe via something as simple as a primitive voltage divider the current draw from infrequent use would allow that.

Or is it battery powered? The 14.8V suggests a 4S Lipo if that is the case then try a 3S, 11.1V.

Lower V in is going to give a lower V out unless there is a voltage amplifier in the box.

What is the input voltage spec on the box (reverser) eg 10V - 16V?
Click to expand...


Reducing the input voltage will cause one of the functions to operate too slowly. The R/C unit I have is feed by 14.8 volts from a Li-on battery. The battery powers the receiver and also the end functions, i.e. hook and lateral travel of the crane along the beam. This is one of those Playmobil MAN overhead cranes that I have motorized. So 14.8 volts causes the hook to raise and lower too quickly.
 
Madman said:
Reducing the input voltage will cause one of the functions to operate too slowly. The R/C unit I have is feed by 14.8 volts from a Li-on battery. The battery powers the receiver and also the end functions, i.e. hook and lateral travel of the crane along the beam. This is one of those Playmobil MAN overhead cranes that I have motorized. So 14.8 volts causes the hook to raise and lower too quickly.
Click to expand...

So use the control output to switch relays.. Feed the relays with the output of a 'buck' regulator to drive the hook motor.. You can then tweak the 'buck' to give you the voltage / speed you want.
 
Madman said:
Basically, it uses one of those R/C motor reversers to operate two functions on a crane I am working on, the one that powers and reverses the hook motor.
Click to expand...

Have to confess when I re read the above back still confused!

Does it read, one reverser, that is operating two functions, ie Up & Down, Backwards & Forwards, or two separate reversers, one to control Up & Down and the other controlling Backwards & Forwards?

If two separate reversers, the diagram in POST 8 can be used to control the input voltage of one of the reversers, hence lowering the output voltage to make the the Up and Down of the hook more manageable, the other reverser can be left as is.

If only one reverser?, that has one voltage input, but multiple outputs, then it's a case of identifying the correct output, and adapting the diode scenario.

As ever a picture, and details of the individual items, would help enormously.........
 
Madman said:
Reducing the input voltage will cause one of the functions to operate too slowly. The R/C unit I have is feed by 14.8 volts from a Li-on battery. The battery powers the receiver and also the end functions, i.e. hook and lateral travel of the crane along the beam. This is one of those Playmobil MAN overhead cranes that I have motorized. So 14.8 volts causes the hook to raise and lower too quickly.
Click to expand...

Ive just looked at the youtube videos of the crane and there appears to be 2 motors under grey covers one for traverse and one for hook. The hook motor end is visible, if this is the one to have reduced voltage I think the simplest method is to use the bridge and extra diodes in Greg Elmassian's link and, as the text says, instead of feeding AC into it just feed DC into it and it will drop the voltage and be reversible. Couple of cheap readily available components that can be mounted any where on the crane out of sight and connected into the motor wire (yellow one seen moving when traversing?).
 
Arthur Aardvark said:
Have to confess when I re read the above back still confused!

Does it read, one reverser, that is operating two functions, ie Up & Down, Backwards & Forwards, or two separate reversers, one to control Up & Down and the other controlling Backwards & Forwards?

If two separate reversers, the diagram in POST 8 can be used to control the input voltage of one of the reversers, hence lowering the output voltage to make the the Up and Down of the hook more manageable, the other reverser can be left as is.

If only one reverser?, that has one voltage input, but multiple outputs, then it's a case of identifying the correct output, and adapting the diode scenario.

As ever a picture, and details of the individual items, would help enormously.........
Click to expand...

OK, here is the motor reverser I have.

http://www.ebay.com/itm/Two-DC-moto...884060?hash=item3aac6c555c:g:iWwAAOxySy9SQoai

As you can see there are four relays. Two of the relays control lateral movement of the boom and cab. The other two control the up and down moment of the hook. The power supplying power to the receiver is also powering the boom and hook.

As mentioned, 14.8 volts works well for the boom but is too much for the hook, causing it to raise and lower too quickly.

One of the reasons I am using a 14.8 volt Li-on battery is because I have an electromagnet hanging on the hook that needs to be powered and the Li-on battery pack seems to have enough reserve power to handle the magnet.

I will try to take some photos tomorrow which may help clear things up.
 
Madman said:
OK, here is the motor reverser I have.

http://www.ebay.com/itm/Two-DC-moto...884060?hash=item3aac6c555c:g:iWwAAOxySy9SQoai

As you can see there are four relays. Two of the relays control lateral movement of the boom and cab. The other two control the up and down moment of the hook. The power supplying power to the receiver is also powering the boom and hook.

As mentioned, 14.8 volts works well for the boom but is too much for the hook, causing it to raise and lower too quickly.

One of the reasons I am using a 14.8 volt Li-on battery is because I have an electromagnet hanging on the hook that needs to be powered and the Li-on battery pack seems to have enough reserve power to handle the magnet.

I will try to take some photos tomorrow which may help clear things up.
Click to expand...

A stack of diodes in series between terminal B and the motor will drop the voltage (I would not use a resistor in series as will get extremely hot) those motors should not draw more than 1 amp so any Part No 1N4001 (1A, 50V) or 1N5400 (3A 50V) diodes should do.
7 of them in series will drop about 4.2V (7 x 0.6V) which will give you about 10.6V, if you want a lower voltage just add more diodes.
They are not very big so will be easy to conceal.
This article by Gregh explains it quite well. http://www.members.optusnet.com.au/satr/motor control.htm
I feel this is the simplest and most cost effective (10 for $1.00) way to achieve what you are after and saves a whole lot of mucking about.
 
Actually it is not the most cost effective. If if you cannot buy a full wave bridge for less than the cost of 4 diodes, the circuit I show would use 7 diodes, as opposed to the article you show that would use 10 diodes. (the example you linked to shows a total of 5 diode drops... a full wave bridge gives you 2 drops, and 3 diodes in the "middle" of the schematic below (on the right) gives 5 drops also, thus 4 + 3 = 7.

dropper.jpg

(the picture above has 6 voltage drops (0.7 volts each), the "GregH" circuit would require 12 diodes, and as you can see this circuit requires 8.

(so this drops 4.2 volts to the motor in either direction)

But perhaps it is easier to comprehend... definitely not cheaper or smaller.

But these are 2 options that Madman can take easily and slow down his motor.

Greg
 
Some photos. The motor in the clear plastic housing near the cab is the one I want to reduce power to. It's one of those tiny gear motors from China.

IMG_5024.jpg IMG_5025.jpg IMG_5026.jpg IMG_5027.jpg IMG_5028.jpg IMG_5029.jpg IMG_5030.jpg
 
Greg Elmassian said:
Actually it is not the most cost effective. If if you cannot buy a full wave bridge for less than the cost of 4 diodes, the circuit I show would use 7 diodes, as opposed to the article you show that would use 10 diodes. (the example you linked to shows a total of 5 diode drops... a full wave bridge gives you 2 drops, and 3 diodes in the "middle" of the schematic below (on the right) gives 5 drops also, thus 4 + 3 = 7.

dropper.jpg

(the picture above has 6 voltage drops (0.7 volts each), the "GregH" circuit would require 12 diodes, and as you can see this circuit requires 8.

(so this drops 4.2 volts to the motor in either direction)

But perhaps it is easier to comprehend... definitely not cheaper or smaller.

But these are 2 options that Madman can take easily and slow down his motor.

Greg
Click to expand...


Greg, I built the voltage reducer as per your sketch. I am please to say that it did the trick nicely. Thanks for taking the time to respond to my question.
 
You must log in or register to reply here.
Back
Top Bottom