DC Motor Modification

Look for a DC-DC converter, step down converter to get 12Vdc from you 24Vdc power supply.

It is not recomended you tie your 12Vdc motor to 24Vdc power supply.

Keep in mind the current your DC motor will demand from your DC-DC converter.

Regards!

You could use a BJT or FET and burn the extra power in it. Set the BJT up as a linear voltage regulator and have the output be 12 V. How much Current does your DC motor draw? If it's not much then you may not need a heat sink but you probably will. The power dissipated in the BJT with be (24v-12v)*I, where I is the current of the motor. This will be much less expensive than a DC-DC converter but a lot more wasteful.

The simplest way to do this is to place a number of diodes in series with the field. Each one will drop 0.7 of a volt so if you want to drop 12 volts = 12/.7 = 17 or 18 diodes. Use a diode rated for the current. a 1N4000 is good for 1 amp. They cost about 4 cents each = 72 cents. Of course a resistor to do the same will work as well.

Now if your motor draws 50 amps, you should make a power voltage dropper because a 50 amp dropping 12 volts = 600 watts, so dieodes are not the choice for high currents. Diodes are good as their voltage drop is constant and not current dependant

If you know someone adept enough at electronic design, you can build a pulse width modulator, using the field inductance to integrate the current to the same value it would be at 12V. That is not hard to do using components as simple as a 555 timer, power FET, and clamp diode. Be certain to use plenty of bypass capacitance at the 24V input to the circuit. Parts, exclusive of blank circuit board cost, should be less than $5 to $6 in quantities of 50-100 for field currents of less than about 3A to 5A, less cost if the field current is lower.

Failing that, use of a DC-DC convertor as stated in at least one other post would be the best bet. Unquestionably that will be the best bet if this is a one-off application rather than doing your own design. Lowest cost DC-DC convertors under the Cincon name can be found at Mouser.com, if your current requirement is not too high. If the field draws more than an amp or so at 12V, I would most respectfully shun the multiple series-connected diode voltage-drop approach.

The only real disadvantage of PWM (pulse width modulation) is EMI generation, which can generally be handled if needed by inductance at the PWM output and use of shieldied wire (shield made of copper braid, not foil, terminated at both ends) for the motor leads. PWM frequency will be a tradeoff between PWM induced ripple in the field current (sawtooth waveform) and eddy current losses as heat in the stator iron. If you can get a look at the stator laminations, you should be OK at 10 KHz to 20 KHz with .014" thick laminations. If they are considerably thicker, a lower frequency is indicated, but the sawtooth component of field current may be excessive. In that case, a DC-DC convertor again is the best alternative for field currents in excess of an ampere.

Feel free to contact me privately if you wish. I can help regarding additional no-cost guidance in PWM design, but will need to know field inductance.

Bernie Katz

for low currents diodes work well and avoid complex circuits and drop the voltage no matter what the current is. Resistor voltage drops are a function of current. One could also use a 7812 three terminal regulator, good for ~3 amps if heat sinked and costs about $2 with heat sink and a couple of capacitors to give stability and it can be used to operate a pass transistor for higher current regulation. Pulse width modulation would save wasting watts in various IR drops but adds complexity

Dear Aurizon: I stated respectfully that at currents in excess of 1 ampere, series diodes are not the preferred solution due to reasons stemming from heat dissipation. At 3A, use of a linear regulator suffers from the same shortcoming, except to a greater extent. Dropping 12V at 3A dissipates 36W as heat, requiring a large heat sink, not to mention any considerations relating to power budget.

PWM solutions need not be complex. A buck controller such as the old National standby LM2576 can be configured to serve as a PWM controller using the motor inductance as the buck inductance, with only a few more parts added to the basic design. That aside, a 24V to 12V stepdown convertor based upon the same only requires 5 parts, is a "stock" design, and approaches 80% efficiency. Later higher frequency versions are 90% or better. The circuit simplicity is evident on page 1 of the data sheet:

http://cache.national.com/ds/LM/LM2576.pdf

Although a 5V version is shown, a 12V part is available as indicated in the table on page 2 of the data sheet.

Again respectfully,

Bernie Katz

I used the devices in order of complexity, with diodes being cheap and simple, yet effective at low current ratings. Resistors of course waste heat, but they are still fairly simple and a non tech can wire one in. A 3 terminal regulator and any attached pass transitor also waste heat and are more complex, but have the ability to be precise and maintain voltage at varied currents.

A PWM system is indeed the most efficient, but is also the hardest for anyone to design, source and build. That small package for $79 seems useaful

TANSTAFL !!! Diodes dissipate as much heat as resistors! PWM at 50% duty cycle and fast switching speeds is much more efficient.

The original question does not state what current is required.

Go Bernie!

yes, however the silicon diodes have a .7v voltage drop from 1-2 ma to 1 amp in the case of a 1 amp diode. A resistor will have an IR drop. Different diodes work with other currents and they will bring a constant voltage drop and not one that is a function of the current. The diodes within their rating dissipate the heat in a string and do not get very hot.

for a one off voltage adjustment, what can be simpler?

At some point, if you have to make hundreds of these, and must have high efficiency a PWM design can be optimized

Look at the IR curve for a diode . . . The diode has a non-linear curve whereas a linear resistor has a straight line "curve." Nonetheless, they both dissipate the same amount of heat for the same voltage drop at the same current. I dare you to prove otherwise.

I never said they dissipated any less? why would I say that? All I said was they give you a standard voltage drop over a wide current range. dissipation remains a I*I*R or E*I

A very simple method would be to connect two 12V automobile batteries in series and connect your 24 V source to this package. Then use 12V from one battery at a time regularly changing from one to the other using each one equally. Your 24 V source will keep them both charged. Simple, strong and CHEAP.

Depending how the motor is wound you may be able to re-connect the field coils in series. It is quite common to have parallel field coils (especially with higher voltage fields). You would need to open the motor and take a look. Double the volts and half the amps! (Has same power loss).

Traditional "field trim" method was to use an adjustable resistor equal to the ohms of the existing field. Wattage rating of resistor needs to be a minimum of the I^2xR losses (or 12V x If amps you have). Conservative use says wattage of resistor is approx 2x what you burn....gives long life and lower temperature.

Fancy new circuits with converters have lower power loss at expense of complexity.

What is the motor size and the expected field current? Is this for commercial mass production or a one-off backyard project? What is your 24V supply made up of (batteries? existing system power supply? spare 24V power supply you had lying around and wanted to use?, etc). Not enough information is provided!!

Please elaborate, what is the application? If it's small (less than 1A of field current) how about a simple 7812 12V regulator chip and a small heatsink. You cannot get much simpler and cheaper than that.