24V-> 12V reduction for dc motor

Much closer to 80 A, initially. I=E/R If only the only change is a doubling of E, and the initial R remains unchanged, what is I gonna do?

Could you use a 24V battery drill or the trigger control for this application?

--Winston

Go here..

These are power diodes that you can mount on a heat sink in series to reduce our voltage.

THey are the 220 style case so you can use a stripe of aluminum across a row of them with screw hole between each one that can thread into a back plane heat sink. just solder them in series and pass the 24+ volts in one end (Anode string side) and use the last Cathode end to supply the motor..

40 AMP DIODES 24+ IN 12+ OUT with load

->|+-------+>|+------+>|+------+>|+--------+

To Motor circuit

These diodes are specified to have a forward drop of around 1.9 to

2.5 volts.. You may need to experiment a little so get more than 3 units.

PS. You would place a small cap across the whole string of diodes to help suppress HV energy coming back from the motor..

A .1 uf 100 ohm snubber is a good choice.

Jamie

Could you use instead a seperate 12 volt supply and just use the 24 volts to actuate a relay?

"Greg Russell" schreef in bericht news: snipped-for-privacy@mid.individual.net...

Suppose the 1A current thru the motor applies for full load, the most simple solution is a series resistor 12R/15W. It will become hot.

You can do it electronically but you'll have to deal with the huge inrush current. For a short time but nevertheless. Once you're using electronics you can as well add a current limiter of - let's say - 10A to make it more manageable. As you actually build a 12V power supply - lineair or switching - you can incorporate that current limiter in it. A lineair supply will become about as hot as the series resistor.

A third possibillity is building a pwm regulator. You'll need a square wave oscillator with a duty cycle 25% and a frequency of 20-100Hz depending on the motor. You'll also need the current limiter in this case.

petrus bitbyter

(...)

Here you go. Just place a resistor in series with the variable control on this part to limit maximum duty cycle to 50%:

--Winston

With a PWM regulator. In most cases you can use the motor inductance, which means that you just need to switch the 24V. It won't be simple.

Nope. If the motor draws 40A when stopped in a 12V system, then it has a resistance of about 0.3 ohm - slap 24V on there and it'll draw 80A, as noted.

Yes, but only when the current is 10A.

Where did you get 48 ohms? Assuming that things would work at all with just a series resistor, then you want to drop 12V at 1A, which works out to 12 ohms.

Nor do you know if you need to in the first place, or if just putting resistors in series will get you the required system behavior.

Permanent-magnet DC motors are somewhat mechanically "stiff" when you drive them from a constant-voltage supply. They get much less so when the supply is not constant-voltage (in fact, you can generate a pretty good constant torque from the right sort of DC motor and a constant- current supply). You can test this if you have a spare motor handy: give it a spin by hand with the motor terminals open. Now short the motor terminals and give it a spin again.

So a 24V supply in series with a 12V resistor and your motor would, indeed, cause the motor to draw 1A at 12V when it is going at the same speed that it does now. But depending on your mechanical arrangement it may not go the same speed -- I can't even tell you if it'd go faster or slower than it does now, nor could I tell you if that would be better or worse behavior than your system exhibits now.

If you really want to replicate the current behavior, you either need to get an equivalent motor to the one you have now, only wound for 24V, or you need to drop a reliable 12V somehow, probably with either a linear supply or a switching one. Because the motor's stall current is so high, you will either need to limit the supply current and live with the reduced torque at startup, or you will have to have a supply that's sized in nearly every way to supply 20A at 12V, and live with the fact that it's a honkin' big thing compared to a 1A, 12V supply.

--
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.
Why am I not happy that they have found common ground?

Tim Wescott, Communications, Control, Circuits & Software
http://www.wescottdesign.com

Greg Russell used his keyboard to write :

At the following link you might find some good information. I realize the company is in australia, but the info might help you.

Hope the link works, you may have to type it.

--
John G.

Isn't that what capacitors are good at - supplying large current for a short period of time.

| Isn't that what capacitors are good at - supplying | large current for a short period of time.

Yes, when discharging.

Charging from a voltage source you'll get a very high inrush current when there is no or only a small series resistance.

A stalled motor represents a small inductance and a very small resistance hence you'll get a huge inrush current of ten, hundred or even more times the steady state current. For a short time so (slow) fuses usually survive but to MOSFETs it's fatal. Once the motor is spinning up, it produces a magnetic field opposite to the one that makes the motor spin and the current drops.

petrus bitbyter