24V-> 12V reduction for dc motor

[Greg Russell]

Hello all,

I have a 12VDC motor that I wish to intermittently run in a 24VDC system,
and am wondering how I might efficiently reduce the voltage, please.

The motor draws 40A at initial torque in a 12V system, but quickly drops
to ~1A. If my beginner's knowledge serves, that would mean 20A initially
in the 24V system.

If E=IR (voltage=amperage*resistance) then does an inline resistor of 1.2
ohms drop the voltage to 12V? What about after initial torque, does the
required resistance change to 48 ohms, and if so, how might I accomplish
that?

I don't know if those color-banded solid-state resistors can handle such
an initial amperage, or how to wire the circuit for the varying
resistance required.

 

[Winston]

Greg Russell wrote:

Hello all,

I have a 12VDC motor that I wish to intermittently run in a 24VDC system,
and am wondering how I might efficiently reduce the voltage, please.

The motor draws 40A at initial torque in a 12V system, but quickly drops
to ~1A. If my beginner's knowledge serves, that would mean 20A initially
in the 24V system.

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?

If E=IR (voltage=amperage*resistance) then does an inline resistor of 1.2
ohms drop the voltage to 12V? What about after initial torque, does the
required resistance change to 48 ohms, and if so, how might I accomplish
that?

I don't know if those color-banded solid-state resistors can handle such
an initial amperage, or how to wire the circuit for the varying
resistance required.

Could you use a 24V battery drill or the trigger
control for this application?
http://tinyurl.com/79qtsdg

--Winston

 

[Jamie]

Greg Russell wrote:

Hello all,

I have a 12VDC motor that I wish to intermittently run in a 24VDC system,
and am wondering how I might efficiently reduce the voltage, please.

The motor draws 40A at initial torque in a 12V system, but quickly drops
to ~1A. If my beginner's knowledge serves, that would mean 20A initially
in the 24V system.

If E=IR (voltage=amperage*resistance) then does an inline resistor of 1.2
ohms drop the voltage to 12V? What about after initial torque, does the
required resistance change to 48 ohms, and if so, how might I accomplish
that?

I don't know if those color-banded solid-state resistors can handle such
an initial amperage, or how to wire the circuit for the varying
resistance required.
Go here..

http://www.mouser.com/ProductDetail/NXP-Semiconductors/BYC20-600127/?qs=sGAEpiMZZMtvcUztdGSumHqLbuom3z1bwT5sy1Abu38%3d

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

 

[petrus bitbyter]

"Greg Russell" <me@invalid.org> schreef in bericht
news:9j7m4eFfvjU1@mid.individual.net...

Hello all,

I have a 12VDC motor that I wish to intermittently run in a 24VDC system,
and am wondering how I might efficiently reduce the voltage, please.

The motor draws 40A at initial torque in a 12V system, but quickly drops
to ~1A. If my beginner's knowledge serves, that would mean 20A initially
in the 24V system.

If E=IR (voltage=amperage*resistance) then does an inline resistor of 1.2
ohms drop the voltage to 12V? What about after initial torque, does the
required resistance change to 48 ohms, and if so, how might I accomplish
that?

I don't know if those color-banded solid-state resistors can handle such
an initial amperage, or how to wire the circuit for the varying
resistance required.

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

 

[Phil Allison]

"Jamie the Radio Ham Idiot "

http://www.mouser.com/ProductDetail/NXP-Semiconductors/BYC20-600127/?qs=sGAEpiMZZMtvcUztdGSumHqLbuom3z1bwT5sy1Abu38%3d

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

** Huh ??

That is a hyper fast diode pack.

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

** At a chip temp of 150C and HIGH current !!

Read the fucking specs - dickwad.

The OP only needs 1 amp average current.

Slow diodes ( rated at about 10 amps) are all he needs and about 10 of them
in series would do.

Plus a 15 amp breaker to cover a stalled motor.


...... Phil

 

On 24 Nov 2011 20:01:19 GMT, Greg Russell <me@invalid.org> wrote:

Hello all,

I have a 12VDC motor that I wish to intermittently run in a 24VDC system,
and am wondering how I might efficiently reduce the voltage, please.

The motor draws 40A at initial torque in a 12V system, but quickly drops
to ~1A. If my beginner's knowledge serves, that would mean 20A initially
in the 24V system.

If E=IR (voltage=amperage*resistance) then does an inline resistor of 1.2
ohms drop the voltage to 12V? What about after initial torque, does the
required resistance change to 48 ohms, and if so, how might I accomplish
that?

I don't know if those color-banded solid-state resistors can handle such
an initial amperage, or how to wire the circuit for the varying
resistance required.
Could you use instead a seperate 12 volt supply and just use the 24

volts to actuate a relay?

 

[Winston]

Winston wrote:

(...)

Could you use a 24V battery drill or the trigger
control for this application?
http://tinyurl.com/79qtsdg

Here you go. Just place a resistor in series
with the variable control on this part to limit
maximum duty cycle to 50%:
<http://www.ereplacementparts.com/switch-vsr-forward-reverse-lever-sold-seperately-p-64884.html>

--Winston

 

[Tim Wescott]

On Thu, 24 Nov 2011 20:01:19 +0000, Greg Russell wrote:

Hello all,

I have a 12VDC motor that I wish to intermittently run in a 24VDC
system, and am wondering how I might efficiently reduce the voltage,
please.

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.

The motor draws 40A at initial torque in a 12V system, but quickly drops
to ~1A. If my beginner's knowledge serves, that would mean 20A initially
in the 24V system.

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.

If E=IR (voltage=amperage*resistance) then does an inline resistor of
1.2 ohms drop the voltage to 12V?

Yes, but only when the current is 10A.

What about after initial torque, does
the required resistance change to 48 ohms, and if so, how might I
accomplish that?

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.

I don't know if those color-banded solid-state resistors can handle such
an initial amperage, or how to wire the circuit for the varying
resistance required.

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

 

[John G]

Greg Russell used his keyboard to write :

Hello all,

I have a 12VDC motor that I wish to intermittently run in a 24VDC system,
and am wondering how I might efficiently reduce the voltage, please.

The motor draws 40A at initial torque in a 12V system, but quickly drops
to ~1A. If my beginner's knowledge serves, that would mean 20A initially
in the 24V system.

If E=IR (voltage=amperage*resistance) then does an inline resistor of 1.2
ohms drop the voltage to 12V? What about after initial torque, does the
required resistance change to 48 ohms, and if so, how might I accomplish
that?

I don't know if those color-banded solid-state resistors can handle such
an initial amperage, or how to wire the circuit for the varying
resistance required.

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

http://secure.oatleyelectronics.com/files/K275%20WEB%20NOTES.pdf
Hope the link works, you may have to type it.

http://secure.oatleyelectronics.com//product_info.php?products_id=840&osCsid=78c9516a506712be955423b51a509ac6

--
John G.

 

[fungus]

On Nov 25, 1:32 am, "petrus bitbyter" <petrus.bitby...@hotmail.com>
wrote:

You can do it electronically but you'll have to deal with the huge inrush
current.

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

 

[petrus bitbyter]

"fungus" <openglMYSOCKS@artlum.com> schreef in bericht
news:29f2ac7a-3d54-4cec-91c0-6ca58edacdc9@i8g2000vbh.googlegroups.com...
On Nov 25, 1:32 am, "petrus bitbyter" <petrus.bitby...@hotmail.com>
wrote:

You can do it electronically but you'll have to deal with the huge inrush
current.

| 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