PDA

View Full Version : Discussion Micro Alternators


Garak
Apr 30, 2006, 02:12 PM
I'm looking for details on building an Alternator based power system. I hate batterys and I want to build a plane that can fly without them(ofcourse I'll still have one onboard as a backup for the RX and servos). I'm also going to be putting lots of different electronics and stuff onboard that are going to require power.

I've been searching on google but I really can't find anything on alternator design. What limits the current an alternator will produce? What is the effect of more or less turns? Smaller or bigger wire? etc...

I've been doing a few experiments with a stepper motors I had kicking around driven by my high speed rotary tool. Unloaded I'm getting around 120VAC accross both coils, I put a small 50ohm load on there and it drops right off to around 3VAC.

I know there are a few commerical solutions out there but I'm but a poor student and I'm looking to build this from scrap.

My target is around 50W at 12VDC. Physically small and as light weight as possible. Could I get that much power from a rewound cdrom motor?

Would it be feasable to mount the can from the cdrom motor right on the engions shaft behind the propeller?

I'm guessing that I'm going to need some high speed diodes for the rectifier? Any suggestions...

robe_uk
Apr 30, 2006, 03:36 PM
http://runryder.com/helicopter/t190393p1/

try the above link, based on a heli system but could be used on a plane

Miami Mike
Apr 30, 2006, 08:10 PM
The TAM-5 (http://tam.plannet21.com/) model that flew the Atlantic in August of 2003 used a brushless motor as an alternator. You might be able to find technical details on the web with a search (http://www.google.com/search?sourceid=navclient-ff&ie=UTF-8&q=tam-5).

vintage1
May 01, 2006, 05:35 AM
Any efficient brushless motor will be an efficient alternator. Use a 6 diode bridge of Schottky diodes.

andrew b
May 02, 2006, 10:26 AM
Problem is that the faster you spin the alternator then the higher the frequency of the output AC, starting at quite low frequencies you start to incur losses, I believe above 400Hz, maybe someone will input some more to this fascinating thread.

Garak
May 03, 2006, 09:44 AM
Any efficient brushless motor will be an efficient alternator. Use a 6 diode bridge of Schottky diodes.

I'm well aware, what I'm interested in is what limits the current and the overall power output.


Problem is that the faster you spin the alternator then the higher the frequency of the output AC, starting at quite low frequencies you start to incur losses, I believe above 400Hz, maybe someone will input some more to this fascinating thread.

I'm thinking the limiting factor is going to be the speed of the diodes. I know I've had problems using regular power diodes in PWM motor control circuits running at a few kHz.

Transmission losses are going to be very low because its still not that high of a frequency and the distance by wire from the coils to the diodes is going to be pretty short.

What I really need at this point is X number of turns, of Y gauge wire on Z number of poles can produce A watts at B RPM... I just need some numbers to use to guesstimate a place to start.

vintage1
May 03, 2006, 09:59 AM
I'm well aware, what I'm interested in is what limits the current and the overall power output.


Power output is what the power in is, less a bit of efficiency :D

what limits the current, is nothing much. If you don't put a load on that is self limiting, it will simply burn out the diodes, the motor or both..


I'm thinking the limiting factor is going to be the speed of the diodes. I know I've had problems using regular power diodes in PWM motor control circuits running at a few kHz.



Schottky diodes.
They will function at several tens of kilohertz.

Frequencies - a 10 pole motor at 10KRPM is about 100K/60 = 16Khz.

Well within Schottky operating parameters.

T

What I really need at this point is X number of turns, of Y gauge wire on Z number of poles can produce A watts at B RPM... I just need some numbers to use to guesstimate a place to start.

Any eklectric mnotor will have a KV..that works the same way backwards spin it at a given RPM, divide by the KV and thats the output voltage.


As far as current rating goes, its I squared R more or less. Calculate from total length of wire and cross sectional area..efficiency will be similar for motoir or generator both..so expect aroudn 70%, and the wasted power will be generator heat.
.

ray foley
Jul 26, 2006, 11:08 PM
There is or was a product called "genesys" which involved a disc with magnets being attached to an engine behind the prop, and a magnetic pickup of some sort that produced current pulses as the magnets passed by. There was also a filter and rectifier board and some voltage regulation. I don't recall the manufacturer for sure.

ciao -rjf

rev: as noted below, Sullivan is the mfgr of the Genesys system. Thanks P4. -r

Zeta Phoenix
Jul 27, 2006, 01:17 AM
The Genesys was made by sulivan if i rember correctly.

Yep, here you go: http://www.sullivanproducts.com/GenesysContent.htm

-Jon

Lakc
Jul 27, 2006, 01:11 PM
Getting back to your origional question, an automobile alternator (we have to call them generators again btw) uses an electromagnet in place of permenant magnets. Simply varying the current will change the output. However, as it has brushes it becomes noisy (electrically) and thats not a good thing in a radio controlled model, but not a total showstopper either.
If you use permenant magnet brushless motors you loose the easy controllability of the field strength, thus the power output.

vintage1
Jul 29, 2006, 08:25 AM
No need for a alternator type thing..the CDROM stuff will do what you need if you can turn it fast enough. 50W is very doable from a 2-oz bell motor.

Trouble is, without some battery, the thing might not work if the engine cuts. Or at idle..

Chris Moon
Aug 15, 2006, 02:07 PM
Isn't it true that you can not gain more than you put into something? I mean, with loss, you would never be able to sustain power, correct? I am thinking that every cycle will have a loss due to inefficiency, therefore making the power output decrease exponentially.

GlowFly
Aug 15, 2006, 03:37 PM
I think you misunderstand. I think the OP is planning to attach an alternator onto a fuel engine.
--
Steve

Chris Moon
Aug 15, 2006, 07:19 PM
Ahhh... gotcha. sorry for the confusion. Interesing thread anyhow.


Chris