On Sun, 05 Oct 2008 23:30:36 -0700, Kevin McMurtrie
put finger to keyboard and composed:
In article ,
"Richard Crowley" wrote:
"EADGBE" wrote ...
The anode of the diode is attached directly to three of the resistors
as shown.
I have tried to re-create the circuit board traces on the other side.
The cathode connects to what N_Cook has described as an inductor.
Take a look at this photo:
http://i240.photobucket.com/albums/f...e123/DIODE.jpg
Thanks to everyone for your help so far.
Any chance of finding a replacement for this diode?
Depending on what the actual diagram looks like,
it is quite possibly a simple, generic silicon diode
like a 1N4001 (power rectifier) or maybe even a
popular signal diode like a 1N4148, etc. These
are both very common parts that likely cost only
a few cents in production quantities.
I believe it takes at least two transistors. It's a positive feedback
system to compensate for losses in the motor, creating a virtual perfect
motor that doesn't change speeds under load. And that brings about the
second problem...
The circuit has only one pot so it's not going to work on a worn motor.
The loss compensation is a fixed resistor so it's only good for a motor
in perfect condition.
My understanding is that a real motor can be represented as a perfect
motor with a small series resistance. To maintain a constant speed
under all load conditions, one needs to add an external series
resistor of equal value and then add the voltage drop across this
resistor to the motor voltage.
o---- Rloss ----- perfect motor -----o o--- Rext ---o
|--- voltage at motor terminals ---| |-- loss --|
Here is a circuit I found in "303 Circuits" by Micro-Tech/Elektor
Electronics.
|-------------------------|
| |
| |\ |
| | \ |
|---|- \ | Vm - Im
| \ -----|--- R3 ---|----o o--- MOTOR ---o GND
V1 o--- R1 ---|---|+ / |
| | / |
| |/ |
| |
|------ R2 ----|
Vm = V1 + Im x [(R2/R1) x R3]
Effectively you have a voltage source with a negative output
resistance of R3 x (R2/R1).
- Franc Zabkar
--
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