A
Arie de Muynck
Guest
On 2020-06-07 18:02, Bob Engelhardt wrote:
That\'s more or less it. Divide that voltage drop by the total DC
resistance in the circuit (assume the mains is zero ohms), and you know
the DC current flowing. Simulate that with a DC supply connected to the
motor, rotate the motor manually, and check how much breaking action
occurs. I\'ve been amazed at it (and used it as a controllable motor
torque load).
Arie
On 6/7/2020 9:23 AM, Arie de Muynck wrote:
On 2020-06-07 14:07, Bob Engelhardt wrote:
On 6/6/2020 3:35 PM, Pimpom wrote:
This should work:
https://www.dropbox.com/s/ablk0cc0iqfhsta/Load%20presence%20indicator.png?dl=0
Diodes D1-D4 must be able to handle the load, including startup
current.
I like it. KISS. A small point: only need one of the D3/D4 pair.
That would produce a small amount of DC in the pump voltage. Shaded
pole motors are very sensitive to that - with some DC you can easily
use them as brake. Better \"Keep It Symmetrical [and] Safer\".
Arie
That\'s because the AC voltage in one direction would be a diode drop
different from the other direction? So the dc voltage would be equal to
the diode drop, or is it more complicated than that?
That\'s more or less it. Divide that voltage drop by the total DC
resistance in the circuit (assume the mains is zero ohms), and you know
the DC current flowing. Simulate that with a DC supply connected to the
motor, rotate the motor manually, and check how much breaking action
occurs. I\'ve been amazed at it (and used it as a controllable motor
torque load).
Arie