Re: [CAD_CAM_EDM_DRO] Re: PWM Back Emf
Posted by
Jim Register
on 2007-01-13 22:06:08 UTC
Frank,
The current sense resistor lets you get at the back emf of the motor
while power is applied, rather than when the motor is coasting.
First, model the motor as a generator (back emf) in series with a
resistor (the motor's internal resistance). Now, add the current sense
resistor from the motor to ground. We'll call the junction between the
motor and the current sense resistor "A".
The voltage at point "A" = (applied voltage - back emf) x voltage
divider ratio.
To cancel out the applied voltage, add a second voltage divider, with
the same ratio, connected from the applied voltage to ground. We'll
call the output of this voltage divider "B". If you use an opamp and
subtract voltage "A" from "B" with a gain equal to the voltage divider
ratio, the output is back emf, all by itself. You will want a trimpot
in this second voltage divider so you can set the ratio precisely. I
just applied a small voltage to the motor while holding the shaft from
rotating and adjusted the trimpot for 0 volts out of the opamp.
The times I have done this, the drive amps were analog, not PWM - I have
not dealt with sampling the back emf signal at a particular time. I've
only done this with small motors - 1 amp drive or less.
Hope this helps,
Jim
The current sense resistor lets you get at the back emf of the motor
while power is applied, rather than when the motor is coasting.
First, model the motor as a generator (back emf) in series with a
resistor (the motor's internal resistance). Now, add the current sense
resistor from the motor to ground. We'll call the junction between the
motor and the current sense resistor "A".
The voltage at point "A" = (applied voltage - back emf) x voltage
divider ratio.
To cancel out the applied voltage, add a second voltage divider, with
the same ratio, connected from the applied voltage to ground. We'll
call the output of this voltage divider "B". If you use an opamp and
subtract voltage "A" from "B" with a gain equal to the voltage divider
ratio, the output is back emf, all by itself. You will want a trimpot
in this second voltage divider so you can set the ratio precisely. I
just applied a small voltage to the motor while holding the shaft from
rotating and adjusted the trimpot for 0 volts out of the opamp.
The times I have done this, the drive amps were analog, not PWM - I have
not dealt with sampling the back emf signal at a particular time. I've
only done this with small motors - 1 amp drive or less.
Hope this helps,
Jim
On Sun, 2007-01-14 at 03:47 +0000, Frank wrote:
> That doesn't really help me understand. It sounds like if I try to
> modify the speed by changing the PWM, it won't; it will return to
> preset speed. I think I am going to have to put a circuit together
> and try and understand it.
>
> Thanks for trying to explain it.
>
>
>
> --- In CAD_CAM_EDM_DRO@yahoogroups.com, "Mariss Freimanis"
> <mariss92705@...> wrote:
> >
> > That's where the trimpot comes in. Too much and the motor speeds up
> > with increasing load. Too little and it slows down. Just right and
> the
> > speed doesn't change at all. Kind of like Goldilocks, the 3 bears
> and
> > the beds, the chairs, the porridge or whatever.:-)
> >
> > Mariss
> >
Discussion Thread
Frank
2007-01-13 02:12:41 UTC
PWM Back Emf
Mariss Freimanis
2007-01-13 07:35:38 UTC
Re: PWM Back Emf
Dennis Schmitz
2007-01-13 14:43:29 UTC
Re: [CAD_CAM_EDM_DRO] PWM Back Emf
Dennis Schmitz
2007-01-13 14:49:13 UTC
Re: [CAD_CAM_EDM_DRO] Re: PWM Back Emf
Frank
2007-01-13 15:04:35 UTC
Re: PWM Back Emf
Mariss Freimanis
2007-01-13 16:08:37 UTC
Re: PWM Back Emf
Frank
2007-01-13 17:54:19 UTC
Re: PWM Back Emf
Mariss Freimanis
2007-01-13 18:58:06 UTC
Re: PWM Back Emf
Frank
2007-01-13 19:48:15 UTC
Re: PWM Back Emf
Jim Register
2007-01-13 22:06:08 UTC
Re: [CAD_CAM_EDM_DRO] Re: PWM Back Emf
Frank
2007-01-13 23:50:32 UTC
Re: PWM Back Emf
Alan Marconett
2007-01-14 11:24:32 UTC
Re: [CAD_CAM_EDM_DRO] Re: PWM Back Emf