On Sat, Oct 23, 2010 at 4:22 AM, Oliver Seitz <[email protected]> wrote:

>
> > The constant current leads to a constant voltage at the
> > external resistor. In fact you're subtracting a known
> > voltage from the EMF voltage you want to measure.
>
> About like this:
> http://johannes.eventify.de/kiste-temp/Voltage_shifter.jpg


That's the same thing I described except the 100K in series with the motor
(which is the V+ source in this case).

 Which is the output impedance on "to analog input" line? I think 100K...

Replace the BC547 with a 500mA darlinghton, 0-2.5V with a variable reference
voltage from PIC, and 10K resistor from the emiter with a smaller resistor,
100K with 0 ohm. Forget about the Hz to GND line since you're able to switch
off the reference, but add 10K from Vref to ground to avoid false current
generation if Vref becomes by accident an input.  If the brake time  is less
than 50-100uS, probably that will not alter too much the speed, nor the
movement since there is a movement conservation (inertia). When the motor is
generator, any load will decrease the EMF voltage in the same way any
mechanical load on a motor is increasing the current sourced (and decrease
the power supply voltage if that hasn't enough power). BTW, I'm wandering
why nobody filter than EMF noise.

Seb, I think you have two software programmable solutions, choose one and
experiment.

 Vasile

>
>
> Output voltage is about Vmotor-((control voltage/10kOhm)*100kOhm)
>
> Output voltage is always higher than control voltage. Rise the control
> voltage only to a level where output voltage stays about 1V above control
> voltage.
>
> Greets,
> Kiste
>
>
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