Re: Simple 555 PWM - disappointing performance



Terry Given wrote:
> Terry Pinnell wrote:
> > "Herbert Blenner" <a1eah71@xxxxxxx> wrote:
> >
> >
> >>Replace Q1 with a PNP transistor in common collection configuration.
> >>Now this circuit can take advantage of the sizable current sink
> >>capability of the 555 timer. Further since you are now switching
> >>voltage across the motor, D1 is not needed.
> >
> >
> > Thanks, Herbert, but according to the LM555 data***, a 555 "output
> > can source or sink 200 mA."
> >
> >>>--
> >>>Terry Pinnell
> >>>Hobbyist, West Sussex, UK
>
> and besides, he is completely wrong about the "dont need the diode",
> which of course you still do. The diode is placed to ensure the motor
> current has somewhere to commutate to when the switch turns off. for a
> pnp switch, the pnp collector goes to the motor and the cathode of the
> diode; the other end of the motor and the diode anode are at zero volts.
>
> advice like that you can live without.

Connecting the emitter to the motor places the pnp transistor in a
common collector configuration. Hence the transistor approximates a
voltage switch. The current however will slew between on and off due to
the inductance of the motor.

>
> BTW, the point Ban made is quite correct. The motor has some inductance,
> Lmotor. This is what "forces" the current to keep flowing when the
> 2n3055 switches off. When the switch is on, current ramps up, current
> slope = dI/dt = Vbattery/Lmotor. Eventually this current gets so high
> the voltage drop across the winding resistance gets in the way (eg in a
> relay, Rwinding sets Irelay). When the transistor switches off, the
> voltage on the winding reverses (Lenz' law) forward biasing the diode
> to, say, 1V. The inductor current now ramps down, dI/dt = Vdiode/Lmotor

In common emitter configuration the 2N3055 approximates a current
switch. Now switching the current produces voltage spikes and
necessitates the diode.

Herbert

>
> Because Vdiode is small (1V cf 15V) the current ramps down slowly. This
> will limit the performance you get from the machine. A simple fix for a
> low power single-ended circuit is to add a zener in series with the
> diode (or a few more diodes), but beware the losses!
>
> Cheers
> Terry

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