Re: Simple 555 PWM - disappointing performance

Herbert Blenner wrote: 

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.

thats a fairly piss-poor description, indicative of a deep lack of understanding.


            ---+-- +4V
               |
            [motor]
               |
               |
               |
             |/e
555--[R]-----|    pnp
             |\c
               |
            ---+-- 0V


Dont forget the 555 runs from +15V.

When the 555 o/p pulls low, the motor turns on, but where does the substantial base current required come from? As the emitter pulls down, the voltage across Rbase drops, so Ibase drops, so PNP turns off. Besides, even if we magically required almost no base current (eg darlington, Ibase = 3A/10,000 = 300uA) the emitter will *always* be Vbe above the base, losing a large chunk of the available 4V supply. The reducing base drive "feature" of this POS *forces* the transistor to dissipate far more power than it would if switched properly.

When the 555 o/p goes to +15V, where does the motor current flow to? by Lenz' law, the emitter voltage will rise, until the base-emitter diode is forward biased (IOW about 15.7V) at which point the pnp will turn on, shunting the motor current to 0V. Of course Vce = 15.7V, with 3A across it, ramping down to zero in time t = Lmotor*Ipeak/15.7V. dissipation acros the pnp is thus 23.5W for however long it takes. The motor current ramps down a lot faster than it ramps up, as it has almost 3x the voltage across it.

If D1 is left connected across the motor, this turn-off behaviour does not happen, as motor current commutates thru D1, keeping about 0.7V across the motor (so current downslope is about 5-6 times *slower* than the up-slope)

I did a simple simulation in Simetrix, using an MJD2955 and a 1mH inductor (no winding R), chopped at 100Hz with 15V square-wave base drive. Peak current is controlled *entirely* by Rbase (ergo by Hfe, so all over the show with time, temperature and type of transistor). To get 2.5A, I needed Rb = 50 Ohms, and Vcesat actually pulls right up to the supply rail! The turn-off current commutation behaviour is as described.

In conclusion, this is a *terrible* idea, and will not work.



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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