Re: Simple 555 PWM - disappointing performance



Terry Given wrote:
> Herbert Blenner wrote:
> >
> > Terry Given wrote:
> >
> >>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?
> >
> >
> > If my memory is correct, the output stage of the 555 has an open
> > collector and grounded emitter. In this case a second resistor from
> > the 4 volt supply to the base of the pnp transistor would ensure prompt
> > turn off.
> > Alternately if the output stage of the 555 has active source and sink
> > stages then you must reconnect the timer to a lower voltage and omit
> > the second resistor.
>
> Incorrect, its always been a totem-pole output stage (which is why 555's
> draw 300mA current spikes on each switching edge, as both totem-pole
> transistors momentarily conduct, hence the need for a hefty bypass cap
> right across the pins of the device). The totem-pole output can sink
> much more than it can source, for much the same disappearing Ibase
> reason as this terrible idea.

Since the output sinks more than it sources, the turnon base current in
a pnp transistor would be higher than with a npn unit.

>
>
> I am not discussing turn off here at all, merely turn on. As the pnp
> turns on, the emitter voltage falls, therefore so does the base,
> therefore so does the voltage across the 555-to-base resistor.

Adjust the resistor to compensate for the change in voltage. The bottom
line is to get the base current up to the proper value.

>
> my simulation shows this clearly, and I use an ideal voltage source (ie
> one that can sink and source many trillions of amps)
>
> >
> >
> >>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.
> >
> >
> > You forget that bipolar transistors are not the only kind.
>
> *you* said "Replace Q1 with a PNP transistor"
>
> please explain to me how I can have a pnp transistor that is *not* bipolar?

After you objected to the higher Vce of the pnp transistor, I reminded
you that not all transistors are bipolar.

>
>
> >
> >
> >>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.
> >
> >
> > You should never allow Vbe of a pnp transistor to go positive. This is
> > especially true for power transistors, whose base-emitter junctions
> > have low reverse breakdown voltages.
> >
>
> read it again. I said "until the base-emitter diode is forward biased"
> which of course means Ve > Vb for a pnp. This is implicit in the "about
> 15.7V" statement immediately thereafter.

Driving the base resistor from a 15 Volt output of a 555 reverse biases
the base-emitter junction of the pnp transistor and causes reverse
breakdown.

>
> although it is possible, depending on stray capacitances and the 555
> dV/dt, to apply a momentarily high reverse voltage across the
> base-emitter junction. yet another reason not to do it this way.
>
> > If the output stage of the 555 has active source and sink stages then
> > connecting the timer supply to four or five volts ensures proper
> > biasing of the transistor.
>
> I should mention that, unless the 555 has output clamp diodes (or a FET
> output stage) the pnp wont work at all, as all base current has to flow
> *into* the 555 output when it is high.

When the output of the 555 is high, the pnp transistor is off and no
base current flows.

> An LMC555 will do that just fine
> (FETs dont care which way the current goes) but I cant recall OTTOMH
> what the vanilla 555 output stage looks like. They probably do have
> clamp diodes, and its probably *not* a good idea to bung more than a few
> mA thru them, lest you trigger the internal SCR which sits across the
> device supply rails (thereby shorting out the 15V supply thru the 555,
> and cooking it quick-smart).
>
> changing the supply voltage to 4V does not change the circuit behaviour,
> only the voltage at which the pnp turns on again (Ve = 4.7V or so), and
> of course the rate-of-change of motor current. Its still a rat***
> circuit, which explains why *nobody* does this.
>
> the magical disappearing base current alone is enough to consign this
> POS to the wastepaper basket, where it truly belongs :)
>
>
> >
> >
> >>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.
> >
> >
> > I suggest doing the circuit analysis. You will find that switching
> > current in an inductance produces voltage spikes. By contrast switching
> > voltage across the inductance gives a continuous change in current with
> > slope discontinuity. Voltage switching in inductive circuits is not an
> > idea, instead it is standard practice in electromechanical controls
> > systems.
> >
> > Herbert
>
> I'd love to see your circuit analysis of this. Admittedly I dont know
> much about pissant little motors (definition: anything a human can lift
> unaided), but my calculus is OK. yes, V=dLambda/dt = LdI/dt (if and only
> if L is constant, otherwise add in an IdL/dt term) so dI/dt = V/L.

The response, i(t), of a series connection of R and L to a voltage
source V u(t) is V / R ( 1 - e ^ ( - R / L t) ).

Now a current source, I u(t), driving the series connection of R and L
develops a voltage v(t) = I R u(t) + L I du(t) / dt = a step plus an
impulse.

>
> all that happens in this case (apart from the appallingly low
> efficiency, due to piss-poor base drive and linear operation) is that
> the pnp transistor functions as the clamp diode, but in a much more
> lossy fashion. For Vcc(555) = 4V, 0.7V is dropped across the motor
> (exactly the same as with D1) *but* the C-E voltage of the pnp
> transistor is 4.7V, so it dissipates 4.7/0.7 = 6.7 times more power than
> D1 would in the same situation.

When Vce is 4.7 V the pnp transistor is off and the current is zero.
During turn on Vce is about 1 volt, which entails three times the power
dissipation of a saturated common emitter transistor. However this
additional dissipation is merely cause to fet.

Herbert

>
>
> >
> >
> >>>
> >>>>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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