Alt-BEAM Archive
Message #03276
To: "'Steven Bolt'" sbolt@xs4all.nl,
From: Wilf Rigter Wilf.Rigter@powertech.bc.ca
Date: Fri, 14 May 1999 20:23:48 -0700
Subject: [alt-beam] Re: My Pagerbot
Hello Steven,
SEs are not my specialty but I have an affinity to all simple circuits which
I can "deeply" understand. Those are circuits for which I have a reasonable
complete model in my head on which I can make reasonably accurate
predictions about the behaviour of the circuit based on the model and first
principles. There are always counter intuitive surprises ( Now that's
strange!) which I treasure. So I would be quite interested if you can shed
some light on the difference in experimental results and what your comments
are with regard to isolating the feedback paths for base resistor.
As I mentioned in my email, I tested the solution I posted, with a diode and
a real motor and using an oscilloscope, I observed the switchoff voltage of
1.5V which I define simply as the voltage that remains on the capacitor at
the end of the discharge and the starting voltage for the next charge cycle.
I don't understand the need for a 1381 voltage detector to make this
measurement, a voltmeter should probably be sufficient.
I indicated in my email and schematic that my circuit uses a 1K base
resistor instead of the conventional 2.2K in part to compensate for the
lower voltage drop across that resistor and to improve Vsat. If you look
more closely at my schematic you will notice that I separated the FLED load
resistor and feedback resistor functions. That was the improvement I
referred to. At the 2N3906 base, a 10K (or larger resistor) base emitter
resistor shunts less current away from the base than the original 2.2K and
should make the circuit more sensitive and less likely to "hang". Moreover,
simply adding the 1N4148 in series with the base of the 2N3906 as you
suggested would appear sub-optimal since it requires an even higher turn-on
current ie the voltage drop across the 2.2K resistor (and shunt current) at
turn-on must be greater (Vbe+Vf).
In the original FLED SE design simply changing the 2.2K resistor to 10K
would reduce turn-on current but also reduce available base currents and for
larger motors this would cause the 2N3904 to come out of saturation.
Adding the diode at the NPN collector and isolating feedback paths by using
a separate FLED larger load resistor for low turn-on current and a separate
smaller feedback resistor for low Vsat during would appear to give the best
of both possible worlds.
regards
Wilf Rigter mailto:wilf.rigter@powertech.bc.ca
tel: (604)590-7493
fax: (604)590-3411
> -----Original Message-----
> From: Steven Bolt [SMTP:sbolt@xs4all.nl]
> Sent: Friday, May 14, 1999 7:48 AM
> To: Wilf Rigter
> Cc: beam@corp.sgi.com
> Subject: RE: My Pagerbot
>
> On Thu, 13 May 1999, Wilf Rigter wrote:
>
> > I have tested the XFLED idea and the theoretical dropout voltages are
> (note
> > Steve Bolt) spot on.
>
> It takes about 1mA before you see a Vf of 0.6V across a 1N4148, and
> 5mA before Vf rises to 0.7V. Since we are talking about the current
> at the point of switch-*off*, I submit that your 3906 (which is not
> the driver) is getting excessive base current. It seems safe to
> predict that a LED in that position will actually flash :)
>
> I tested the solution I posted:
>
> If you want to stay with a switch-off level (instead of a
> switch-off time), increasing that level is probably the most
> efficient solution. You might try a 1N4148 diode in series with
> the 2K2 resistor (anode towards the 3906 base). That will make
> the SE switch off at about 1V instead of less than 0.8V.
>
> And with a 1381C as voltage detector, I measured switch-on at 2.15V,
> switch-off at 0.78V. Adding the diode increased the switch-off
> level to 1.06V, indicating an appropriate base current.
>
> Best,
>
> Steve
>
> ----------------------------------------------------------------------
> # sbolt@xs4all.nl # Steven Bolt # popular science monthly KIJK #
> ----------------------------------------------------------------------
>
>
>
>
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