Re: 74LS05 driving Relay

From: Fritz Schlunder (me_at_privacy.net)
Date: 01/28/05 

Date: Thu, 27 Jan 2005 21:46:50 -0700

"Bradley1234" <someone@yahoo.com> wrote in message
news:iO9Kd.1089$lg5.828@trnddc06...

> Okay, so putting the damper across the relay is better? I stand
corrected.
> I dont read hobbiest schematics on the internet.

Yes much better.

> > In a TV the damper diode is needed to provide a conduction path for the
> yoke
> > coil current which is kind of in parallel with the horizontal output
> > transistor. In your standard transistor-drives-relay-coil circuit, the
> > relay coil (inductive device) is in series with the transistor. This is
> an
> > important topological difference.
> >
>
> kind of in parallel? I thought it was one of 2 yokes and the flyback, in
> series and thats no yoke.

I'm not an expert on TVs, but I'll refer to the schematics of figure one and
two of this document:

http://www.alphacron.de/download/miscellaneous/horiz-tv.pdf

They show the yoke coil being in series with a capacitor which together is
in parallel with the horizontal output transistor (HOT).

> But to really stop the spikes to make "no voltage" wont it also require a
> capacitive and resistive load?

For the relay coil, no. Inductors obey E=L*dI/dt

Where E is the voltage generated (or applied to the inductor), L is the
inductance in Henries, and dI/dt is the rate of change of current in the
inductor measured in amps per second. Current never changes instantaneously
in an inductor. This means current doesn't reverse direction
instantaneously either. It must first slowly ramp down to zero, and then
change direction slowly ramping back up to a higher value for AC
applications. The voltage measured accross the inductor can instantly
change, and does instantly change at the moment of transistor turn off, but
the current keeps flowing in the same direction.

With the diode antiparallel to the relay coil, and the transistor turns off
(pretty much instantaneously), then a new current path is needed. The diode
provides this current path. The relay coil thus produces an external
voltage of only one diode drop or about 0.7V. Hardly enough transient to
even call it a spike at all.

> And when an arc takes place across a relay contact? Doesnt that mean
> thousands of volts?

If the relay contacts were fully open, then yeah it would likely take
incredibly high voltage. If the relay contacts have just closed, but
"bounce" open again slightly, then the air gap will be exceedingly small.
It doesn't take much voltage to breakdown this tiny air gap. If you closely
study a little two AA cell flashlight (say using 500mA or more lamp) switch
in action, you should be able to observe tiny electric arcs (assuming you do
this at night, turn off all the lights and cover the lightbulb with
something very dark) as the contacts open and close. The voltage is very
low at 3V or less, but the gap is small. The effect is much more pronounced
and much more observable if you use higher voltages and higher currents (say
12V at 5A). The load need not be inductive.

If the relay contacts are fully closed, and an inductive load is being
powered, then at the moment of contact opening an arc will likely form
(though this assumes the inductive load doesn't have an antiparallel diode
across it for DC applications). Again the voltage produced when the arc is
first initiated is very small since the gap is very small. As the relay
contacts get farther and farther from each other a higher and higher voltage
is needed to sustain the arc. Eventually the inductive energy will all be
dissipated in the arc and parasitic coil/wiring resistance and the arc will
extinguish.

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