How about a thread how the sawtooth sweep waveform is generated..?
 

In the same vein as Miniman82's thread on chroma processing/demodulation, it'd be cool to have a discussion on -
how the Horizontal and Vertical sawtooth waveforms are actually generated and appear across the yoke windings.

The sawtooth waveform is one of the 'givens' that's pretty much taken for granted, without considering just how sweep circuits generate it.

It's easy to visualize how a neon bulb (or gas-discharge tube) relaxation oscillator creates a sawtooth (like for H sweep in very old 'scopes). But it's a bit more esoteric in TV sets.

Just a thought to toss out there, inspired by Miniman82's excellent interactive.

The problem is that there are many different methods of
generating these things, and as many different ones for syncing them.

Somewhere buried in one of the sites dedicated to early color is one of the best descriptions of the synchroguide circuit I've seen.

Indeed. When my CT100 had a horizontal intermittent I looked
at scads of those, all the URLs now forgotten. That and lots
of testing with a scope never found the problem, the circuit is
too complicated and feedback-y. Nor did doing simulations. The problem
was solved by replacing ALL the resistors in or directly adjacent
to that circuit. One was bad at high temps and/or voltages. Testing
all the old ones with an ohmmeter and heat gun didn't find a bad one.
The caps had already long been replaced, along with two
resistors (replaced again).

I have a pretty good explanation of the synchroguide in one of my manuals, I'll scan it when I get home so we can discuss.

I believe Pete's site says it best, no need to rehash it here so read what's in the link:

http://www.earlytelevision.org/Deksn...horiz_osc.html

Actually what I was tryin' to express in the OP was not about synchronization, but rather would pertain to a free-running vert or horiz sweep amplifier without any sync input to the oscillator.
What design factors combine to produce a nice linear sawtooth wave (as opposed to a sine wave or 'humpy' wave) across the yoke winding (or to the deflection plates in an electrostatic set)?

I gotta confess to having spent a whole 'career' in the trade without having heard a good synopsis on just how the sawtooth wave gets produced. It was always one of those 'givens' that's always taken for granted without question. Sorta analogous to an auto repair techie not pondering the chemical minutiae of fuel/air combustion.

IIRC, (I'm a little hazy on this) the sawtooth is produced by using the inductance of the deflection yoke and transformers. The output tube is turned on hard, and that starts an increasing current into the inductance, and that increases the magnetic field that deflected the electron beam inside the CRT. At some point (middle of the screen IIRC) the tube turns off, but the current in the inductance keeps going, so there's the damper diode that turns on. This pumps the energy from the inductance into a boosted B+, and keeps the magnetic field going.

What's hard to understand about an oscillator? Wa2ise pretty much has it, did you want to see schematic explanations?

wa2ise is explaining, in the post above, what happens to the output
of the horizontal output tube.

The question is the oscillator itself. As I said, there are several different
forms of this. The usual suspects are either based on a transformer, usually
with just one oscillator tube, or a two tube multivibrator circuit. The latter
comes in two basic forms: one has each plate connected to the other grid through
a capacitor, the other has the plate of tube 1 connected to the grid of tube 2 by
a capacitor, and the two cathodes connected together and connected
to ground by a 1K resistor. In the latter case the grid of tube 1 has to have a low
resistance to ground (typically 10k) and the plate of tube 2 is connected to
B+ (or in an electrostatic set, the CRT high voltage) by a high resistance , 1 to 40
megohms, in parallel with a capacitor to ground.

Just to be careful:

Vertical and horizontal are two very different animals.

For horizontal sweep, the oscillator does not generate a linear ramp. It generates basically a rectangular wave to drive the horizontal output hard on and off. The waveform is modified somewhat to get the fastest possible switching speed of the horizontal output device, because the major power dissipation is during the switching time. In the case of an NPN horizontal output transistor, this means starting the off pulse with a strong negative spike to pull stored charge out of the junction. As stated above, it's the yoke / flyback inductance that makes the sawtooth current out of a rectangular voltage.

For vertical sweep, generating a linear waveform is required because the output is working mainly against the resistance of the yoke. Because of the inductance, you will also see a pulse component in the voltage across the yoke, but the vertical output is not specifically a switched-mode stored-energy recovery system like the horizontal.

I would note that some Zenith designs had feedback that forced the yoke current to be a linear sawtooth so that no vertical linearity adjustment was required.

In typical tube designs the vertical oscillator circuit does not generate
a ramp either: it generates something approximating, not terribly well,
a short "on" current pulse and a long "off" period. It is then integrated
at the grid of the vertical output tube to make a sawtooth.

Didnt some of the early electrostatic sets use a gas tube oscillator like some oscilloscopes?

These are all illustrations of the duality of current and voltage. You can integrate a constant current in a capacitor to get a voltage ramp, or integrate a constant voltage in an inductor to get a current ramp.

Most certainly some pre-war 405 line British sets did (thyratrons).
I believe that some early magnetic ones did too.

E. G.

http://www.earlytelevision.org/ekco_...technical.html



Doug McDonald