Good point Albert, I see what you mean and I just checked it. With a 1KHz signal (1mS) there are about 11.5 cycles when I turn the horizontal position from left to right. So when I "compress" the picture, in simple words, there are two possibilities;
- the length of the visible trace becomes shorter till we have 10 divisions but without changing the individual time length of one cycle. Thus still having the problem. or;
- the total length of the trace is more compressed (it is now about 12.5 div in display length) till there are exact 10 divisions. And also "compressing" the individual cycle time wich is now 20% to much.
I think indeed that the second option is the way to the solution.
Good point from you!
I have just changed the two 2N3906 with two BC556b (Vceo 65V) because they are not adekwate for this voltage, as some of you said.
In the future I search for a permanent replacement that can withstand a higher voltage but until now the BC556b?s doing their work perfect.
Now I have to dig in the schematics and try to understand what the possible cause can be why the sweeplength is so long.
grtz and thanx for all the help, Ren¨¦
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--- In TekScopes@..., "Albert" <aodiversen@...> wrote:
Hi Ren¨¦,
Not sure what you mean with "too long".
If you display 1 kHz signal at 1 ms/div, then
- each period is 1.2 div and there are about 10 periods visible if you rotate Hor Pos to see start and end.
In this case the gain of the Hor Output Amplifier seems too high for some reason.
-- or --
- each period is 1.2 div and there are only about 8-8.5 periods visible, even if you rotate Hor Pos.
This would indicate too fast timing of the sweep generators. This seems not realistic when both A and B suffer from this by such a big percentage.
Albert
The only thing what?s left is the timing from the sweep rates, they are about 20% to long. All of them on both timebases, so if somebody knows...
grtz and thanx so far for all the help, Ren¨¦
