I found a possible error in a classic book. I have a printed version of the Seely's "Electron Tube Circuits", and searching for cathode follower data, and in the problem section of one of the chapters I found a mistake what is observed in the rectangle. Unless the tube is driven the grid positive, and taking into account that RC must be a time constant, not a simple grid stopper, C must be connected to the grid, say, between R and the grid. Any further idea? Am I wrong?
Thanks guy. I don't know why it hadn't been loaded as a pic so it can be seen directly as an image as you could do.
The problem has two items, imho. One is a error in the drawing. I believe that the resistor labelled R must be wired between grid and input, and the C at the joining of R and grid, because as it appears in the drawing, if the gid isn't driven positive, it simply does nothing. And second, and perhaps the basic idea of the problem presented, is that the cap is between grid and cathode, it appears multiplied by 1-K with K the gain of the stage. As Seely says in the text, K is less than one, and near 0.9. So in practice, C is demultiplied (Perhaps a neologism) and acts as a smaller value than its true real value. Then the stage has no sense.
1) the original image is not shown in the page because it´s a BMP ; should be converted (what PRR did) into a gif/JPG/PNG which is "understood" by HTML
2) don´t think the book is actually *wrong* , there must be further explanation in the text or another image, but in principle an integrator implies some *current* source feeding a capacitor, which creates both phase shift or a time delay (to ways to describe the same phenomenon) , the simplest current source being a resistor between generator and capacitor.
Unless the generator is a current source by itself.
You will have an integrating time constant calculated from R and C
What is confusing in these diagrams is that the "R" shown is NOT the "integrating R".
*Maybe* this is a trick question, Author expects Students to use the visible R as the integrating one ... which it is not, OR use raw C capacitance value in the second example ... while its effective value is much higher, because it´s bootstrapped.
And if cathode follower gain is, say, 0.9 ; capacitor is not reduced to 0.9 of its raw value but divided by (1-0.9) , so about 10X larger
We would actually need the full page.
2) don´t think the book is actually *wrong* , there must be further explanation in the text or another image, but in principle an integrator implies some *current* source feeding a capacitor, which creates both phase shift or a time delay (to ways to describe the same phenomenon) , the simplest current source being a resistor between generator and capacitor.
Unless the generator is a current source by itself.
You will have an integrating time constant calculated from R and C
What is confusing in these diagrams is that the "R" shown is NOT the "integrating R".
*Maybe* this is a trick question, Author expects Students to use the visible R as the integrating one ... which it is not, OR use raw C capacitance value in the second example ... while its effective value is much higher, because it´s bootstrapped.
And if cathode follower gain is, say, 0.9 ; capacitor is not reduced to 0.9 of its raw value but divided by (1-0.9) , so about 10X larger
We would actually need the full page.
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OK. I am fan of DOS, still using the 7.10 Version.
2) don´t think the book is actually *wrong* , there must be further explanation in the text or another image, but in principle an integrator implies some *current* source feeding a capacitor, which creates both phase shift or a time delay (to ways to describe the same phenomenon) , the simplest current source being a resistor between generator and capacitor.
Unless the generator is a current source by itself.
You will have an integrating time constant calculated from R and C
What is confusing in these diagrams is that the "R" shown is NOT the "integrating R".
*Maybe* this is a trick question, Author expects Students to use the visible R as the integrating one ... which it is not, OR use raw C capacitance value in the second example ... while its effective value is much higher, because it´s bootstrapped.
And if cathode follower gain is, say, 0.9 ; capacitor is not reduced to 0.9 of its raw value but divided by (1-0.9) , so about 10X larger
We would actually need the full page.
No, dear. The cap is between cathode and grid, and is demultiplied. It would be enlarged if it is between plate and grid, adding to the Miller cacpacitance, also called capacitance multiplier, like Phantastron and Sanatron ramp generators. If K is .9, then the cap seen in the circuit as it is presented in ten times less (0.1 * C).
Yes, I got three problems/schematics. But 8-8 was checked and boxed, so I (stupidly?) assumed it was the question.
MS-BMP is a VERY "fat" uncompressed image file, and hardly ever used on the Web. There is no reason browsers "can't" support BMP. I believe I have seen it happen. However GIF JPEG and PNG (all of these are compressed files) are much-much more common on the Internet. The *specific* problem here may be a forum-software prejudice against BMP. It "accepts" it but does not presume the reader's browsers can render it, so only offers to download (which my browser does by opening in a MS viewer).
Correct, if the bias voltage between cathode to grid is still negative. The 6N7 is unique in that it requires positive grid voltage to drive with cathode at zero volts. Regards.
Still with negative bias above -1V or grid positive respect to cathode, will draw grid current. In this way, the grid will act as a small plate collectin' electrons. If in this condition, the external circuit permits the plate and grid power dissipation be in their limits, the tube can perform as well as with negative bias, given the previous stage can support the grid current at low impedance.
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