All fairly sensitive to grid treatment and gas content, so the operarting point is not well-defined, but with proper choice of RP is is real sure to work for small signals. It is fairly stable because if plate-cathode current increases, grid-bump current increases, grid goes more negative, plate-cathode current decreases. This is a useful bias and operating point for *small* signals. 10Meg seems to give 0.5V of grid voltage. If the tube tends to suck more, plate voltage drops, it sucks less. See Radiotron Designer's Handbook, 4th Edition, section 2.2 (snip attached).Īlso notice that the plate resistor has a major effect on operating bias. A 0.050V uncertainty on a 1-2V bias is not important. What is the grid-cathode voltage needed for normal operation? Often 1V or 2V. Put that in 1Meg, we have 50mV or 0.050V. Most meters will leak as much as a grid.īut for a small triode at small plate current, pencil '50 nA'. It will normally be "high" when you take it out of the box and decline over a few days of hot operation. It varies between different tubes, batches, brands, and vintages. And I disagree we can casually measure it and get a full picture. Therefore we can use a "large" resistor, which does not load-down most sources, and still have the grid voltage "zero" for practical purpose. Grid current is normally much smaller than any other current in an amplifier. I wonder if the electron buildup at the grid, when open or grid-leak is too high, essentially forms another electron cloud which affects/reduces the field between the cathode electron cloud and positive plate. ![]() An electric field is then created from the electron cloud to the positively charged plate. My question is, does the charge buildup at the grid actually flow through the grid resistor making the grid more negative or do the electrons get stuck there at the grid and effectively reduce the electric field the electrons heading toward the plate are traveling through.īase on physics, a space charge electron cloud is created off the cathode from due to the heating of the cathode by the filament. I'm guessing it makes the grid more negative as well. The electrons at the grid repel electrons coming off the cathode. From what I understand, charge buildup occurs on the grid. The slight bias is due to the electrons heading toward the plate and smashing into the grid along the way, correct? Īlso where do you find the datasheets for Mullard tubes? They appear to be owned by New Sensor Corporation (Mike Matthews of EHX) now.Īnother interesting point is if the grid-leak resistor is left off or too high. This would bias the tube only slightly negative, correct? Effectively near the saturation and clipping region (Eg = 0) of the tube from my understanding. The cathode was tied to ground and the grid had a 5 Meg resistor to ground. On an early model of the Fender Deluxe, the first tube (6SC7) used a grid-leak bias. ![]() Is this the same principle for the grid-leak bias? If this is true, the grid to cathode bias voltage is technically the cathode voltage (Ic*Rc) minus the small negative voltage at the grid (produced by electrons hitting the grid and the resulting current flow to ground through the grid-leak resistor). So when a tube is cathode biased, the grid voltage is slightly below ground (due to electrons smashing into the grid and the small grid-leak voltage) and a number of volts below the cathode (due to the current through the cathode resistor), correct? Thanks for the replies, this is starting to make sense.
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |