sorenj07 said:
hopefully "this" isn't any of the circuits i've recommended so fari should remove the pic links to not tempt people
By the way, if to modify your circuit a little bit, it may sound like a violin. You may use it's input for a vibrato control.
mm. Here's another shot in the dark before I actually learn what I'm doing. I'm so busy nowadays 
. Anyway this is supposed to drive a pair of 6P14P's. Gain is .94 but most CD players can already drive an amp like this to full power so I'm not so worried. I've got 10 of those Russian tubes coming so I'll probably make a new thread as I try to design (ouch) a working layout for it.
. Anyway this is supposed to drive a pair of 6P14P's. Gain is .94 but most CD players can already drive an amp like this to full power so I'm not so worried. I've got 10 of those Russian tubes coming so I'll probably make a new thread as I try to design (ouch) a working layout for it. An externally hosted image should be here but it was not working when we last tested it.
sorenj07 said:mm. Here's another shot in the dark before I actually learn what I'm doing. I'm so busy nowadays
An externally hosted image should be here but it was not working when we last tested it.
Hmm... What current will go through the tube with 9K in cathode?
What will be the voltage on it? Is it enough for the swing?
An externally hosted image should be here but it was not working when we last tested it.
sorenj07 said:
Don't you want to solder and try it this way? 😉
If you'll discover that the current is not enough, or the swing is too low to drive 6P14P, you always may solder one more resistor in series with catode load, and connect a grid resistor between them. 😉
You really, really REALLY need to do your homework, starting with being able to interpret plate curves and drawing a load line. We already had a discussion of this sort over in the solid state forum, with a kid that was trying to learn basic things by using a simulator. NEVER forget that a simulator is very much a garbage in, garbage out type of thing. Keeping this in mind, you will never learn how things work from it, unless you already know the basics beforehand.
There is an obvious error in your circuit. Sanity check: if plate current = cathode current, and this is supposedly 5mA, what would the grid voltage be, referred to the cathode?
Well, since grid is at zero, and cathode is at supposedly 5mA * 9000 ohm = 45V, this means grid is -45V relative to cathode.
Also, since B+ is 250V, and 45V is burned off on the plate resistor, Vkp is 250 - 45 - 45 = 160V. Supposedly.
Now, look at the plate curves. Look for a curve for -45V grid... there isn't one. If there was one, it would intersect 5mA at Vkp well over 1000V, so obviously this circuit fails the 'sanity check'.
Here is a simplified calculation to help you to do what you are trying to do. First, you need to split your cathode resistor into two. Their sum should still be 9k. Now, Find the point 160V, 5mA (this is your actual Vkp and Ip chosen as a DC working point) on the plate curves, and see which grid voltage you need for it. It comes out as approximately -1.6V. In order to drop that voltage at 5mA, you need a resistor of 320 ohms. Therefore, you need to split your cathode resistor of 9k into 320 ohms and 8.68k. The 320 ohms goes closer to the cathode and you need a grid leak resistor from the grid, to the junction of the two resistors, say 1M. Now, because your grid is at 45 - 1.6V = 43.4V DC positive, you also need an input coupling capacitor.
There is an obvious error in your circuit. Sanity check: if plate current = cathode current, and this is supposedly 5mA, what would the grid voltage be, referred to the cathode?
Well, since grid is at zero, and cathode is at supposedly 5mA * 9000 ohm = 45V, this means grid is -45V relative to cathode.
Also, since B+ is 250V, and 45V is burned off on the plate resistor, Vkp is 250 - 45 - 45 = 160V. Supposedly.
Now, look at the plate curves. Look for a curve for -45V grid... there isn't one. If there was one, it would intersect 5mA at Vkp well over 1000V, so obviously this circuit fails the 'sanity check'.
Here is a simplified calculation to help you to do what you are trying to do. First, you need to split your cathode resistor into two. Their sum should still be 9k. Now, Find the point 160V, 5mA (this is your actual Vkp and Ip chosen as a DC working point) on the plate curves, and see which grid voltage you need for it. It comes out as approximately -1.6V. In order to drop that voltage at 5mA, you need a resistor of 320 ohms. Therefore, you need to split your cathode resistor of 9k into 320 ohms and 8.68k. The 320 ohms goes closer to the cathode and you need a grid leak resistor from the grid, to the junction of the two resistors, say 1M. Now, because your grid is at 45 - 1.6V = 43.4V DC positive, you also need an input coupling capacitor.
to sorenj:
go there : http://teodorom.atspace.com
there is numerous site like this on net,and this is just first one I grabbed from my bookmarks.
read,learn basics,and only THEN simulator will be tool,not a master for you 😉
go there : http://teodorom.atspace.com
there is numerous site like this on net,and this is just first one I grabbed from my bookmarks.
read,learn basics,and only THEN simulator will be tool,not a master for you 😉
Warning: the first of the English links that I tried resulted in an ad for atspace that wouldn't go away until I closed the window. The Italian links seem OK but I can only make out bits and pieces (the little Italian I have is all food words).
Wot - no spaghetti?
.
No ads or pop-ups for me, even with the blocker turned off.
Hey SY, are you still using a windoze computer? It's probably infected..
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No ads or pop-ups for me, even with the blocker turned off.
Hey SY, are you still using a windoze computer? It's probably infected..
Ah yes, I'll give you that one, but no problems with it. I can click back OK.
It just seems to be a broken link that defaults to the host.
It just seems to be a broken link that defaults to the host.
edit: the light at the end of the tunnel is approaching! (my friend helped me order Valve Amplifiers, 3rd Edition!!)
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