|22nd November 2008, 08:11 PM||#1|
Join Date: Sep 2008
Understanding Simple SE design choices
Hi, so I was reading this thing:
I'm trying to figure out how to design an amp myself, so I started following along with the reasoning presented.
So, I graphed myself a load line using a 5k output transformer and 450 volts total B+. I marked the Y axis point at 90 mA (450/5k). I marked the X axis point at 450 volts. Cool. This should be the line on which my tube can operate along, right?
So, say I'm aiming for 30 watts of dissipation. I drew the line for that, representing all the points that would cause 30 watts of dissipation. Alright. Now, for me to find a place where I can dissipate 30 watts, I need to look for an intersection with my load line, right? This doesn't seem possible with this loadline.
Here's my graph. The red line is a 30 watt line. Purple line is 28 watts. Yellow line is 10 watts. It seems to me like the only operating points I'll get will be near 10 watts...
I must be doing this wrong. Help!
|22nd November 2008, 08:28 PM||#2|
Join Date: Jan 2005
Re: Understanding Simple SE design choices
Your loadline is a resistor load load, not a transformer load. Transformers do not drop so many volts.
What you need are the tube curves. Say, for instance, that there is a -20V curve that hits 280V at 110mA. Use that as a starting point, bias the grid to -20V, apply a B+ of 280V, ground the cathode. Your load line will be parallel to the one you drew, but it will run through that point. Then you will be dissipating 30W.
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