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Old 22nd June 2012, 09:55 PM   #11
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Originally Posted by tomchr View Post
Does this make sense?
Yes it does make sense but if I do that the maximum current is about 240+ mA!!
So if the quiscient current is 80 mA the Q-point is not at the middle... right? and if this is right why does it happend? I think that in class A amplifiers the Q-point must be at the middle..
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Old 23rd June 2012, 12:28 AM   #12
12E1 is offline 12E1  United Kingdom
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If your bias point is with the current at 80mA then the current can swing plus / minus 80mA maximum (and in reality a bit less than that). So the peak current flow would be just under 160mA.

I think you assumed that the grid would go to 0 volts. To swing the current from 0 to 160mA the grid will not need to go that far - with a 2.5k load line biased for 80mA idle, it would go no more positive than about -30 volts for 160mA current.

If you do take the grid more positive than -30 then you will simply introduce lots of distortion since the bias point would then not be at the middle of the current swing.

Last edited by 12E1; 23rd June 2012 at 12:31 AM.
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Old 23rd June 2012, 09:53 PM   #13
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Originally Posted by 12E1 View Post
If you do take the grid more positive than -30 then you will simply introduce lots of distortion since the bias point would then not be at the middle of the current swing.
Ok I see what you mean.. But how can I be sure that the grid voltage wil not reach 0 or -10 volts?
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Old 26th June 2012, 09:06 PM   #14
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anyone??
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Old 1st July 2012, 07:28 AM   #15
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Originally Posted by greg_e46 View Post
so after reading some articles I have more questions about the schematic.
I found EH 300B datasheet and tried to draw the load line and calculate the bias point. in the picture below I set the plate voltage at 360v (80% of Vplate max). With 880Ω resistor I have about 69volts dropped across cathode resistor and the bias current is ~80 mA. so the max current is 150mA. But if I draw the load line I don't have 2.5kΩ load as the schematic shows. So I suppose I am thinking-doing something wrong... Right?
Not correct. If you drop 69 vols on the cathode resistor, this means that your HV (high voltage) will be 360-69 V, because ths is what the 300B "sees" between anode and cathode. So the center of the load line will be at the intersection of 291 volts with the 69Volts grid curve.
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