THD prediction

Boris_The_Blade · 16th October 2007, 07:55 PM

I have been wondering this for a little while....

For a single-ended, resistance loaded amplifier, is there a simple equation that will estimate the THD produced by knowing:

that ratio of the plate voltage excursion in one direction vs the opposite.

i.e. for a grid input of +1 and -1, the plate voltage may 'reach' from vp-8 to vp+10, respectively, where vp is the DC plate voltage. This is probably easiest to picture by thinking of pentode plate curves where operation may span unequally spaced grid voltage lines.

I can see this being applied to other active devices as well by just knowing a loadline..i.e. a JFET.

I suppose that, if no one knows, i may be able to come up with something with MATLAB, but i wanted to check first.

Mark Kelly · 16th October 2007, 10:04 PM

According to RDH4 the % 2HD is given by 100 x EQ - QF / 2(EQ + QF) where Q is the quiescent point, E is the point of highest current and F the point of lowest current.

In your example EQ = 10, QF = 8 so 2HD = 100 x 2 / 2 x 18 or 5.5%

Boris_The_Blade · 17th October 2007, 01:55 PM

exactly what i was looking for, thanks

16th October 2007, 07:55 PM	#1
Boris_The_Blade diyAudio Member Join Date: Mar 2007	THD prediction I have been wondering this for a little while.... For a single-ended, resistance loaded amplifier, is there a simple equation that will estimate the THD produced by knowing: that ratio of the plate voltage excursion in one direction vs the opposite. i.e. for a grid input of +1 and -1, the plate voltage may 'reach' from vp-8 to vp+10, respectively, where vp is the DC plate voltage. This is probably easiest to picture by thinking of pentode plate curves where operation may span unequally spaced grid voltage lines. I can see this being applied to other active devices as well by just knowing a loadline..i.e. a JFET. I suppose that, if no one knows, i may be able to come up with something with MATLAB, but i wanted to check first.

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16th October 2007, 10:04 PM	#2
Mark Kelly diyAudio Member Join Date: Feb 2006 Location: Geelong	According to RDH4 the % 2HD is given by 100 x EQ - QF / 2(EQ + QF) where Q is the quiescent point, E is the point of highest current and F the point of lowest current. In your example EQ = 10, QF = 8 so 2HD = 100 x 2 / 2 x 18 or 5.5%

17th October 2007, 01:55 PM	#3
Boris_The_Blade diyAudio Member Join Date: Mar 2007	exactly what i was looking for, thanks

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