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How van these 2 curves shown below map to each other: So where would point B or point A on the IDD vs VDD curve be on the power output vs power input curve?

IDD vs VDD curve

Power output vs Power input in a power amplifier

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1 Answer 1

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The lower curve is the power product of an AC coupled signal. The Psat is the maximum power due the knee or saturation limit defined by a distorted peak of the entire sine waves limited by the "knee" of the upper curve on the left and clipped by Vdd on the right.

Point A is out of reach at max current because there are resistors in the path Rd,Rs to achieve a gain and limit current.

Point B is beyond the sine wave if the input were increased to cutoff current then Vds=Vdd when the JFET resistance is high.

Imagine a small signal input on the top curve (around Q on the load line) This is a just an RMS power point near the origin on the lower linear power curve.

Note: On chart 1 when Vds=2.1V, Id=3.9mA when Vgs = -0.1V peak swing. This implies Rd+Rs = (12V-2.1) / 3.9mA= 2.54 K then Vds/Id=Ron

Now can you figure out gm the inverse or ΔI/ΔV=gm from Vds and Id?

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  • \$\begingroup\$ so if a waveform is large enough to swing beyond the knee on the left hand side and beyond point B on the right hand side it would be mapping somewhere past the P1dB point on the lower curve, is that right? \$\endgroup\$
    – ali khalil
    Dec 20, 2019 at 19:00
  • \$\begingroup\$ Also what does operating in the "ohmic" region mean? Does it mean hitting the knee 'vertical" line on the first plot Ids vs Vds? and what influence does the gate biasing have on the gain or the output power level if the amplifier is operating in the ohmic region? \$\endgroup\$
    – ali khalil
    Dec 20, 2019 at 19:04
  • \$\begingroup\$ The vertical slope in top left is when the the FET is operating like a low ohmic switch. You can't go any more left than this. Gate bias determines symmetry of large swing. Here with a cutoff of around Vgs(t) = -0.85V the optimum Q point is about 1/3 of Vgs(t) = -0.3 \$\endgroup\$ Dec 20, 2019 at 20:02
  • \$\begingroup\$ Would different quiscent points on the same Vgs line, say -0.6 V. represent the same gain (assuming the line is quasi horizontal)? \$\endgroup\$
    – ali khalil
    Dec 20, 2019 at 21:15
  • \$\begingroup\$ To answer your first question a 0.2 V change in gate bias created a change in Ids of 3.9-2.2= 1.6 A so gm = 1.7/0.2= 8.5 , is that right? \$\endgroup\$
    – ali khalil
    Dec 20, 2019 at 21:26

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