At the point where the load line intersects the plate voltage
axis, the plate current is zero and there is no voltage drop across
the load resistor, RL . Therefore, the voltage drop across the tube
must equal to the supply voltage, Ebb: 300 volts.
8.
a -- para 50d(2)
= rp x gm = 9,000 ohms x 3,000 micromhos
= 9,000 ohms x .003 mhos
= 27
9.
c -- para 52a, 52c(2), 56a
10.
c -- para 52c, d; Attached Memorandum, fig.
3-5
In figure 3-5 a horizontal line drawn from point C (Ecc = -4V)
to the plate current axis shows the plate current to be 4.6 mA.
11.
d -- para 52c, d
When the signal voltage swings 4 volts positive, the grid
voltage will be -2 volts (Ecc + es = eg, or -6 + 4 -2) and the
corresponding point of operation will be point B.
When the signal
voltage swings 4 volts negative, the grid voltage will be -10 volts
and the point of operation will be point F.
12.
b -- para 52d(2), 56d(1); fig.
65
13.
b -- para 52c(1)
When the load line intersects the plate current axis, the
voltage across the tube is zero (eb = 0).
Thus, at this point the
current in the plate circuit is limited only by RL and is found to
be:
Ib
Ebb/RL = 250/50K = 5 mA.
14.
a -- para 53g, 54b
15.
c -- para 56a; fig.
65
16.
a -- para 56c; fig.
65
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