Example 5.1

In the previous chapter we calculated the gain of a 12AX7 amplifier to be 65.2. The tube manual gives the input capacitance of the 12AX7 to be 1.6 pf and grid to plate as 1.7 pf. What is the effective input capacitance of this amplifier.

Solution.

Cinp = Cin + Cgp ( Av + 1 )

Cinp = 1.6 pf + 66.2 x 1.7 pf = 114 pf.

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Example 5.2

A 6AU6 has a stated transconductance of 3900 micro mhos and a plate resistance of 0.5 meg ohms. These values are given for a g2 voltage of 100 volts. Now we look at the transfer function graphs and find the g1 voltage that gives 3900 micro mhos on the g2 = 100 volts curve. Now if we look at the same g1 voltage the point on the g2 = 50 volts curve gives a transconductance of 2200 micro mhos. What will be the gain of a resistance coupled amplifier using a 220 k ohm plate resistor and the following stage grid resistor of 470 k ohms. The cathode resistor is 1000 ohms. What is the gain of this amplifier (a) with a cathode bypass capacitor and (b) without a cathode bypass capacitor?

Solution.

(a)

Rbc = Rb Rcf / (Rb + Rcf)

Rbc = 220 e+3 x 470 e+3 / (220 e+3 + 470 e+3 ) = 150 e+3

Av = Gm rp Rbc / ( rp + Rbc )

Av = 2200 e-6 x 500 e+3 x 150 e+3 / ( 500 e+3 + 150 e+3 ) = 254

(b)

Av = Gm rp Rbc / [ rp + Rbc + Rk ( Gm rp + 1 )]

Av = 2200 e-6 x 500 e+3 x 150 e+3 / [ 500 e+3 + 150 e+3 + 1 e+3 ( 2200 e-6 x 500 e+3 + 1 ) ] = 94.2

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Example 5.3

The plate resistance of the pentode section of a 6U8, triode - pentode, is given as 0.4 meg ohms. The resistance coupled amplifier chart gives the following data.

Ebb = 250 v
Rb = 270 k ohms
Rg2 = 820 k ohms
Rcf = 1 meg ohm
Rk = 1200 ohms
Ib = 0.72 mA
Ig2 = 0.26 mA
Ec1 = -1.0 v
Ec2 = 37 v
Eb = 55 v

Esig = 0.1 vAC
Eout = 22 vAC
Gain = 220
%Dist = 0.82 %

Calculate (a) the operating transconductance of the tube, (b) the gain without a cathode bypass capacitor.

First we must calculate Rbc = Rb Rcf / ( Rb + Rcf ) = 270 k x 1 M / (270 k + 1 M ) = 213 k ohms. Now we calculate the operating transconductance.

Gm = Av ( rp + Rbc ) / ( rp Rbc )

Gm = 220 ( 0.4 e+6 + 213 e+3 ) / ( 0.4 e+6 x 213 e+3 ) = 1583 micro mhos.

Now we calculate the gain without a cathode bypass capacitor as follows.

Av = Gm rp Rbc / [ rp + Rbc + Rk ( Gm rp + 1 )]

Av = 1583 e-6 x 0.4 e+6 x 213 e+3 / [ 0.4 e+6 + 213 e+3 + 1200 ( 1583 e-6 x 0.4 e+6 + 1 )]

Av = 98.2

Any volunteers to breadboard and test the circuit? Maybe I'll get around to it someday but I wouldn't bet the farm on it if I were you.

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Example 5.4

Using the full gain of the pentode section of a 6U8 above calculate the effective input capacitance. Capacitance values for a shielded tube are as follows. Cin = 5 pf. Cgp = 0.006 pf.

Solution.

The effective input capacitance is given by,

Cinp = Cin + Cgp ( Av + 1 )

Cinp = 5 pf + 0.006 pf x ( 220 + 1 ) = 6.33 pf.

That's a long way from the value for a triode.

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