33

Output impedance:

The output impedance of this stage will become the source impedance for

whatever circuit it feeds, so it is necessary to know what it is in order to select a

suitable input impedance of the following stage. This is particularly important if the

stage is expected to drive a heavy load, such as a tone stack. Fig. 1.26 shows a

Thévenin equivalent circuit, which is how the circuit in fig. 1.18 appears to AC only.

Because the power supply is connected to ground via very large smoothing

capacitors, as far as AC signals are concerned, the power supply and ground are one

and the same: they appear to be shorted together. Additionally, rather than show a

valve with unknown properties, it has been replaced by a perfect signal generator in

series with a resistance, ra, which

represents the anode resistance

of the valve. Such diagrams are

mathematical simplifications of

the real circuit and are invaluable

for solving AC circuit problems

such as input and output

impedances. If the cathode

resistor is bypassed it is

effectively short circuited as far

as AC is concerned, and so it is

missing from circuit a. There it

can be seen that the total

resistance through which current

has travelled to reach the output,

is the parallel combination of Ra

and ra, so the output impedance is:

Zout (Rk bypassed) = Ra || ra………………………….…………………………..…...VIII

It was calculated earlier that ra is in this case about 70kΩ, so the output impedance

is:

Zout = 100k || 70k

100k x 70k

Zout

100k + 70k

=

= 41kΩ

If the cathode resistor is unbypassed, which it will be at very low frequencies, then

the circuit appears as in fig. 1.26b, and Rk is now in series with ra. Because Rk is

below the cathode its value appears multiplied by μ+1. Therefore the output

impedance when the cathode is unbypassed becomes:

Zout (Rk unbypassed) = Ra || (ra + Rk(μ+1))………………………………

If Rk is 1.5kΩ and μ is 100 then:

Zout (Rk unbypassed) = 100k || (70k + 1.5k x (100 + 1))

Zout (Rk unbypassed) = 100k || 251.5k

(Rk unbypassed)

100k x 251.5k

100k + 251.5k

Zout =

= 72kΩ

This clearly indicates that an unbypassed cathode resistor considerably increases the

output impedance. If the stage is only partially bypassed it will therefore have a

higher output impedance at low frequencies than at high frequencies, and this will

somewhat increase the degree of bass attenuation caused by the output coupling

capacitor