QThe noise figure f( Vgn) is given for a 20dB/V scaling ( cf page 5 of the DS
). What will it be for a 40dB/V ??
AI don't have any measured results to give you, but the basic idea is that the
AD605 has roughly constant output noise voltage, whatever the gain you select
(page 5 figure 12). Usually, amplifiers are modelled with a constant input
noise voltage, so the output noise voltage is proportional to gain (more gain,
more the input noise voltage is amplified at the ouput). The AD605 does not
work like this! Since the output noise stays constant with increasing gain, the
input referred noise must be decreasing with increasing gain (page 5 figure 13).
Noise figure is the ratio of the output noise of a real amplifier (AD605 in
this case) with an "ideal" noiseless amplifier driven from the same source
impedance. Put simply, the noise of the ideal amplifier increases with gain
while the output noise of the AD605 does not increase with gain. Therefore the
ratio (AD605 output noise)/(ideal output noise) will become smaller as the gain
So, what would figure 18 on page 9 look like,if the gain scaling factor was
changed from 20dB/V to 40 dB/V. Basically, it would look the same, only you
would take all the VGN values and divide them by 2. The change from 20dB/V to
40dB/V only affects the gain control interface, not the signal path. For the
same gain, you have the same noise figure. However, you need only 1/2 the
voltage on VGN input for 40dB/V compared with 20dB/V scale to get a particular
There may be other secondary effects, related to the internal architecture of
the AD605, but in the first order analysis, there is no change in the
Gain/noise figure characteristic.