--- In "John Crockett" wrote:
>
> I think I have answered my questions to my own satisfaction.
>
> John
>
Hmmn, I should have typed the following faster then :)
E.I.N. (Equivalent Input Noise)
The EIN of a system is a convenient way of lumping together any noise volta=
ges generated within that system, and presenting it as though it were an Eq=
uivalent Noise voltage at the Input to a noisless system.
e.g. Assume you had a black box amplifier with a gain of 40dB
If you short circuit the input to ensure that no signal, either wanted or n=
oise is able to enter the box, then you could reasonably expect that the ou=
tput should remain at zero. i.e. minus infinity dB, afterall, (40dB gain =
=3D x100) ... (zero x100 =3D zero !)
In reality however, due to the various noise mechanisms within the amplifie=
r electronics, an output noise voltage will indeed exist.
Now, if we measure that output voltage at say 1mV (-60dBV) and we recall th=
at our amplifier has a gain of 40dB, then we can say that black box has an =
EIN of (-60dBV minus 40dB), i.e. -100dBV.
In other words, if we had applied a noise voltage of -100dBV to the input o=
f a noisless 40dB amplifier then we would get -60dBV out. Black box has an =
EIN of -100dBV.
Suppose we input a noisless signal of say -30dBV to the above black box.
Well, it would come out +10dBV, i.e. gained by 40dB.
The EIN (-100dB) would come out -60dBV
We would have a signal to noise ratio of +10 minus -60 i.e. SNR 70dB
Important : Although we used the gain of the amplifier to calculate the EIN=
, the gain itself has no bearing on the EIN figure. i.e. for the above box,=
the EIN is always -100dBV.
If its gain were reduced to 20dB then the output noise would be -80dBV i.e.=
-100dBV plus 20dB
If you take a second box (err, white box ?) with a better EIN performance o=
f -120dBV with a gain of say 35dB, and we input our same noisless signal of=
-30dBV to it, then the output would be +5dBV, i.e. gained by 35dB and our =
output noise would be -120dBV gained by 35dB i.e. -85dBV
We would have a signal to noise ratio of +5dBV minus -85dBV i.e. SNR 90dB
So we can see that a 20dB improvement of EIN equates directly to a 20dB SNR=
improvement in output, irrespective of gain settings.
If you were to combine these two systems such that the better performing wh=
ite box preceded the black box and adjusted their individual gains to achie=
ve an overall gain of 40dB to replicate the original black box setup, then =
...
white box (gain +35dB, EIN -120dBV) ... black box (gain +5dB, EIN -100dBV)
Input a -30dBV signal, comes out of white at +5dBV, noise at -85dBV.
Now, at this point we've got an interesting situation. We have a signal lev=
el of +5dBV, but we have two noise voltages!! one is the -85dBV white box o=
utput and the second is the black box EIN, -100dBV
So ... all three of these signals now go into the +5dB gain black box ampli=
fier, and come out as +10dBV signal, -80dBV white box noise and -95dBV blac=
k box noise. In reality the three outputs are summed and don't exist as sep=
arate entities. The two noise sources actually sum as the square root of th=
e sum of their squares, and become
-79.87dBV
So, finally, we end up with a 40dB amplifier with an input of -30dBV and an=
output of +10dBV signal and -79.87dBV noise i.e. a SNR of 89.87dB very nea=
rly the full 20dB improvement of the white box vs black box EIN individual =
figures.
Caveat : Although I said that the gain of an amplifier has no bearing on it=
s EIN, practically, due to the topology of common input stages, generally s=
peaking the EIN performance of an amplifier will improve with increasing ga=
in.
Rob
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