---+++ NoiseContraints The diagram below demonstrates what the feedback topology will look like.

* FeedbackDiagram.jpg:

From this figure we can derive some noise gain equations for both the bolometer voltage noise source and the opamp voltage noise source.

Considering the bolometer noise source only (The amplifier noise voltage =0) This can easily be derived if it is assumed that all the current flowing through the bolometer flows through Rf and just calculating the voltage across Rf. Note that this expression can also be derived if we assume a non-zero input impedance to the amplifier. Effectively a summing amplifier is created and we get the same expression.

Vout = (Rf/Rb) * VBoloNoise

Considering the amplifier noise source. (The Bolometer noise voltage = 0V)

For a given Vout, the voltage across the primary of the transformer is as follows:

Firstly defining Zin as the Bolometer resistance in parallel with the Transformer input impedance. (Ideally the transformer input impedance = inf, but having a magnetizing inductance and a winding capacitance reduces this)

Zin = Rbolo // Ztran

A signal at the output of the amplifier is coupled to the input of the transformer via the feedback resistor which is in series with Zin. The voltage across Zin is equal to the voltage across the primary. Note that Rf and Zin form a potential divider.

VPrimary=Vout * Zin /( Zin + Rf)
Vsecondary = -N*VPrimary = - N*Vout * Zin / (Zin + Rf)
VAmpIn= VAmpNoise - N*Vout*Zin/(Zin+Rf)
G=1+R2/R1
Vout=VAmpIn*G
Vout=G* ( VAmpNoise - N*Vout*Zin/(Zin+Rf) )

solving for Vout gives :

Vout = VAmpNoise/ (1/G + N*Zin/(Zin+Rf) )
Vout = VAmpNoise *   ( G   //  (Zin+Rf)/(N*Zin) )

for Zin<<Rf (since Rbolo << Rf)

Vout = VAmpNoise * ( G   //  Rf/(N*Zin) )

If we divide the output voltage contribution due to the bolometer by the output voltage contribution of the amplifier noise, we kind of have a S/N ratio.

S/N = (Rf/Rbolo) * VBoloNoise  /   ( VAmpNoise * ( G   //  Rf/(N*Zin) ) )
S/N = (VBoloNoise/VampNoise) * (Rf/Rbolo) / (G // Rf/(N*Zin))

Now, if we assume that Rf/(N*Zin) << G

S/N = (VBoloNoise/VampNoise) * (Rf/Rbolo)  /  (Rf/(N*Zin))
S/N = (VBoloNoise/VampNoise) * N *Zin / Rbolo

If Zin=Rbolo i.e Ztran=Inf (input inductance is huge and capacitance is insignificant)

S/N = (VBoloNoise/VampNoise) * N

Given a VBoloNoise =5pV/sqrt(Hz) and a VAmpNoise =4nV/sqrt(Hz)

We need a transformer turns ratio of 800 for the S/N to be equal to 1 With an AmplifierNoise of 1nV/sqrt(Hz) however this reduces to 200.

These values are only applicable to an ideal transformer with no input impedance. For the squids (and Trevors thesis) at 1Mhz, the input impedance of the Squid was 1Ohm. This would therefore reduce Zin from 1Ohm to 0.707 Ohms (assuming squid Z is inductive). This reduction requires the transformer ratio to be at least 280

-- KevinMacDermid - 04 Mar 2010

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This topic: ColdFeedback > WebHome > ColdTransformerOpampSpecifications > NoiseContraints Topic revision: r1 - 2010-03-04 - KevinMacDermid

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