What is the optimum choice of Rbolo?

We will assume for these calculations that we are operating in the middle of our useful bandwidth, i.e in the flat region. Also assuming that the -3dB on the S/N occur at 50kHz and 500kHz.

Output voltage due to amplifier noise (1)

VoutNoise=Vn*Rf/(N1^2*Rbolo*N2)

Output voltage due to bolometer (2)

VoutBolo=Rf*sqrt(4KT) /(N1*sqrt(Rbolo))

Signal to Noise (3)

StoN=VoutBolo/VoutNoise= N1*N2*sqrt(Rbolo*4KT)/Vn

Minimum primary inductance of second transformer (4)

Lmin=N1^2*Rbolo/(2*pi*fmin)

Maximum primary capacitance of second transformer (5)

Cmax=1/(2*pi*fmax*N1^2*Rbolo)

Minimum number of turns on primary of second transformer (6)

AL=Henries/turn^2;
N2primMin=sqrt(Lmin/AL)=N1*sqrt(Rbolo/(AL*2*pi*fmin))

Minimum number of turns on secondary of second transformer (7)

AL=Henries/turn^2;
N2secMin=N2primMin*N2=N2*N1*sqrt(Rbolo/(AL*2*pi*fmin))

Capacitance on secondary of second transformer (8)

AC=Farads/turn
Csec=N2primMin*N2*AC=N2*AC*sqrt(N1^2*Rbolo/(AL*2*pi*fmin))

Capacitance on primary of second transformer (9)

Cprim=Csec*N2^2=N2^3*AC*sqrt(N1^2*Rbolo/(AL*2*pi*fmin))

Ratio between actual capacitance on primary of second transformer to maximum allowed capacitance on primary of second transformer (10)

Cratio= (N1^2*N2^2*Rbolo)^(3/2) * AC*fmax*sqrt(2*pi/(AL*fmin))

Minimum opamp gain allowed (11)

Rin=Rf/(G*N2) << N1^2*Rbolo
Rin=Rf/(Gmin*N2) = N1^2*Rbolo/20     (since 1Ohm/50mOhms=20)
Gmin=20*Rf/(N1^2*Rbolo*N2)

Maximum allows capacitance of secondary of second transformer (12)

Cmaxsec=Cmax/N2^2= 1/(2*pi*fmax*N2^2*N1^2*Rbolo)

Maintaining S/N ratio and output voltage due to bolometer current

If we aim for a constant S/N ratio, whatever we do, from (3), N1*N2*sqrt(Rbolo) needs to remain constant. From (7) this implies that the number of turns on the secondary will remain constant, additionally from (12) it also means that the maximum allowed capacitance on the transformer secondary remains constant.

If we also aim for a constant output voltage due to the bolometer from (2) we need Rf /(N1*sqrt(Rbolo)) to remain constant

The only way to reduce the capacitance is to improve the winding technique (minimise AC), and increase AL, so that less turns are required on the transformer secondary in the first place.

At the limit, if N2=1, from (3),

N1=SN*Vn/sqrt(Rbolo*4KT)

With a 1Ohm bolometer, N1=190 and Vn=1nV/sqrt(Hz) gives an SN of 1 If we achieve this then we don't actually need a second transformer and the design objective becomes ensuring that the secondary inductance of the transformer satisfies (4) and (9). Unfortunately this would require that we have less than 8.8pF on the secondary of our first transformer, and that (the difficult bit) the secondary of the first transformer has an inductance of 115mH.

So, we definitely need another transformer in there

Trying out some combinations of Rbolo, N1 and N2

If we use Rbolo=1, N1=10, N2=19.1 We need Cpmax=3.18n and Lpmin=0.318m This implies that our first transformer must have at least this (or much more) which is still very hard

If we consider Lmin=N1^2*Rbolo/(2*pi*fmin) to be our constraint, we need Lmin as small as possible which implies Rbolo should be as small as possible. If N1=1, and Rbolo=1, we need N2=190 => Cmax=0.32uF, Lmin=3.2uH. If Al=240nH/turn, we need 4 turns on the primary, with 4 on the primary we need 760 on the secondary. The capacitance/turn on the secondary will have to be less than 0.012pF/turn to achieve a 500kHz upper frequency. Different winding strategies, and a larger core will certainly make achieving this more reasonable.

If we assume that the secondary of N1 has the same inductance as the primary of N2 and if we assume that each is 50uH (I believe that this could be built for the first transformer) If N1=28 and Rbolo=10mOhms, Lmin=25uH (which is two 50uH in parallel), and Cmax=41nF. N2 has to be 68 to satisfy the S/N, and Rf has to be 5.3kOhms to give 10nVs out due to the bolometer only. With Al=240nH/turn^2, we need N2primary=15turns, and N2secondary=1029 turns. Assuming all the capacitance is due to the secondary winding of the second transformer we get a secondary capacitance of 8.9pF which therefore gives a capacitance/turn of 8.6f F/turn, given that the secondary inductance will be 0.23H, the resonant frequency of the second transformer must be higher than 110kHz. Which sounds reasonable?

If Rbolo=1mOhm, and N1=33 turns, this gives N1^2*Rbolo=1Ohm, which gives Lmin=3.2uH and Cmax=0.32uF. Lets assume our first transformer has 50uH on the secondary, this is way bigger than Lmin so its the second transformer that limits us. With Lmin=3.2uH, with 240nH/turn^2 we need 4 turns on the primary. To achieve our S/N of 1 or more, lets have a gain of 200 on the second transformer, this means that we have 800 turns on the secondary. Assuming that all our capacitance is due to the secondary side of the second transformer, the highest capacitance we can have on the secondary is 8pF. With 8pF we need less than 0.01pF/turn. The resonant frequency of the second transformer would be 160kHz.

Finally, if we buy a 100kHz self resonant transformer with 10uH input inductance (see http://www.cmr-direct.com/images/products/global/ltthbrochure.pdf) and use the 1:300 connection. Assuming N1^2*Rbolo=1Ohm, we have easily achieved the Lmin spec (3.2uH), and with a self resonance of 100kHz, the capacitance on the primary must be less than 0.25uF, which has also satisfied our 0.32uF maximum. The S/N would be 1.6, i.e better performance than we need. Additionally, from simulation, the S/N=1 at 13kHZ and at 800kHz.

-- WinterlandUser - 12 Mar 2010

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This topic: ColdFeedback > WebHome > RboloOptimum Topic revision: r2 - 2010-03-12 - WinterlandUser

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