SPICE transfer function for RIAA testing - Page 2 - diyAudio
 SPICE transfer function for RIAA testing
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diyAudio Member

Join Date: Aug 2002
Location: Germany
...the plot

edit: bad pic, well, the range is -30dB at 1Hz till +77 dB at 20kHz. It rather should be at 0dB
Rüdiger
Attached Images
 spiceplot.jpg (35.4 KB, 563 views)

 2nd March 2006, 12:47 PM #12 diyAudio Member     Join Date: Jun 2002 Location: 3RS Did you check the "resources tab" at the Beige Bag web site ? www.beigebag.com The formula is given in that article about the RIAA correction. Application might be slightly different in that GUI but you get the idea. __________________ AM
 2nd March 2006, 01:04 PM #13 diyAudio Member     Join Date: Aug 2002 Location: Germany Hi, The formula itself seems to be ok, my problem is the actual implementation. I'm sure, the beigebag-spice software is good, but I shall get it working with ltspice, Rüdiger
 2nd March 2006, 05:08 PM #14 diyAudio Member     Join Date: Nov 2002 Location: Netherlands Hi Rudiger, In plain pspice text as a sub circuit this works for me: .SUBCKT I_RIAA IN OUT PARAMS: tau_1=3180u tau_2=318u tau_3=75u +tau_4=3.18u * E1 OUT 0 LAPLACE {V(IN)}={(1+tau_1*S)*(1+tau_3*S) +/((1+tau_2*S)*(1+tau_4*S))} R1 OUT 0 1meg * .ENDS For a normal invers RIAA make tau_4=0 and for a corrected RIAA incorporating the 3.16 us time constant, make tau_4=3.18u Maybe that works in your pspice? Cheers
diyAudio Member

Join Date: Nov 2002
Location: Netherlands
Quote:
 Originally posted by Onvinyl E_RIAA out 0 LAPLACE {1117*V(in)}={(1+0.00318*S)*((1+0.000075*S)/(1+0.000318*S))}
The plus for extending the expression on the next line is missing. It must be:

E_RIAA out 0 LAPLACE +{1117*V(in)}={(1+0.00318*S)*(1+0.000075*S)/(1+0.000318*S)}

Btw. Why that factor "1117"?

diyAudio Member

Join Date: Aug 2002
Location: Germany
Quote:
 Originally posted by Pjotr The plus for extending the expression on the next line is missing. It must be: E_RIAA out 0 LAPLACE +{1117*V(in)}={(1+0.00318*S)*((1+0.000075*S)/(1+0.000318*S))}
Pjotr,
there is no next line. however, the expression computes without error warning, but shows strange behaviour. This is not related to the gain modifier (1117*)
I give up for today,
Rüdiger

diyAudio Member

Join Date: Nov 2002
Location: Netherlands
This is what I get with the sub circuit mentioned.

Attached Images
 i_riaa.gif (4.9 KB, 574 views)

diyAudio Member

Join Date: Apr 2002
Location: Llanddewi Brefi, NJ
In beigebag and multisim there is a transfer function utility -- just multiply and apply the constants:
Attached Images
 inverse_riaa.gif (19.4 KB, 552 views)

diyAudio Member

Join Date: Nov 2002
Location: Netherlands
Hi,

That sub circuit should work in any standard pspice compliant simulator. Since we are most time interested in the gain at 1 kHz I have modified the model so it gives 0 dB gain at 1 kHz:

.SUBCKT I_RIAA IN OUT PARAMS: tau_1=3180u tau_2=318u
+tau_3=75u tau_4=3.18u CORR=9.897
E1 OUT 0 LAPLACE {V(IN)}={(1+tau_1*S)*(1+tau_3*S)
+/((1+tau_2*S)*(1+tau_4*S)*CORR)}
R1 OUT 0 1meg
.ENDS

Cheers
Attached Images
 i_riaa_2.gif (4.0 KB, 510 views)

diyAudio Member

Join Date: Jun 2004
Location: Richmond CA
Here's an LTSpice example of Laplace for inverse RIAA. This inverse RIAA includes the "Neumann" 3.18us or 50kHz lead in addition to the usual 50/2122Hz lags and 500Hz lead. The laplace works well in LTSpice only for AC analysis, for transient analysis you should use an RC network. Of course you need the RC network or a test record for frequency response verification if you built something.

The example uses the "El Cheapo" feedback network with an ideal 1-pole opamp, results are similar to OPA637 model.

I have built and tested frequency response of both tube and solidstate phono preamps modeled in LTSpice, agreement between simulation and measurement from 100Hz and 10kHz normalized at 1kHz is usually within 0.2dB, 0.7 to 1dB from 20Hz to 20kHz. I doubt it is possible to verify frequency response this closely with a test record and phono cartridge, if you have done this please let me know.

The E source with the Laplace formula used adds 19.9dB gain at 1kHz, so for 5mV on the preamp input the AC source (V3 in this example) should be 505.8uV.

The example circuit for Laplace:
Attached Images
 elcheapo_1a.png (10.7 KB, 589 views)

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