Tube DIY Asylum: How to design a passive single-network RIAA (two tubes), 2nd draft by Kurt Strain

To calculate the correct values for this phono preamp RIAA equalizer, using 6922 triodes, I developed the following formulae for anyone wishing to implement this single network two-tube topology.

The things you need to know beforehand (gather the data):

For the 6922 in the datasheet:
mu = 33, rp = 2.9K, gm = 11.5 ma/V, all at Vp = 90V and Ip = 12mA.
Cgp = 1.4 pF

Now in this case I am running the 6922 away from its published operating points, where it states that Vp = 90V and Ip = 12mA. Here I am running it (with a B+ = +130V) at Vp = +80V and Ip = 2.5mA. This is because I think this is a better operating point for lower noise and small signals. You can get a better number for rp in this case by following the following formula:

rp = rpo*[(Ipo/Ip)^(1/3)]

where Ipo is the published operating current, rpo is the published plate resistance.

In this case:

rp = 2.9K*[(12mA/2.5mA)^(1/3)] = 4.9K, a substantial difference.

But mu remains very constant, 33, and gm now shifts to mu/rp = 33/4.9K = 6.7 mA/V.

These numbers are also calculable from the plate curves for the 6922, if you know the technique. These numbers above are sufficient for us to use.

The amplifier is not loaded with a CCS, but a 20K plate resistor. For simplicity I bypassed the cathode resistor. If you unbypass it then the plate resistance will go up considerably and needs to be recalculated. If you battery bias it with a small resistor left unbypassed in the cathode, again recalculate the effective plate resistance of the new circuit. This version creates an output resistance, Ro, from V1, equal to:

Ro = rp || 20K plate resistance = 4.9K || 20K = 3.9K ohms.

The output resistance of V1, Ro, is something you need to add to R1 in order to get the total series resistance from a modeled voltage source, called Rs, that precedes the shunt capacitor node on the other side of R1.

Now that the preliminaries are out of the way, Iâ€™m not going to derive the complete transfer function here because you donâ€™t need to know. You want to plug and chug and get a real good passive RIAA going.

First choose C1, a value that makes sense. The only major criterion that C1 should meet is that it should be much larger than [3*Cgp*(mu+1)] + 20pF using V2â€™s parameters, a good rule of thumb to avoid inaccuracy due to input capacitive loading from V2 and some stray capacitance. For the 6922, this value is [3*1.4pF*(33+1)]+20pF = 163pF. â€œMuch largerâ€ means 50 times larger to accept a small error of about 2%, or you can simply add the amount of capacitive loading to C2 later for greater accuracy. Neglect that in this example.

Iâ€™m going to try one I know works here. Set C1 = 0.033uF, which is 200 times more capacitance than my rule of thumb says, 163pF. I can forget about adding the loading capacitance to C2 with that much room to spare.

Now calculate the exact value for R2. It happens that R2 calculates directly from the transfer function when C1 is known:

R2 = 318usec / C1 = 318E-6 / 0.033E-6 = 9636.36 ohms, or about 9.64K. Notice I am carrying everything to a large 6 significant digits during the calculation process.

This one gets the 318 usec breakpoint, a frequency of 500.5 Hz. Only two more values left!

Rs = series resistance talked about earlier = 6.87736*R2. Thatâ€™s a known constant I calculated out that works every time, derived by iterative process:

Rs = 6.87736*R2 = 6.87736*9636.36 = 66272.7 ohms, or about 66.3K.

From here calculate R1 necessary to get Rs = 66.3K:

R1 = Rs â€“ Ro = 66.2727K â€“ 3.9K = 62.4K ohms. I lost calculation precision by the less accurate figure for Ro.

Now we need to check if this value for R1 makes sense. It needs to be less than 500K to avoid isolating the drive of V1 output to the input of V2 and even less to overcome the effect of the 1M grid resistor on V2. Maximize this grid resistance! Even 1M has an effect on the bass response accuracy and the higher Rs is, the worse this situation gets. For accuracy, itâ€™s really best to have Rs much less than this 1M grid resistor. It also needs to be larger than Ro so that V1â€™s importance is not critical, as this varies over time and with different 6922â€™s. Also, R1 determines the lowest impedance load on V1 as the frequency goes up, so itâ€™s best to keep it a drivable load from V1 . Well, R1 is 25 times that of Ro and 16 times less than the 1M grid resistance, which is not bad.

The last value to pick is C2:

C2 = C1 / 2.91600 (where 2.91600 is another calculated constant that works all the time, to 6 significant digits)

And so:

C2 = 0.033uF / 2.91600 = 0.0113169uF.

In summary we have our values already as:
R1 = 62.4K
R2 = 9.64K
C1 = 0.0330uF
C2 = 0.0113uF

â€¦ to 3 significant figures. If you choose Caddock values for resistors and standard capacitor values, the following choices are close:

R1 = 64.9K
R2 = 9.76K
C1 = 0.033uF
C2 = 0.01uF + another 0.001uF in parallel = 0.011uF.

Choose close tolerance parts, +-1 percent or better for the resistors and +-5 percent or better for the capacitors should be sought. But really itâ€™s better to select measured capacitors for close values for better precision than 5%, 2% being much more desirable.

With this circuit you can get the standard RIAA to be within +-0.5 dB without trying hard, or trim values to get them closer for better accuracy.

For further enhancement, itâ€™s possible to add in a rumble filter of -6dB/octave at around 20 Hz and maybe even a â€œhidden breakpointâ€ at 3.18usec where it stops attenuating above about 50KHz. For the 20Hz rumble filter, set the coupling cap values to do this job. For the 3.18usec hidden breakpoint (where recordings on average stop increasing treble boost), you can add a small resistor, R3, in series with C2, where R3*C2 = 3.18usec. In this case, that would make R3 = 3.18E-6 / 0.011E-6 = 289 ohms, and a 301 ohm Caddock would be close enough. The difference is a small approximately 0.6 dB boost at 20KHz.