Passive RIAA Equalization

Let’s now look at some passive RIAA equalization networks. Passive RIAA equalization requires as little as two resistors and two capacitors. The classic configuration consists of a series resistor shunted by the combination of a capacitor and a capacitor and resistor in series. Below is the classic configuration and its values nicely represent the underlying equations for its design, as

R₁C₁    =  2187µS

R₁C₂    =  750µS

R₂C₁    =  318µS

C₁/C₂   =  2.916

Testing this equalization network is simple enough in B² A/D Spice. First we add a voltage source to provide an AC signal source that can be made to sweep from 10Hz to 100kHz.

Second we add the network itself; and, finally, we add a voltage meter to the output to see the results. The following circuit was drawn in B² A/D Spice and the file is downloadable here ( Passive RIAA Inversion Eq 1.ckt Ver 4.2) ( Passive RIAA Inversion Eq 1.ckt Ver 2001) 

The graph follows the desired curve closely, as can be seen below.

Still as was mentioned earlier, comparing several tens of points to the table is tedious and error prone. One workaround is to have B² A/D Spice create a table instead of the graph, which would allow comparing tables to tables, but there is still an easier way. We can place the RIAA inverse transfer function, which we have already covered earlier in this article, in series with the output and look for the deviations from a flat output.

Now, if we look at just the second voltage meter’s output, we will see how close the network matches the desired equalization curve.

Not bad. If you think otherwise, be sure to check the units used on the two axes, 10Hz to 100kHz, not 20-20kHz, and milli-decibels (one-thousandth of a decibel or one-tenth of a mB), not dbs (decibels). If we change the Y-axis increments to dBs, then the overwhelming flatness is apparent. (Decibels differ from ohms and watts and coulombs and meters and kilos in that they do refer to any fix quantity, but rather they represent a relationship between two fixed units, such as volts and watts.)

Unfortunately, this equalization network seldom used with an infinite load impedance or a zero-ohm source impedance. Well, two FET-input op-amps do come close, but even in this case a coupling capacitor and ground resistor must be added to prevent the first op-amplifier’s DC offset from being amplified. For example, 1mV of input offset becomes 1V of DC offset at the output of a +40dB gain preamp, as the DC gain is 20dB higher than the gain at 1kHz.

How much influence will the added capacitor and resistor make to the equalization curve? The answer can be found by adding these two components to the B² A/D Spice circuit.  

 Why didn’t we add the actual op-amp models to the circuit? One the things you quickly learn when using SPICE is that simple is better. If you examine the actual models provided by the op-amps’ manufacturers, you might be shocked to find that the models don’t look anything like the op-amps’ actual schematics, with the models consisting of only a few transistors and a few resistors. What is going on here? Where are all the cascoded input stages and current mirrors? The answer is that the model only has to give the same results as the actual op-amp, not define all of its hundreds of parts, which would only slow down or even possibly halt the SPICE engine’s execution, particularly if the schematic held hundreds of op-amps.  

The preamp’s results are not too bad, being only 1dB off at 10Hz. By the way, if you think the capacitor is the culprit here, your blaming the wrong part. The .1µF capacitor and the 1M resistor define a –3dB cutoff point of 1.59Hz, which is still far enough away from 10Hz to make no real difference. If you are still unconvinced, just remove the coupling capacitor or make its value 100µF. No, the real culprit is the 1M resistor, as it effectively reduces the 75k resistor’s value to 69k, which throws the network’s time constants off. Overcoming this problem requires using a source impedance of 6k or adding the 6k to the 75k to yield a 81k resistor. In the case of the op-amps, the output impedance is usually well under 100 ohms, so the latter method works better, but with tube circuits, the source impedance might be even higher than 6k, which would require using a grid resistor even lower than 1M or reducing resistor R₁’s value. For example, a 12AX7 used in a grounded-cathode configuration with a 150k plate-load resistor results in an output impedance of roughly 44k (62k in parallel with 150k). When this value is added to equalization network’s 75k, the error becomes more noticeable and moves in the opposite direction, giving a positive boost at 10Hz (+2.74dB). 

If we reduce resistor R₁’s value to 37k, the output returns to close to flat, but with a slight bump in low bass. What would be the optimal value for resistor R₁? One way to find out is to use B² A/D Spice’s parameterized AC sweep test. Here we can enter a beginning value for R₁ and incrementally add 100 ohms to it until 39k is reached. After we run the test, we look for the flattest graph plot.

In this case, the bold green line looks about as good as we can hope for. This line represents the 14^th sweep, so the value of R₁ must be 14 times 100 ohms plus 36k (37.4k).   

We can increase the accuracy of our circuit analyses by adding the grid-stopper resistor and the Miller-effect capacitance. Because the grid-stopper resistor’s value is so low compared to R₁’s value, this resistor’s effect is minor at audio frequencies. And the Miller-effect capacitance just adds to C₂’s value, so using a C₂ value that accounts for the added capacitance is a good idea. Of course, some readers are going to want more than just the modeling of the equalization network; they will want the whole circuit. Well, here it is.