PULTEC EQUALIZER SCHEMATIC HOW TO
PULTEC EQUALIZER SCHEMATIC SERIES

It's the same formula as above, but now the 69.8 is replaced with an impedance including the reactance of the cap, and then before taking the log you find the magnitude of the complex number that's the voltage ratio at the divider.ДЇrom looking at the circuit, this appears to be about 7dB above the maximum attenuation, so it's not the 3dB point.Дoing the math on the low cut turns up about the same numbers: max possible cut around 20 or a little more, actual cut at the specified frequency around 15.

PULTEC EQUALIZER SCHEMATIC SERIES

Unfortunately, if you plug and chug the cap values you find out that with 47nF cap in series with the 75 ohm resistor and the max cut dialled in, the actual attenuation at 20kHz is about -35.5dB, about a 15dB below nominal. The 3dB point would be 3dB above this, at about -40dB (19dB below nominal). That's about 22dB lower than the nominal attenuation of 20.8 db, so it looks like a 22dB maximum cut.

Then at maximum cut the 75 ohm resistor would be parallel with the 1K pot for a total resistance to ground of: The biggest cut this could possibly make is if the cap was a short. Now you can see that the high cut allows high frequencies to go around the 1K pot down to ground through a 75 ohm resistor and a selectable cap in series. If you neglect a few things you can assume the eq's basic attenuation is -20.8 db: It's a little weird trying to figure out exactly why they choose the numbers that they did. Usually when you specify the frequency of a shelving response you're talking about the 3dB point, the point where the response is 3dB less than the 'maximum'. So the basic attenuation of the passive eq here is about 20dB (a little more). The 10K low boost pot looks like it should be in there as well, but it's bypassed by a cap at all but just the low frequencies, so from the 'flat' perspective you can think of it as 0 ohms. In this case the basic flat attenuation is provided by the 10K (series) high boost pot, and the 1K (shunt) high cut pot. Of course the idea of a passive eq is to cut all frequencies by the same amount with a pair of resistors, and then to boost by going around the series resistor, or to cut by going around the shunt resistor. I'm sure its an easy concept, but its got my mind spinning. If through the 100n cap, the reactance would be almost 80k at the frequency of 20h.This doesn't seem right, and then there is only 11k to ground.ДЄnyone that could shed some light on this would be very benevolent. So theoretically, when the bypass switch has the filters engaged the signal in the lowcut section would travel through one of the caps and then through the 1k and 10k to ground? Where do you get the resistance to ground to calculate the total resistance, is it here? I am assuming that you are looking for the resonant frequencies between the cap reactance and the resistance? There are also 3 additional reisistors(1k, 10k, and 10k). Now i understand the theory behind this, but when looking at the schematic it seems much more complicated.ДЇor instance: There are three pots(10k, 10k, and 1k) that act to send a certain amount of signal to the filter networks.

On the low end, as well as the high cut it is employing a cap/resitance network that, depending on the routing either seems to work as a low pass or hight pass filter. On the highs it is simply calculated through the resonant frequency between the inductors and the caps.

PULTEC EQUALIZER SCHEMATIC HOW TO

I am in the process of making the Gyraf pultec clone.( ) The build is going fine but for some reason i cannot wrap my head around how to formulate the low end frequencies.

Pultec equalizer schematic