If you wanted to splice in some extra frequency bands into your equalizer, how would you know what values to use for the caps and inductors??
First: Pick your frequency that you want to add.
Second: Find the reactance constant of the values of the caps and inductors in the stock circuit.
Third: Use the formulas for inductive and capacitive reactance to figure out your new values of your cap and inductor.
Fourth: Verify your math by plugging in the new values into the formula for resonance.
Fifth: Breadboard the filter to make sure it works the way you want it to in real life.
For example, here is the stock treble boost section of a Pultec EQP-1A.
Reactance values in ohms are shown next to their respective components.
Notice that the values for reactance are not the same!
Why? A couple of reasons: One, Pultec wanted nice round numbers on their faceplate dials. But they were limited by the fact that caps do not come in every value, unless you parallel them, which was not too practical for economic and space reasons. So, they choose some values that got them close and lied about their true resonant frequency. A second reason might have been that the values they choose made the equalizer have smoother transistions between frequencies than the math had predicted. This could be due to inductors having dc resistance and other factors. Also, Pultec used some of the same coil sections for two different frequencies, so the math will not come out perfect.
OK, so lets say we want to insert a new band at 6.5 khz in order to have smoother steps on our frequency selector. How do we do it?
Pick an average value for reactance according to all the other stock values. Notice that there are a few values that lie outside the norm. We will throw out the 3 khz and 16 khz reactance values since they do not match up with the others.
Lets use 2200 ohms for our new cap and inductor reactance values.
So we want our cap and inductor to both have a reactance of 2200 ohms at 6.5 khz.
Lets figure out the inductor value first, since the formula is a little easier to manipulate.
Inductive reactance is equal to XL=2 pi ? L
We want to find out the value for L, so we will isolate it from the other variables:
L=XL/2 pi ? we can just use 6.28 as an aproximation for 2 pi:
L=XL/6.28 ? now sub in your value for reactance which is 2200:
L=2200/6.28 ? our new frequency band is 6.5khz, which is 6500 hz, so
L=2200/40820 so our inductor needs to be 0.0539 henries, or about 54 millihenries. A quick glance at the schematic seems to verify our math, as the inductor values above and below 6.5 khz are 47 and 68 millihenries.
Now, we need to figure out the cap value. We could do this two ways, use the formula for resonance, or use the formula for reactance like we did for the inductor. Lets use the reactance method, and verify our results with the resonance formula.
Capacitive reactance is XC=1/(2 pi ? C)
Isolating C we have C=1/(XC 2pi ?) now just plug in our values like we did for the inductor:
C=1/( 2200 X 6.28 X 6500) C=1/89,804,000
So C=0.000000011135 Why does that number seem so small?
Because the formula gives capacitance in Farads, so we need to move the decimal up six places to get the answer in microfarads.
So our answer is 0.011 microfarads, or 11 nanofarads if you like.
Looking at the schematic verifies that we are in the ballpark.
Lets check both values by plugging them into the formula for resonance:
The formula for resonance is ?=1/(2 pi
L C)
So we have ?=1/(6.28 X
0.00000000060129)
so ?=1/(6.28 X 0.00002452) , ?=1/0.00015399 which works out to
6493 hertz, or about 6.5 khz! Pretty close to want we wanted.
So there ya go.
Backing up a little, what is reactance? A simple answer would be the resistance a cap or inductor offers to an AC signal. So in order to calculate the reactances of the caps and inductors in the above circuit, we just use the frequency and cap and inductor values and plug them into the formulas. Then we try and match that with any new values we want to add.
What about shelving curves you say? It's even easier than the boost section, as you only have to find the cap values. You would use the same approach. Find the capacitive reactance at the designated frequency and repeat the process we just used on the boost section, only you won't have to compute the inductor value or use the resonance formula.
Not to good at math, you say? Fret not! A seat of the pants approach would be to just pick cap and inductor values that are in between the values that lie above and below the new frequency band that you want.
Then download a free program that does electronics math to check your picks.
:guinness:
cj
First: Pick your frequency that you want to add.
Second: Find the reactance constant of the values of the caps and inductors in the stock circuit.
Third: Use the formulas for inductive and capacitive reactance to figure out your new values of your cap and inductor.
Fourth: Verify your math by plugging in the new values into the formula for resonance.
Fifth: Breadboard the filter to make sure it works the way you want it to in real life.
For example, here is the stock treble boost section of a Pultec EQP-1A.
Reactance values in ohms are shown next to their respective components.
Notice that the values for reactance are not the same!
Why? A couple of reasons: One, Pultec wanted nice round numbers on their faceplate dials. But they were limited by the fact that caps do not come in every value, unless you parallel them, which was not too practical for economic and space reasons. So, they choose some values that got them close and lied about their true resonant frequency. A second reason might have been that the values they choose made the equalizer have smoother transistions between frequencies than the math had predicted. This could be due to inductors having dc resistance and other factors. Also, Pultec used some of the same coil sections for two different frequencies, so the math will not come out perfect.
OK, so lets say we want to insert a new band at 6.5 khz in order to have smoother steps on our frequency selector. How do we do it?
Pick an average value for reactance according to all the other stock values. Notice that there are a few values that lie outside the norm. We will throw out the 3 khz and 16 khz reactance values since they do not match up with the others.
Lets use 2200 ohms for our new cap and inductor reactance values.
So we want our cap and inductor to both have a reactance of 2200 ohms at 6.5 khz.
Lets figure out the inductor value first, since the formula is a little easier to manipulate.
Inductive reactance is equal to XL=2 pi ? L
We want to find out the value for L, so we will isolate it from the other variables:
L=XL/2 pi ? we can just use 6.28 as an aproximation for 2 pi:
L=XL/6.28 ? now sub in your value for reactance which is 2200:
L=2200/6.28 ? our new frequency band is 6.5khz, which is 6500 hz, so
L=2200/40820 so our inductor needs to be 0.0539 henries, or about 54 millihenries. A quick glance at the schematic seems to verify our math, as the inductor values above and below 6.5 khz are 47 and 68 millihenries.
Now, we need to figure out the cap value. We could do this two ways, use the formula for resonance, or use the formula for reactance like we did for the inductor. Lets use the reactance method, and verify our results with the resonance formula.
Capacitive reactance is XC=1/(2 pi ? C)
Isolating C we have C=1/(XC 2pi ?) now just plug in our values like we did for the inductor:
C=1/( 2200 X 6.28 X 6500) C=1/89,804,000
So C=0.000000011135 Why does that number seem so small?
Because the formula gives capacitance in Farads, so we need to move the decimal up six places to get the answer in microfarads.
So our answer is 0.011 microfarads, or 11 nanofarads if you like.
Looking at the schematic verifies that we are in the ballpark.
Lets check both values by plugging them into the formula for resonance:
The formula for resonance is ?=1/(2 pi
So we have ?=1/(6.28 X
so ?=1/(6.28 X 0.00002452) , ?=1/0.00015399 which works out to
6493 hertz, or about 6.5 khz! Pretty close to want we wanted.
So there ya go.
Backing up a little, what is reactance? A simple answer would be the resistance a cap or inductor offers to an AC signal. So in order to calculate the reactances of the caps and inductors in the above circuit, we just use the frequency and cap and inductor values and plug them into the formulas. Then we try and match that with any new values we want to add.
What about shelving curves you say? It's even easier than the boost section, as you only have to find the cap values. You would use the same approach. Find the capacitive reactance at the designated frequency and repeat the process we just used on the boost section, only you won't have to compute the inductor value or use the resonance formula.
Not to good at math, you say? Fret not! A seat of the pants approach would be to just pick cap and inductor values that are in between the values that lie above and below the new frequency band that you want.
Then download a free program that does electronics math to check your picks.
:guinness:
cj
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