Help can some explain this circuit

[Amagic]

Hi

Can somene explain how this circuit works

Also Can somebody help me draw this circuit on switcherCAD so i can view the results of this circuits, as i dont know how to use this program,

Many Thanks

[SY]

Remember the equation for the gain of an inverting amp? That's what this is, except the two resistors have been replaced by two complex impedances. Just for simplicity, pick the midpoints in the adjustments and draw this with fixed components. Then you can have the fun of grinding through the math to get an expression for the ratio of the impedances, which will magically show a pole and a zero on either side of a center frequency.

To check your work, look at the low frequency limit, where the caps can be approximated by an open circuit, and the high frequency limit, where the caps can be approximated by a short.

[I_Forgot]

Originally posted by Amagic

Also Can somebody help me draw this circuit on switcherCAD so i can view the results of this circuits, as i dont know how to use this program,

Many Thanks

The best way to learn how to use the program is to start simulating some simple circuits. Your tone control is a good starter.

You can see how to set up the simulation in the circuit you posted in the other thread. The pot is split into two parameterized resistors. You step one of the two values and use total resistance minus the stepped resistance for the other, that way the total resistance remains constant, like a pot.

Use an AC simulation if you want to see frequency and phase response of the circuit. The program will set the stepped resistance value then sweep the frequency. The frequency and phase response curves will be plotted in different colors for each of the stepped resistance values. Frequency response is shown in solid lines, phase response in dotted lines.

The schematic is drawn by inserting the required components and connecting them together with wires. You must have at least one grounded node in a circuit, and every node must have a DC path to ground (through non capacitive components).

You can select a real op-amp from the library or use the "universal" opamp (which is what I used). In that case you must include the ".lib opamp.sub" statement on the schematic.

In the plot window you must specify the the node voltage or component current you want to plot (in this case, output voltage). This is done at time of simulation or after the fact by simply clicking on a node or component on the schematic.

I_F

[I_Forgot]

Here it is now...

I_F

[I_Forgot]

Here's the bass control circuit- it decided to work today...

I_F

[Amagic]

Hi I would just like to say a big thankyou to I_forgot for HOOKING me up,

I really the appreciate help

Many thanks

[Amagic]

Hi

I have begun to understand how to work out the max and min gain using R2/R1 when taking out the capacitor.

How you incorporate the capacitor in the equation. for example if i was to in put 500hz into one of the circuits, is there a formula i can use to work out the gain when 500hz has put input in one of the circuits.

i tried by working out the impedance of the capacitor at 500hz
1/(2*Pi*f*R) the added the answer to R2, the R2/R1 , but the answer i got was well off

[cliffforrest]

I find it always best to analyse the extremes to understand what is going on - and no maths required!

For some high frequence, the impedance of the cap will be effectively zero, so the pot is shorted out and R3 is in parallel with R1 and R5 in parallel wirh R2.

This gives a fixed gain which is easily calculated.


At th other end of the spectrum, at some low frequency (DC will do!) , the imedance of the cap will be effectively infinite so can be ignored. The gain can then be calculated with the pot at either end by adding 500K to first the input resistance and then the feedback resistance.

Doing this should give you a feel for the transfer function and you should be able to sketch the curves - except you don't yet know the turnover frequency.

Mathmatical types will then derive the equations and solve it.

Any easier way is to calculate the frequency where Xc = 500K, the resistance of the pot. your formula is correct, just make sure you plug in the right values.

Remember 82nF is 82 * 10^-9 F !

[cliffforrest]

I find it always best to analyse the extremes to understand what is going on - and no maths required!

For some high frequence, the impedance of the cap will be effectively zero, so the pot is shorted out and R3 is in parallel with R1 and R5 in parallel wirh R2.

This gives a fixed gain which is easily calculated.


At th other end of the spectrum, at some low frequency (DC will do!) , the imedance of the cap will be effectively infinite so can be ignored. The gain can then be calculated with the pot at either end by adding 500K to first the input resistance and then the feedback resistance.

Doing this should give you a feel for the transfer function and you should be able to sketch the curves - except you don't yet know the turnover frequency.

Mathmatical types will then derive the equations and solve it.

Any easier way is to calculate the frequency where Xc = 500K, the resistance of the pot. your formula is correct, just make sure you plug in the right values.

Remember 82nF is 82 * 10^-9 F !

[Amagic]

are the same formulas used to work out band pass filters (like the one attached)