The circuit around U2.1 won't work.
If you are a perfectionist, you have to add a second-order Q = sqrt(2)/2 all-pass filter in the upper path to make its phase shift equal to that of the other two paths.
If you are a perfectionist, you have to add a second-order Q = sqrt(2)/2 all-pass filter in the upper path to make its phase shift equal to that of the other two paths.
Resistor R7 should go to the negative input of U2.1,
instead of to the output of the previous stage.
instead of to the output of the previous stage.
The way you have drawn it, there is practically no feedback around U2.1 when U1.1's closed-loop output impedance is very low (as it will be).
Rayma assumes that you want to add the input signals, but I have the impression you want to subtract them to make a balanced input - at least the terminal numbers match the terminal numbers of a normal XLR connector. If so, then you need something like the figure at the top of https://en.wikipedia.org/wiki/Instrumentation_amplifier Choosing R1 = 0 and Rgain -> infinity, the first stage matches your U1.1 and U1.2, but the second part doesn't match your U2.1.
He said "Summing L and R " which would be the circuit in post #6.
In any event the feedback resistor is wrongly connected.
In any event the feedback resistor is wrongly connected.
He, she or it indeed wrote about summing left and right, I had overlooked that. Then it's indeed just one wrongly connected resistor.
A lot of boards from China also have inputs with a common ground, and so they only have 3 pins,
even on RIAA phono circuit boards. Not a good idea.
BTW, don't forget the local decoupling capacitors for each supply pin of each IC package, 0.1uF ceramic.
Those should be on the schematic too.
even on RIAA phono circuit boards. Not a good idea.
BTW, don't forget the local decoupling capacitors for each supply pin of each IC package, 0.1uF ceramic.
Those should be on the schematic too.
Do you thing i really need it ? i do not know if i will ear the difference, but also i could ad it later.
Yes i will not forget
Yes summing from an vintage jukebox preamp to a new active loudspeaker
Two more comments, I'll get back to your question:
-With the 20 kohm potmeters at their maximum, the time constant of 1 uF with a 20 kohm potmeter in parallel with a 10 kohm resistor results in a cut-off at 23.87 Hz.
-If you use op-amps with a nonnegligible bias current flowing into their inputs, like NE5532, make sure not to use carbon track potmeters. According to an old Panasonic application note, carbon track (trimming) potmeters can't handle DC current drawn out of their wipers for very long, as it causes anodization.
-With the 20 kohm potmeters at their maximum, the time constant of 1 uF with a 20 kohm potmeter in parallel with a 10 kohm resistor results in a cut-off at 23.87 Hz.
-If you use op-amps with a nonnegligible bias current flowing into their inputs, like NE5532, make sure not to use carbon track potmeters. According to an old Panasonic application note, carbon track (trimming) potmeters can't handle DC current drawn out of their wipers for very long, as it causes anodization.
Theoretically, not having the all-pass will result in a 50 degree phase error and a resulting dip of -0.8545 dB at 2 kHz if the driver phase responses are perfect and their acoustic centres are perfectly aligned. It could be audible in a direct AB comparison, but chances are it will be small compared to aberrations you get anyway from driver and room imperfections.
If you should want to try it with an all-pass filter, there is a useful document on the Analog Devices site called MT-202, https://www.analog.com/media/en/training-seminars/tutorials/MT-202.pdf Figure 5 and the equations below it show how to design one, except that the equation for R4 is wrong, it should have been R4 = Q/(2k).
In your case, that leads to this circuit:
with values
C = 33 nF
R1 = 5683 ohm
R2 = 11366 ohm
R3 = 5683 ohm
R4 = 2841.5 ohm
to be inserted between U7.2 and C38. The circuit attenuates by three so you will have to make up for that somewhere else in the chain, for example by reducing R40 of your original circuit to 2 kohm.
In your case, that leads to this circuit:
with values
C = 33 nF
R1 = 5683 ohm
R2 = 11366 ohm
R3 = 5683 ohm
R4 = 2841.5 ohm
to be inserted between U7.2 and C38. The circuit attenuates by three so you will have to make up for that somewhere else in the chain, for example by reducing R40 of your original circuit to 2 kohm.
Thanks, i will try this filter for sure. For my personal information, what is Q and k in the formula ? and why 33nF ?
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A fourth-order Linkwitz-Riley filter consists of a cascade of two second-order filter stages with Q = 0.5 √2. An all-pass filter has twice the phase shift of an ordinary filter, so one second-order, 600 Hz, Q = 0.5 √2 all-pass section has the same phase shift as a fourth-order, 600 Hz Linkwitz-Riley filter.
The bass and midrange signals in your original circuit go through a 2 kHz Linkwitz-Riley filter and through a 600 Hz Linkwitz-Riley filter, while the treble only goes through a 2 kHz filter. You therefore need to pass the treble through an all-pass section to get everything back in phase.
The bass and midrange signals in your original circuit go through a 2 kHz Linkwitz-Riley filter and through a 600 Hz Linkwitz-Riley filter, while the treble only goes through a 2 kHz filter. You therefore need to pass the treble through an all-pass section to get everything back in phase.
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