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 Chip Amps Amplifiers based on integrated circuits

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 6th July 2011, 04:15 AM #11 sregor   diyAudio Member   Join Date: Jan 2006 Location: massachusetts You can modify Q on Sallen and Key circuits by varying the gain of the opamp - K (the gain) is 3 - 1/Q for a circuit with both Rs equal and both Cs equal. For Q = .5, K = 1 : for Q= .7 , K = 1.59. Reference here (there many more references if you search.) __________________ Steve
crazifunguy
diyAudio Member

Join Date: Aug 2008
Quote:
 Originally Posted by sregor You can modify Q on Sallen and Key circuits by varying the gain of the opamp - K (the gain) is 3 - 1/Q for a circuit with both Rs equal and both Cs equal. For Q = .5, K = 1 : for Q= .7 , K = 1.59. Reference here (there many more references if you search.)
I actually had that site already bookmarked and was using it as a refrence. Thank You though. Any idea as to how the origional gain is calculated? I value any input on the other questions as well. Thank You

 6th July 2011, 09:41 AM #13 AndrewT   diyAudio Member   Join Date: Jul 2004 Location: Scottish Borders i can't read clearly but both input opamps appear to be inverting. The gain of an inverting stage like that is Rfb/Rs Rfb = 2.50k? Rs = 10.0k? + 332r + Rs (of source equipment) Gain is ~ 2.5times (~+8dB) but will vary with different source components and with different source impedances eg capacitor coupling will vary the gain with frequency. You really do need a well buffered output stage in your source equipment to ensure that the 10k swamps the variations in cables and sources. The upper inverter has a compound feedback arrangement. I don't have the skills to analyse that for you. I could research it, but I leave that as your homework. __________________ regards Andrew T.
 6th July 2011, 10:57 AM #14 crazifunguy   diyAudio Member   Join Date: Aug 2008 So essentially that second stage is acting as a buffer? I will do somemore homework and try to calculate the initial gain. Is there any benefit to invering 2 times? To me that seems like you end up where you started so...why invert? Last edited by crazifunguy; 6th July 2011 at 11:00 AM.
 6th July 2011, 01:15 PM #15 crazifunguy   diyAudio Member   Join Date: Aug 2008 Using WYE to Delta conversions I found the total resistances for the first gain segment to be 7281/11385 This turns out to be a gain of 1.42/2.22 Those gains seem low to me so can someone check my math x---R1---o---R2---y .............| ............R3 .............| .............z R1=6800 R2=482 & 10482 Low Gain/High Gain (Potentiometer) R3=482 Delta Converison Ra=R2+R3+R2*R3/R1 Rb=R1+R3+R1*R3/R2 Rc=R1+R2+R1*R2/R3 After delta conversion Rxy=(Ra+Rb)*Rc/(Ra+Rb+Rc) Last edited by crazifunguy; 6th July 2011 at 01:20 PM.
sregor
diyAudio Member

Join Date: Jan 2006
Location: massachusetts
Quote:
 Originally Posted by crazifunguy I actually had that site already bookmarked and was using it as a refrence. Thank You though. Any idea as to how the origional gain is calculated? I value any input on the other questions as well. Thank You
Start with equation 6. (Q of generalized sallen and key) Then make all the Rs the same, and all the Cs the same. (r1=r2, c1=c2) and it reduces to Q = 1/(3-K)
__________________
Steve

 6th July 2011, 06:42 PM #17 crazifunguy   diyAudio Member   Join Date: Aug 2008 I rechecked my math a few times and I had an error. The values seem to be 15011/34698. This gives gains of 2.9 and 6.8. When adding gains are they simply added or multiplied?
 7th July 2011, 08:40 AM #18 AndrewT   diyAudio Member   Join Date: Jul 2004 Location: Scottish Borders multiplied. __________________ regards Andrew T.
crazifunguy
diyAudio Member

Join Date: Aug 2008
I have a final question. In this schematic in the phase adjustment it looks like there is a 10uf POLARIZED cap. If the audio signal is AC shouldnt there be 2 caps or a non ploarized cap used?
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 8th August 2011, 09:27 AM #20 AndrewT   diyAudio Member   Join Date: Jul 2004 Location: Scottish Borders that final 10uF capacitor is the DC blocker to prevent damage downstream if something goes wrong with this unit or in a prior stage. 10uF is quite large and can drive good extended bass into >=10k input of the next stage. It cannot drive extended bass into the 4k7 that is fitted on the output. If you know the Rin of your amps then you can scale that 10uF to an appropriate value to maintain or curtail the lowest bass signal. Multiply the C value in Farads by the resistor value in ohms to arrive at the RC time constant in seconds for a passive RC or CR filter, provided Rs= zero ohms and Rload is near infinity (>4M7). Aim for ~100ms for extended bass in a home system. 10uF & 10k = 100ms 1uF & 100k = 100ms. Cutting that bass filter by an octave or even two, is preferred by some builders/listeners. Worth trying RC time constants of that bass filter in the range from 10ms to 200ms. PA generally does not try to reproduce extreme bass and they use filters quite a bit higher, 5ms to 20ms are probably quite common. They also tend to use active filters with 2 or more poles to protect their equipment from damage. Note that the input is filtered with 47ms (@ min volume). Increasing volume to maximum, changes that bass input filter to ~15ms. As the operator turns the volume up the bass is progressively more filtered. Once that extended bass is thrown away at the input you cannot get it back by using extended flat response elsewhere. The fact that the designer (LC Audio) has chosen >=15ms @ input and <47ms @ output tells me this bass filtering system is not intended for home listening. It is more likely targeted at a PA or Disco duty. In summary, that 10uF could be a value between 470nF and 22uF. Choose to taste. Select a film cap if you can fit it into the case and it can fit your budget. __________________ regards Andrew T. Last edited by AndrewT; 8th August 2011 at 09:35 AM.

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