Newbie 312 questions (output atten., DI, mic/line in)

[vertiges]

[quote author="Harpo"][quote author="vertiges"]I thought the cap in the feedback loop acted like a low pass filter and was facultative. I thought that if its value was reduced, a lost of hi frequencies will occur.

I don't understand why you have to change the value of the components around if you change the "gain pot".[/quote]
Hi vertiges,

the voltage gain of this non-inverting amp is 1+(Rfb / Rshunt) where Rshunt is the resistor to gnd in series with your potentiometer wired as rheostat. For 4,7k feedback resistor, 47R at gnd and 4,7k pot your voltage gain is varying between 101 (pot value 0R=max.gain) and 1,99 (pot value 4,7k=min.gain*) or 40dB to 6dB.

The cap across the feedback resistor forms a lowpass filter.
The corner frequency where the signal is dropped by -3dB is calculated by 1 / (2xPi() x Rfb x Cfb) with R in Ohm and C in Farad.
If you fill in the alternative parts couple from Bo's quote (4,7k/470pF, 10k/220pF, 22k/100pF...) your formula looks like 1/(2*Pi()* 4700 * 0,000000000470) resulting in about 72kHz.
This 72kHz is the frequency at wich your signal has dropped -3dB (voltage gain of 0,707). You most probably wont hear the 72kHz, but this filter only has a 6dB/oct. slope, so your interest might by, what is the loss at 20kHz.
Voltage gain at 20kHz for this is 1 / (squareroot (1+ (20000Hz / Lpf-frequency)^2)), giving 0,963. This loss in dB is log10(0,963)*20, giving -0,32dB. The introduced phase error at 20kHz for this Lpf is at Arctan(20000Hz / Lpf-freqency) * 180 / Pi(), giving 15,5°. (phase at 72kHz is 45°).

Same goes for the cap between gnd and Rshunt, but this forms a highpass filter.
The -3dB cutoff frequency for 47R and 1000uF is (same formula as above 1/(2*Pi()*R*C) 3,4Hz. (The highpass filter asumes worst case condition (value of potentiometer = 0). If you add up the 4,7k from the pot to the 47R (giving 4747R), this hpf goes down to 0,03Hz)
Your interest may be the loss at 20Hz, giving 1 / (squareroot (1+ (Hpf-frequency / 20Hz)^2)), giving voltage gain of 0,986 or -0,12dB. Phase error for this Hpf at 20Hz is -Arctan(Hpf-freqency / 20Hz) * 180 / Pi(), giving -9,6°. (phase at 3,4Hz is -45°).

edit: corrected example pot value from 22k to 4,7k[/quote]

I have another question... :roll:

Those caps have an other function ? I mean like blocking DC ?

I'm asking because I just don't understand why using a Hi-pass and a low-pass in this circuit... So the answer is maybe we need to block some DC current, and we have to calculate the values of the caps for being sure they don't cut anything between 20 and 20000 Hz... am I wrong ?

And about the cap on the Feedback loop... It's calculated for giving +/- 0 dB loss at 20 Hz and 20kHz. Does this circuit induce something we don't wanna hear like weird hi/low frequencies phenomenons or something like that ?

Thanks a lot !

 

[Harpo]

[quote author="vertiges"]Those caps have an other function ? I mean like blocking DC ?

I'm asking because I just don't understand why using a Hi-pass and a low-pass in this circuit... So the answer is maybe we need to block some DC current, and we have to calculate the values of the caps for being sure they don't cut anything between 20 and 20000 Hz... am I wrong ?[/quote]
The cap in the shunt arm for the hpf is for blocking DC, the cap in the feedback loop is to limit the opamp for picked up HF garbage and keep it from oscillating. As any R/C circuit it will affect the frequency response of your amp. For usual you try to design it (in this particular case for linear response, not for a filter or effect unit), that your audible range (maybe 20Hz-20kHz) is least affected by these filters. Its a compromise. The largest cap to gnd you can get will have nearly ideal frequency response (on paper), but might/will induce some EMI trouble, acting like an antenna and Bo's already quoted 'scraping/thumbing gain pot'. A too small - if any cap in the feedback loop will burden your opamp with amplifying signals in the HF range it can't do up to a specific level, depending on type of your opamp, resulting in audible or inaudible oscillation. There is no such thing like an ideal opamp/DOA with unlimited power and frequency range.
[quote author="vertiges"]And about the cap on the Feedback loop... It's calculated for giving +/- 0 dB loss at 20 Hz and 20kHz. Does this circuit induce something we don't wanna hear like weird hi/low frequencies phenomenons or something like that ?[/quote]
Sorry, couldn't find my '20Hz' quote in the lpf. The only 20 I wrote was part of voltage gain to dB conversion [value in dB=log10(voltage gain)*20].
You don't have +/- 0 dB loss at 20Hz and 20kHz, you try to get close to this. From the 4k7||470pF lpf example, the -3dB cutoff frequency is at 72048 Hz. At 20kHz this loss is still at voltage gain of 0,963 (=-0,32dB), not voltage gain of 1 (=0 dB) with phase error of 15,5°. At 10kHz this loss is still at voltage gain of 0,99 (=-0,08dB) with phase error of 7,9°. At 6303Hz your phase error from the above 72kHz filter has dropped below 5° (still with loss -0,03dB, not 0dB). If you want this below 5° phase response happen not at 6303Hz but at 20kHz, you have to decrease your feedback cap from 470pF to 150pF. This will set the -3dB lpf from 72kHz to 225kHz. Now look, what your opamp/DOA can handle.

 

[vertiges]

[quote author="vertiges"]
[quote author="vertiges"]And about the cap on the Feedback loop... It's calculated for giving +/- 0 dB loss at 20 Hz and 20kHz. Does this circuit induce something we don't wanna hear like weird hi/low frequencies phenomenons or something like that ?

Sorry, couldn't find my '20Hz' quote in the lpf. (1) The only 20 I wrote was part of voltage gain to dB conversion [value in dB=log10(voltage gain)*20].
You don't have +/- 0 dB loss at 20Hz and 20kHz, you try to get close to this (2) [/color]. From the 4k7||470pF lpf example, the -3dB cutoff frequency is at 72048 Hz. At 20kHz this loss is still at voltage gain of 0,963 (=-0,32dB), not voltage gain of 1 (=0 dB) with phase error of 15,5°. At 10kHz this loss is still at voltage gain of 0,99 (=-0,08dB) with phase error of 7,9°. At 6303Hz your phase error from the above 72kHz filter has dropped below 5° (still with loss -0,03dB, not 0dB). If you want this below 5° phase response happen not at 6303Hz but at 20kHz, you have to decrease your feedback cap from 470pF to 150pF. This will set the -3dB lpf from 72kHz to 225kHz. Now look, what your opamp/DOA can handle.[/quote]

(1) Actually, I was thinking about the two caps : the C-Shunt and the C on the feedback loop. My question wasn't very well formulated. (I was tired...)

You answered to this question by responding to the previous one :

- The cap in the Fb loop limits the opamp for picked up HF garbage and keeps it from oscillating.
- The cap in the shunt arm blocks DC.

I didn't know about the EMI trouble neither... :roll:

(2) Yes, yes ! That's what I've had understood. :wink: "+/- 0dB" meant "about" or "approaching"

Thanks for your help Harpo ! :thumb:

eD)))

 

[sr1200]

I got a nublet question regarding the 312 project (didnt wanna start a new thread for something probably obvious)

i have every part for the pres EXCEPT for the switches for the pad/PP/Ph.  Suggestions for something cool (illuminated would be nice) 

What has everyone else used?

 

[Luca]

[quote author="vertiges"]I thought the cap in the feedback loop acted like a low pass filter and was facultative. I thought that if its value was reduced, a lost of hi frequencies will occur.

I don't understand why you have to change the value of the components around if you change the "gain pot".

Hi vertiges,

the voltage gain of this non-inverting amp is 1+(Rfb / Rshunt) where Rshunt is the resistor to gnd in series with your potentiometer wired as rheostat. For 4,7k feedback resistor, 47R at gnd and 4,7k pot your voltage gain is varying between 101 (pot value 0R=max.gain) and 1,99 (pot value 4,7k=min.gain*) or 40dB to 6dB.

The cap across the feedback resistor forms a lowpass filter.
The corner frequency where the signal is dropped by -3dB is calculated by 1 / (2xPi() x Rfb x Cfb) with R in Ohm and C in Farad.
If you fill in the alternative parts couple from Bo's quote (4,7k/470pF, 10k/220pF, 22k/100pF...) your formula looks like 1/(2*Pi()* 4700 * 0,000000000470) resulting in about 72kHz.
This 72kHz is the frequency at wich your signal has dropped -3dB (voltage gain of 0,707). You most probably wont hear the 72kHz, but this filter only has a 6dB/oct. slope, so your interest might by, what is the loss at 20kHz.
Voltage gain at 20kHz for this is 1 / (squareroot (1+ (20000Hz / Lpf-frequency)^2)), giving 0,963. This loss in dB is log10(0,963)*20, giving -0,32dB. The introduced phase error at 20kHz for this Lpf is at Arctan(20000Hz / Lpf-freqency) * 180 / Pi(), giving 15,5°. (phase at 72kHz is 45°).

Same goes for the cap between gnd and Rshunt, but this forms a highpass filter.
The -3dB cutoff frequency for 47R and 1000uF is (same formula as above 1/(2*Pi()*R*C) 3,4Hz. (The highpass filter asumes worst case condition (value of potentiometer = 0). If you add up the 4,7k from the pot to the 47R (giving 4747R), this hpf goes down to 0,03Hz)
Your interest may be the loss at 20Hz, giving 1 / (squareroot (1+ (Hpf-frequency / 20Hz)^2)), giving voltage gain of 0,986 or -0,12dB. Phase error for this Hpf at 20Hz is -Arctan(Hpf-freqency / 20Hz) * 180 / Pi(), giving -9,6°. (phase at 3,4Hz is -45°).

edit: corrected example pot value from 22k to 4,7k
[/quote]

Guys I'm not sure if it is that simple...
Because there are two caps in the same network that interferee each other!
So I really don't know if is it enogh to take the two case separately, I don't think so.
Maybe I'm wrong but in these cases the best practice is to calculate the LAPLACE TRANSFER FUNCTION.
Basically the net under the opamp is a band-cut filter.
Calculating LTF there are two poles and two zeros.
First pole at 0,00347Hz and first zero at 0,0038Hz
Second zero at 80KHz and second pole at 87,5KHz.
Assuming all original values and pot at 0R (CW).
Obviously that diagram is the OUT of the cut band filter that goes into - of the opamp.
At the out of the opamp the zeros becomes the poles and vice versa. The central band is where the opamp amplifies. The left and right bands where the gain is always 1.
Important: the action of the pot change all the spectrum not only the near 0Hz part...

 

[ruffrecords]

Guys I'm not sure if it is that simple...
Because there are two caps in the same network that interferee each other!
So I really don't know if is it enogh to take the two case separately, I don't think so.
[/quote]

If the turnover frequencies are far apart it is perfectly reasonable to consider each capacitor in the network separately because there will be no significant interaction. In this case they are very far apart - at least three decades.

Cheers

Ian

 

[SV.L]

Chris the boards were never meant for DI but it can be done.
For the DI you Need 4 Diodes, 2 caps and 3 resistors (1M included)
Diodes are 12V zeners.!

Chris if you see my board. the RL 150K can be soldered at RC to CC and use RL to solder the 1M!

About the SC-A12 i supposed its an atenuator before the DOA.

Do i need a switch between DI and Mic?

I ask because i don't know how to put it in the circuit. Without it seems easy to get Bo's DI Mod in circuit.

 

[SV.L]

Chris that's Fabios DI version wich is fine i'd choose Bo Hansen version over Fabio.

Bo protects the DOA with 2x 1N4148 from + in DOA to - In DOA

i would use a switching jack i would use Fig# 2...so i would go from Input transformer to DOA but when inserting a jack it would break into a DI directly to the DOA.!

Look at this: (sorry its big)

I think my question would be answered with this picture. Can you post it again?