pstamler
Well-known member
Okay, this is going to brand me as a naif in some ways, but here goes.
I was looking at output stages for mic preamps a couple of days ago, and was thinking about what a preamp should be expected to drive. It doesn't seem unreasonable to me for a professional preamp to be expected to drive 100' (30 meters) of cable. Sometimes we're a good distance away from wherever the recorder is, and in many studios we're looking at running wire through floors, walls, ceilings, etc.. 100' is probably extreme but not unheard of.
Okay, what is the load a 100' cable puts onto an amplifier? Looking at cable specs for something like Gotham GAC-2. It has a conductor-to-conductor capacitance of 38pF/ft (well, less than that, but they don't say how much less, so this is the specified maximum) and a conductor-to-shield capacitance of 58pF/ft (same deal). That adds up to 96pF/ft -- and since this is one of the lower-capacitance pro-audio cables out there, let's round that to 100pF/ft to make the arithmetic easier. (See question #1.) A 100' cable will therefore have about 10,000pF of capacitance, or 10nF.
What kind of signal am I looking at? Let's assume a standard pro-audio output, nominally +4dBu with 20dB of headroom. If this is an electronically-balanced circuit, each leg will be putting out a nominal -2dBu; the maximum output would then be +18dBu on each leg. That works out to about 6.156Vrms, or (assuming a sine-wave) about 8.7Vpk.
According to Walt Jung's concervative rule of thumb, a circuit should have 1V/us of slew rate for every 1V of peak output. So that means our output amplifiers should have a slew rate of 8.7V/us or greater. Not too hard to do with 5534s, OPA604s, etc.. Into a resistive circuit.
But this ain't a resistive circuit. It's got 10nF of capacitance in parallel with the load. So the output amplifier has to drive the capacitance without causing slew rate issues.
There's another rule of thumb for slew rate calculations:
V / us = mA / nF
That means that to get a certain slew rate in a capacitative load, you need a certain amount of current to charge a capacitor. The units compute right if you set the equation up as above. You can also rearrange it, if you want to figure out how many mA of charging current you'll need:
(V / us) * nF = mA
So for a slew rate of 8.7V/us and a capacitance of 10nF you'll need 87mA of current.
It's a bit worse than that, because you'll also need a bit of current to drive the resistive part of the load. Say that's 2.5k (half of a 5k load, not standard but if you check a Studer recorder that's the input impedance). That will take another 3.5mA or so. Figure the output amp has a 10k feedback network, and you need another another 0.9mA. So to drive your Studer recorder back there in the machine room, plus 100' of cable, youl'll need about 91.4mA of current.
Shit.
That's a *lot* of current, guys. Almost nobody these days builds preamps that can put that much out, period, never mind in staying in Class-A or staying clean or anything. It just ain't there. (For comparison, a 5534 opamp maxes out at 38mA.)
So that's what I'm looking at, and I have some thoroughly naive quesiotns:
1) Am I calculating the capacitance properly? Is it reasonable to add the conductor-to-conductor and conductor-to-shield capacitances? Or not?
2) Am I being unreasonable in applying Jung's strictest criterion to this application? Would I be more reasonable applying his looser rule (0.5V/us per peak output volt)?
3) What effect, if any, will there be from the 100-200 ohm resistor usually placed in series with outputs?
4) Have I just re-discovered why real pro equipment is big and heavy and runs hot? And why real pro output stages were made with discrete transistors and stepdown transformers (less voltage traded for more current)?
5) Is there something real obvious that should be staring me in the face but isn't?
6) And if not...what do y'all think is the best strategy for building an output stage that will drive a long cable? All-discrete? Op-amp with power transistors? Op-amp with high-current buffer a la Jung-Martell? BIG vacuum tubes and a whopper transformer?:
Peace,
Paul
PS This train of thought got started because I have a 78 transcription preamp in the living room and the computer is here in the workroom, with a 75' cable between them. The problem is a lot less demanding; my soundcard in this computer is set for -10dBV nominal sensitivity, and the input clips at about +8dBu, so the maximum voltage drive is about 2.75Vpk. Cable capacitance is, I'm guessing, about 75pF/foot (unbalanced), or about 5.6nF, so this would take about 15.4mA to drive to the appropriate slew rate, and even the TL071 in the preamp's output can manage that -- just.
I was looking at output stages for mic preamps a couple of days ago, and was thinking about what a preamp should be expected to drive. It doesn't seem unreasonable to me for a professional preamp to be expected to drive 100' (30 meters) of cable. Sometimes we're a good distance away from wherever the recorder is, and in many studios we're looking at running wire through floors, walls, ceilings, etc.. 100' is probably extreme but not unheard of.
Okay, what is the load a 100' cable puts onto an amplifier? Looking at cable specs for something like Gotham GAC-2. It has a conductor-to-conductor capacitance of 38pF/ft (well, less than that, but they don't say how much less, so this is the specified maximum) and a conductor-to-shield capacitance of 58pF/ft (same deal). That adds up to 96pF/ft -- and since this is one of the lower-capacitance pro-audio cables out there, let's round that to 100pF/ft to make the arithmetic easier. (See question #1.) A 100' cable will therefore have about 10,000pF of capacitance, or 10nF.
What kind of signal am I looking at? Let's assume a standard pro-audio output, nominally +4dBu with 20dB of headroom. If this is an electronically-balanced circuit, each leg will be putting out a nominal -2dBu; the maximum output would then be +18dBu on each leg. That works out to about 6.156Vrms, or (assuming a sine-wave) about 8.7Vpk.
According to Walt Jung's concervative rule of thumb, a circuit should have 1V/us of slew rate for every 1V of peak output. So that means our output amplifiers should have a slew rate of 8.7V/us or greater. Not too hard to do with 5534s, OPA604s, etc.. Into a resistive circuit.
But this ain't a resistive circuit. It's got 10nF of capacitance in parallel with the load. So the output amplifier has to drive the capacitance without causing slew rate issues.
There's another rule of thumb for slew rate calculations:
V / us = mA / nF
That means that to get a certain slew rate in a capacitative load, you need a certain amount of current to charge a capacitor. The units compute right if you set the equation up as above. You can also rearrange it, if you want to figure out how many mA of charging current you'll need:
(V / us) * nF = mA
So for a slew rate of 8.7V/us and a capacitance of 10nF you'll need 87mA of current.
It's a bit worse than that, because you'll also need a bit of current to drive the resistive part of the load. Say that's 2.5k (half of a 5k load, not standard but if you check a Studer recorder that's the input impedance). That will take another 3.5mA or so. Figure the output amp has a 10k feedback network, and you need another another 0.9mA. So to drive your Studer recorder back there in the machine room, plus 100' of cable, youl'll need about 91.4mA of current.
Shit.
That's a *lot* of current, guys. Almost nobody these days builds preamps that can put that much out, period, never mind in staying in Class-A or staying clean or anything. It just ain't there. (For comparison, a 5534 opamp maxes out at 38mA.)
So that's what I'm looking at, and I have some thoroughly naive quesiotns:
1) Am I calculating the capacitance properly? Is it reasonable to add the conductor-to-conductor and conductor-to-shield capacitances? Or not?
2) Am I being unreasonable in applying Jung's strictest criterion to this application? Would I be more reasonable applying his looser rule (0.5V/us per peak output volt)?
3) What effect, if any, will there be from the 100-200 ohm resistor usually placed in series with outputs?
4) Have I just re-discovered why real pro equipment is big and heavy and runs hot? And why real pro output stages were made with discrete transistors and stepdown transformers (less voltage traded for more current)?
5) Is there something real obvious that should be staring me in the face but isn't?
6) And if not...what do y'all think is the best strategy for building an output stage that will drive a long cable? All-discrete? Op-amp with power transistors? Op-amp with high-current buffer a la Jung-Martell? BIG vacuum tubes and a whopper transformer?:
Peace,
Paul
PS This train of thought got started because I have a 78 transcription preamp in the living room and the computer is here in the workroom, with a 75' cable between them. The problem is a lot less demanding; my soundcard in this computer is set for -10dBV nominal sensitivity, and the input clips at about +8dBu, so the maximum voltage drive is about 2.75Vpk. Cable capacitance is, I'm guessing, about 75pF/foot (unbalanced), or about 5.6nF, so this would take about 15.4mA to drive to the appropriate slew rate, and even the TL071 in the preamp's output can manage that -- just.