What is the benefit of a bipolar supply for opamps? I'm going to toy with non-inverting amplifiers for taking guitar level to balanced line level (Gain factor of 10-ish?)
Depending on the circuit, you can end up requiring fewer parts. Though one obviously has to distinguish single-ended/bipolar from actual supply voltage, when people ask your question, they are generally referring to something where the bipolar supply involves greater overall supply voltage (e.g., +/-9v vs +9V, as opposed to +/-6v vs +12V). Greater overall supply voltage will provide more headroom, assuming you have that much gain in yor circuit.
Quote from: Mark Hammer on November 17, 2011, 04:38:19 PM
Depending on the circuit, you can end up requiring fewer parts. Though one obviously has to distinguish single-ended/bipolar from actual supply voltage, when people ask your question, they are generally referring to something where the bipolar supply involves greater overall supply voltage (e.g., +/-9v vs +9V, as opposed to +/-6v vs +12V). Greater overall supply voltage will provide more headroom, assuming you have that much gain in yor circuit.
Thank you. One other question (probably a dumb one). When you see opamp circuits with a single single supply you'll see the input biased with a voltage divider. I'm assuming this voltage divider puts your "zero reference" at half of the supply voltage so the output can swing in either direction, correct? R1 and R2 in this schematic:
http://www.generalguitargadgets.com/pdf/ggg_icbuf_sc.pdf?phpMyAdmin=78482479fd7e7fc3768044a841b3e85a
One more thing. Is there a golden rule for the amount of headroom you should provide?
> Is there a golden rule for the amount of headroom you should provide?
"Never clip!"
Except when you're doing it on purpose, of course :icon_lol: E.g. Sansamp GT2, comparator fuzz and the like.
Quote from: therecordingart on November 17, 2011, 04:08:17 PM
What is the benefit of a bipolar supply for opamps? I'm going to toy with non-inverting amplifiers for taking guitar level to balanced line level (Gain factor of 10-ish?)
The main benefit is that you can have a ground that does not carry supply current, so you do not get the output modulated by ground currents. I used to design preamps for ultra-low noise infrared scanners that were used to detect aircraft exhaust plumes (3 to 5 micron using indium antimonide photovoltaic sensors) and thermal imaging (8 to 14 micron using mercury-cadmium-telluride sensors). All of these used ±15 volt supplies for the photovoltaic InSb sensors and the same for the photoresistive HgCdTe sensors. There would have been complete chaos with supply currents multiplied by ground resistance combining their noise into the extremely small signals if the power supply had been unipolar. Even with pedals, where the signals are much higher, we frequently find problems with ticking LFO's in phasers and delay / echo / reverb pedals that could be reduced substantially by using dual supplies. I also designed regulators for the preamps I built and I measured the power supply at 33 nV/(Hz)
0.5. The amplifiers for the photovoltaic units were proprietary units from an outside vendor, but my photoresistive amps were rated at 0.64 nV/(Hz)
0.5 as bare amplifiers and still less than 1.0 nV/(Hz)
0.5 with the current supply added in.
Another benefit is that ground is always within the permissible DC input level of the op amps. Some op amps invert the output when you go beyond the allowable input range. It pays to have DC coupling from the guitar to the preamp because AC coupling using capacitors in series require a resistance to ground (or a Vcc/2 level) that decouples the source from the input at low frequencies and permits the resistor noise to dominate. Also, since signals are centred on ground, switch pops will be minimal.
The only disadvantage I can think of is that output capacitors, if they are electrolytic, have to be bipolar because they swing above and below ground.
Just when I think I'm getting pretty crafty, I read a post that I understand about every third word of. :icon_rolleyes:
I'm reminded on a pretty much daily basis to be humble. Not a bad thing by any means.
Quote from: Gordo on November 18, 2011, 12:58:15 PM
Just when I think I'm getting pretty crafty, I read a post that I understand about every third word of. :icon_rolleyes:
I'm reminded on a pretty much daily basis to be humble. Not a bad thing by any means.
Agreed.
The more I learn the more I'm coming to the realization that stompbox designers cut a lot of corners or under design "because it's only a stompbox." I'm too nit picky to just let things slide so I'm going all out with the stuff I build.
> It pays to have DC coupling from the guitar to the preamp because AC coupling using capacitors in series require a resistance to ground ... ... permits the resistor noise to dominate.
NOT if the capacitor is picked very large. Comparable to source impedance, not load impedance. 1K source, 100uFd, 100K load, the hiss above 10Hz is roughly that of a 1.01K-1.1K resistor, not a 100K resistor. 10uFd will be fine above 200Hz, where we hear hiss.
Which means no low-cut in hiss-critical input networks.
Bipolar supplies are often useful but their "popularity" stems from analog computers where DC of either polarity must be computed. In straight audio the benefits are less. In simple audio, maybe not worth the trouble. In larger systems like mixers, getting power off of ground helps. You would think you could eliminate coupling caps, but the high gain of large systems brings-up significant DC error, and large systems are full of pots which scratch with small DC; often there are _more_ coupling caps in a large bipolar system than a large single-supply system (at least no fewer).
This is fine when coupling caps cost a dime. In loudspeaker outputs the required cap costs several dollars, and harms bass damping. In mono single-side you need a power cap and an output cap. Mono bipolar needs two power caps, about the same. But stereo unipolar is three big caps, stereo bipolar is two big caps... nobody has made cap-output stereo speaker amps in decades because bipolar winds up a buck cheaper, a buck that can be put to a sexier faceplate or a lower selling price.
Quote from: PRR on November 18, 2011, 04:00:37 PM
> It pays to have DC coupling from the guitar to the preamp because AC coupling using capacitors in series require a resistance to ground ... ... permits the resistor noise to dominate.
NOT if the capacitor is picked very large. Comparable to source impedance, not load impedance. 1K source, 100uFd, 100K load, the hiss above 10Hz is roughly that of a 1.01K-1.1K resistor, not a 100K resistor. 10uFd will be fine above 200Hz, where we hear hiss.
Which means no low-cut in hiss-critical input networks.
If you have a linear system, low frequency noise is not that much of an issue if it is below the audible threshold. But suppose you have a guitar going into a high-gain fuzz pedal that uses either diode clipping or non-linear amplification. That low frequency noise that you cannot hear modulates the amplifier DC operating point and if you don't have AC coupling to the clipping diodes, it modulates the signal and varies the even order harmonic content, even if the modulating signal cannot be heard. I currently have a rather elaborate fuzz in work that uses a ±12 VDC wall wart as its power supply.
Agreed that low-cut does not belong at an input. Anything that determines frequency response belongs between buffers to avoid having interface impedances affect the response.
Quote from: PRR on November 18, 2011, 04:00:37 PM
NOT if the capacitor is picked very large. Comparable to source impedance, not load impedance. 1K source, 100uFd, 100K load, the hiss above 10Hz is roughly that of a 1.01K-1.1K resistor, not a 100K resistor. 10uFd will be fine above 200Hz, where we hear hiss.
Which means no low-cut in hiss-critical input networks.
I didn't believe this at first, but I re-read a few times and now I think I get it.
1. The biasing resistor generates a thermal noise current. If there was nothing attached to the input (no coupling cap or source) then all of this current would be turned into a voltage by V=IR.
2. However, attaching the input coupling capacitor and source impedance provides a second path for the noise current to flow. Now the noise voltage is V=IZ where Z is [biasR||(couplingCap+sourceImpedance)]. Since Z is lower the noise voltage is lower.
3. Making the coupling cap impedance as small as possible allows the source impedance to "soak up" the most noise current. If the coupling cap presents a higher impedance, the source can not "soak up" as much of the noise current and so it becomes a larger voltage.
Is this correct?
> source impedance to "soak up" the most noise current
Yes.
Noise Power is an important concept too. The Noise Power in an ideal resistor at given temp is a constant. The way this Power is distributed between Voltage and Current is set by resistance value.
What if you put two physical resistors together? You might think twice the noise power. However each one sucks-up some of the noise power of the other. Two physical 100r in parallel gives the same noise power as one 50r alone.
A full analysis of the R-C input is tedious. Assuming C is very-large (relative to resistances and frequency range) we may call it a short. A 1K source and a 10K load combine to make a 909 ohm noise resistance. Noise voltage is down by square-root of R, about 0.95. However we really want signal-to-noise. The 1K is signal, the 10K is just loss. Loss is around 0.91. Compared to no input network at all, we have 0.95 hiss and 0.91 of signal. S/N is 0.957 of perfection or a 0.4db Noise Figure.
Excellent info, thanks!