What's the point in a preamp?
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I'm talking in the context of guitar amps, but I assume that this question is relevant for any type of audio amplifier.
Very often in amplifier schematics I see two stages of amplification -- first, the signal is amplified a smaller amount by a preamp circuit and then amplified again by a power amp circuit.
This seems redundant to me. What's the point in amplifying a signal in two small steps rather than just one greater-gain amplification?
My first thought was: does this multi-stage amplification help to reduce unwanted noise from the signal? But the more I think about that, the less it makes sense, since surely the second stage would be amplifying any noise as well.
amplifier
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add a comment |
$begingroup$
I'm talking in the context of guitar amps, but I assume that this question is relevant for any type of audio amplifier.
Very often in amplifier schematics I see two stages of amplification -- first, the signal is amplified a smaller amount by a preamp circuit and then amplified again by a power amp circuit.
This seems redundant to me. What's the point in amplifying a signal in two small steps rather than just one greater-gain amplification?
My first thought was: does this multi-stage amplification help to reduce unwanted noise from the signal? But the more I think about that, the less it makes sense, since surely the second stage would be amplifying any noise as well.
amplifier
$endgroup$
$begingroup$
There is also the problem of gain bandwidth product. For a given amplifier, more gain means less bandwidth. If you use too much gain in one stage, then you limit the bandwidth of that stage. This can lead to distortion - it takes gain and bandwidth for negative feedback to compensate for distortion.
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– JRE
2 hours ago
1
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You don't want the high currents (to the loudspeaker) anywhere near the input signal from the guitar pickup, or the vinyl-record signals.
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– analogsystemsrf
2 hours ago
1
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The first amplifier in any signal path usually is the one that adds all the noise to the signal. So the pre-amp must be designed so as to avoid adding more noise to the signal than necessary. Generally low noise devices and design techniques are incompatible with high power devices and design techniques.
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– mkeith
1 hour ago
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@mkeith I think your comment is the best general answer I've seen yet on this. Combined with Dave Tweed's answer, it all makes sense in terms of guitar amplification.
$endgroup$
– Todd Wilcox
23 mins ago
add a comment |
$begingroup$
I'm talking in the context of guitar amps, but I assume that this question is relevant for any type of audio amplifier.
Very often in amplifier schematics I see two stages of amplification -- first, the signal is amplified a smaller amount by a preamp circuit and then amplified again by a power amp circuit.
This seems redundant to me. What's the point in amplifying a signal in two small steps rather than just one greater-gain amplification?
My first thought was: does this multi-stage amplification help to reduce unwanted noise from the signal? But the more I think about that, the less it makes sense, since surely the second stage would be amplifying any noise as well.
amplifier
$endgroup$
I'm talking in the context of guitar amps, but I assume that this question is relevant for any type of audio amplifier.
Very often in amplifier schematics I see two stages of amplification -- first, the signal is amplified a smaller amount by a preamp circuit and then amplified again by a power amp circuit.
This seems redundant to me. What's the point in amplifying a signal in two small steps rather than just one greater-gain amplification?
My first thought was: does this multi-stage amplification help to reduce unwanted noise from the signal? But the more I think about that, the less it makes sense, since surely the second stage would be amplifying any noise as well.
amplifier
amplifier
asked 3 hours ago
Jacob GarbyJacob Garby
27610
27610
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There is also the problem of gain bandwidth product. For a given amplifier, more gain means less bandwidth. If you use too much gain in one stage, then you limit the bandwidth of that stage. This can lead to distortion - it takes gain and bandwidth for negative feedback to compensate for distortion.
$endgroup$
– JRE
2 hours ago
1
$begingroup$
You don't want the high currents (to the loudspeaker) anywhere near the input signal from the guitar pickup, or the vinyl-record signals.
$endgroup$
– analogsystemsrf
2 hours ago
1
$begingroup$
The first amplifier in any signal path usually is the one that adds all the noise to the signal. So the pre-amp must be designed so as to avoid adding more noise to the signal than necessary. Generally low noise devices and design techniques are incompatible with high power devices and design techniques.
$endgroup$
– mkeith
1 hour ago
$begingroup$
@mkeith I think your comment is the best general answer I've seen yet on this. Combined with Dave Tweed's answer, it all makes sense in terms of guitar amplification.
$endgroup$
– Todd Wilcox
23 mins ago
add a comment |
$begingroup$
There is also the problem of gain bandwidth product. For a given amplifier, more gain means less bandwidth. If you use too much gain in one stage, then you limit the bandwidth of that stage. This can lead to distortion - it takes gain and bandwidth for negative feedback to compensate for distortion.
$endgroup$
– JRE
2 hours ago
1
$begingroup$
You don't want the high currents (to the loudspeaker) anywhere near the input signal from the guitar pickup, or the vinyl-record signals.
$endgroup$
– analogsystemsrf
2 hours ago
1
$begingroup$
The first amplifier in any signal path usually is the one that adds all the noise to the signal. So the pre-amp must be designed so as to avoid adding more noise to the signal than necessary. Generally low noise devices and design techniques are incompatible with high power devices and design techniques.
$endgroup$
– mkeith
1 hour ago
$begingroup$
@mkeith I think your comment is the best general answer I've seen yet on this. Combined with Dave Tweed's answer, it all makes sense in terms of guitar amplification.
$endgroup$
– Todd Wilcox
23 mins ago
$begingroup$
There is also the problem of gain bandwidth product. For a given amplifier, more gain means less bandwidth. If you use too much gain in one stage, then you limit the bandwidth of that stage. This can lead to distortion - it takes gain and bandwidth for negative feedback to compensate for distortion.
$endgroup$
– JRE
2 hours ago
$begingroup$
There is also the problem of gain bandwidth product. For a given amplifier, more gain means less bandwidth. If you use too much gain in one stage, then you limit the bandwidth of that stage. This can lead to distortion - it takes gain and bandwidth for negative feedback to compensate for distortion.
$endgroup$
– JRE
2 hours ago
1
1
$begingroup$
You don't want the high currents (to the loudspeaker) anywhere near the input signal from the guitar pickup, or the vinyl-record signals.
$endgroup$
– analogsystemsrf
2 hours ago
$begingroup$
You don't want the high currents (to the loudspeaker) anywhere near the input signal from the guitar pickup, or the vinyl-record signals.
$endgroup$
– analogsystemsrf
2 hours ago
1
1
$begingroup$
The first amplifier in any signal path usually is the one that adds all the noise to the signal. So the pre-amp must be designed so as to avoid adding more noise to the signal than necessary. Generally low noise devices and design techniques are incompatible with high power devices and design techniques.
$endgroup$
– mkeith
1 hour ago
$begingroup$
The first amplifier in any signal path usually is the one that adds all the noise to the signal. So the pre-amp must be designed so as to avoid adding more noise to the signal than necessary. Generally low noise devices and design techniques are incompatible with high power devices and design techniques.
$endgroup$
– mkeith
1 hour ago
$begingroup$
@mkeith I think your comment is the best general answer I've seen yet on this. Combined with Dave Tweed's answer, it all makes sense in terms of guitar amplification.
$endgroup$
– Todd Wilcox
23 mins ago
$begingroup$
@mkeith I think your comment is the best general answer I've seen yet on this. Combined with Dave Tweed's answer, it all makes sense in terms of guitar amplification.
$endgroup$
– Todd Wilcox
23 mins ago
add a comment |
4 Answers
4
active
oldest
votes
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Quick and dirty answer:
Buffering is one reason. Interconnects between things can have a lot of capacitance and require a lot (comparatively) of current to drive.
Noise immunity is another. Think about this scenario: Send a signal through a wire where it picks up, say, 10mV noise, then amplify it by 100x: total noise, 1000mV. But if you instead amplify it by 10x, then send it through the wire where it gets 10mV noise, then amplify by another 10x, your total signal amplification is still 100x, but your total noise is only 100mV.
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Are you saying the noise picked up inside the amp chassis would be equal to or greater than the noise picked up by the guitar pickups out in the world? That doesn't seem right to me. In the case of electric guitars, the part of the signal chain most prone to noise is the source (the pickups), not an interconnect (the cables or flywires or the traces on a PCB).
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– Todd Wilcox
31 mins ago
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@ToddWilcox I did say this was a quick and dirty answer, and may not apply quite so well to the specific scenario the asker is asking about. It is not the best answer and needs a lot of work but I don't have the time or energy to work on it right now, and frankly I'm amazed it got as many upvotes as it did. That said, the definition of "noise" that I'm using here is implicitly assuming that the signal you want is exactly what the transducer outputs, that the signal as it exists on the terminals of the transducer is noise-free by definition.
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– Hearth
27 mins ago
add a comment |
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In audio gear, it is useful to do most of the signal manipulation at a standard level, known as "line level". This includes mixing, equalization, compression, etc.
Some signal sources (microphones, guitar pickups, etc.) do not inherently produce line level outputs, so a preamplifier is used to boost the signal to that level. Some signal sources (record players) require not only a boost, but also a special equalization to flatten the frequency response.
Then, after all of the signal processing is done, a second, "power" amplifier is used to drive the speaker(s).
This kind of modularity allows signal sources, processing stages, and different kinds of speakers to be mixed and matched freely.
$endgroup$
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In case anyone needs this broken down to the simplest level for electric guitar amps: the preamp gets the signal ready for the tone controls, then after the tone controls the power amp makes it ready for the speaker.
$endgroup$
– Todd Wilcox
37 mins ago
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Oh, you're right. I did not notice he was talking about amps in the same unit which was implied by the "in the same schematic part" bit.
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– Toor
35 mins ago
add a comment |
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A major reason for separate boxes for preamps and poweramps is the GROUND currents and also magnetic coupling. [there is numeric example, at 20KHz and 6 amps to the speakers, at end of this answer, with the Preamp only 10cm from the Power amplifier]
Suppose you built the preamp and the poweramp on the same PCB. Why not?
Some of the loudspeaker current will be flowing around on the GROUND, and end up combining with the input signal.
To minimize this "combining", make that PCB long and thin, so the PowerAmp Grounds are far away from the PreAmp Grounds.
How to improve on this? use long thing regions between the Preamp and the Poweramp.
In the extreme, a coax cable provides a long-thin-region, to ensure very small combining of input and output.
Give low millivolt signals from a vinyl record Moving Magnet cartridge, or even 0.5 millivolt from Moving Coil cartridges, that become near-100-volt audio outputs, the entire system needs 100,000:1 isolation.
simulate this circuit – Schematic created using CircuitLab
=============================================
How bad can crosstalk be? assume output current is 6 amps peak at 20,000Hz. The dI/dT is 6* d(sin(2*pi*20,000*time))/dT = 6 * 2*pi*20,000*cos(2pi20000T)
or dI/dT = 700,000 amps per second.
Assume the preamp input (remember that 1 millivolt signal from the cartridge, and you want at least 10,000:1 SNR or tonal feedback, thus 0.1 microvolt feedback is the desired floor) is 0.1 meter from the Speaker output.
V_magnetic_induce = (2.0e-7 * Area/Distance) * dI/dT
and we'll assume the input loop area (signal to ground) is 1cm by 4cm.
Now run the math; remember we need 0.1 microvolt feedback.
Vinduce = 2e-7Henry/meter * 1cm * 4cm /10cm * 700,000
Vinduce = 2e-7 * 0.0004meter/0.1meter * 700,000
Vinduce = 2e-7 * 0.004 * 7e+5
Vinduce = 2e-7 * 4e-3 * 7e+7 = 56 e-3 = 56 milliVolts.
The magnetic feedback, caused by having the Poweramplifer near the Preamplifier, is 56mV / 0.1 microvolt or 560,000X stronger than what "clean" music can tolerate.
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add a comment |
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To minimize the noise factor, which is the SNR of the output divided by the SNR of the input. An ideal amplifier should keep the SNR constant, since the input noise is amplified by the same amount as the input signal. A real amplifier, however, adds extra noise. The noise factor is given by
$$ F = 1 + frac{N_mathrm{additional}}{N_mathrm{input}G}.$$
If you cascade a series of amplifiers the total noise factor is given by Friis’ equation
$$F_mathrm{total} = F_1 + frac{F_2 - 1}{G_1} + frac{F_3 - 1}{G_1 G_2} + frac{F_4 - 1}{G_1 G_2 G_3} + dots.$$
Where $F_n$ is the noise factor of the nth stage and $G_n$ is the gain of the nth stage. This is because the additional noise of the first stage is amplified by the second and subsequent stages but the additional noise of the second stage is amplified by only the third and subsequent stages etc.
As you can see, the the noise factor of a given stage is divided by the gain product of all previous stages. So the first stage is the most important when it comes to noise. That’s why you have a low noise pre-amp stage as your very first component in the signal chain. This configuration has the added benefit of not having to worry about the noise figure of the power amplifier.
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And this is true from DC to daylight, as they say. The first amplifier sets the noise figure is what they say in RF (as a rule of thumb).
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– mkeith
1 hour ago
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Quick and dirty answer:
Buffering is one reason. Interconnects between things can have a lot of capacitance and require a lot (comparatively) of current to drive.
Noise immunity is another. Think about this scenario: Send a signal through a wire where it picks up, say, 10mV noise, then amplify it by 100x: total noise, 1000mV. But if you instead amplify it by 10x, then send it through the wire where it gets 10mV noise, then amplify by another 10x, your total signal amplification is still 100x, but your total noise is only 100mV.
$endgroup$
$begingroup$
Are you saying the noise picked up inside the amp chassis would be equal to or greater than the noise picked up by the guitar pickups out in the world? That doesn't seem right to me. In the case of electric guitars, the part of the signal chain most prone to noise is the source (the pickups), not an interconnect (the cables or flywires or the traces on a PCB).
$endgroup$
– Todd Wilcox
31 mins ago
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@ToddWilcox I did say this was a quick and dirty answer, and may not apply quite so well to the specific scenario the asker is asking about. It is not the best answer and needs a lot of work but I don't have the time or energy to work on it right now, and frankly I'm amazed it got as many upvotes as it did. That said, the definition of "noise" that I'm using here is implicitly assuming that the signal you want is exactly what the transducer outputs, that the signal as it exists on the terminals of the transducer is noise-free by definition.
$endgroup$
– Hearth
27 mins ago
add a comment |
$begingroup$
Quick and dirty answer:
Buffering is one reason. Interconnects between things can have a lot of capacitance and require a lot (comparatively) of current to drive.
Noise immunity is another. Think about this scenario: Send a signal through a wire where it picks up, say, 10mV noise, then amplify it by 100x: total noise, 1000mV. But if you instead amplify it by 10x, then send it through the wire where it gets 10mV noise, then amplify by another 10x, your total signal amplification is still 100x, but your total noise is only 100mV.
$endgroup$
$begingroup$
Are you saying the noise picked up inside the amp chassis would be equal to or greater than the noise picked up by the guitar pickups out in the world? That doesn't seem right to me. In the case of electric guitars, the part of the signal chain most prone to noise is the source (the pickups), not an interconnect (the cables or flywires or the traces on a PCB).
$endgroup$
– Todd Wilcox
31 mins ago
$begingroup$
@ToddWilcox I did say this was a quick and dirty answer, and may not apply quite so well to the specific scenario the asker is asking about. It is not the best answer and needs a lot of work but I don't have the time or energy to work on it right now, and frankly I'm amazed it got as many upvotes as it did. That said, the definition of "noise" that I'm using here is implicitly assuming that the signal you want is exactly what the transducer outputs, that the signal as it exists on the terminals of the transducer is noise-free by definition.
$endgroup$
– Hearth
27 mins ago
add a comment |
$begingroup$
Quick and dirty answer:
Buffering is one reason. Interconnects between things can have a lot of capacitance and require a lot (comparatively) of current to drive.
Noise immunity is another. Think about this scenario: Send a signal through a wire where it picks up, say, 10mV noise, then amplify it by 100x: total noise, 1000mV. But if you instead amplify it by 10x, then send it through the wire where it gets 10mV noise, then amplify by another 10x, your total signal amplification is still 100x, but your total noise is only 100mV.
$endgroup$
Quick and dirty answer:
Buffering is one reason. Interconnects between things can have a lot of capacitance and require a lot (comparatively) of current to drive.
Noise immunity is another. Think about this scenario: Send a signal through a wire where it picks up, say, 10mV noise, then amplify it by 100x: total noise, 1000mV. But if you instead amplify it by 10x, then send it through the wire where it gets 10mV noise, then amplify by another 10x, your total signal amplification is still 100x, but your total noise is only 100mV.
answered 2 hours ago
HearthHearth
5,11511339
5,11511339
$begingroup$
Are you saying the noise picked up inside the amp chassis would be equal to or greater than the noise picked up by the guitar pickups out in the world? That doesn't seem right to me. In the case of electric guitars, the part of the signal chain most prone to noise is the source (the pickups), not an interconnect (the cables or flywires or the traces on a PCB).
$endgroup$
– Todd Wilcox
31 mins ago
$begingroup$
@ToddWilcox I did say this was a quick and dirty answer, and may not apply quite so well to the specific scenario the asker is asking about. It is not the best answer and needs a lot of work but I don't have the time or energy to work on it right now, and frankly I'm amazed it got as many upvotes as it did. That said, the definition of "noise" that I'm using here is implicitly assuming that the signal you want is exactly what the transducer outputs, that the signal as it exists on the terminals of the transducer is noise-free by definition.
$endgroup$
– Hearth
27 mins ago
add a comment |
$begingroup$
Are you saying the noise picked up inside the amp chassis would be equal to or greater than the noise picked up by the guitar pickups out in the world? That doesn't seem right to me. In the case of electric guitars, the part of the signal chain most prone to noise is the source (the pickups), not an interconnect (the cables or flywires or the traces on a PCB).
$endgroup$
– Todd Wilcox
31 mins ago
$begingroup$
@ToddWilcox I did say this was a quick and dirty answer, and may not apply quite so well to the specific scenario the asker is asking about. It is not the best answer and needs a lot of work but I don't have the time or energy to work on it right now, and frankly I'm amazed it got as many upvotes as it did. That said, the definition of "noise" that I'm using here is implicitly assuming that the signal you want is exactly what the transducer outputs, that the signal as it exists on the terminals of the transducer is noise-free by definition.
$endgroup$
– Hearth
27 mins ago
$begingroup$
Are you saying the noise picked up inside the amp chassis would be equal to or greater than the noise picked up by the guitar pickups out in the world? That doesn't seem right to me. In the case of electric guitars, the part of the signal chain most prone to noise is the source (the pickups), not an interconnect (the cables or flywires or the traces on a PCB).
$endgroup$
– Todd Wilcox
31 mins ago
$begingroup$
Are you saying the noise picked up inside the amp chassis would be equal to or greater than the noise picked up by the guitar pickups out in the world? That doesn't seem right to me. In the case of electric guitars, the part of the signal chain most prone to noise is the source (the pickups), not an interconnect (the cables or flywires or the traces on a PCB).
$endgroup$
– Todd Wilcox
31 mins ago
$begingroup$
@ToddWilcox I did say this was a quick and dirty answer, and may not apply quite so well to the specific scenario the asker is asking about. It is not the best answer and needs a lot of work but I don't have the time or energy to work on it right now, and frankly I'm amazed it got as many upvotes as it did. That said, the definition of "noise" that I'm using here is implicitly assuming that the signal you want is exactly what the transducer outputs, that the signal as it exists on the terminals of the transducer is noise-free by definition.
$endgroup$
– Hearth
27 mins ago
$begingroup$
@ToddWilcox I did say this was a quick and dirty answer, and may not apply quite so well to the specific scenario the asker is asking about. It is not the best answer and needs a lot of work but I don't have the time or energy to work on it right now, and frankly I'm amazed it got as many upvotes as it did. That said, the definition of "noise" that I'm using here is implicitly assuming that the signal you want is exactly what the transducer outputs, that the signal as it exists on the terminals of the transducer is noise-free by definition.
$endgroup$
– Hearth
27 mins ago
add a comment |
$begingroup$
In audio gear, it is useful to do most of the signal manipulation at a standard level, known as "line level". This includes mixing, equalization, compression, etc.
Some signal sources (microphones, guitar pickups, etc.) do not inherently produce line level outputs, so a preamplifier is used to boost the signal to that level. Some signal sources (record players) require not only a boost, but also a special equalization to flatten the frequency response.
Then, after all of the signal processing is done, a second, "power" amplifier is used to drive the speaker(s).
This kind of modularity allows signal sources, processing stages, and different kinds of speakers to be mixed and matched freely.
$endgroup$
$begingroup$
In case anyone needs this broken down to the simplest level for electric guitar amps: the preamp gets the signal ready for the tone controls, then after the tone controls the power amp makes it ready for the speaker.
$endgroup$
– Todd Wilcox
37 mins ago
$begingroup$
Oh, you're right. I did not notice he was talking about amps in the same unit which was implied by the "in the same schematic part" bit.
$endgroup$
– Toor
35 mins ago
add a comment |
$begingroup$
In audio gear, it is useful to do most of the signal manipulation at a standard level, known as "line level". This includes mixing, equalization, compression, etc.
Some signal sources (microphones, guitar pickups, etc.) do not inherently produce line level outputs, so a preamplifier is used to boost the signal to that level. Some signal sources (record players) require not only a boost, but also a special equalization to flatten the frequency response.
Then, after all of the signal processing is done, a second, "power" amplifier is used to drive the speaker(s).
This kind of modularity allows signal sources, processing stages, and different kinds of speakers to be mixed and matched freely.
$endgroup$
$begingroup$
In case anyone needs this broken down to the simplest level for electric guitar amps: the preamp gets the signal ready for the tone controls, then after the tone controls the power amp makes it ready for the speaker.
$endgroup$
– Todd Wilcox
37 mins ago
$begingroup$
Oh, you're right. I did not notice he was talking about amps in the same unit which was implied by the "in the same schematic part" bit.
$endgroup$
– Toor
35 mins ago
add a comment |
$begingroup$
In audio gear, it is useful to do most of the signal manipulation at a standard level, known as "line level". This includes mixing, equalization, compression, etc.
Some signal sources (microphones, guitar pickups, etc.) do not inherently produce line level outputs, so a preamplifier is used to boost the signal to that level. Some signal sources (record players) require not only a boost, but also a special equalization to flatten the frequency response.
Then, after all of the signal processing is done, a second, "power" amplifier is used to drive the speaker(s).
This kind of modularity allows signal sources, processing stages, and different kinds of speakers to be mixed and matched freely.
$endgroup$
In audio gear, it is useful to do most of the signal manipulation at a standard level, known as "line level". This includes mixing, equalization, compression, etc.
Some signal sources (microphones, guitar pickups, etc.) do not inherently produce line level outputs, so a preamplifier is used to boost the signal to that level. Some signal sources (record players) require not only a boost, but also a special equalization to flatten the frequency response.
Then, after all of the signal processing is done, a second, "power" amplifier is used to drive the speaker(s).
This kind of modularity allows signal sources, processing stages, and different kinds of speakers to be mixed and matched freely.
answered 2 hours ago
Dave Tweed♦Dave Tweed
124k10153268
124k10153268
$begingroup$
In case anyone needs this broken down to the simplest level for electric guitar amps: the preamp gets the signal ready for the tone controls, then after the tone controls the power amp makes it ready for the speaker.
$endgroup$
– Todd Wilcox
37 mins ago
$begingroup$
Oh, you're right. I did not notice he was talking about amps in the same unit which was implied by the "in the same schematic part" bit.
$endgroup$
– Toor
35 mins ago
add a comment |
$begingroup$
In case anyone needs this broken down to the simplest level for electric guitar amps: the preamp gets the signal ready for the tone controls, then after the tone controls the power amp makes it ready for the speaker.
$endgroup$
– Todd Wilcox
37 mins ago
$begingroup$
Oh, you're right. I did not notice he was talking about amps in the same unit which was implied by the "in the same schematic part" bit.
$endgroup$
– Toor
35 mins ago
$begingroup$
In case anyone needs this broken down to the simplest level for electric guitar amps: the preamp gets the signal ready for the tone controls, then after the tone controls the power amp makes it ready for the speaker.
$endgroup$
– Todd Wilcox
37 mins ago
$begingroup$
In case anyone needs this broken down to the simplest level for electric guitar amps: the preamp gets the signal ready for the tone controls, then after the tone controls the power amp makes it ready for the speaker.
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– Todd Wilcox
37 mins ago
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Oh, you're right. I did not notice he was talking about amps in the same unit which was implied by the "in the same schematic part" bit.
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– Toor
35 mins ago
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Oh, you're right. I did not notice he was talking about amps in the same unit which was implied by the "in the same schematic part" bit.
$endgroup$
– Toor
35 mins ago
add a comment |
$begingroup$
A major reason for separate boxes for preamps and poweramps is the GROUND currents and also magnetic coupling. [there is numeric example, at 20KHz and 6 amps to the speakers, at end of this answer, with the Preamp only 10cm from the Power amplifier]
Suppose you built the preamp and the poweramp on the same PCB. Why not?
Some of the loudspeaker current will be flowing around on the GROUND, and end up combining with the input signal.
To minimize this "combining", make that PCB long and thin, so the PowerAmp Grounds are far away from the PreAmp Grounds.
How to improve on this? use long thing regions between the Preamp and the Poweramp.
In the extreme, a coax cable provides a long-thin-region, to ensure very small combining of input and output.
Give low millivolt signals from a vinyl record Moving Magnet cartridge, or even 0.5 millivolt from Moving Coil cartridges, that become near-100-volt audio outputs, the entire system needs 100,000:1 isolation.
simulate this circuit – Schematic created using CircuitLab
=============================================
How bad can crosstalk be? assume output current is 6 amps peak at 20,000Hz. The dI/dT is 6* d(sin(2*pi*20,000*time))/dT = 6 * 2*pi*20,000*cos(2pi20000T)
or dI/dT = 700,000 amps per second.
Assume the preamp input (remember that 1 millivolt signal from the cartridge, and you want at least 10,000:1 SNR or tonal feedback, thus 0.1 microvolt feedback is the desired floor) is 0.1 meter from the Speaker output.
V_magnetic_induce = (2.0e-7 * Area/Distance) * dI/dT
and we'll assume the input loop area (signal to ground) is 1cm by 4cm.
Now run the math; remember we need 0.1 microvolt feedback.
Vinduce = 2e-7Henry/meter * 1cm * 4cm /10cm * 700,000
Vinduce = 2e-7 * 0.0004meter/0.1meter * 700,000
Vinduce = 2e-7 * 0.004 * 7e+5
Vinduce = 2e-7 * 4e-3 * 7e+7 = 56 e-3 = 56 milliVolts.
The magnetic feedback, caused by having the Poweramplifer near the Preamplifier, is 56mV / 0.1 microvolt or 560,000X stronger than what "clean" music can tolerate.
$endgroup$
add a comment |
$begingroup$
A major reason for separate boxes for preamps and poweramps is the GROUND currents and also magnetic coupling. [there is numeric example, at 20KHz and 6 amps to the speakers, at end of this answer, with the Preamp only 10cm from the Power amplifier]
Suppose you built the preamp and the poweramp on the same PCB. Why not?
Some of the loudspeaker current will be flowing around on the GROUND, and end up combining with the input signal.
To minimize this "combining", make that PCB long and thin, so the PowerAmp Grounds are far away from the PreAmp Grounds.
How to improve on this? use long thing regions between the Preamp and the Poweramp.
In the extreme, a coax cable provides a long-thin-region, to ensure very small combining of input and output.
Give low millivolt signals from a vinyl record Moving Magnet cartridge, or even 0.5 millivolt from Moving Coil cartridges, that become near-100-volt audio outputs, the entire system needs 100,000:1 isolation.
simulate this circuit – Schematic created using CircuitLab
=============================================
How bad can crosstalk be? assume output current is 6 amps peak at 20,000Hz. The dI/dT is 6* d(sin(2*pi*20,000*time))/dT = 6 * 2*pi*20,000*cos(2pi20000T)
or dI/dT = 700,000 amps per second.
Assume the preamp input (remember that 1 millivolt signal from the cartridge, and you want at least 10,000:1 SNR or tonal feedback, thus 0.1 microvolt feedback is the desired floor) is 0.1 meter from the Speaker output.
V_magnetic_induce = (2.0e-7 * Area/Distance) * dI/dT
and we'll assume the input loop area (signal to ground) is 1cm by 4cm.
Now run the math; remember we need 0.1 microvolt feedback.
Vinduce = 2e-7Henry/meter * 1cm * 4cm /10cm * 700,000
Vinduce = 2e-7 * 0.0004meter/0.1meter * 700,000
Vinduce = 2e-7 * 0.004 * 7e+5
Vinduce = 2e-7 * 4e-3 * 7e+7 = 56 e-3 = 56 milliVolts.
The magnetic feedback, caused by having the Poweramplifer near the Preamplifier, is 56mV / 0.1 microvolt or 560,000X stronger than what "clean" music can tolerate.
$endgroup$
add a comment |
$begingroup$
A major reason for separate boxes for preamps and poweramps is the GROUND currents and also magnetic coupling. [there is numeric example, at 20KHz and 6 amps to the speakers, at end of this answer, with the Preamp only 10cm from the Power amplifier]
Suppose you built the preamp and the poweramp on the same PCB. Why not?
Some of the loudspeaker current will be flowing around on the GROUND, and end up combining with the input signal.
To minimize this "combining", make that PCB long and thin, so the PowerAmp Grounds are far away from the PreAmp Grounds.
How to improve on this? use long thing regions between the Preamp and the Poweramp.
In the extreme, a coax cable provides a long-thin-region, to ensure very small combining of input and output.
Give low millivolt signals from a vinyl record Moving Magnet cartridge, or even 0.5 millivolt from Moving Coil cartridges, that become near-100-volt audio outputs, the entire system needs 100,000:1 isolation.
simulate this circuit – Schematic created using CircuitLab
=============================================
How bad can crosstalk be? assume output current is 6 amps peak at 20,000Hz. The dI/dT is 6* d(sin(2*pi*20,000*time))/dT = 6 * 2*pi*20,000*cos(2pi20000T)
or dI/dT = 700,000 amps per second.
Assume the preamp input (remember that 1 millivolt signal from the cartridge, and you want at least 10,000:1 SNR or tonal feedback, thus 0.1 microvolt feedback is the desired floor) is 0.1 meter from the Speaker output.
V_magnetic_induce = (2.0e-7 * Area/Distance) * dI/dT
and we'll assume the input loop area (signal to ground) is 1cm by 4cm.
Now run the math; remember we need 0.1 microvolt feedback.
Vinduce = 2e-7Henry/meter * 1cm * 4cm /10cm * 700,000
Vinduce = 2e-7 * 0.0004meter/0.1meter * 700,000
Vinduce = 2e-7 * 0.004 * 7e+5
Vinduce = 2e-7 * 4e-3 * 7e+7 = 56 e-3 = 56 milliVolts.
The magnetic feedback, caused by having the Poweramplifer near the Preamplifier, is 56mV / 0.1 microvolt or 560,000X stronger than what "clean" music can tolerate.
$endgroup$
A major reason for separate boxes for preamps and poweramps is the GROUND currents and also magnetic coupling. [there is numeric example, at 20KHz and 6 amps to the speakers, at end of this answer, with the Preamp only 10cm from the Power amplifier]
Suppose you built the preamp and the poweramp on the same PCB. Why not?
Some of the loudspeaker current will be flowing around on the GROUND, and end up combining with the input signal.
To minimize this "combining", make that PCB long and thin, so the PowerAmp Grounds are far away from the PreAmp Grounds.
How to improve on this? use long thing regions between the Preamp and the Poweramp.
In the extreme, a coax cable provides a long-thin-region, to ensure very small combining of input and output.
Give low millivolt signals from a vinyl record Moving Magnet cartridge, or even 0.5 millivolt from Moving Coil cartridges, that become near-100-volt audio outputs, the entire system needs 100,000:1 isolation.
simulate this circuit – Schematic created using CircuitLab
=============================================
How bad can crosstalk be? assume output current is 6 amps peak at 20,000Hz. The dI/dT is 6* d(sin(2*pi*20,000*time))/dT = 6 * 2*pi*20,000*cos(2pi20000T)
or dI/dT = 700,000 amps per second.
Assume the preamp input (remember that 1 millivolt signal from the cartridge, and you want at least 10,000:1 SNR or tonal feedback, thus 0.1 microvolt feedback is the desired floor) is 0.1 meter from the Speaker output.
V_magnetic_induce = (2.0e-7 * Area/Distance) * dI/dT
and we'll assume the input loop area (signal to ground) is 1cm by 4cm.
Now run the math; remember we need 0.1 microvolt feedback.
Vinduce = 2e-7Henry/meter * 1cm * 4cm /10cm * 700,000
Vinduce = 2e-7 * 0.0004meter/0.1meter * 700,000
Vinduce = 2e-7 * 0.004 * 7e+5
Vinduce = 2e-7 * 4e-3 * 7e+7 = 56 e-3 = 56 milliVolts.
The magnetic feedback, caused by having the Poweramplifer near the Preamplifier, is 56mV / 0.1 microvolt or 560,000X stronger than what "clean" music can tolerate.
edited 1 hour ago
answered 2 hours ago
analogsystemsrfanalogsystemsrf
16.1k2822
16.1k2822
add a comment |
add a comment |
$begingroup$
To minimize the noise factor, which is the SNR of the output divided by the SNR of the input. An ideal amplifier should keep the SNR constant, since the input noise is amplified by the same amount as the input signal. A real amplifier, however, adds extra noise. The noise factor is given by
$$ F = 1 + frac{N_mathrm{additional}}{N_mathrm{input}G}.$$
If you cascade a series of amplifiers the total noise factor is given by Friis’ equation
$$F_mathrm{total} = F_1 + frac{F_2 - 1}{G_1} + frac{F_3 - 1}{G_1 G_2} + frac{F_4 - 1}{G_1 G_2 G_3} + dots.$$
Where $F_n$ is the noise factor of the nth stage and $G_n$ is the gain of the nth stage. This is because the additional noise of the first stage is amplified by the second and subsequent stages but the additional noise of the second stage is amplified by only the third and subsequent stages etc.
As you can see, the the noise factor of a given stage is divided by the gain product of all previous stages. So the first stage is the most important when it comes to noise. That’s why you have a low noise pre-amp stage as your very first component in the signal chain. This configuration has the added benefit of not having to worry about the noise figure of the power amplifier.
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$begingroup$
And this is true from DC to daylight, as they say. The first amplifier sets the noise figure is what they say in RF (as a rule of thumb).
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– mkeith
1 hour ago
add a comment |
$begingroup$
To minimize the noise factor, which is the SNR of the output divided by the SNR of the input. An ideal amplifier should keep the SNR constant, since the input noise is amplified by the same amount as the input signal. A real amplifier, however, adds extra noise. The noise factor is given by
$$ F = 1 + frac{N_mathrm{additional}}{N_mathrm{input}G}.$$
If you cascade a series of amplifiers the total noise factor is given by Friis’ equation
$$F_mathrm{total} = F_1 + frac{F_2 - 1}{G_1} + frac{F_3 - 1}{G_1 G_2} + frac{F_4 - 1}{G_1 G_2 G_3} + dots.$$
Where $F_n$ is the noise factor of the nth stage and $G_n$ is the gain of the nth stage. This is because the additional noise of the first stage is amplified by the second and subsequent stages but the additional noise of the second stage is amplified by only the third and subsequent stages etc.
As you can see, the the noise factor of a given stage is divided by the gain product of all previous stages. So the first stage is the most important when it comes to noise. That’s why you have a low noise pre-amp stage as your very first component in the signal chain. This configuration has the added benefit of not having to worry about the noise figure of the power amplifier.
$endgroup$
$begingroup$
And this is true from DC to daylight, as they say. The first amplifier sets the noise figure is what they say in RF (as a rule of thumb).
$endgroup$
– mkeith
1 hour ago
add a comment |
$begingroup$
To minimize the noise factor, which is the SNR of the output divided by the SNR of the input. An ideal amplifier should keep the SNR constant, since the input noise is amplified by the same amount as the input signal. A real amplifier, however, adds extra noise. The noise factor is given by
$$ F = 1 + frac{N_mathrm{additional}}{N_mathrm{input}G}.$$
If you cascade a series of amplifiers the total noise factor is given by Friis’ equation
$$F_mathrm{total} = F_1 + frac{F_2 - 1}{G_1} + frac{F_3 - 1}{G_1 G_2} + frac{F_4 - 1}{G_1 G_2 G_3} + dots.$$
Where $F_n$ is the noise factor of the nth stage and $G_n$ is the gain of the nth stage. This is because the additional noise of the first stage is amplified by the second and subsequent stages but the additional noise of the second stage is amplified by only the third and subsequent stages etc.
As you can see, the the noise factor of a given stage is divided by the gain product of all previous stages. So the first stage is the most important when it comes to noise. That’s why you have a low noise pre-amp stage as your very first component in the signal chain. This configuration has the added benefit of not having to worry about the noise figure of the power amplifier.
$endgroup$
To minimize the noise factor, which is the SNR of the output divided by the SNR of the input. An ideal amplifier should keep the SNR constant, since the input noise is amplified by the same amount as the input signal. A real amplifier, however, adds extra noise. The noise factor is given by
$$ F = 1 + frac{N_mathrm{additional}}{N_mathrm{input}G}.$$
If you cascade a series of amplifiers the total noise factor is given by Friis’ equation
$$F_mathrm{total} = F_1 + frac{F_2 - 1}{G_1} + frac{F_3 - 1}{G_1 G_2} + frac{F_4 - 1}{G_1 G_2 G_3} + dots.$$
Where $F_n$ is the noise factor of the nth stage and $G_n$ is the gain of the nth stage. This is because the additional noise of the first stage is amplified by the second and subsequent stages but the additional noise of the second stage is amplified by only the third and subsequent stages etc.
As you can see, the the noise factor of a given stage is divided by the gain product of all previous stages. So the first stage is the most important when it comes to noise. That’s why you have a low noise pre-amp stage as your very first component in the signal chain. This configuration has the added benefit of not having to worry about the noise figure of the power amplifier.
answered 1 hour ago
user110971user110971
3,4041717
3,4041717
$begingroup$
And this is true from DC to daylight, as they say. The first amplifier sets the noise figure is what they say in RF (as a rule of thumb).
$endgroup$
– mkeith
1 hour ago
add a comment |
$begingroup$
And this is true from DC to daylight, as they say. The first amplifier sets the noise figure is what they say in RF (as a rule of thumb).
$endgroup$
– mkeith
1 hour ago
$begingroup$
And this is true from DC to daylight, as they say. The first amplifier sets the noise figure is what they say in RF (as a rule of thumb).
$endgroup$
– mkeith
1 hour ago
$begingroup$
And this is true from DC to daylight, as they say. The first amplifier sets the noise figure is what they say in RF (as a rule of thumb).
$endgroup$
– mkeith
1 hour ago
add a comment |
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$begingroup$
There is also the problem of gain bandwidth product. For a given amplifier, more gain means less bandwidth. If you use too much gain in one stage, then you limit the bandwidth of that stage. This can lead to distortion - it takes gain and bandwidth for negative feedback to compensate for distortion.
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– JRE
2 hours ago
1
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You don't want the high currents (to the loudspeaker) anywhere near the input signal from the guitar pickup, or the vinyl-record signals.
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– analogsystemsrf
2 hours ago
1
$begingroup$
The first amplifier in any signal path usually is the one that adds all the noise to the signal. So the pre-amp must be designed so as to avoid adding more noise to the signal than necessary. Generally low noise devices and design techniques are incompatible with high power devices and design techniques.
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– mkeith
1 hour ago
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@mkeith I think your comment is the best general answer I've seen yet on this. Combined with Dave Tweed's answer, it all makes sense in terms of guitar amplification.
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– Todd Wilcox
23 mins ago