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Question about line level input
This thread has 20 replies. Displaying posts 1 through 15.
Post 1 made on Saturday July 25, 2015 at 14:41
crosen
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Suppose you have an amplifier that takes a standard, unbalanced, analogue line level audio signal via RCA terminals.

What I'd like to understand is what determines the maximum strength of this signal that the amplifier can handle before the amplification is clipped. Also, is this max oriented on voltage? or amps? or wattage? or ...?

The question is not in connection with a particular project. It's just something I'd like to understand better to the end of better designied audio systems. Much appreciate any input.
If it's not simple, it's not sufficiently advanced.
Post 2 made on Saturday July 25, 2015 at 15:28
fcwilt
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The design and settings of the amp - from input all the way to the output.

The rating will either be in volts, dBv, dBu or perhaps dBm.

If you desire more info:

[Link: en.wikipedia.org]
Regards, Frederick C. Wilt
Post 3 made on Saturday July 25, 2015 at 15:50
buzz
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crosen,

For line level inputs, since their input impedance is typically very high, current is very, very low, therefore power is low and input voltage is the limit. Mild overload will simply sound bad, major overload, such as a static discharge, could result in physical destruction of circuit elements.

On another level, lets say that the gain of our gadget is 10x. If the power supply voltage is 10V, the output cannot go beyond 10V. Therefore 1V would be the maximum input before clipping. (Note that it is not generally possible for output to exactly reach the power supply "rail".)

From a system design standpoint, the line level input will have a maximum input voltage spec that should be respected. The output from this line level device should match whatever the follow-on device needs. If the follow-on device is a loudspeaker, then significant current is involved and power and cooling should be discussed.
Post 4 made on Saturday July 25, 2015 at 16:53
Ernie Gilman
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Buzz has A LOT to say about this, and it's all correct. There's just so much to it!

On July 25, 2015 at 14:41, crosen said...
Suppose you have an amplifier that takes a standard, unbalanced, analogue line level audio signal via RCA terminals.

What I'd like to understand is what determines the maximum strength of this signal that the amplifier can handle before the amplification is clipped. Also, is this max oriented on voltage? or amps? or wattage? or ...?

First, the line input. A line input samples the input voltage. There's negligible current into a line input. The following amplifier stages amplify this signal.

Now, if there's an amplification stage right at this input, then the line input will cause distortion if, when amplified, the output level of that stage is too high of a voltage for that stage to cleanly pass. For instance, if that stage can output 3 volts but the line level is so high that, when amplified, 4 volts would come out of that stage, that last volt will not come through cleanly and will appear mostly as distortion.

Next, we have the output stages where voltage and current are delivered. If all the previous stages are clean but the signal presented by those stages to this output stage makes it try to output more voltage than it can do cleanly, then there's distortion.

There can also be distortion due to current limits. If the amp is connected to a low impedance load and the power supply voltage of the amp does not stay up at its normal voltage on peaks; then those peaks won't accurately represent the input signal. Not matching the input is the definition of distortion.

The question is not in connection with a particular project. It's just something I'd like to understand better to the end of better designied audio systems. Much appreciate any input.

There's a horrible term in this industry that is very helpful if you understand what it means and don't take it literally. That term is gain structure. If your gain structure is incorrect, you will have excess noise, excess distortion, or both.

Let's look at a system that's a variable output source; a preamp or mixer; and a trimmable power amplifier.

Let's say you have an adjustable output CD player (a pro model) in that system. If you turn the volume way down on the CD, you'll have to turn the volume way up on the mixer to hear it. Turning the mixer volume way up, you'll start to hear the hiss normal in all electronic circuits, that we mask by having the audio much louder than the hiss. Once this hiss can be heard, it's always there.

But if we turn up the CD player, we can then turn down the mixer, which will turn down the amount of hiss in the signal. See? Two components, different performance depending on where we set their volume controls.

Now, if the CD, at full volume, outputs so much volume that it causes distortion in the mixer, then you've got distortion and turning down the mixer won't fix it. Again, two components, and adjustments that can cause problems.

What you want to do is turn the CD up enough so that its level hides the noise but keep it down enough that it doesn't create distortion in the mixer. All this is called having proper gain structure.

Now add the power amp. If its volume controls are SET WAY THE HELL UP, then you might not be able to light more than one or two LEDs on the mixing board without driving people out of the place. And residual hiss in the mixer can be audible. Turn the power amp down some, you can turn up the volume control on the mixer, and things will work better. Gain structure is the balancing act of all inputs and outputs for minimum distortion and minimum noise.


The reason you can't take term "gain structure" literally? It's because every mixer, preamp, or power amplifier has gain blocks, circuits where a certain non-adjustable amount of gain occurs. Let's say we have a power amp with an input control and a maximum voltage gain* of 30. 1 volt of line level in equals an output voltage of 30. 30 volts AC is about 110 watts into 8 ohms.

What happens when we turn down the input control? We don't adjust the gain of the amp. It would still output 30 volts if fed 1 volt. We now feed it less than 1 volt, so its gain of 30 now outputs less than 30.

No volume control or slider adjusts the actual gain of any stage. It attenuates the signal from its input value all the way down to zero. And no gain ever changes.

However, it's convenient to talk as though we're changing the gain. To begin with, gain is only one syllable! Next, we're amplifying, so it's easier to talk about voltages increasing than talking about how we might minimize their decreasing. So we talk about changing gain as though that's what we're doing.



*going beyond talking about voltage gain is another subject
A good answer is easier with a clear question giving the make and model of everything.
"The biggest problem in communication is the illusion that it has taken place." -- G. “Bernie” Shaw
OP | Post 5 made on Saturday July 25, 2015 at 17:16
crosen
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Thanks very much for all the great info. I'm going to set aside some time to fully digest this.
If it's not simple, it's not sufficiently advanced.
Post 6 made on Saturday July 25, 2015 at 18:31
alihashemi
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This can be made much simpler. IDEAL amplifiers swing rail to rail. This means they can take a small signal and take it all the way down to their lowest voltage or all the way up to their highest voltage (these are the +/- voltage supplies that are connected to the amplifier). Now, that's the ideal case. Transistors can never actually output EXACTLY that highest or lowest voltage due to leakage current. For example, we can just say this: If we have an amplifier that is connected to a +/-15V supply, it'll be able to output up to +/-14.7V. When your amplified signal tries to go past the 14.7V limit, clipping occurs. Generally, that extra voltage drop is due to something called threshold voltage.

As Ernie said, these amplifiers have fixed gain, so the max value you can input will always be in terms of voltage. Also, most Audio Amplifiers have FET inputs meaning they have infinite input impedance and no current flows in (ideally no current, there is always leakage), therefore this value will never be a maximum in terms of current.

Ernie is right, however, on loading the amplifier with a low impedance. Ideal amplifiers have 0 output impedance. Since you're always loading an amplifier in parallel, whatever load will be in parallel with 0 ohms which can then be approximated as ~0 load. This value is actually closer to ~10 Ohms for good amplifiers and ~100 Ohms for the crappy ones. If you load an amplifier with something that is less than its output impedance, you'll generally get distortion due to higher current draw.

The conversation past here gets very complicated. Audio amplifiers (talking down at the IC level here) have multiple stages, each designed for a specific purpose. The Class A,B,C, AB, etc. amplifiers we all know of represent the output stage, etc.
Ali Hashemi
Post 7 made on Saturday July 25, 2015 at 23:47
Ernie Gilman
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While this is simpler it brings up some things I left out on purpose.

On July 25, 2015 at 18:31, alihashemi said...
This can be made much simpler. IDEAL amplifiers swing rail to rail. This means they can take a small signal and take it all the way down to their lowest voltage or all the way up to their highest voltage (these are the +/- voltage supplies that are connected to the amplifier). Now, that's the ideal case. Transistors can never actually output EXACTLY that highest or lowest voltage due to leakage current.

I think it's not leakage current, but minimum voltage drop across the transistor when it's on full. With NPN and PNP transistors, that's in the range of 0.6 volts, germanium 0.3 volts. But we don't use germanium for output devices because they can't pass as much current as the other technologies. I do't know what it is for other kinds of devices.

For example, we can just say this: If we have an amplifier that is connected to a +/-15V supply, it'll be able to output up to +/-14.7V. When your amplified signal tries to go past the 14.7V limit, clipping occurs. Generally, that extra voltage drop is due to something called threshold voltage.

I must go look up threshold voltage, though that description doesn't include any "extra voltage drop...." wait a minute: that's probably the standard voltage drop across a junction.

There's another issue that makes this kind of amp look like it outputs less than rail to rail. If it could swing completely from rail to rail, that's its instantaneous voltage. If the signal is a sine wave, its RMS voltage is 0.707 times its peak voltage. So in this particular perfect world, an amp that outputs 10 volts peak to peak, with a plus and minus 10 volt power supply (20 volts rail to rail), outputs 7.07 volts RMS. I threw in the 20 volt number because it's true that rail to rail is twice the peak voltage of the RMS voltage, and I didn't want someone coming back to say "but wait, there's 20 volts there! How does it become 7.07 at full volume?"

As Ernie said, these amplifiers have fixed gain, so the max value you can input will always be in terms of voltage. Also, most Audio Amplifiers have FET inputs meaning they have infinite input impedance and no current flows in (ideally no current, there is always leakage), therefore this value will never be a maximum in terms of current.

I haven't investigated actual input circuits for years, but I doubt that most inputs are FET inputs. Back in plain old transistor days, some line inputs went directly to an input buffer, some went through switching, volume and tone without any amplification at all! In other words, things varied.

Another high input impedance device is the op amp. The inverting input is a short to ground and the noninverting input is (ideally) an open circuit! Both of those facts dramatically change the details of how the line level signal is handled.

Ernie is right, however, on loading the amplifier with a low impedance. Ideal amplifiers have 0 output impedance. Since you're always loading an amplifier in parallel, whatever load will be in parallel with 0 ohms which can then be approximated as ~0 load. This value is actually closer to ~10 Ohms for good amplifiers and ~100 Ohms for the crappy ones. If you load an amplifier with something that is less than its output impedance, you'll generally get distortion due to higher current draw.

The output impedance of an amp is an impedance in series with the output, not in parallel. An amp with zero output impedance will not decrease in output voltage due to low impedance loads. Those are the kind of amps that blow up or weld when their active outputs are shorted with a signal coming out of them.

Output impedance limits the amount of current that can be delivered, I agree. But that means it can't be in parallel with the load. An output impedance of zero ohms per that concept is a short circuit across the output. Look up buildout resistance for an example of adding resistance in series with an output in order to get a better result.

The conversation past here gets very complicated. Audio amplifiers (talking down at the IC level here) have multiple stages, each designed for a specific purpose. The Class A,B,C, AB, etc. amplifiers we all know of represent the output stage, etc.

Yes.

crosen, maybe you can ask another question if all this has been so much as to obscure the thing you wanted to ask about!
A good answer is easier with a clear question giving the make and model of everything.
"The biggest problem in communication is the illusion that it has taken place." -- G. “Bernie” Shaw
Post 8 made on Sunday July 26, 2015 at 02:16
Ernie Gilman
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I just ran across a GREAT white paper on headphone amps. The latest trend in headphone amps is toward zero ohms source impedance. The paper talks about the drawbacks of buildout resistance, which is electronically a series resistance. I have not had the time to read it yet, but the frequency response curves tell that they're looking at how source impedance affects frequency response.

The discussion with a link is at [Link: proaudiodesignforum.com]. I could not get the first paper to open, though. It bears investigating.
A good answer is easier with a clear question giving the make and model of everything.
"The biggest problem in communication is the illusion that it has taken place." -- G. “Bernie” Shaw
Post 9 made on Sunday July 26, 2015 at 12:45
alihashemi
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On July 25, 2015 at 23:47, Ernie Gilman said...
While this is simpler it brings up some things I left out on purpose.

I think it's not leakage current, but minimum voltage drop across the transistor when it's on full. With NPN and PNP transistors, that's in the range of 0.6 volts, germanium 0.3 volts. But we don't use germanium for output devices because they can't pass as much current as the other technologies. I do't know what it is for other kinds of devices.

I must go look up threshold voltage, though that description doesn't include any "extra voltage drop...." wait a minute: that's probably the standard voltage drop across a junction.

Close. That voltage drop is only valid for BJT's. There are down drops. One across the Base Emitter junction, VBE, and one across the Collector Emitter junction, VCE. For a BJT to be in "amplification" or active mode, it requires a VBE of ~0.7V and a VCE>VCEsat. VCEsat is generally ~0.2V. For MOSFETS this is different. Most ICs mostly consist of FETS now a days. One of the few drawbacks of using MOSFET's instead of BJT's is power, thus BJTs still exist in almost all amplification circuit. One of the largest benefits of MOSFETs is that no current flows into their gate and also, the gate is pretty much a capacitor thus it has infinite input impedance. Like BJT's, FET's also have different operating modes. Additionally, they have a threshold voltage. This is actually a built in chemical potential that must be overcome for the device to turn "on".

There's another issue that makes this kind of amp look like it outputs less than rail to rail. If it could swing completely from rail to rail, that's its instantaneous voltage. If the signal is a sine wave, its RMS voltage is 0.707 times its peak voltage. So in this particular perfect world, an amp that outputs 10 volts peak to peak, with a plus and minus 10 volt power supply (20 volts rail to rail), outputs 7.07 volts RMS. I threw in the 20 volt number because it's true that rail to rail is twice the peak voltage of the RMS voltage, and I didn't want someone coming back to say "but wait, there's 20 volts there! How does it become 7.07 at full volume?"

The only way to actually get the supply voltage at the output of a transistor is to put it into cutoff mode. Ideally, this means that there is zero current flowing through the transistor, which is the path to ground, thus allowing for the supply voltage to appear at the output node. The problem is that due to leakage current, the current in cutoff mode will never be zero, causing a voltage drop and thus making the output voltage less than the supply. There are a lot of other factors that can cause this issue like frequency,etc. I am not sure if I understand your question, though?

I haven't investigated actual input circuits for years, but I doubt that most inputs are FET inputs. Back in plain old transistor days, some line inputs went directly to an input buffer, some went through switching, volume and tone without any amplification at all! In other words, things varied.

Yes, this is exactly true. Input stages are usually buffer stages and that is to ensure that input impedance is nearly infinity. Most buffers, or in other words source followers, are made of MOSFETs entirely.

Another high input impedance device is the op amp. The inverting input is a short to ground and the noninverting input is (ideally) an open circuit! Both of those facts dramatically change the details of how the line level signal is handled.

Op amps are high input impedance generally because of their buffered inputs (either source followers or emitter followers). The inverting input is definitely not a short to ground. It depends on the design. A bare op-amp without anything connected to it outside of the IC is a differential amplifier with almost infinite gain. Negative feedback allows you to set gain, lower output impedance, and raise the input impedance.

The output impedance of an amp is an impedance in series with the output, not in parallel. An amp with zero output impedance will not decrease in output voltage due to low impedance loads. Those are the kind of amps that blow up or weld when their active outputs are shorted with a signal coming out of them.

This is not true. First, if the output was a series resistance with the load, a load would never change the amount of current drawn because it is in series. Additionally, if you look at a transistor level circuit diagram of any basic differential amplifier (not even a full blown op amp) you will see that the output impedance always falls in parallel with the load resistance and that is why low impedance loads change the actual gain (by changing the circuit) and thus make for a lower output. Look at the picture below:

 photo CMOS OpAmp.png

This is a very simple two stage op amp. If you look at the output node, you will see that the output impedance is the impedance seen at the drain of Q6 in parallel with the resistance seen at the drain of Q7. A load resistance will fall in parallel with the resistance seen at the drain of Q7 and thus it wall fall in parallel with the final output impedance.


Output impedance limits the amount of current that can be delivered, I agree. But that means it can't be in parallel with the load. An output impedance of zero ohms per that concept is a short circuit across the output. Look up buildout resistance for an example of adding resistance in series with an output in order to get a better result.

You are right, the value is never actually zero, but ideally, it should be. There are a couple of things I've learned in amplifier designs and that is that a perfect amp has the following:

1. infinite differential gain
2. zero common mode gain and thus infinite CMRR
3. infinite input impedance
4. zero output impedance
5. infinite bandwidth
6. zero input offset voltage

Yes.

crosen, maybe you can ask another question if all this has been so much as to obscure the thing you wanted to ask about!
Ali Hashemi
Post 10 made on Sunday July 26, 2015 at 12:51
alihashemi
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On July 26, 2015 at 02:16, Ernie Gilman said...
I just ran across a GREAT white paper on headphone amps. The latest trend in headphone amps is toward zero ohms source impedance. The paper talks about the drawbacks of buildout resistance, which is electronically a series resistance. I have not had the time to read it yet, but the frequency response curves tell that they're looking at how source impedance affects frequency response.

The discussion with a link is at [Link: proaudiodesignforum.com]. I could not get the first paper to open, though. It bears investigating.

Frequency response is very dependent on impedance. Two very simple method used for finding high and low cutoff frequencies are the open and short circuit time constant methods. The way this works is that at low frequencies, internal capacitance of all transistors, etc. can be assumed to be open circuits. If one can then find the resistance seen by each external capacitance, one can find each pole in the transfer function. Additionally, at high frequencies, all external capacitances can be seen as shorts. If one then finds the resistance seen by each internal capacitance, one can find the other poles in the transfer function. Pretty much any resistance added within the circuit will affect the cutoff frequencies and thus the bandwidth.
Ali Hashemi
Post 11 made on Sunday July 26, 2015 at 16:58
Ernie Gilman
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The poor guy just asked a simple question about line level inputs..................
A good answer is easier with a clear question giving the make and model of everything.
"The biggest problem in communication is the illusion that it has taken place." -- G. “Bernie” Shaw
Post 12 made on Sunday July 26, 2015 at 17:50
fcwilt
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On July 26, 2015 at 16:58, Ernie Gilman said...
The poor guy just asked a simple question about line level inputs..................

Which is why I gave him a simple answer. :)

Last edited by fcwilt on July 27, 2015 03:58.
Regards, Frederick C. Wilt
Post 13 made on Sunday July 26, 2015 at 20:51
amirm
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Ali, there are some mistakes in your explanation. The voltage drop across a bipolar transistor in an amplifier is Vce, not Vbe. There are two junctions there and therefore the drop is 1.2 volts, not 0.6 and certainly not some small leakage voltage. FETs are used in small minority of power amplifier designs. Vast majority use bipolar transistors.

The impedance as seen by the load is also in series as Ernie explained. This is shown in the equiv. circuit of any amplifier:



As you see, Ro (output impedance) is in series with the load.
Amir
Founder, Madrona Digital, http://madronadigital.com
Founder, Audio Science Review, http://audiosciencereview.com
Post 14 made on Monday July 27, 2015 at 02:00
alihashemi
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On July 26, 2015 at 20:51, amirm said...
Ali, there are some mistakes in your explanation. The voltage drop across a bipolar transistor in an amplifier is Vce, not Vbe. There are two junctions there and therefore the drop is 1.2 volts, not 0.6 and certainly not some small leakage voltage.

There are three voltage drops in a transistor: Vbe, Vce, Vcb, which are you referring to? The only one that you can actually get off the top of your head is Vbe because for forward active mode, or in other words amplification mode, it HAS to be ~0.7V. The voltage drop across Vce has to be GREATER THAN Vcesat which is ~0.3V but that isn't a set number. Depending on how much current your drawing and the circuit itself that value will be different. If you take a look at any document explaining the operating regions of a BJT you'll see that there are 4 modes and that only forward active can be used for amplification. The two things I mentioned above are required for the BJT to be in that mode.

|FETs are used in small minority of power amplifier designs. Vast majority use |bipolar transistors.

This is true, and I mentioned that when I said FET's drawback is in power, but FET's are common for input stages. In fact, they're almost better in every case except for power and that is usually only one stage in an amplifier. Even the simplest op amps have 4-5 stages of amplification.

The impedance as seen by the load is also in series as Ernie explained. This is shown in the equiv. circuit of any amplifier:



As you see, Ro (output impedance) is in series with the load.

Also, sorry, I should've been more clear. Yes you can model it that way at DC. However, at AC, all DC sources are shorted, thus putting load resistances in parallel with the output resistance. Please look at page two of the document below. Input and output impedances as well as gain are all small signal values which are all given at AC, not DC.

[Link: whites.sdsmt.edu]
Ali Hashemi
Post 15 made on Monday July 27, 2015 at 08:06
Mario
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I was going to get deep into it... I'm glad you guys beat me to it.
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