I need a Gain of 1 in my feedbackloop. But which values are better, lower or higher? R1=120k, R2 120k or R1=10Ω and R2=10Ω?
Best regards.
Best regards.
Inverting, or non-inverting? (I'm guessing it's inverting.)
What happens if you make the feedback resistors ten ohms? The answer is super simple, and glaringly obvious.
What happens if you make the feedback resistors ten ohms? The answer is super simple, and glaringly obvious.
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Generally you use kohm (k = 1000) resistors in a feedback network. You wouldn't use a value like 10 ohms unless you had a specific reason.
For audio the general procedure is to keep to lower values, say under 50k. But any values will work.
After that, "gain of one" can be a confusing way to put it. It's better to say unity gain, or else state the gain figure, as shown below.
Again referring to the illustration below, R2 is generically called Rf, for feedback resistor. R3 is generically called Rg, for gain resistor. So the gain formula is generically stated as Gain = 1 + (Rf / Rg).
But in a real schematic these components would always be numbered like any other, not called Rf or Rg.
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Generally you use kohm (k = 1000) resistors in a feedback network. You wouldn't use a value like 10 ohms unless you had a specific reason.
For audio the general procedure is to keep to lower values, say under 50k. But any values will work.
After that, "gain of one" can be a confusing way to put it. It's better to say unity gain, or else state the gain figure, as shown below.
Again referring to the illustration below, R2 is generically called Rf, for feedback resistor. R3 is generically called Rg, for gain resistor. So the gain formula is generically stated as Gain = 1 + (Rf / Rg).
But in a real schematic these components would always be numbered like any other, not called Rf or Rg.
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Attachments
Thanks for the explanations. Its a non-inverting Amp. I read about Johnson Noise and thought about the right selection of the gain resistor values. then i will use Rf=Rg=1k to obtain a gain of 2.
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Right idea. But wait, you thought it would be that simple? C'mon.
Don't panic, things are going to get back to simple right away. But first I have to tell you that you need to consider the load-driving ability of your op amp. Data sheets generally state this as "drives 600 ohm loads," or words to that effect.
But what's a "load"? There's never--or rarely--just one resistor connected to an op amp's output, and the calculations can get complex.
Complex calculations are for engineers. We mere humans just use rules of thumb. So getting back to simple, the rule of thumb is never require an op amp to drive less than 2k unless it's the final output (to speakers or headphones).
In this particular case Rf would be 2.2k (the closest standard value), and Rg would be sized to match.
You're right to consider Johnson noise, but broadly speaking this is not something to worry about until you get up around 100k, in that area.
Addendum: remember that rules of thumb are, of necessity, very broad statements. You'll see this one--and all of them--broken from time to time. Not a problem, it'll work fine.
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Right idea. But wait, you thought it would be that simple? C'mon.
Don't panic, things are going to get back to simple right away. But first I have to tell you that you need to consider the load-driving ability of your op amp. Data sheets generally state this as "drives 600 ohm loads," or words to that effect.
But what's a "load"? There's never--or rarely--just one resistor connected to an op amp's output, and the calculations can get complex.
Complex calculations are for engineers. We mere humans just use rules of thumb. So getting back to simple, the rule of thumb is never require an op amp to drive less than 2k unless it's the final output (to speakers or headphones).
In this particular case Rf would be 2.2k (the closest standard value), and Rg would be sized to match.
You're right to consider Johnson noise, but broadly speaking this is not something to worry about until you get up around 100k, in that area.
Addendum: remember that rules of thumb are, of necessity, very broad statements. You'll see this one--and all of them--broken from time to time. Not a problem, it'll work fine.
.
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Hmmm. You're totally winging it, aren't you? Well, nothing wrong with that. When in doubt, take a shot, see what happens.
Most op amps won't drive a 40 ohm load. You might be well served to take a look at the very well known Cmoy, which is here:
A Pocket Headphone Amplifier | HeadWize
Enclosures don't have to be an Altoids tin, or on the other hand something elaborate. I personally have been known to use tuna fish cans, or if you or your neighbor has a cat you're home free.
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Hmmm. You're totally winging it, aren't you? Well, nothing wrong with that. When in doubt, take a shot, see what happens.
Most op amps won't drive a 40 ohm load. You might be well served to take a look at the very well known Cmoy, which is here:
A Pocket Headphone Amplifier | HeadWize
Enclosures don't have to be an Altoids tin, or on the other hand something elaborate. I personally have been known to use tuna fish cans, or if you or your neighbor has a cat you're home free.
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I have a Buffer lt1010 on the output. Do you know wich Bias resistor Value i should take? And i have 30mA in the output, is it to much for headphones?
The data sheet implies the bias resistor will set the quiescent current, which if I'm understanding it correctly means you can decide whether you essentially want it to run in class A or not. You would need to know the peak voltage that you will see across the load resistance to make an informed choice. My own tests when I designed my headphone amp were that around 0.8 volts pk/pk were more than enough to drive the headphones to any level I would want. So that would suggest around 20 ma peak.
From the data sheet,
For the TO-220 package, the output stage quiescent current can be increased by connecting a resistor between the BIAS pin and V+. The increase is equal to the bias terminal voltage divided by this resistance.
From the data sheet,
For the TO-220 package, the output stage quiescent current can be increased by connecting a resistor between the BIAS pin and V+. The increase is equal to the bias terminal voltage divided by this resistance.
Thank you for your explanation, and wich resistance value is it for 0.8 volts peak voltage? Or which chart I have to look at?
Lol, I knew you would ask that 😀
0.8 volts was the peak/peak voltage that I found covered all listening with my own Sony MDR-V7 headphones. So its just a ballpark figure when other headphones are considered because all are different.
To select a resistor to bias the IC you will have to follow the formula given in the data sheet (that I quoted above in the text). If power consumption isn't an issue then try something that will give say 60ma quiescent current in the output stage. That would cover all practical headphone use. I ran my single ended class A headphone amp at 50ma.
0.8 volts was the peak/peak voltage that I found covered all listening with my own Sony MDR-V7 headphones. So its just a ballpark figure when other headphones are considered because all are different.
To select a resistor to bias the IC you will have to follow the formula given in the data sheet (that I quoted above in the text). If power consumption isn't an issue then try something that will give say 60ma quiescent current in the output stage. That would cover all practical headphone use. I ran my single ended class A headphone amp at 50ma.
I have another question. how much current through the headphones ist ok? My output have 25mA it is to much?
I'm not quite sure what you mean by that tbh.
The headphones current depends totally on the applied voltage (the signal). The more you turn it up, the more signal voltage appears across the 'phones and so the more current they draw. 1 volt across 40 ohms would draw 25 milliamps. That would be loud.
The headphones current depends totally on the applied voltage (the signal). The more you turn it up, the more signal voltage appears across the 'phones and so the more current they draw. 1 volt across 40 ohms would draw 25 milliamps. That would be loud.
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I don't mean to be smarty mouth, but this thread is nagging at me. Obviously npn never heard of Ohm's law, so I'm going to jump back in because I speak newbie. Translation is required and I'm going to take my shot.
Amps is a slave, not a master. You speak of your output having 25ma, but this means there are 25ma available, and the critical key word is "available." It does not mean that 25ma flows at all times, or that it ever flows.
Amperage (current) flows in obedience to voltage, and to the resistance of whatever load is present. Which is to to say that amperage, compelled by voltage, flows through a resistive load.
Voltage is a force, it doesn't move, it compels motion. Amperage has physical existence (or close enough), it's compelled by voltage to flow. Resistance is whatever load you design into a circuit, or whatever you hook to the output.
All of which was elegantly summed up in 1827 by a German researcher named Georg Simon Ohm. Hence we have Ohm's Law, which in electricity/electronics rules all:
E=IR. As well stated as I=E/R, or R=E/I.
Getting to cases, we use voltage amplifiers. A small voltage is applied to the input, a larger voltage appears at the output. The voltage is applied across (it's correct to say "across," not "to") a load, which is a speaker, headphones, a following circuit, whatever.
Voltage is applied across a load, and amperage flows in obedience to both voltage and load. Usually the only question about amperage is "do I have enough available?" This is why you're using an LT1010, because it can carry/supply more current. Pretty much any op amp can apply enough voltage across your headphones to make them work, but can't necessarily supply enough current without making that unpleasant "snap" that indicates catastrophic failure.
After all of which I'm force to mention impedance. Your headphones don't actually have 40 ohms resistance, they have an impedance of 40 ohms.
Impedance is the AC version of what in DC is resistance. We use a different term because impedance changes with frequency, wheras resistance does not. Generally, DIYers probably shouldn't worry too much about impedance, we just stick with rules of thumb and build it the way the schematic says. Don't feel bad about not having mastery of impedance, there are engineers who don't fully catch on.
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I don't mean to be smarty mouth, but this thread is nagging at me. Obviously npn never heard of Ohm's law, so I'm going to jump back in because I speak newbie. Translation is required and I'm going to take my shot.
Amps is a slave, not a master. You speak of your output having 25ma, but this means there are 25ma available, and the critical key word is "available." It does not mean that 25ma flows at all times, or that it ever flows.
Amperage (current) flows in obedience to voltage, and to the resistance of whatever load is present. Which is to to say that amperage, compelled by voltage, flows through a resistive load.
Voltage is a force, it doesn't move, it compels motion. Amperage has physical existence (or close enough), it's compelled by voltage to flow. Resistance is whatever load you design into a circuit, or whatever you hook to the output.
All of which was elegantly summed up in 1827 by a German researcher named Georg Simon Ohm. Hence we have Ohm's Law, which in electricity/electronics rules all:
E=IR. As well stated as I=E/R, or R=E/I.
E is voltage (Electromotive force). I is current ("Intensité de courant," amperage). R is Resistance (load).
One volt applied to a resistive load of 1 ohm will cause 1 amp of current to flow through the load.
One volt applied to a resistive load of 2 ohms will cause 1/2 amp of current to flow through the load.
Current always flows in obedience to voltage applied, and load present.One volt applied to a resistive load of 1 ohm will cause 1 amp of current to flow through the load.
One volt applied to a resistive load of 2 ohms will cause 1/2 amp of current to flow through the load.
Getting to cases, we use voltage amplifiers. A small voltage is applied to the input, a larger voltage appears at the output. The voltage is applied across (it's correct to say "across," not "to") a load, which is a speaker, headphones, a following circuit, whatever.
Voltage is applied across a load, and amperage flows in obedience to both voltage and load. Usually the only question about amperage is "do I have enough available?" This is why you're using an LT1010, because it can carry/supply more current. Pretty much any op amp can apply enough voltage across your headphones to make them work, but can't necessarily supply enough current without making that unpleasant "snap" that indicates catastrophic failure.
After all of which I'm force to mention impedance. Your headphones don't actually have 40 ohms resistance, they have an impedance of 40 ohms.
Impedance is the AC version of what in DC is resistance. We use a different term because impedance changes with frequency, wheras resistance does not. Generally, DIYers probably shouldn't worry too much about impedance, we just stick with rules of thumb and build it the way the schematic says. Don't feel bad about not having mastery of impedance, there are engineers who don't fully catch on.
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But my LTspice simulation show me a current of 25mA through the Load Resistor(headphones). Does it mean I have on full power the 25mA?
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First thing: get over amps. Forget about amps. You don't care about amps.
Second thing, remember how headphones don't have resistance, they have impedance? You're not simulating your headphones in LTspice, you just have a resistance, which is not the same thing.
Third thing, forget about amps. Now then, going on from there...
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First thing: get over amps. Forget about amps. You don't care about amps.
Second thing, remember how headphones don't have resistance, they have impedance? You're not simulating your headphones in LTspice, you just have a resistance, which is not the same thing.
Third thing, forget about amps. Now then, going on from there...
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To begin with, all you want to know is how to hook up your LT1010. So why am I telling you all this about Ohm's law, and voltage is a force, and blah blah?
Because you suffer from two very common newbie misconceptions: that voltage flows, and that amperage enters things on its own, neither of which is true. So pardon my speaking frankly, but you don't know enough to ask questions that have answers. For instance, "i have 30mA in the output, is it to much for headphones" is simply meaningless. I'm not trying to be a smart guy, I'm just stating the fact.
But even so I probably overdid it. After all, you didn't ask for a course in basic electronics. And even if you did, probably all you really need to know is:
And maybe not even that.
So let's attack from a different angle. It seems all you really want to know is: 1) How much gain should your circuit have, and: 2) How should the LT1010 be connected to give that much gain?
To find out the answer to #1, state your audio source. If it's a CD player, a computer, or similar, then a gain of around 5 will do fine. No need to be precise, just get as close to 5 as standard component values might allow, err on the low side.
To find out the answer to #2...well...is there some reason you're not just using the schematic in the data sheet? The "Typical Application" on page one? Data sheet here:http://www.linear.com/docs/1683
Or else, as was already asked, what circuit are you using? That has to be known. You don't have to post the circuit itself, if that's hard to do, but provide a link, something.
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To begin with, all you want to know is how to hook up your LT1010. So why am I telling you all this about Ohm's law, and voltage is a force, and blah blah?
Because you suffer from two very common newbie misconceptions: that voltage flows, and that amperage enters things on its own, neither of which is true. So pardon my speaking frankly, but you don't know enough to ask questions that have answers. For instance, "i have 30mA in the output, is it to much for headphones" is simply meaningless. I'm not trying to be a smart guy, I'm just stating the fact.
But even so I probably overdid it. After all, you didn't ask for a course in basic electronics. And even if you did, probably all you really need to know is:
And maybe not even that.
So let's attack from a different angle. It seems all you really want to know is: 1) How much gain should your circuit have, and: 2) How should the LT1010 be connected to give that much gain?
To find out the answer to #1, state your audio source. If it's a CD player, a computer, or similar, then a gain of around 5 will do fine. No need to be precise, just get as close to 5 as standard component values might allow, err on the low side.
To find out the answer to #2...well...is there some reason you're not just using the schematic in the data sheet? The "Typical Application" on page one? Data sheet here:http://www.linear.com/docs/1683
Or else, as was already asked, what circuit are you using? That has to be known. You don't have to post the circuit itself, if that's hard to do, but provide a link, something.
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