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|Sostratos forex peace||Can you advise what my options are to correct so i get the right ohms out for the headphones. However, even if the theory were correct for loudspeakers, its applicability to headphones is suspect. I wouldn't spend a bunch of money to upgrade your set up. If you really want the Shures, I'd get the headphones and try them out with your devices and see if you're happy. While a few headphones might sound better to some listeners with more resistance between the headphones and the amp, most just sound worse.|
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|Merrill edge it investing streamlined process||But then again, 6moons doesn't publish any measurements or test results, and they also defend the improvements of ALO audio's cables on several headphones. As of well over million iPods had been sold. Just atleast theoretically by looking at the advertised changes for an IMOD do you think its worth it? It's nearly all "zero ohm" gear as they're after accuracy. I was wondering about the power transfer theorem. Oh by the way your blog is one of those "internet treasure chests".|
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|Capital dollar||Will that be like a short? Drew Ralph May 28, at AM. Hi NwAvGuy, great write-up like always. I know planars don't need to worry about FR changes and they always require mechanical and acoustic damping which will mitigate any loss in electrical damping but that still leaves the article you just linked. I'm not sure where your "by all accounts" comes from, but if it's mainly Head-Fi, you have to remember Schiit is a major sponsor there and there's a strong trend on Head-Fi to have lots of favorable posts about sponsor products--even when they have serious problems such as Schiit, NuForce and AudioGD.|
|Investing op amp gain 10lbs||Thanks Kevin. David G. I will assume that the Line out of the IPod is a safer option than a headphone out. But, regardless, Asus either doesn't understand, or doesn't care source, the benefits of a low headphone output impedance. I was wondering about this because I tried out a new planar headphone and I heard bloated bass and recessed treble. In practice, for a variety of reasons, I doubt it does. Subjectively the deviation is more audible on the SE because of it flat frequency response against the U shaped response of the UM3x in which the bump in the mids still leaves them way below the bass end treble levels.|
In real life, however, this is not easily attained. The performance of a real op-amp in this regard is most commonly measured in terms of its differential voltage gain how much it amplifies the difference between two input voltages versus its common-mode voltage gain how much it amplifies a common-mode voltage. The ratio of the former to the latter is called the common-mode rejection ratio , abbreviated as CMRR:. An ideal op-amp, with zero common-mode gain would have an infinite CMRR.
Real op-amps have high CMRRs, the ubiquitous having something around 70 dB, which works out to a little over 3, in terms of a ratio. Because the common mode rejection ratio in a typical op-amp is so high, common-mode gain is usually not a great concern in circuits where the op-amp is being used with negative feedback.
If the common-mode input voltage of an amplifier circuit were to suddenly change, thus producing a corresponding change in the output due to common-mode gain, that change in output would be quickly corrected as negative feedback and differential gain being much greater than common-mode gain worked to bring the system back to equilibrium. Sure enough, a change might be seen at the output, but it would be a lot smaller than what you might expect. A consideration to keep in mind, though, is common-mode gain in differential op-amp circuits such as instrumentation amplifiers.
We should expect to see no change in output voltage as the common-mode voltage changes:. Aside from very small deviations actually due to quirks of SPICE rather than real behavior of the circuit , the output remains stable where it should be: at 0 volts, with zero input voltage differential. Our input voltage differential is still zero volts, yet the output voltage changes significantly as the common-mode voltage is changed.
More than that, its a common-mode gain of our own making, having nothing to do with imperfections in the op-amps themselves. With a much-tempered differential gain actually equal to 3 in this particular circuit and no negative feedback outside the circuit, this common-mode gain will go unchecked in an instrument signal application.
There is only one way to correct this common-mode gain, and that is to balance all the resistor values. Suppose that all resistor values are exactly as they should be, but a common-mode gain exists due to an imperfection in one of the op-amps. With the adjustment provision, the resistance could be trimmed to compensate for this unwanted gain.
One quirk of some op-amp models is that of output latch-up , usually caused by the common-mode input voltage exceeding allowable limits. In JFET-input operational amplifiers, latch-up may occur if the common-mode input voltage approaches too closely to the negative power supply rail voltage. On the TL op-amp, for example, this occurs when the common-mode input voltage comes within about 0. Such a situation may easily occur in a single-supply circuit, where the negative power supply rail is ground 0 volts , and the input signal is free to swing to 0 volts.
Latch-up may also be triggered by the common-mode input voltage exceeding power supply rail voltages, negative or positive. As a rule, you should never allow either input voltage to rise above the positive power supply rail voltage, or sink below the negative power supply rail voltage, even if the op-amp in question is protected against latch-up as are the and op-amp models.
At worst, the kind of latch-up triggered by input voltages exceeding power supply voltages may be destructive to the op-amp. While this problem may seem easy to avoid, its possibility is more likely than you might think.
Consider the case of an operational amplifier circuit during power-up. If the circuit receives full input signal voltage before its own power supply has had time enough to charge the filter capacitors, the common-mode input voltage may easily exceed the power supply rail voltages for a short time.
If the op-amp receives signal voltage from a circuit supplied by a different power source, and its own power source fails, the signal voltage s may exceed the power supply rail voltages for an indefinite amount of time! Another practical concern for op-amp performance is voltage offset. That is, effect of having the output voltage something other than zero volts when the two input terminals are shorted together.
When that input voltage difference is exactly zero volts, we would ideally expect to have exactly zero volts present on the output. However, in the real world this rarely happens. Even if the op-amp in question has zero common-mode gain infinite CMRR , the output voltage may not be at zero when both inputs are shorted together.
This deviation from zero is called offset. A perfect op-amp would output exactly zero volts with both its inputs shorted together and grounded. However, most op-amps off the shelf will drive their outputs to a saturated level, either negative or positive. In the example shown above, the output voltage is saturated at a value of positive For this reason, offset voltage is usually expressed in terms of the equivalent amount of input voltage differential producing this effect. In other words, we imagine that the op-amp is perfect no offset whatsoever , and a small voltage is being applied in series with one of the inputs to force the output voltage one way or the other away from zero.
Offset voltage will tend to introduce slight errors in any op-amp circuit. So how do we compensate for it? Unlike common-mode gain, there are usually provisions made by the manufacturer to trim the offset of a packaged op-amp.
These connection points are labeled offset null and are used in this general way:. On single op-amps such as the and , the offset null connection points are pins 1 and 5 on the 8-pin DIP package. Inputs on an op-amp have extremely high input impedances. We analyze the circuit as though there was absolutely zero current entering or exiting the input connections. This idyllic picture, however, is not entirely true. Op-amps, especially those op-amps with bipolar transistor inputs, have to have some amount of current through their input connections in order for their internal circuits to be properly biased.
These currents, logically, are called bias currents. Under certain conditions, op-amp bias currents may be problematic. The following circuit illustrates one of those problem conditions:. At first glance, we see no apparent problems with this circuit.
In other words, this is a kind of comparator circuit , comparing the temperature between the end thermocouple junction and the reference junction near the op-amp. The problem is this: the wire loop formed by the thermocouple does not provide a path for both input bias currents, because both bias currents are trying to go the same way either into the op-amp or out of it. In order for this circuit to work properly, we must ground one of the input wires, thus providing a path to or from ground for both currents:.
Another way input bias currents may cause trouble is by dropping unwanted voltages across circuit resistances. Take this circuit for example:. We expect a voltage follower circuit such as the one above to reproduce the input voltage precisely at the output. But what about the resistance in series with the input voltage source? But even then, what slight bias currents may remain can cause measurement errors to occur, so we have to find some way to mitigate them through good design.
One way to do so is based on the assumption that the two input bias currents will be the same. In reality, they are often close to being the same, the difference between them referred to as the input offset current. If they are the same, then we should be able to cancel out the effects of input resistance voltage drop by inserting an equal amount of resistance in series with the other input, like this:.
With the additional resistance added to the circuit, the output voltage will be closer to V in than before, even if there is some offset between the two input currents. In either case, the compensating resistor value is determined by calculating the parallel resistance value of R 1 and R 2. Why is the value equal to the parallel equivalent of R 1 and R 2? This gives two parallel paths for bias current through R 1 and through R 2 , both to ground.
A related problem, occasionally experienced by students just learning to build operational amplifier circuits, is caused by a lack of a common ground connection to the power supply. This provides a complete path for the bias currents, feedback current s , and for the load output current. Take this circuit illustration, for instance, showing a properly grounded power supply:. The effect of doing this is profound:.
The feedback can be frequency dependent, or flat as required. The two simplest examples of op amp circuits using feedback are the formats for inverting and non-inverting amplifiers. The circuit for the inverting op-amp circuit is shown below. The op amp circuit is quite straightforward using few electronic components: a single feedback resistor from the output to the inverting input, and a resistor from the inverting input to the input of the circuit.
The non-inverting input is taken a ground point. This op amp circuit uses only two additional electronic components and this makes it very simple and easy to implement. It is easy to derive the op-amp gain equation. This means that any current flowing into the chip can be ignored. From this we can see that the current flowing in the resistors R1 and R2 is the same, because no current is flowing out of the junction between the two resistors. Hence the voltage gain of the circuit Av can be taken as:.
As an example, an amplifier requiring a gain of ten could be built by making R 2 47 k ohms and R 1 4. The circuit for the non-inverting op-amp is shown below. It offers a higher input impedance than the inverting op amp circuit. Like the inverting op amp circuit, it only requires the addition of two electronic components: two resistors to provide the required feedback.
The non-inverting amplifier also has the characteristic that the input and output are in the same phase as a result of the signal being applied to the non-inverting input of the op amp. The gain of the non-inverting circuit for the operational amplifier is also easy to determine during the electronic circuit design process. The calculation hinges around the fact that the voltage at both inputs is the same.
This arises from the fact that the gain of the amplifier is exceedingly high. If the output of the circuit remains within the supply rails of the amplifier, then the output voltage divided by the gain means that there is virtually no difference between the two inputs. We can assume that for the purpose of our calculation, the input to the operational amplifier draws no current as the impedance of the chip inputs will be well above the resistor values used.
This means that the current flowing in the resistors R 1 and R 2 is the same. The voltage at the inverting input is formed from a potential divider consisting of R 1 and R 2 , and as the voltage at both inputs is the same, the voltage at the inverting input must be the same as that at the non-inverting input. Hence the op amp gain equation for the voltage gain of the circuit Av can be taken as:. As an example, an amplifier requiring a gain of eleven could be built by making R 2 47 k ohms and R 1 4.
Op-amp gain is very easy to determine. The calculations for the different circuits is slightly different, but essentially both circuits are able to offer similar levels of gain, although the resistor values will not be the same for the same levels of op amp gain. It is normal to use operational amplifiers in linear applications with negative feedback, although this is not always the case.
This utilises the very high gain of the open loop amplifier to provide repeatable performance governed by the external components. However it is also possible to use operational amplifiers with other forms of feedback to produce other effects. One of the applications of using positive feedback within an op amp circuit to provide switching, for which comparators provide much better performance as they operator much faster and do not suffer from latching issues, but that does not mean that the basic principles of positive feedback do not apply.
However the basic principles of feedback and gain still apply to this type of IC or circuit block. That said, negative feedback is by the most widely used form of feedback for analogue, linear applications. Check out our video on op-amp gain. Read more about. Shopping on Electronics Notes Electronics Notes offers a host of products are very good prices from our shopping pages in association with Amazon.
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