# Tag Info

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The opamp in both circuits is just a voltage follower with a gain of 1, so it is irrelevant for the purpose of calculating gain. The left circuit is a simple R-C high pass filter, and the right circuit a simple R-C low pass filter. The gain of each of these is 1 well into the passband. Well into the stopband, the gain will decrease 6 dB/octave or 20 ...

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Opamps are optimized for linear operation, in which the voltage difference between the input terminals is kept very small via feedback. As a consequence, the performance when using them in a nonlinear or open-loop application tends to be poor. In particular, charge storage on internal nodes tends to cause opamps to respond very slowly when coming out of an ...

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But also how do I calculate where the -3 dB point is of both circuits? In general, you would find the magnitude of the transfer function, set it equal to $\dfrac{1}{\sqrt{2}}^*$, and solve for the frequency. For a simple 1st order filter, this is almost trivial. In the 2nd circuit, the voltage at the inverting input of the op-amp is: $$V_+ = ... 4 Consider conservation of energy. Very roughly, 1.2 V * 50 mAh is about 215 J. 5 V * 500 mA is .25 W. At a .25 W rate of use, 215 J will be consumed in about 450 seconds, or 7.5 minutes. Edit Let's look at the chips you mention. TPS61200 is a boost converter. It will "boost" the power supply voltage from 1.2 V to some higher voltage --- 5 V in your ... 3 This seems to be a problem I've seen a few times on stack exchange. Consider an op-amp with localized negative feedback - The manufacturer designs the op-amp so that under the very worst case situations it is stable. The worst case situation is unity gain - this has the biggest chance of being unstable. Anyway, each year the boundaries get pushed a bit more ... 3 I found that it's possile to go from 1.2v to 5v with the TPS61200 and go from 50mA to 500mA with the L272 but what if they're used together ? Will it preserve my battery lifespan? Let's just say the answer is "yes". Say you had a 1.2V battery which can now provide 5V, and where it could provide 50mA, it could now provide 500mA, with no reduction in ... 3 Welcome to the world of real opamps. Real opamps are not ideal. For this particular case - the output does not swing exactly to rails. Event for opamps that are advertised as "rail to rail" output will be several mV from the real rail. Output swing is usually specified in the datasheet. It's called output swing, min/max output or something like that, look ... 3 If something has an infinite impedance then any frequency cannot affect it however, it's never going to be infinite and as frequency rises the small amount of capacitance associated with the input will start to become significant even if the input resistance stays in the Giga ohm range. The frequency won't affect the capacitance but the amplitude of the ... 3 First, many InAmps are integrated into one IC, also the three-opamp versions. Then the cost of the extra opamps isn't that high. Consider the extra cost of less than a mm^2 real estate against the total cost to create an IC. Next, I've never seen a single-opamp inamp. There are two-opamp and three-opamp versions, and there's a reason why you can't make one ... 3 An op-amp is not required. Given the minimal requirements given, a voltage divider will do. Gain = 0.5. simulate this circuit – Schematic created using CircuitLab If you need to buffer the output, you can just place a unity gain op-amp at the output of the voltage divider. 3 Generally, input resistance for commercial CMOS opamps are high enough (in the 100M to G Ohms range at DC) that optimizing the circuit for it are futile, considering the optimal resistors will have mismatch themselves. This is caused by the differential input stage for the opamp being a (or two in the case for rail to rail opamps) MOS differential pair ... 2 I don't see either why it is a requirement. However, you can achieve a 2nd order LPF with attenuation without an an op-amp, and without introducing power supply noise, dealing with input bias current noise, and the other few limitations of op amps. From TI's Design Methodology for MFB Filters Using an active filter for a Q < 0.5 (two real poles) ... 2 In case you meant the voltage between the two inputs as "swing" take care on this. Some opamps are quite happy to have the full supply rails separating the inputs whilst other opamps would smoke under the same circumstances because they have an input circuit that will only tolerate maybe a volt or so. Usually this is due to input protection diodes. Common ... 2 Getting a gain less than 1 requires an inverting amplifier where $A=-\dfrac{R_f}{R_i}$. It requires cascading two stages. Use one for the inversion, or -1 gain, and another with your desired gain, -0.5 fro, your example. simulate this circuit – Schematic created using CircuitLab 2 You are designing a linear circuit and the first rule with opamps is that the voltage at both inputs is identical in dc level and ac level. Negative feedback ensures this but, only if you set up things reasonably. Your circuit will try and adjust it's output to make both input pins exactly the same but it can't because the output cannot go sufficiently ... 2 Although I can't tell from your plots what is going on, I would suggest the following. Any time you build an amplifier circuit, such as yours, it is extremely wise to provide some high frequency feedback. To do this, place a small cap across R21. Typically 10 pF is enough. This will create a low pass response at Omega = 1/(R21*C), but that typically isn't a ... 2 Add power supply decoupling caps to all IC's as near to its power pins as possible. A good value is usually 100nF, but you may want to double check the datasheet for the various chips you used. Saving money by leaving decoupling caps out is a bad design choice. Caps are cheap and troubleshooting/redesign time is expensive. 2 Looks like mains hum. (I'm assuming the annotations on the graph show you are changing the input gain during acquisition, so that the sinewave is not continuous). Now we don't know: the period of the observed "sine" wave the ADC sampling rate your mains frequency but you do, so you can work out if that's a possible hypothesis. For example if you are ... 2 741D is a variant of the 741, you would need to read the manufacturer's own datasheet to find what the D signifies, but it is probably the package (e.g. plastic 8-pin DIP as opposed to ceramic DIP or SOIC). However there is a chance that it is a variant in supply voltage range or some other parameter like input offset voltage, so that is worth checking if ... 2 Theoretically, or more precisely, ideally, the answer is no. If$$Z_{in}(\omega) \rightarrow \infty$$for an ideal op-amp, then there is zero input current for any input voltage of any frequency. But, this isn't surprising or, for that matter, very interesting. Ideal op-amps are an abstraction that, in the appropriate context, are a good approximation ... 1 If you are looking for a specific OpAmp chip I'm sorry, I'm not familiar with any 3.5 V ones. About other things, if it will be a battery driven device make sure the input impedance is high ( to consume less current ). Usually (extra) high input impedance OpAmps are doing worse with noise characteristics, but if that's not of terrible importance the trade ... 1 Local feedback: Feed the opamp's output back to that opamp's own input. Global feedback: Feed the (multistage) amplifier's output back into the (multistage) amplifier's input. The question is about how the amplifier properties (like bandwidth, maybe other properties that I can't think of right now) change with changing the feedback architecture. Notice ... 1 You should look at COTS telephone line interface modules (TLI). They take all of the headache out of connecting an MCU to a phone line. Typical services include on/off hook signalling, ring detection, AC-coupled line in/out, sometimes a DC output derived from the DC present on the phone line (the 42.7Vdc you measured), and decent isolation (1-2000V) to ... 1 There are a number of different issues in play here. Assuming you're talking about a three-op-amp instrumentation amplifier and a one-op-amp differential amplifier. First, lets talk about hand-made instrumentation-amps and differential amplifiers vs. purpose-built integrated circuits. Hand-made amplifiers of both classes will not be as good as the IC's, ... 1 What you are missing is that the pot is wired such that it produces a ratio of the input signal with the ratio varying from 0 to 1 accross the full sweep of the pot. This is why the 5 kΩ and 10 kΩ pots resulted in the same full volume. The pot achieves this by being a resistor divider. It does not work by adding resistance in series with a ... 1 I don't see that it makes much sense for a device to define a common-mode input range which is larger than the allowable input swing; my guess is that the "input swing" and common mode range both extend 300mV beyond the rails, but describing the input and output swings together as being "rail-to-rail" made it clear that they both extend at least that far ... 1 Input swing not a very specific term, so it's hard to tell what it means exactly. Common mode input range, however, is well defined. If you were to tie both inputs together, the common mode input range is the voltage range over which you can drive the two inputs with the opamp still working as intended. Look closely at the sentence you quoted, and you ... 1 The common mode voltage of a differential signal $V_+$ and $V_-$ is defined as:$$ V_{cm} = (V_+ + V_-)/2 $$It is the common voltage your differential signals are superimposed upon, and as the differential signal (the input swing) is defined as:$$ V_{dif} = V_+ - V_-  For example, I could define a differential signal, $V_{dif}$ with 900mV ...

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This is called a virtual ground circuit. And this one will not work in all applications - it's very limited in terms of the amount of current it can source or sink from the virtual ground. For a simple audio amplifier running on a 9 volt battery, it's fine. But for more demanding applications, it likely won't be sufficient. A very simple improvement ...

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You should have no issues running multiple amplifiers from the same virtual ground. However, the virtual ground circuit itself will depend on what your circuit is doing. If you will be driving DC levels, you can't use a simple voltage divider without any buffering, for example. This is a good page that gives an overview of virtual ground techniques: ...

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