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9

No, there is no way to make an inverting buffer with just an op-amp that does not depend on the resistor values. You can get resistors with very fine accuracy and stability (at an equally impressive price) or you can get networks with matched (in value and in temperature coefficient) where the absolute accuracy may not be so impressive but the ratio is ...


9

You can generate a negative voltage with a small current quite simply with a source of AC, and a voltage doubler rectifier. (This shows a doubler, reverse the diodes and capacitors for an inverter) See here for some circuit diagrams. The AC would come from your Arduino, by toggling a single pin regularly. You must add a series resistor too. It might be ...


8

Yes, an op-amp can act like a precision rectifier but not in power applications - if you have a signal that you want to rectify (maybe in order to convert to a peak value) then use the circuit below: - The negative part of the signal is accurately reproduced as a positive going signal at Vout. Two op-amps can be combined to produce a full wave rectifier ...


6

An op-amp's output pin can only swing between its power rails - many op-amps can only get within a volt or so of the rails, while 'rail to rail output' amps can reach voltages very close indeed to their rails. They can't output voltages outside that range, though - the power still has to come from somewhere! I suspect your simulation used a simplified op-amp ...


5

The label in the diagram refers to the wiper of the potentiometer, which is connected to Vee. The two outer terminals are connected to the offset-null of the Op Amp. The offset null terminals allow you to calibrate an op-amp to ensure the output is exactly zero at exactly equal inputs, despite manufacturing tolerances, by altering the position of the ...


5

An ideal op-amp wants to make its two inputs equal in voltage through the negative feedback path. Look at a very simple "precision diode": Notice that the inverting input (-) is also Vout. Vin is the other input. Remember, the op-amp wants to make its inputs the same voltage. Let's assume that Vout starts at 0V, and Vin is 5V. Because the inverting input ...


5

There are probably better switchable gain op-amp configurations if you use a non-inverting gain topology. For a start, the impedances around the feedback and grounding resistors can be much lower and hence the leakage currents produced by the analogue switch produce significantly smaller offset voltage errors. One further advantage is that the analogue ...


5

Attenuation is pretty simple: use a potentiometer. If the output resistance isn't acceptable, then you can follow that with a buffer: simulate this circuit – Schematic created using CircuitLab There are not a lot of gotchas with this approach: the op-amps you have are probably unity gain stable (check the datasheet). Of course the problem is ...


5

It's fine as long as you don't go outside the common mode input range. Remember that the op-amp thinks only in terms of V+ and V- applied. It doesn't care what you call ground.


4

I guess by "complete the feedback loop" you mean "hold the inverting and noninverting inputs at the same voltage". This is basically the op-amp's only goal in life, and given suitable negative feedback, it will accomplish it. If it can't, then it will drive the output into one supply rail or the other attempting to do so. So, why can IC1 accomplish this, ...


4

Since there is no resistor in series with the - input, you really are using the opamp as a comparator and the feedback resistor can be removed. Your circuit is a inverting comparator, BUT, its threshold level is 0 V since that's what you are feeding into the + input. There are several things you need to do to fix this circuit: Set the threshold voltage ...


4

Actually the resistor divider is providing a virtual ground. The difference is that unlike other filter topologies that need a stiff ground to absorb current through network or feedback resistors, the Sallen-Key low pass filter only needs a high impedance bias voltage (its only load is the opamp's non-inverting input, which should draw negligible bias ...


4

Series (non-inverting) diode limiter. This is the simplest circuit of an "ideal diode" connected in series to the load RL (the load does not belong to the circuit): In this circuit, the op-amp raises its output voltage with VF to overcome the undesired (here) forward diode voltage drop VF. Look also at this Wikibooks story to see how my students converted ...


4

I don't see how a serious transient that would affect a relatively slow op-amp is being caused directly by that switch, nor do I see any problem the way you're using it in general. Do you have a resistor directly across the piezo (not from the non inverting input to ground)? If not, then leakage may be causing the voltage to rise across the piezo while it ...


4

The top circuit is better in some ways, but I would prefer to see the switch right on the op-amp input and the resistors on the other side, which will greatly reduce the nonlinearity (distortion). If the input voltage can exceed the mux power supply range you have to be careful about the 'disconnected' resistors conducting current through the chip's internal ...


4

It depends on the opamp. In the old days when we had to trudge to school barefoot in the snow uphill both ways, opamps inputs drew enough current to matter to many circuits. For such opamps, the imbalance in current between the two inputs, called the input offset current was lower than the total current drawn by each input. You could cancel out the common ...


4

First let's remove the fluff. The 20K and 6.8K are there so that if the input becomes disconnected, the output will be approximately zero (because the effective input will be 1.268V, which is fairly close to presumed mid-scale of 1.25V. The gain of this circuit is 1 + 27K/(13K || 5.6K) = +7.90 For the offset, assume the input is mid-scale of 1.25V and ...


4

There's no reason in general to be afraid of unequal supplies. You might have to calculate some resistor values for the regulators or whatever, but that's no big deal. You do have to be concerned a bit about headroom- if you're using an inverting amplifier, for example, the output can only swing to (say) -9V reliably, so you can't have a gain of -1 and a ...


4

A zener diode connected from op-amp output to inverting input (possibly with a series std diode) and NOT switched by S1 plus a resistor from Vsense to inverting input will limit Vout+ excursion. If this is dual supply then back to back zeners will do the same thing symmetrically. When Vout approaches Vzener negative feedback is provided. The resistor from ...


4

Your formula is only true for an active differentiator, which includes (most simply) an op-amp (figure below). In this situation, there's a power supply, which sets the maximum output voltage, which can be higher than the input voltage. The properties of the op-amp will determine what happens if your formula predicts higher output voltages, but most likely ...


4

The equation given is a low frequency approximation that holds only when $$\omega RC \ll 1 $$ The exact phasor equation for an RC 'differentiator' circuit is $$V_{out} = \frac{j\omega RC}{1 + j\omega RC}V_{in}$$ Note that $$V_{out} \le V_{in}$$ for all frequencies. However, when \$\omega RC \ll 1\$, we have that $$V_{out} \approx j\omega RC\;V_{in} ...


3

As @circuit fantasist said in his now deleted answer @iggy is right - this is exactly an offset null adjustment. Look at the end of this article The page he cited shows the circuit below. (where the term '-Vcc' is synonymous with '-Vee' in the original question). FWIW - that is an excellent Op Amp introduction page - but much or all of the ...


3

I do not think this is intended to vary Vee. I think this is simply a way to zero out the OPAMP offset voltage. Can you give us the source of the picture?


3

I'm almost afraid to tackle this one, but here goes. Stop. Just stop. Forget EKGs. Get a cheap DMM, find a web page that deals with really simple op amp circuits, and start building really basic stuff, like a voltage follower, then a non-inverting amp with a gain of 2, and work up from there. The LM358 data sheet has some very basic circuits that you can ...


3

Whoever told you that was mistaken. The papers you cite are talking about "latched comparators" that are specifically designed to have internal storage of their state. These are specifically designed to support ADCs. The reason most ADCs require sample-and-hold circuits is that they require a certain amount of time to make a conversion, and the analog ...


3

If you tested the part correctly to see the output saturated at 1.3 to 1.5V below the positive supply rail then that certainly does not live up to the claim for an LMV324 opamp being a rail to rail output part. It could very well be that the parts you purchased are cheap junk LM324 opamps that were remarked as LMV324 to make them appear to be better parts. ...


3

Suggest you set it up as a gain of two non-inverting amplifier to test it: simulate this circuit – Schematic created using CircuitLab If you are seeing it saturate at less than about V1 - 0.1V it's not a real LMV324


3

"So why do we only assume negative feedback takes place" It is not correct that "we assume negative feedback" only. Who say this? The shown two-opamp circuit (introduced by A. Antoniou) can be interpreted as a combination of two "Negative Impedance Converter (NIC)" circuits. There are two basic NIC types (current-inversion - INIC, and voltage-inversion - ...


3

If there's two different input voltages, does the op-amp amplify the difference between them ? Yes. Because op-amp is a differential amplifier. The output voltage for an ideal op-amp is 0 if the input voltages are equal. But isn't the voltages of the two inputs the same anyway in the ideal op-amp ?! Not always. The voltages at the input ...


3

It it to mitigate the input leakage current of the OPAMP The OPAMP will attempt to keep the difference between + & - to equal zero. The input leakage current of the OPAMP in conjunction with gain resistors will produce a voltage at the - pin. If the + pin was tied to 0V there would now be an offset error voltage By providing a resistance to 0V (usually ...



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