# Tag Info

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Can An Operational Amplifier Circuit Be Made Entirely Out Of Diode Nand And Nor Gates? This apparently simple-enough question is somewhat ambiguous and can be answered several ways. Spehro has assumed that you convert the input to a digital value and perform digital arithmetic on it. So he says the answer is yes. ScottMcP takes your question at face ...

4

This is basically a classic window comparator, with stuff around it to make is actually useful in the particular application. PIR sensors report changes in IR accross the sensor area. C2 removes the DC bias, and the circuit around IC1D amplifies the result and also does some frequency filtering. This is probably in part to reduce frequencies that aren't ...

4

The heart of the operational amplifier is the difference amplifier, where a current is divided between a pair of highly matched input transistors. This is what allows high gain combined with low offsets. Logic gates simply don't have the geometry to do this. So the answer is no. That said, logic inverters (and by extension almost any inverting gate such ...

3

Since the op-amp inputs are (ideally) open circuits, case 1 is correct. Case 2 is wrong conceptually. In the actual circuit, resistors $R_3$ and $R_4$ are series connected - all of the current through $R_4$ is through $R_3$. However, in your case 2 schematic, this is not the case; there can be a non-zero current through the wire connected the two ...

3

Well, if that circuit does in some way function it's well outside of the normal operation of a 741. The common mode range of a 741 only goes to within a couple volts of the negative rail, below that the current sinks that bias the differential front end will no longer function. Even with a single-supply op-amp the circuit will not function properly ...

3

You only need to be concerned about the voltage that the op-amp will actually see, both when negative feedback is intact, and when negative feedback is broken. If you use a "typical" transimpedance amplifier (i.e., current to voltage converter), shown below, for example, so long as you choose the resistor such that the output voltage never exceeds the ...

3

This is a kind of silly question, but not in a bad way, it can be used as a teachable moment. The issue you have isn't so much that resistors can be poorly matched, it is that you're assuming op-amps are ideal. That is, "the only source of error is from the resistors, so if I get rid of those I will reach nirvana". Even if you get matched resistor made ...

3

The equation is correct, although the unsimplified form would be clearer. The first stage sums in the inverting input, which means that those terms should be negative when summed in the second stage. And then it inverts in the second stage. -(-(a + b) + c + d) = a + b - c - d

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When I teach introductory op-amp analysis techniques, I emphasize the following to start with: (1) Check for the presence of negative feedback. This means that the output of the op-amp is connected via some network to the inverting input of the op-amp. (2) If negative feedback is present, assume the inverting and non-inverting input voltages are equal ...

3

Sure, if you have digital numbers for the input voltages an op amp merely does the calculation Vout = Av(Vin+ - Vin-). Easily done with gates (and either NAND or NOR gates would be sufficient). NAND and NOR gates require more than just diodes, they require transistors, of course. OR and AND gates are not sufficient. If you do the calculation ...

3

One of the simplest op-amps looks like this: simulate this circuit – Schematic created using CircuitLab And you can build it using NAND gates: simulate this circuit simulate this circuit So the answer is YES. Keep in mind that this will be crap, and will operate only over avery narrow range. The primary tool for operation that ...

2

That configuration is an instrumentation amplifier. The point you ask about is where you can tap to get the common mode voltage, which can be fed back to the body in what would be called "driven leg" if this were an ECG amplifier. It effectively reduces electrode impedance for common mode signals, reducing noise.

2

The circuit is an instrumentation amplifier with the normal Rg split into two. The center point will have a potential equal to the average of the two input electrodes. That voltage can be buffered and fed to a third electrode to establish a common-mode voltage range at the patient with respect to the circuit ground. Ideally, the common-mode voltage will be ...

2

Your problem may be that the LM358 is not designed to drive a speaker as a load. You will need to add an output stage with a low output impedance of ~4 to 32 ohms or less, depending on your speaker's impedance. Here is an example of a typical configuration using BJTs as the output stage: Note how the feedback loop has been modified to incorporate the ...

2

You will need to apply a 2.5 volt bias to the op-amp positive (+) INPUT terminal. You can use a resistor divider (something like 4.7K and 4.7K from positive to ground at your positive terminal. This will cause the output to be at 2.5 volts when no signal is present at the microphone. Also, you will need a capacitor in series with you speaker (headphone), ...

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Just divide the voltages down to something reasonable. If you use a CMOS-input comparator, or op-amps as a comparator, you can use very high value resistors without significant loss of accuracy. BTW, I presume you are talking about DC here, otherwise the problem is a bit different. If you used a 10M + 101K resistors you would have a 1.8V to 2.3VDC ...

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The zenner limits the input voltage to the resistor ladder to 6.2V. The resistor ladder divides the voltage up into discrete steps for comparison. For each step sum the resistors above and below, then use the 6.2V as the input voltage ($V_{IN}$) to the standard voltage divider formula: $$V_{OUT} = \frac{R_2}{R_1 + R_2}V_{IN}$$ $R_1$ is the sum of the ...

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The opamp is rail-to-rail and it can run from 3.3 V, so that's OK. One thing to check is that Rl is small enough. At 1 mA, the emitter of of Q1 will be at 1 V. That means the collector should be 1.5 V minimum, preferably 2 V, for good regulation. That leaves 16 V for the load. This can't be more than 16 kΩ, else there isn't enough available ...

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The opamp is only rail-to-rail for outputs. It is not rail-to-rail for inputs. Its' common-mode input range is Vdd - .9 Your schematic is completely unreadable, so I'm guessing you have a 5-volt supply. The results of overdriving the input are unpredictable. Why you would get the same result with 3.3 volts I don't know, but the fact that you seem to get ...

2

Your schematic shows the op-amps being powered from a 3.3V supply. The most you'll get at the op-amp outputs is 0 to 3.3V. (Even the data sheet confirms this on line 3 of the features list - "Rail-to-Rail Swing at Output"). With the op-amp configured as a unity gain block this means that the input cannot range outside of 0 to 3.3V. If you try to drive the ...

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The amount of boost or cut for any given frequency is based on the ratio of the levels of signal applied to the boost and cut busses. For flat response, both busses get the same amount of signal. If you eliminated the cut bus, anything that wasn't boosted would be at "full cut", which is not very useful. You want to be able to vary the boost/cut smoothly ...

2

No, you can't flip the + and - inputs of a opamp and still expect the circuit to work. Think about it. Draw out a circuit that uses a opamp, then flip the two inputs and analyze the circuit again. It's not going to work the same as the original circuit.

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The REAL op amp output is G*(V+ - V-). "G", here, is the internal gain of the op amp, and is very high. In fact, when we assume G is infinite, and that zero current can enter the input terminals (both of which are APPROXIMATELY true), lovely things start to happen so long as the op amp is in a NEGATIVE FEEDBACK configuration. What sort of things? Well, ...

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An operational amplifier must have gain. There is no way to get gain out of diodes.

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It looks like the op amp symbol you dropped in your schematic is not mapping to the correct pins on the device. What symbol did you use? You can make a new symbol and make sure the pin names match those in the .lib file...

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What you have is probably two strain-gauges in each package forming a half-bridge. The wire colours could mean anything so what you have to do is take a multimeter and measure the resistance. You should be able to figure out what you measure from this: - simulate this circuit – Schematic created using CircuitLab Between two of the three wires ...

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The problem is the CM input voltage range- for a voltage follower you need rail-to-rail input and output if you want the output to be able to swing rail-to-rail. Note that the maximum input voltage of Vdd-0.9V (for functionality) is only guaranteed at 25°C, so caution would suggest you allow more headroom that than. Also, you should not assume that ...

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Just think of AMP1 as a summing amplifier; its two inputs are the outputs of AMP0 and AMP2. If AMP2 is functioning as an inverting integrator or low-pass filter, this makes the output of AMP1 the difference between the original signal and the low-pass version, which leaves the high-pass signal.

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Weird, I don't know how that first circuit worked with a 741. Maybe there was enough current leaking out of the inverting input to bias the the input coupling cap to some value that let it run. Biasing the non-inverting input to mid supply was the right thing to do. To get rid of the clipping (low volume) maybe you could reduce the gain of the first stage ...

1

Best in class input/output rail-to-rail op. amps.: OPA350 (includes DIP-8) OPA333 (Non DIP). Also, a lot of other great features. See the datasheets.

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