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10

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 ...


5

At first, the principle of "virtual ground" can be applied during DESIGN of opamp-based amplifiers. This simplifies calculations - and the error is in most cases acceptable. Error? Yes - because there is always a differential voltage between both opamp inputs, which is exactly Vdiff=Vout/Aol. (Aol=open-loop gain of the opamp). Because of the large values for ...


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

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

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


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

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

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

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 ...


3

Now if this is the case, and the inverting and non-inverting inputs are brought to the same potential, won't the op-amp stop amplifying? The op-amp amplifies the voltage across (difference) the input terminals regardless. For the ideal op-amp, the voltage gain \$A\$ is 'infinite' (arbitrarily large) thus, the voltage difference can be arbitrarily ...


2

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 ...


2

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 ...


2

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 ...


2

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

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


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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, ...


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.


2

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...


2

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 ...


1

Other answers have touched upon how an op amp works, but may not be clear on why the simulated circuit works as it does. A couple key observations to start: If an op amp is wired using infinitely-fast components such that the potential difference (+ minus -) on the two inputs at some moment in time is a uniformly-decreasing function of the output which ...


1

An op-amp will produce a positive output if the '+' input is more positive than the '-' input. And will produce a positive output if the '-' input is more positive than the '+' input. In all the circuits where we say the '+' and '-' pins are at the at the same potential such as: Think about what would happen if the '-' terminal were not at the same ...


1

The shorthand way of analyzing a (linear) op-amp circuit is to assume that the inverting and non-inverting inputs are at the same potential. That assumes it has been wired properly (and that the gain is essentially infinite). You could think of the output being at the bottom of a steep valley- any changes to the output are multiplied by the open-loop gain ...


1

I don't think there is any way** that does not reduce to having a tightly matched pair of resistors somewhere. Of course you can buy those precision-trimmed resistors as part of a chip and pretend they're not there (for example, use a dual instrumentation amplifier with G=1, inputs connected in inverse). You can buy a pair of 0.05% ratio matched ...


1

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 ...


1

No. In the circuit configuration shown, the input impedance of the first opamp stage is dominated by the values of the resistors, such as the 10K tied to +4.5V. The transistor isolates the source from this load.


1

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.


1

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 ...



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