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10

Simulated oscillators usually don't start on their own, try setting an initial condition to break the feedback loop during the bias point calculation. I can do this with the Pulsonix (SIMetrix) SPICE simulator by adding an initial condition with a value of zero, you should be able to do something similar with the simulator you are using - see the ...


9

Well, the LM339 is a comparator, not an op-amp. It is intended to compare the two input voltages and put out one of two levels depending on the result of the comparison. It is not intended to be used with feedback, except possibly positive feedback for hysteresis. The output is an open collector, and there is a pull-up to 5V, meaning that when the ...


8

Duncan, rather than try to answer the question, I'm going to go on a tangent, inspired by my overwhelming belief that you are going about this the hard way. I speak as one who, many years ago, took a course in just this sort of thing under "Doc" Edgerton. The first thing you need to do is replace the straw/dart combination with something with a consistent ...


5

The most obvious problem in the circuit is that the + input of the amplifier is floating; it is not biased to any DC voltage level. You need to convey a reference voltage to this input. (This has to be done through a reasonably large resistor; if you just tie the input to a stiff reference voltage, the input will have no impedance to work against.) ...


4

The resistor is in parallel to the battery, and for that reason you can remove it from the circuit. It doesn't influence the opamp at all. You can easily redraw the circuit with the resistor in parallel to the 5V battery without changing anything to the remainder of the circuit. Rule of thumb: An ideal opamp with negative feedback will try to set both ...


4

The main reason for using open-collector is so that several comparators can have their output connected together in a wired OR gate. All the open collectors can be tied together to a single resistor without any conflicts between comparators. This would not have been easy to do when your comparator has also the capability to source current at its output. ...


4

There is no current flowing into the inverting (negative) input, because the input impedance of an ideal op-amp is infinite, therefore the current flowing through the feedback 1k\$\Omega\$ resistor (let's call it \$i_3\$) is the sum of the two currents flowing through the two other resistors (\$i_1\$ and \$i_2\$). The inverting input is a virtual ground, as ...


3

jippie's right. D2 is your problem. On the reverse cycle, you've got no regulation on the voltage/current from D2.L3's top is at "virtual ground". L3's bottom is at positive 23 volts or so. This forward biases D2. The output is the cathode side of D2 I assume. That means you've only got regulation on 1/2 of your cycle (through D1). Your regulator sees the ...


3

Usually we're concerned with how much loading can be put on the output and still have it function properly. The output swing specification will be at a specific load resistance, usually load is between the output and ground on both dual and single-supply op-amps. For example, the dual-supply OP-07: You can see that over the whole temperature range with ...


3

For reference, here is information for real-world protection units, meeting sets of safety regulation specification, as used in industrial plant settings, says, sensors and remote control wiring running hundreds meters on outdoor paths, subject to lightning strike. As original poster suspects, cascaded protection is typically used inside these unit. You ...


3

To gain insight into what is happening, replace the op-amp with an ideal voltage amplifier model (we assume the gain \$A \rightarrow \infty\$): simulate this circuit – Schematic created using CircuitLab Now it's easy to see two important points \$R\$ can only change the current through the 5V source - it has no other effect there is no path ...


2

Regarding R10, the virtual ground is driving stray capacitance. If you loaded the output of U2 directly with that capacitance it would probably oscillate. R8 actually hurts stability of U4. It would improve it only if C4 was connected to pin 1, however in most cases users will not load the output capacitively (eg. by attaching a long cable) so it's ...


2

This will function okay, particularly if you use good resistors, and don't need more than some % DC accuracy. The 78L05 is not a very good regulator and its a worse reference so it's almost surely going to limit the overall DC system accuracy if not dealt with. You could measure the voltage with one channel of your ADC and deal with it digitally, but power ...


2

This idea will work providing you don't need to supply more than (say) 20 mA of current at 2V to other devices because it's likely that the op-amp wouldn't be able to do this - some op-amps will supply (say) 50mA such as the AD8605 (from memory) of course. DC offset is an issue if you want to use an op-amp so check the data sheet. Why not find an ...


2

That drawing isn't very well laid out - the LED should have been drawn to the LEFT of the photodiode to make it a little more clear. The LED generates light full-time. The photodiode may or may not pass current, depending upon whether it can "see" the LED, so the voltage available between the photodiode and its resistor will rise and fall as the photodiode ...


2

An opamp is driving an analogue voltage. To be useful, it needs to control the voltage swing completely. A comparator is a binary switch. Its output is effectively in only two states. It can work correctly with an external resistor. It doesn't need to drive the signal high when used with an external pull-up resistor. An NPN transistor can pull down when it ...


2

To see the difference between the parts, have a look at the electrical characteristics table in the datasheet: The 'A' types have lower specs for input offset voltage and current. This probably means they are cherry-picked parts from the same production run. The 'A' types also have lower specs for offset drift over temperature. This might mean they are ...


2

I always thought of comparators as the simplest A-D converter. You may want a different supply voltage on the analog side (maybe +/-15V or something) and the output going to a different digital voltage. (+5 V perhaps.) The open collector lets you easily adjust the output voltage reference.


2

You are trying to create a detailed model of a sloppy process. For one thing, the gain-bandwidth product is a minimum spec, or sometimes even just "typical". The actual gain-bandwidth can be quite different for any one part, and I wouldn't count on the phase margin either. The raw DC gain can also vary a great deal, as can other parameters. Getting this ...


2

Actually not the voltages add up but the currents. All currents flowing to the negative input (which is assumed to be high impedance, i.e. "open" and at ground level) must add up to 0. Since currents are proportional to the voltage across resistors, the voltage of the output must be equal in magnitude (but opposite in sign) to the SUM of the voltages at the ...


2

Just to make the point clear and actually put an answer to this question. (see all the comments by Dave, Jippie et al.) They are totally different ICs. The 358 is is dual op amp (it has two separate op amps in the same 8 pin package). The 386 is an audio amplifier. They have different pinouts and cannot be interchanged.


1

In this configuration you are missing an all-important resistor known as the inverting input resistor. This allows the inverting input pin node to be dragged down to match the non-inverting pin voltage: - simulate this circuit – Schematic created using CircuitLab I'm also assuming that the brown wire in your diagram is meant to be ground ...


1

No. As shown your circuit will tend to not do what you want. BUT an extension of this method will. What happens in your circuit is that when NOT1_Out is high D1 will cause the capacitor to charge faster when NOT2_output is high, and discharge slower when NOT2_output is low. If R3 is somewhat larger than R2 this will have the effect of changing the mark ...


1

Just a suggestion... use a pair of metal-film resistors as a voltage divider between +5 and ground. MFRs are a lot quieter than carbons. R1 should be exactly 150% of R2.


1

The DC currents through R1 and R5 are flowing through R3, causing the virtual ground to increase in voltage. According to the OP the gain is working correctly now.


1

"What are the tradeoffs between using a dual supply as opposed to single supply and virtual ground?" There are two important aspects connected with such a trade-off: 1.) AC applications for single supply configurations always are connected with a lower cut-off frequency (high-pass effect due to the coupling capacitor). 2.) You must ensure unity dc gain ...


1

In my application input can be AC but the output needs to be a DC referenced level (being fed into an ADC, 0-2V5) If your application is AC then you can couple the signal via a capacitor to a 1.25V reference voltage provided by resistors - this sets the average voltage at 1.25 volts i.e. half the ADC range. This strikes me as the easiest way to deal ...


1

It's usually called Output Current. Parameter "Io" on this TI part, page 4. ADI calls it Maximum Output Current or "Iout", page 3. Occasionally, it is called Short Circuit Current or "Isc" like in this Linear part which is 5-9mA depending on operating conditions, pages 4 and 6. It is a slightly roundabout way of saying even if you short the output with 0 ...


1

The "5th" opamp is a virtual ground so that you other amps have a DC offset. This is done because you're using a single supply (just 'positive' volts). No matter what input voltage you pick, this circuit will take half and use it as the negative rail for the op-amps. In that sense, this circuit is robust enough to handle the voltage change well. However, ...



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