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14

The terminology can be confusing for a newbie, actually. The term "virtual short circuit" refers to the fact that in an opamp circuit with negative feedback the circuit is arranged in a way that (ideally) makes the voltage across the two opamp inputs zero. Since one of the properties of a short circuit between two points is that the voltage across those ...


6

The Elenco 9425 is a breadboard. ("Circuit board" usually means a PCB.) Breadboard holes are 0.1 inches apart. Through-hole DIP packages commonly use this same spacing, with the distance between the pin rows being 0.3 or 0.6 inches. I'll walk you through how to pick a package on Digi-Key. The key is to use the filters one at a time to narrow down the list ...


6

This is a fun problem, but thankfully, you aren't the first person to run into this. High-speed ADCs tend to be picky about input drive, and high-speed ADC drivers are often made in low voltage processes, which means that accepting a high-voltage input requires some gymnastics, for which resistor dividers aren't an option due to high-speed impedance control ...


5

Very good question indeed. I think much of this can be answered by looking at the equivalent circuit of an op amp. For an ideal op amp, the current flowing into V+ and V- is zero, so this means Rin must be infinite. When an ideal op amp is setup in a feedback arrangement(Vout is connected to V+ or V- in some way), the voltage at V+ will equal V-. The ...


4

Step 1: PC speakers are crappy at best. Who knows what they call ground, how they relate it and where exactly it comes from. Thus, Step 2: You need to have ONE ground point in your system. Which follows to Step 3: You need AC coupling in your schematic, because you do not know what will be at the output, in terms of power-loops through whatever other ...


4

The crater and magic smoke release is the best clue here. I've had parts returned with a crater hole in a past life as an applications engineer. This is a very strong indicator that you are causing a latchup condition in your opamp. This can happen by either a voltage spike on its power rail, or more commonly, by driving one of the input pins above its ...


4

There are many break-out boards available for virtually every size of small-outline package. They typically have 0.1 inch pins for plugging into a breadboard. For example...


4

IC's have different standard "packages", with defined component size, pin pitch, etc. For any given IC, you can generally see which packages are available at the end of the datasheet. The capabilities of the IC aren't determined by the package (much), so if you find a specific IC that meets your specs it will work no matter which package you choose. This ...


3

It is just an AC coupling capacitor. It prevents any DC component of the input signal from being amplified. Ra and Rb serve to set the DC input voltage to halfway between the rails. It is common in any application where the DC component is unimportant, and especially in single supply circuits such as this one, where the "mid rail" voltage is +Vs/2 rather ...


3

The circled diode provides a -0.7Vdc negative rail or reference. Note that the anode of the diode is connected to circuit ground (mostly hidden by your red circle - but visible at high magnification). This negative rail allows the (+) input of the 2nd op-amp to operate near or below ground. Note that I would connect pin 4 of the op-amp to the cathode of ...


3

Just to clear the air. If an op-amp is NOT being used as a comparator, in other words it has a negative feedback resistor, then it will output the difference between the (+) and (-) input times the gain to keep the (+) and (-) inputs at the same voltage. In the real world, the input impedance of an op-amp can never be infinite or zero ohms, but is somewhere ...


3

In short, there's a difference between the input impedance of the op amp, and the input impedance of the overall amplifier circuit. Even in terms diff amp you show, there is no current actually entering the op amp, which (ideally) has infinite input impedance. As an aside, note that the difference amp inputs see different input impedances, which is a built ...


3

For those not familiar with 4 - 20 mA sensors and control loops, they are commonly used in industrial control systems. They have several advantages and these have been listed in Why is zero represented by 4mA in 4 - 20 mA industrial control systems?. Suffice it to say that the author is building a sensor which is powered by the loop. It has to work when less ...


3

Well, using your much improved FFT you could approximate THD to the amplitude of the 2nd harmonic - it appears to be about -42 dBV and your main signal is at -15 dBV. The 2nd harmonic is therefore 27 dB lower. As a ratio that is 22.4 times lower or about 4.5 % THD. If you included the 3rd harmonic at about -58 dBV you have to convert to a real voltage and ...


3

This does not look like the same model I got from TI website. Here is the header: *////////////////////////////////////////////////////////////////////// * (C) National Semiconductor, Inc. * Models developed and under copyright by: * National Semiconductor, Inc. *///////////////////////////////////////////////////////////////////// * Legal Notice: This ...


3

Although the OP says no resistor divider networks, in a comment he says he aims to reproduce the capabilities of a cheapo multimeter. These have typically 10Mohm input resistance (though you can find 1Mohm if you go really cheap). So, to get +/- 15v down to 0/3.3v with 10Mohm input, you could use this 10:1 divider simulate this circuit – ...


3

They differ in the level of the signal that is required to change their state. In a Schmidt trigger, a relatively small, well-controlled change of signal level is needed. The input is designed to accomodate this signal, and the input current will be relatively low. In the case of an HC14 supplied with 5v, the hysteresis, the difference in switching levels, ...


2

The opamp power is 5v relative to its ground, which is not connected to the ground in the circuit above. Well here's one problem - the negative power rail being 0V does need to connect to the ground points. Not being connected to ground means the signal ground can float off relative to power ground and break rules on input common mode range. Using an ...


2

The method on the right describes what happens if you have a balanced circuit with a balanced (differential) input signal. The top right capacitors with the ground in the middle directly represent the input capacitances of the PMOSFETs. Since the middle node is grounded (actually Vcc but for small signal that is ground) the voltage on the node is zero, as ...


2

One way to simplify the problem is to use the Thevenin equivalent circuits for those voltage dividers you have there. That is, you could draw the circuit in this manner: simulate this circuit – Schematic created using CircuitLab Here $$V_{th1}=\frac{R_1}{R_{sensor}+R_1}V_{cc}$$ \$R_1\$ is just your 680 ohm resistor in series with your 500 ohm ...


2

Unless you have the "cold-junction" offset handled by some other method which you did not mention in your question, it is not recommended to use a simple op-amp to "condition" a thermocouple signal. Analog Devices makes a range of specialty "Thermocouple Interface Amplifiers". Highly recommended to use one of those. There are some "break-out boards" ...


2

I believe you simply do not have the MAX492 biased correctly. Look at the "Common Mode Input Voltage Range" ("CMIVR") in the MAX492 data sheet. It goes from Vee-0.25 to Vcc+.25. This is the legitimate range of voltage you should be applying to either input on the MAX492. Since your L-C circuit is grounded to the power supply common, its sinusoidal voltage ...


2

The circuit is an op-amp with ground as the negative rail. This means Vin cannot go below the negative rail. Now look at the output current formula - it involves Vin and, if Vin cannot go below ground, then the current through the MOSFET HAS to be positive only. In short, you cannot put an AC voltage at Vin unless it is biased to make its most negative ...


2

A simple pull-down resistor (connection to GND via resitor) at the input can fix the problem. It would pull down the floating potential of the open line to 0V. Note: This works if the output impedance of the "voltage input" part (A) is low enough. It wouldn't be good if the "input voltage" part (A) is just a potentiometer because then the pull down ...


2

Look at the datasheets! The LMC6062 datasheet says on page 3: Input common mode voltage range: \$0..V^+\$ − 2.5V i.e. at \$V^+ - 1V\$ you are already 1.5V above the range that is guaranteed to work. The LMC6482 datasheet on the other hand says it is a rail-to-rail input and output OpAmp, i.e. it works with input voltages from GND rail (0V) to up to the \...


1

I'm not certain what your circuit is attempting to do, but when you see a diode in parallel with a base-emitter of a transistor (Q1), then this is usually to add a small voltage drop. The base-emitter is a similar diode junction, so the circuit is usually matching voltage drops. This technique can reduce effects such as crossover distortion in some types ...


1

The answer is: No there is no circuit that just uses op-amps and resistors that can multiply two voltages. You can multiply one voltage times a resistor value. You can do things with log amplifiers and antilog amplifiers, you can do things with PWM, but all of those things (except the resistor value one) involve other (nonlinear) elements.


1

It could be input offset voltage drift with temperature. The data sheet says it is typically 7 uV/degC and with a gain about 680 (10k resistor set to zero), this will cause a shift in output voltage of 4.7 mV for every degC that the LM358 warms up. It should eventually settle down of course. Try touching the LM358 with the tip of a soldering iron to see ...


1

What you show should work. Most likely you have a wiring error arond U4. Check the datasheet and note how two of the opamps in the quad package have mirrored pinouts of each other. That answers your question, but a few other things need to said about your circuit. That's one convoluted way to make a triangle wave oscillator! The slope of the triangle ...


1

simulate this circuit – Schematic created using CircuitLab Figure 1. (a) What you want. (b) What might have happened? Just suggesting you check for silly stuff: if the output from your U4 is in phase with the input it would suggest that you have somehow wired U4 as a unity gain buffer as shown in Figure 1b.



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