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11

What parameters of a real op amp determine the lowest voltage it can amplify? It's all about signal (desired) to noise (not desired) ratio (SNR). The LM358 has an equivalent input voltage noise of 55 nV per sqrt(Hz). Now that probably sounds confusing but it isn't. Let's say your bandwidth is 10kHz\$^1\$ - the total noise will be 55 nV x sqrt(10,000) ...


8

Your assumption "output voltage value is same either if I connect input signal into + port or if I connect input signal into - port." is wrong. Your left circuit does not work as a linear amplifier. Iuses positive Feedback. Depending on the input voltage it saturates close to one of the supply voltages (not shown in your diagramms; probably +10V and -10V; ...


7

No, the LM358 will not work for your application as described. But not due to the input offset voltage. The input offset voltage is the voltage that must be applied across the input terminals for the output to be forced to 0V (or whatever the output is referred to and what you're choosing as ground). That might not be the most helpful description though, ...


6

The resistors at the non-inverting input are for the bias current compensation. In a non-ideal opamp the bias currents flowing into the inputs are non-zero. Parallel resistance of R7 and R8 is nearly equal to the parallel resistance of R2 and R3 (1485 Ohms). The matching in this circuit is not correct, because the output impedance (R1) of the source is not ...


6

Try learn something about allowed Common-mode Input Voltage Range. The Common-mode Input Voltage Range tells you the voltages at the inputs that cause them to work properly or to not work. As for your problem look here And here at question "What other features of op amps should the user know about? " ...


6

A voltage is a difference in potential measured between two points. Traditionally, we label ground as 0V and measure from there, but you can choose any point on your circuit to call zero. If you choose your negative supply, then just by changing your labels, it's back to a single supply! So yes, it is always possible. That said, take a bit of care when ...


6

LM358 is not a rail-to-rail output op-amp, so you should not expect it to produce an output near the positive supply voltage. Unfortunately, the datasheet doesn't go out of its way to make this obvious. The main clues are: Nowhere on the front page does it claim to have rail-to-rail outputs. Figure 10 on page 10: This shows that if only modest currents ...


5

Without a resistor in series to IN1, the current through the Z-diode is not limited. It will still protect from ESD, but any permanent voltage from a low impedance source below -0.7V or above 5.6V will destroy the diode.


5

Yes, it is possible to charge an NiMH battery with that circuit. Just not reliably, and no, it would not be safe. ∆V charge termination is fiddly and unreliable at best. Worse, it is not a state, but a single event, and events can be missed. This is an easily missed event, so there is the definite possibility of uncontrolled overcharging, and it is even ...


5

The ultimate answer is to use SMD components on a PCB - the parts are smaller, and using traces instead of wires will reduce stray capacitance; this can have a large effect in sensitive circuits. I understand your reluctance to use a PCB because it is more work and more expensive initially - however, in the long run it sounds like the only way you will get ...


5

1) Todays high-definition audio equipment that works with 24 bit audio, such as mixers, use 600 ohm differential inputs and internal buss connections (until the signal is digitized) that need a voltage swing of +/- 10 volts to handle the 120dB dynamic range of SACD / DVDV / Blu-Ray audio tracks. The diff driver IC is often a SSM2142 which has supply rails of ...


4

Let's start with a statement: **The output of an op-amp is $$ V_{OUT} = A(V_{IN+} - V_{IN-}) $$ Where \$A\$ is the open loop gain, \$V_{IN+}\$ is the non-inverting input voltage and \$V_{IN-}\$ is the inverting input voltage. This is the way opamps are designed and A (the open loop gain) is a large number, typically in the order of 10\$^5\$. From this we ...


4

Your maths looks fine, your circuit looks fine. However... a) is the opamp really needed? If the MCU ADC input draws any significant current, then you need that opamp there to provide it with a low impedance drive. However, in my experience, most MCU ADCs have a high impedance, and could safely be driven from just the R1/R2 voltage divider with no ...


4

In the older days, amplifiers needed a good amount of margin between the output voltage range and the supply voltages. Older opamps often had 3V of voltage drop per rail. Moreover, having the signal output further away from this limit guaranteed a better supply rejection ratio (older designs often used poorly regulated supplies, using only a zener+NPN as ...


3

You could add a pot as shown. The LM324 is not a good op-amp to use for mV DC signals- offset can be as much as +/-20mV and this will only adjust for +/-5mV offset (you have observed 0.5mV). simulate this circuit – Schematic created using CircuitLab If you don't like having to come up with +/- supplies (or having it drift around by maybe ...


3

If you're up to it, I'd recommend looking into something called a monostable vibrator (it's really just a one shot pulser, each time you trigger it, it generates a single pulse - you can chose how long the pulse is). Monostables can be built using 555 timer chips and there's about a billion 555 example circuits on the internet for doing just about ...


3

Your question doesn't state that you are only allowed use one op-amp. If you are allowed use two then use two inverting summing amplifiers. Use the first to sum all the positive voltages and the second to sum the inversion of those and the negative voltages. simulate this circuit – Schematic created using CircuitLab Figure 1. Two stage ...


3

The assumption that the inverting (-) and non-inverting (+) terminals are at approximately the same voltage, is only valid when the op-amp is configured for negative feedback and its output is not saturated. Negative feedback causes the output to converge such that the inverting terminal and non-inverting terminals are at very close to the same voltage. The ...


3

why don't we consider C3 and C4 for the corner(Cut-off) frequency calculation It's a simplification. Consider the following; all caps the same value and both resistors the same value. The differentially placed capacitor C2 and R2 and R3 form a filter that might have a cut-off of (say) 1 MHz. As individual wires, the cut-off frequency has to be 2 MHz ...


3

The TL081 is not good in this circuit because its inputs only work down to about 3V above the negative supply. You should use an op amp whose common mode input range includes the negative supply, eg. LM358. A 9.6V NiMH battery will go up to about 11.2V (1.4V/cell) at full charge. Therefore your constant current circuit needs to drop less than 0.8V at 1A ...


3

Q1: There is some internal circuitry within opamps (bias, ...) whose current consumption depend on the provided voltage. So, even if it does not change anything to the opamp output, changing its supply voltage can change the current it internally draws (called the quiescent current). Q2: You can't, with this model (see below) Q3: It is indeed inconsistent. ...


3

1) In the case that you do not have R1 in place you would expect the + input node to be floating. But that is not the whole story. All the pins of opamps and in fact every chip have ESD protection diodes between the pins and the supplies. So if you apply a signal at the left of C1 there is a way to charge and discharge C1 if the input signal goes below or ...


3

Yes, you can probably share the divider. The only currents it sees are capacitively coupled and currents that flow through the op-amp input. They will cause small errors, but probably not significant to you. I would suggest reducing the divider resistors to 5K or 10K each and increasing the capacitor to 100uF.


3

These are all loose terms used to describe filters or coupling networks, often it is the -3dB point that is being referred to, but not always. With a single pole RC filter, the -3dB point is convenient as its frequency drops straight out of the sums for the RC time constant. With a Butterworth filter, regardless of the order, the cnventional edge of the ...


3

R2 is often on opamp outouts .Like John D commented it provides some capacitive load isolation .It also isolates RF and is sometimes referred to as a "Stopper resistor".For termimation you assume that the opamp has zero output impedence so in your example the circuit has an output impedance of 33R .The resistor does not waste much output voltage and can ...


3

The resistance range of a potentiometer is fixed at manufacturing. You can't reasonably change it on your own. Lowering the resistance is especially hard since it means adding material. If you wanted to increase the resistance using a opamp, then you could try opening the pot and using the opamp to scrape away some material. However, the result will be ...


3

The data sheet says pin 8 (bias select) needs to be connected as shown below: - A 3V3 power supply is OK as is the way you have connected R2 so ignore the other answer.


3

I'm assuming you're using the DIP package (Vcc- on pin 11). You can use this op-amp with a single sided supply for your application (using it as a comparator between other positive signals). The datasheet shows a number of examples using the op-amps in single sided mode (Vcc- connected to ground). http://www.ti.com/lit/ds/slos081i/slos081i.pdf#page=16 Bop ...


3

Because the input is only 0.5V (-6db). You should divide the output voltage by the input voltage to avoid such problems or even better use 1V when doing an AC analysis.


2

The LMV324 has interesting specifications; in the absolute maximum ratings table, we usually see the maximum input voltages at the IN(+, -) pins relative to Vcc and to ground, but that is not quite the case here: This implies that the input can handle 5.7V to -0.2V on any single supply, but I suspect that is incorrect as the internal ESD diodes will turn ...



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