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New answers tagged operational-amplifier

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A couple options. I'd recommend that you do a math channel and divide your output by your input. You can then turn on a measurement to measure the average of the math channel waveform. You could also do measurements on the waveforms themselves and do the division by hand/in your head.

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The equation that you need is: $V_{out} = A_v (V_{+}-V_{-})$ Av is the open loop gain. The resistances can be neglected if this is an ideal opamp (if it's an opamp with protection diodes, then the resistors do matter) as ideal opamps have infinite input impedance meaning zero current flows into them. If this is a real op amp, an input bias current ...

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You shouldn't just use some rough noise density number for this calculation. The reference fig for noise in your datasheet is Figure 10. From the datasheet you linked to: Noise density figures like this show a knee, below which pink low-freq noise dominates the white noise, and above which, white noise. The important parameters are the frequency of the ...

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Input is signal frequency is 100Hz. I want to use the 6.5nV/sqrt(Hz) but it seems to be a spectral density, and I can't add it to a voltage This is where the bandwidth of the analog system comes in, the max bandwidth determines the noise. Let's say it's an ADC with a low pass filter at 1000Hz. The ADC would see: (1000Hz)^(1/2)*6.5nV/sqrt(Hz) = 2.05uV-rms ...

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The most simple way for explaining the negative feedback effect is to split the whole procedure in several steps: Example: Non-inverting opamp stage with two equal resistors in the negative feedback loop. Supply voltages +/- 10volts. Open-loop gain Aol. Step 1: Apply a DC input of +1V at the non-inv. terminal. Step 2: Because each amplifier has a certain ...

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If $V_+ = V_-$ then, by definition, $V_D = 0$. If you look at the transfer curve you included in your answer, the region of operation where $V_D =0$ is at the center of the graph. In this region, the output voltage is a linear function of $V_D$. Specifically, $V_O = A_{OL}\times V_D$. EDIT: To clarify based on the comments, the output voltage of ...

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Why opamp doesn't behave as datasheet "guaranteed"? It does, but only the values in the tables are guaranteed. The rest is, well, up to anyone's interpretation. How can I achieve 0V output with this opamp? Maybe by loading it with only resistor to ground. 9 mV sounds quite good to me already. If you really need to go to 0 V, use a negative supply rail! ...

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At 2kHz the impedance of C2 is 16kOhm, hence your amplifier acts like if R1 was equal to 300kOhm || 16kOhm = 15kOhm Without the capacitor, your gain is -1/3 but with the capacitor the gain becomes -6.7 and hence you saturate the op amp (1V * 6.7 = 6.7 V pick > 5V pick)

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When a parallel C(is labelled as C2) is added, the output is saturated. What is the explanation of this? At 2 kHz, the 5 nF cap has an impedance of about 16 kohm. Because it is in parallel with a resistor of value 300 kohm, the op-amp gain is increased about twenty times and it is much more likely that the op-amp output will saturate. How can i tackle ...

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Indoor lighting usually has a large amount of ripple caused the AC voltage varying lamp brightness at twice the mains frequency. Depending on the particular bulb the variation might be anywhere from 5% to 100%, and could be many times larger than the signal you are trying to detect. A filter attenuates out of band frequencies that could overload the gain ...

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simulate this circuit – Schematic created using CircuitLab This topology will work. OP1 is a non-inverting schmitt trigger. OP2 is an integrator. OP3 is configured as a comparator. The output of OP1 is a square wave which drives integrator OP2 the output of which is a triangle wave. OP3 compares the triangle wave to a settable threshold to produce ...

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THis will easily work with dual supply . But use Pot for R to change frequency from 1k to 10K using fixed 1k R in series with 10k Pot and C=0.1uF The Positive feedback is simply equal R values from output to Mid-supply (V+ +V-)/2 = 0V in this case. This changes the threshold of the Relaxation Oscillation just like CMOS Schmitt triggers with 50% ...

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The concept of a 'virtual short' is not complete without understanding feedback. An opamp without feedback cannot maintain a virtual ground. In the picture below, Vp is equalized with Vn only through the output changing. If Vn is greater than Vp then the output will be adjusted down until Vp and Vn are equal (and vice versa if Vp is greater than Vn). Very ...

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The notion of a "virtual short" means that the voltages at the inverting and non-inverting op amp inputs are at the same voltage. This is true for an ideal op amp and a reasonably close approximation for a real op amp, provided that they are in a circuit with negative feedback and within their specified limits. The virtual short does not allow current to ...

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simulate this circuit – Schematic created using CircuitLab If it generates more than 30V , consider this simple solution. Op Amps make slow comparators without gain reduction. So be careful in your specs/choices. You probably want a comparator with a pull up R for open collector/drain types.

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You really did not tell us anything about the ADC specs that you require. Various manufacturers like Analog Devices and TI have vast lines of products, including integrated parts that have PGA (programmable gain amplifier) front ends. Here is a copy and paste from A TI webpage with a full scale input range in hundreds of millivolts. "The ADS111x perform ...

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What value to use for resistor (R1 below), and how to calculate it? Depends on input bias current, an opamp with very low bias current can have a R1 very high without affecting the measurement, several Mohms. Higher resistor also means higher noise. How much current will be flowing through the op-amp due to the high voltage spike? if the opamp has ...

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Are there any obvious problems with the circuit, that may lead to the observed drifting behavior? The minimum supply for an LPV821 is 1.7 volts and you are expecting it to work above 1.6 volts. Your AC current is 1 uA and this produces a voltage (across the 20 ohm resistor) of maybe 28 uVp-p. Given that your op-amp is running without feedback AND it has ...

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We're clearly not talking about ideal opamps here, which have infinite gain and infinite bandwidth, among other properties. The G-BW product is a constant for any real opamp that has a dominant pole, and it only applies above the frequency associated with that pole. Below that frequency, the opamp has some finite maximum gain dictated by the limitations ...

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Thanks all for your great answers suitable for a variety of situations. In my case, I am trying to use items already on hand. Hence, the ADC with a restricted range. Also, the ADC works from a 5V supply, so originally I did not have a voltage above 10.1 as reference. For future reference, I see that it is dead simple to chose an ADC with an appropriate ...

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Most likely one company came up with the design first. Then other companies reverse-engineered the design to sell copycat parts. Some customers might even insist that a second source of supply be available for such parts (linear regulators, jelly bean transistors, basic logic gates, etc) so it wouldn't help the original vendor to try to stop the copy-cat ...

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I was going to post a comment(especially since it's an old question), but it became too long. That Application Note National AN-1852 (Which it located at TI now), describes in large detail the reasons for the inclusion of the op-amp to begin with. It provides two totally different services to the circuit. First it provides a low impedance 512mV bias to ...

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One important factor is that the input impedance of your OPAMP is getting very low because of 500 ohms. if you take it to be some 10k or so, your oscillations may sustain. also your loss of signal due to low Q can be compensated. as input impedance falls, loading effects the opamp and hence your oscillations are damping.

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It is a very slow opamp like a 741 so at only 10kHz its open loop voltage gain is 40dB which is 100 times. An OPA134 audio opamp has 10 times more open loop voltage gain at all frequencies but this circuit has no high audio frequencies.

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Yes, it means the open loop gain is 40dB, i.e. ~100 V/V (20 log 100 = 40 dB), up to around 300kHz. It is a typical open loop gain of differential signal. Since you want to amplify the input signal by a factor of 1 to 15 and the output signal is: out = in * (1 + RV1/R1) you have to make sure RV1 will vary from 0 to 14 * R1 = 14 * 3.48 = 48.72 Ohm, so 50 ...

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fatal flaw ---- when tapped hard, your piezo likely will output 20 or 40 or 100 volts, which will destroy the opamp, unless you have a current-limiting resistor in series with VIN+ of OA1.

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The classic 4-resistor difference amplifier will do what you want: simulate this circuit – Schematic created using CircuitLab This circuit keeps the difference between Vout and Vref equal to the difference between V+ and V-, effectively adding Vref to V+ if V- is grounded. The opamp tries to keep its two input terminals at the same potential using ...

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The total resistance for 200 mA at 20V is 100 Ohm. The wires should be quite long or thin. But as answer: I guess you want to make some kind of heating. If you plan PWM control, high enough switching frequency should quarantee the temperature of the wire stays in certain range, no matter the current is alternating being part of the time 0 mA and the rest ...

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This is completely unnecessary, because the LM358 is already rated for continuous short-circuit to ground of any one output. Specifically, Note 5 under "Absolute Maximum Ratings" states: Short circuits from the output to V+ can cause excessive heating and eventual destruction. When considering short circuits to ground, the maximum output current is ...

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Technically, R5 has the same role in both circuits: In amplifier A, R5 plays as an RC low pass filter for the IF source signal. The fc is 1/2*3.14*330k*100u = 0.0048 Hz. The IF signal is normally more than that frequency (3-1000 Hz), so it cannot pass that filter and disturb the 2.5 V DC bias. In amplifier B, we have V+ = V-, so the IF source signal will ...

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Try this circuit configuration: simulate this circuit – Schematic created using CircuitLab Because form what I see you want a DC gain around $5.6 \:[V/V]$ and you want the HF gain to be 0dB (1 V/V). And for this circuit, we have one pole at $$F_P = \frac{1}{2 \pi C_1 R_2 }$$ And the Zero at: $$F_Z = \frac{1}{2 \pi C_1 (R_1||R_2) }$$

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I think I would be looking at low capacitance small MOSFETs like the 2N7000/2N7002 - you don’t want your buffer to add more capacitance than what you are driving. There are others in the BSS series. Look for low Id parts and fairly high voltage rating (like 30-80V). A lot of the physically small parts are low voltage high current designs that can switch ...

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I think that what you are saying is to "add gain" but not "invert" the signal. Your added amplifier is an "inverting" amplifier. You are "going into" the - input. So just use a "non-inverting" amplifier, "going into" the + input. Not sure of your experience level .... do you know how to do that? You can search for "non-inverting op amp amplifier" on ...

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When I touch (grab) the tip of my oscilloscope probe, I get the following waveform. The 'scope is grounded via its power supply, but I'm only touching the probe tip. Note that the AC power in my region is 60 Hz, not 50 Hz: Note that the primary wave has a 60 Hz period, but has a very jagged wave shape....the AC power source is certainly not so "unclean", ...

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Your option b is correct. Your body is mostly insulated from ground, so it acts like a plate of a capacitor. The other plate is the rest of the world, or better, the portion of world somewhat near you. A part of this other plate is, for example, the mains wires in the ceiling or walls. Since the voltage in them oscillates at 50 Hz, and you and them make a ...

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R5 provides DC bias of Vss/2=2.5V to Vin+ used by Amp A. Amp B does not use Vin+ and rather uses 2 stages of inverting gain thru Vin-. Thus R5 does not affect the signal impedance on Amp B. Mainly it has more sensitivity with the extra filtered gain stage with the dual Op Amp then finally a comparator to produce a pulse. Basically Amp A is missing the ...

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Few comments: Your feedback opamps have no compensation. You would end up trying to regulate out the ripple on your output and start oscillating like crazy. Check this link: http://www.ti.com/lit/an/slva662/slva662.pdf it is an article titled "Demystifying Type II and Type III compensators". I'd recommend a Type II compensator for this design (three extra ...

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With the output connected directly to IN- there is about a 30mV difference between the input and the output, where the input is about 400mV. Measuring a high impedance node with a voltmeter can result in a significant error. The input node impedance is around 783 kohm and your meter might have an input impedance of around 10 Mohms so, when you put your ...

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You are correct about the equation's restrictions. The words you are looking for are correlated and uncorrelated noise. That equation is for valid uncorrelated noise sources only. But it's a tall order to have all the noise sources in a circuit to be of the correlated variety, let alone be correlated in such that they match so perfectly and in exactly the ...

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Using these reading from Agilent MSO-X2012a oscilloscope: and How do you calculate the close loop gain using these measurements. You can look in detail at the picture: - And see that channel 1 is about 11 times bigger than channel 2 i.e. gain is 1 + 10k/1k.

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If the op amp is powered from V+ and ground, of course it won't work at all. No matter what the state of the diode, the output will remain at ground in order to best approach the virtual ground state, and the diode will be unbiased. If the op amp is powered from V+ and V-, the output will remain in the vicinity of ground, keeping the diode current at zero ...

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One way to get a low leakage clamp is to employ a BJT emitter-base junction. For example, simulate this circuit – Schematic created using CircuitLab Typical peak current at 4.2V is about 30nA leading to an error of 3uV with a 100 ohm source resistance. It clamps to typically about 5.2V with 12V applied with power dissipation of about 350mW, which ...

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consider something like this simulate this circuit – Schematic created using CircuitLab

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Zeners have a very 'soft' characteristic. As you discovered, they start drawing current well before their advertised voltage. The best way to limit overvoltage into an ADC is to use a silicon diode to a fixed voltage. Usually this will be the ADC rail, especially if you put a series resistor between the diode and the ADC to limit any current drawn, but you ...

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R2 makes the circuit work and sets the "gain" or ratio between output and input voltages. Forget the diode for a moment, it is not necessary for mentally analysing this. The op-amp is in an inverting configuration, so the op-amp output wants to do everything it can to keep the inverting input at the same voltage than the non-inverting input, which is 0V (...

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Like drawn, the circuit is incomplete / unclear as there is no resistor (and/or capacitor) between $V_{in}$ and the - input of the opamp. That means that the feedback circuit (R6, C4) actually doesn't do anything and the opamp will work as a comparator. Without the DC biasing (which isn't needed if we use a symmetric power supply) that would look like: ...

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The current is is equal to the current moving through the resistor. This is why we called a virtual ground, the voltage of the negative terminal is the same as the positive terminal if in negative feedback, but no current flows into the terminal in the case of an ideal op amp such as this one. Because of this the currents are equal and the current flowing ...

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I might be late on this front, but the evaluation board for the ADHV4702-1 shows how to create the +/-110V rails off of a 5V input using a LT8304-1 power management IC: https://www.analog.com/media/en/technical-documentation/user-guides/EVAL-ADHV4702-1CPZ-UG-1444.pdf

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There are probably ESD clamp diodes from the input to the GND rail and power rail of each IC. So the result would have a partially or fully powered IC from the current being provided by Vin. since Vin should be larger than Vth of the MOSFET in order for the switch to be on. This remark doesn't make sense. You are either mistaking the gate drive ...

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Well, we are trying to analyze the following circuit: simulate this circuit – Schematic created using CircuitLab Using KCL, we can write: $$\begin{cases} \text{I}_{\text{R}_1}+\text{I}_{\text{R}_2}=0\\ \\ \text{I}_\text{op}=\text{I}_{\text{R}_2}+\text{I}_{\text{R}_3}\\ \\ \text{I}_{\text{R}_3}=\text{I}_{\text{R}_4} \end{cases}\tag1$$ Using KVL, ...

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