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104 Threads found on Opamp Output Impedance
The impedance at V2 is 5.8k || 580k which is just somewhat less than 5.8k. The output impedance isn't answered by the information provided as it depends on the amplifier, not the resistors shown. An ideal opamp for example would result in zero output impedance for that circuit.
hi guys I want to know how to measure output impedance of opamp in cadence. Regards Arunkumar
An LM3900 is a very weird quad general purpose opamp. I have never seen one and I have seen thousands of audio circuits that use audio opamps. A TL071 single, TL072 dual and TL074 quad opamp are designed for audio with low noise, wide frequency response and wide slew rate, very low distortion and very low output (...)
Icbergn - are you really forced to realize an oscillator with an inductor? As you know - there are many ocillator topologies based on RC combinations only. More than that, did you consider the fact that the active element in the classical Hartley oscillator is a current source (BJT) - whereas the opamp is a voltage source?
Your schematic has nothing to supply the input bias current to the opamp. If it is the signal generator then its resistance will cause the offset voltage. Add a same value resistor in series with the (-) input so that the bias currents cancel.
The output resistance of a typical opamp is maybe 100 ohms. The gain at DC and low frequencies (below 10Hz) is extremely high which reduces the output impedance when negative feedback is used so the output impedance could be much less than 1 ohms at DC and low frequencies.
Hi, it seems you are talking about ASIC design. If so, then please post in the ASIC design section. *** Generally a unity gain buffer is an amplifier with gain of +1. With an opamp you can IN+ use as input and connect IN- to the output. Maybe with a resistor. Then you have high impedance input and low impedance (...)
The voltage follower does two things: 1) It isolates the 1st order RC circuit at trhe input from the second order part. 2) It provides an important extremely low output impedance to drive the second order part. If you put the input RC stage in the feedback of the opamp then the (inverting) opamp inverts the phase shift of (...)
please explain me whether the opamp(two stage) i have used can drive the resistors as shown You can answer the question yourself if you calculate open loop voltage gain and output impedance of your opamp. Of course the answer also depends on the resistor values. But in case of a wideband amplifier, the resistors need to (...)
If you've got the output of one opamp driving the EXTREMELY HIGH INPUT impedance of the next op amp, how much current does that output need to supply? Just about nothing, dynamic or otherwise. If you've got feedback resistors, those will also draw some current, as will capacitors (dynamically!). Without seeing a schematic (...)
Dear all, I am using cadence to simulate an opamp's input referred current noise. However, i have no idea how to build the test circuit for it; since I cannot use a current source as a input noise source in the noise analysis. If i do, the bias condition will be disturbed. can anyone help me? You can use a c
However the bandwidth of the opamp is limited to about 70MHz and also the output impedance increases from there on. A more appropriate description would say, the output impedance is continuously rising beyond 1 MHz. And not explicitely specified in the datasheet, but obvious by circuit analysis basics, (...)
Dear forum, just a simple (?) beginner question: An answer to your question was given above. However, as you are a beginner: Are you aware that using only one single supply voltage (s shown in your diagrams) imposes some severe operational restrictions? The classical opamp application requires dual supply (+/- voltages
I show an opamp but it could be an audio power amplifier. On the left side there are circuits with a positive and negative dual polarity supply. You are trying to make this from a single supply by using brute force to feed the power supply for the amplifiers. On the right side I show amplifiers that have a single supply and are biased at half the
Negative feedback reduces the output impedance of an opamp. It does not increase it. When negative feedback reduces the gain of an opamp to 100 then the output resistance of 75 ohms is reduced to an impedance of much less than 1 ohm at DC and low frequencies.
hi veeru you need make the circhuit with opamp to measure the PH output along with voltage offeset thanks
The opamp has negative feedback that produces an output level exactly the same as the input level. Cf feeds the output level to the junction of R1 and R2. Then R1 has the same signal level at both ends then has no current in it so the input impedance is extremely high.
I think a rail-to-rail opamp should be used because a transistor has a low input impedance especially if it has negative feedback from its output to its input to reduce its voltage gain to be "inverting one".
Hey guys, I was just wondering how i would make a voltage buffer/follower without using an opamp. I have used some class A (emitter follower) and AB amplifiers, but when my output is very low impedance, my output starts to saturate with an offset. I was wondering if you guys had any designs in mind with very high (...)
Hello, Is R14 going to be OK for use as a loop injection resistor for feedback loop measurement? (attached schematic & LTspice simulation). Also, the opamp, U2 , is included purely for ensuring that the feedback loop measurement can be successful. (low impedance of opamp output is needed) That is, R14 and U2 (...)
I know that OTA and opamp differs in that opamp has a output buffer with low output impedance while OTA has high output impedance (ro). Why is it that OTA can't drive a resistor? I get that the gain drops from gm*ro to gm*R, if R is small but isn't this the same for an (...)
I have a very basic question about opamps. Say we have an integrator, using R and C. 98282 The equivalent model for the opamp is, 98283 If I take the second figure and add R and C and ground one of the diff pair, does it still function as an integrator? If so, do people just use the first pi
Hi All, when I study feedback effect, one of its effect is to reduce the output impedance. Then I am curious, if we have an OTA with high output impedance, is it possible to use feedback to reduce this impedance to a very low level, then this OTA will be a opamp? Since we always need (...)
The signal source of the Sallen-Key lowpass filter must be much lower than the input impedance of the circuit. Frequently the very low output impedance of an opamp drives it.
Tricka90, as your desired operating point is at Vcc/2 you must connect the positive opamp input to a Vcc/2 (voltage division using two equal resistors connected to Vcc).
impedance is AC resistance. The input impedance and input resistance of an opamp is 1M ohms to billions of ohms. The output impedance of an opamp is 0.01 ohms at DC and low frequencies to 75 ohms at higher frequencies. Its output resistance is 75 ohms.
I am experiencing strange behaviour of AD8607. It should be pA biass current and uV offset voltage opamp, but simple voltage follower configuration shows this results: 95124 Error seems to be related to source impedance as it dissapears if I connect opamp input directly to voltage reference output (no potentiometer). But
Hello, I am a newbie of IC. And these days, I really confuse about the "input impedance". According to TWO-PORT, for a VCVS, we should open the output to calculate the input impedance, and for a VCCS, we should short the output to calculate the input impedance. But for a opamp or OTA, (...)
why opamp has infinite input impedance, infinite bandwidth, infinite slew rate, infinite CMRR, Zero output impedance, infinite voltage gain. please its urgent. suchita
if you are talking about the standard opamp circuit. Input via a resistor (Ri) to the negative input with a resistor also from this point to the output (Rf). The input impedance is Ri, the output impedance is the actual output impedance of the opamp (100 (...)
Like any opamp, a comparator exhibits a bandwidth limitation characterized by its dominant pole, which in most cases is its output pole. Hence its "built in" 3db bandwidth is approximately 1/2πRC , R being its output impedance, C its load capacitance.
Hi When analyzing an opamp circuit using KVL/KCL/Nodal, how do the + and - signs designating non-inverting and inverting opamp inputs affect the analysis? How do we apply them? Thanks.
I am trying to measure the output impedance of an opamp in the attached configuration. Results are attached. Basically, Zo = Vo/Io = 10mV/40mA = 0.25 Ohms. Without the capacitor in the circuit, I get mostly a large dc current. Is there a better method to measure output impedance? I would like to (...)
Hi all, what we have learnt is that the output impedance of an opamp is very small. I cannot seem the find this information in the datasheet of the opamp I am looking at. May I know how can I measure the output impedance of an opamp using only spice?
This is not mtaching/unmatching issue... Your board can not drive 50 Ohm because of it's high output impedance.You should use a simple buffer amplifier for instance an opamp buffer etc..
- If I have 3.3 in the source - I have 12V on the engine - If I have 0V in the source - I have -12V on the engine(reverse) - If I have 1,65 in the source - I have 0 V on the engine (off) If you want +-12V on the motor then the supply must be more than +-12V. +-15V or +-18V would be good. +-24V is too high; it
Hi dino, two additional remarks: What do you mean * with "ideal amp"? Transistor, opamp, OTA? * with Rout? (The term Rout in your formula primarily consists of the load resistance and - in most cases - is not the output resistance of the amplifier itself).
Hello everyone. For the past few days, I've been struggling with my oscillating OPA549 opamp when I try to drive a simple DC motor with it. Since I do not know the inductance values of the motor, I cannot employ an RC snubber circuit in parallel to the motor since I will not be able to calculate the R and C values - I tried trial and error bu
what is the input and output impedance of an ideal opamp? How can you use an opamp as buffer? which transistor configuration is used for power amplication? How power amplification occurs in such a configuration?
Why is that the output impedance of the OTA is ideally very high? I thought we need a high output current but why is it that the ideal output impedance is very high? Can you please explain it to me? In contrast to the well known opamp (which is a voltage source) the OTA shall act as a (...)
It is a lowpass of 2nd order with unity gain. Advantage: Offset-free (no opamp output) Disadvantage: No low-impedance output (load impedance influences transfer function).
The output of the opamp log converter is buffered, i.e. has a much lower output impedance.
Hi guys, I'm a beginner in electronics and have a problem I can't solve. I have a 2 stage opamp that I use to amplify a sinusoidal input of 40Hz (1v peak to peak). The opamp amp is designed to returned a non-inverted magnified signal. I have a variable RC filter at the end to filter any high frequencies from the input source. The circuit works well
Jon, what is the purpose of the voltage divider R2,R10? Can you reduce the gain of the first opamp instead? The output of the first opamp has low impedance, so you can connect it directly to the ADC. - Nick
An amplifier could be an attenuator when it has a voltage gain less than 1. On the other hand, using an opamp allows to improve drive capability because its output impedance.
HI friends i am designing the Folded cascode opamp. i am getting maximum gain of opamp as 46 dB..! @L=900nm and VDD=1.8....if this is the only gain can be obtained then what is good to implement for increasing gain upto 70 dB 1) gain boosting 2)adding one more stage my GBW requirement is 50MHz and SR is 30v/usec.....what do i do please sug
can you give me some papers about simulation of input/output impedance of opamp both open loop and closed loop. thanks To measure output impedance, regardless closed loop or open loop, u just put ideal current source Idc with ac=1 at the output, then measure the voltage at the (...)
Hi, In non-inverting amplifier Gain (A) = 1 + R1/R2 The impedance of capacitor at given frequency (fr) is Xc = 1/(2*pi*fr*C) In your case R1 = R (in drawing) and R2 = Xc1 = 1/(2*pi*fr*C1) If we write = s then R2 = Xc1 = 1/(s*C1) opamp output is at Vx so Vx= Gain * Vi Vx/Vi = Gain (A) = 1 + R/Xc1 or Vx/Vi = 1 + R/{1/(s*C1)} So Vx/Vi
I couldn't find a datasheet for your device. If the output impedance of the sensor is high then you can use an opamp as unity gain amplifier (buffer) so that the A/D works properly. If your A/D range can't work up to 4.5v then you have to add a voltage divider in the buffer output to lower the voltage before feeding the A/D (...)
Hi, Thank you for watching this post. For a simple feedback, like R1=R2=R, opamp's gain is A. Non-inverting feedback amplifier, the input impedance is (1+AB)*Ri, B is feedback factor. Inverting feed amplifier , the input impedance is R+R/(1+A) I can do some math to get the result. But how to intuitive analysis (...)

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