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99 Threads found on Opamp Output Impedance
Hi, I have a question about output impedance of the opamp. In basic circuit theory class, we say that output impedance of the opamp is very small so that output voltage does not drop within opamp itself. However in transistor level (...)
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 (...)
Hello, I have designed a two stage opamp without the output buffer. Is it possible to drive resistive loads with this block? I'm not getting the expected results when I connected the resistors at the output and fed back the portion of output. So, should I go for a output buffer to drive the resistive (...)
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? Thanks a lot!
Hello all, For a wideband load I'm trying to design the power supply decoupling. The load pulls about 200mA DC current from my power supply with about 100mA of amplitude peak current at 3V supply. This peak current is at frequencies from about DC to 2GHz (multi-carrier signal) but for now I'm simulating with single-tone sine over the mentioned f
Dear forum, just a simple (?) beginner question: What is the difference e.g. having a differential opamp circuit with v=10 and using 1) 10k-resistor and 100k-resistor (opamp_1.png) OR 2) 1k-resistors and 10k-resistor (opamp_2.png) Thanks so far!
I need to design a split power supply starting from a single power supply. The single power supply is supposed to be 24V (at least 2A) so as to have a perfect 12-0-12 split supply, but I'd like to be able to use even one as low as 19V (still 2A minimum). The required output power of the split supply is at least 9V@1A each. I have searched the w
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 (...)
opamps DO NOT HAVE a single-ended collector output in class-A. Their output is produced by very low output impedance complementary emitter-followers in class-AB.
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.
I agree that the (+) input of the opamp must be properly biased at about half the supply voltage. Your existing unbiased opamp is a rectifier. Your opamp is inverting with an input impedance of only 100 ohms that loads down the high impedance output of the photo-transistor. The gain of the (...)
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 As mentioned in post#2, you have listed IDEAL conditions that never can be met. However, to answer your questions: All opamp manufacturers try to
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 (...)
In case of inverting amp: lets take a practical example, your input has 5V R1=R2=1K, output is -5V. Since you have the inverting inputs at virtual gnd, the input signal sees 1K load, simple is that. In case of non-inverting configuration, the input sees only the opamps bias current which is negligible.
The internal topology of the opamp chosen is not important as long as the transit frequency is at least - let's say - 5 MHz. Thus, you have a bandwidth-to-UGF ratio of 100 which is sufficient in most cases. The output impedance of the opamp plays no major role if the resistors connected at the output are (...)
Hello everyone, this is my first "new thread" to the forum :-) Please have a look at this datasheet: it's a typical IF demodulator - or at least I saw several ICs almost The filter opamp has only two pins accessible
Beacause of high output impedance of piezo transducer transformer as an amplifier is not a good option .. Instead consider an opamp based amplifier and more info you can find in this appnote: Signal Conditioning Piezoelectric Sensors IanP
I have a fliter that uses opamp based integrator. I want to make the same filter but using OTA-C instead of opamps. But I'm confused on the process of doing that. Anyone can help me on how to get the same behavior from the transition from one to an other?In other words translating the RC values in the omAmp version into the OTA version with the sam
OTA's are designed to be the gain stages of an opamp. the result is a high output impedance. This can sometimes be overcome by the very high gain. An opamp provides an OTA with a current gain stage to act as a buffer. In some cases the buffer is too expensive or undesirable, making the simple OTA a desirable choice.