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Hello Friends, I am designing a Charge sensitive amplifier and would like to study about the stability. we know, A/1+AB where AB is the open loop gain. If AB is equal to -1 then the system will oscillate. How would I know that AB will be equal -1. Please tell me the process to calculate. A= G sCdRf/ 1+sRf( Cd+Cf) B= (...)
GBW of aux opamps has to be equal to dominant pole of main opamp. In other way You get zero (if GBW is lower) or additional pole if it is higher. It results with auxilliary amps dominant pole location equal to main opamp dominant pole divide by aux opamp gain.
Hello, I had a question in op amp? If we provide equal dc voltages to an op amp it will produce some voltage 'Veq'. The op amp has a finite gain. My understanding is, if we use this circuit in a negative feedback loop of another circuit X, its input voltages come closer to each other. However they do not necessary become exactly equal. I (...)
Equation 23, page 14 , of AN57 by power integrations (below) shows a parameter called ?K TL31? which is the gain of the TL431. From the attached TL431 datasheet , does this gain equal ??.which equals 1/0.0014 = 714 TL431 datasheet: AN57 by Power Integr
For a good understanding of the feedback principle it is necessary, I think, to realize that the voltages at both inputs are NOT equal (due to negative feedback). This is because the opamp always needs an input voltage difference to produce an output voltage. However, due to the large open-loop gain Aol of the opamp this differential input vol
Yes, that means the op amp will be unstable if used as a +1 follower. A minimum closed-loop gain of +2 for a non-inverting amp or a gain of -1 for an inverting amp is the same from a loop-gain point of view since in each case you have two equal value resistors going to the op amp (-) input. The only difference between the (...)
Normally all values are equal for a unity gain differential amp ( 10K) to minimize offset from input bias current. inverting gain= -R7/R8 *7V = (-1) * 7V Non Inv gain = 1+ |inverting gain| * V+in= (1+1) * 5V*R5/(R5+R6)= (1+1) 5v * 1/2= 5v Using superposition rules the output-7+5 =2 the difference. As a (...)
Where do you see doubled current, in which situation? In steady state, the inner loop current setpoint must be obviously equal to output current, are you talking about transient states?
In open loop configuration it's enough to use a single pulse source in series with one of the Vcm bias supplies (small signal stimulus). In closed loop configuration (your application) you better use 2 equal anti-phase pulse sources in series with a common Vcm bias supply and (if so) the gain setting input series impedance d
hello everyone... is it necessary to have the UGB of single ended differential amplifier used as CMFB of main opamp to be equal to the UGB of the main opamp ???? Is there any specific crieteria???
It seems that you should use linear values for every specification. The reason 1 is subtracted from NF is because it is always greater than one in linear scale. Therefore, that performs is just how much better than 1 your design is. The same goes for the gain, if you double the gain while maintaining everything else equal, you should have (...)
The loop gain at DC is 90dB. However, my PSRR at DC (taken at the top of the resistor) is 80dB. Does this make sense? I would have thought the PSRR at DC should equal the loop gain at DC. There's no direct connection between open loop DC gain and PSRR at DC (better: at low frequencies: DC is static, no cha
You can use a positive bandgap reference of any voltage and your negative supply scaled to be equal and opposite. Then your inverting input is now 0V ,( with +Vin=0V) your (-) must now use the same source impedance with a common feedback gain. You may add series RC values for derivative and proportional gain control and common RC (...)
A Sallen-Key filter with equal resistor and capacitor values will have a Butterworth response when its gain is 1.586 and will oscillate when its gain is about 2.0 and higher. Try it with a gain of 1.8 times.
The folded cascode has a higher gain than that of simple amplifier, but its gain is little lower than that of cascode. As far as I remember, the main idea of the folded cascode is that the input and the output common mode points are equal. Why do we need such a thing? in case you are designing multiple stages (cascading) in order to increase (...)
Hi, I am finding Current gain of Common gate amp with gate biased and applying supply at source. If I measure Iout/Iin getting 1 as current gain till mosfet is ON and when mos enters to cut off it is giving gain of sudden spike around 20. and also by varying input sweep in steps 0.01/0.001/0.0001. In this case source current is (...)
It's related to the fact that the gain of an inverting op amp is ideally equal to the Feedback-Impedance divided by the Input-Impedance. As the frequency increases the impedance of C1 goes down. Since a reduction of the input impedance increases the gain of the op amp, the input RC causes the gain to increase with (...)
If the current gain is not equal, a differential output voltage will exist for an equal input current. This is even more important if only a single ended output is used and current mirrors are used with a different Vbe characteristic. Some more info here
Correct - however, does this justify the dimensioning with equal components? I don`t think so - because this has severe disadvantages. Selection of equal component values requires a fixed gain rater close to the stability limit (which is "3") and the filter parameter sensitivity upon tolerances is rather large. A better design is based o
Things are very simple. Whatever gain the RX and TX antennas have and whatever loss is between them, the received signal power level (at RX input) cannot be higher than transmit power level (at TX output). In the best case (and ideal) they can be equal.
Comparing DCM and CCM flybacks, both in current mode, which has the higher gain of the feedback loop?...(assuming all other things equal.)
Hi, i'm trying to calculate small signal gain for common emitter amplifier with bjt diode connection load. For diode connected bjt resistance is equal to parallel connection of Rbe and current source b*Ib. Rbe much smaller then Rcs and therefore Requal ~ Rbe. For gain we have A = -gm1*Rbe2 Am i right? Update: Sorry, (...)
You mean F=SNRi/SNRo but just at certain level of noise? Yes, let's suppose we have a noisy amplifier having gain G. At the input we'll place a resistor having value equal to the input impedance of the amplifier. This resistor will generates a noise power given by N=k*T*B where k is the Boltzmann constant, T the tem
... keep the same stage gain. I did try to use a non inverting op amp stage after it instead of two inverting stages but couldn't seem top get it to work correctly. After the tap of your poti you could do it with one opAmp and 4 equal resistors (say: 100 kΩ) , and a positive reference voltage of 7.
Can inverters, like the ones found on 74LS04 ICs, operate like op-amps? I read somewhere that by connecting the output to the input, it can be made to function like an op-amp. If this is true, then the gain is equal to the feedback resistor value / input resistor value, assuming that that is all there is connected to the
Not quite right. Output impedance of common collector (emitter follower) is measured between the emitter and common, not emitter to collector. The output impedance of the common collector is approximately equal to the source resistance divided by the current gain (hfe).
In simple terms if Noise is calculated using: Noise Power Density = kTB + NF + gain (dBm/Hz) Then for your case with the amplifier and transmission line: Noise Power Density = kTB + NF(amp) + gain(amp) + NF(line) + gain(line) For any passive device, like the transmission line, the Noise Figure is equal to the (...)
Hello, Can anyone recommend a high gain, low power BJT for use at 450MHz? I have been using the BFT25 but I was wondering if there is something out there with more gain. 10LogS21/S12 is roughly equal to 14db. Thank you
Hello everybody ,why do we perform DC and AC analysis in mosfet amplifier ,is the voltage gain equal to the ac output voltage over the ac input voltage or the resultant of DC and AC voltages, and finally is Id the DC drain current or ac drain current such in this equation Id=.5 Kn*(Vov)^2 ?
You must show us your schematic. The easiest way is to use two equal value resistors directly across the 0-10V signal. At their center it will be 0-5V and no op-amp is needed. Brian.
breakfrequencie, 3dB Bandwith and unity gain bandwith are one of many things i am bit unsure of. I know that the breakfrequency is the frequency for where the gain drops 3dB. which means that unity gain should be the same, (which i know is wrong, since formula is different) And at last 3dB bandwith should equal (...)
Could you please show me in any graph th -3 dB frequency of the buffer connected op-amp please Junus - it is very simple: For any first-order circuit the 3dB point is at a frequency where both imaginary and real part of the denominator are equal. This applies also to the simple RC lowpass. The denominator of the
vo = A ( v(+) - v(-) ) that is output voltage vo is equal to the difference in voltage between + and - inputs multiplied by the Open loop gain of the op. amp. In an ideal op. amp the current at the inputs is zero (infinite input impedance) and A is very large (--> infinite). Output resistance is zero.
it is really difficult to get an exact equal voltage at the input of a very high gain comparator, a comparator with a very haigh gain can switch the output with a very small difference. assume the gain is 10000(80dB) and the supply voltage is 3.3 and by asssuming that the offset voltage is zero then the minimum voltage (...)
The Barkhausen criterion requires a loop gain of equal or larger than unity and a phase shift of 360 deg at a certain frequency. In a typical ring oscillator the loop gain always is larger than unity. Each stage has an inherent delay, which can be increased using external capacitors. There is one particular frequency Fo for which this (...)
you can use a differential amplifier 77360 Apply 0.5v to VR, V1 to gnd and the input to V2. If all resistors are equal then you will get 0-2v, if you use RF twice the size of R1 then you can get a gain of 2 and 0-4v output. You can also use a VR of 1v to get some headroom in case the input is <-0.5v and then compens
paths add constructively and destructively with equal probabilities, thus the channel gain varies about the mean channel gain -> 0 dB (gain = 1).
You cannot remove poles. But you can compensate those poles which is not a very efficient solution. A better solution is to make those gain boost amplifiers excess stable as said in Vadim Ivanov's book. If you are compensating you are reducing the speed that you have already paid for with power by using capacitors which is equal to area. Use only o
To summarize (and partly correct) some of the statements made before, I like to point out the following - applicable to each kind of harmonic oscillators (LC or RC): * Each harmonic oscillator must be designed to enable the following condition (formulated in the 1930th by H. Barkhausen): (1) Loop gain magnitude equal to unity for one single frequen
I want to connect between two stage PGA that the overall gain equal i have designed the first stage by telescopic cascode op amp and feedback resistor loop on it SO how i connect between them or any one can give me any materials that talk about this problem..!!
I just came across this- We know that feedback gain is "β". I read that when β=1, we have worst condition. Why is that so? Thanks in advance.:razz:
I am simulating a waveguide in HFSS having 2 ports( rectangular dielectric waveguide) .I want to ask what is real meaning of S(1,2) because in one case i am getting S(1,2) plot showing 5db gain while the propagation constant (beta) is equal to 0. Field overlay also shows wave dying much before even reaching port 2 then how to interprete S(1,2) val
Yes you can do so if your Yagi-Uda balanced feeding is converted to unbalanced with correct impedance, as the whip you have seems to be unbalanced. Plus, you'd need to ensure the PC also can transmit to the same range No. If it result in antenna directional gain is this gain equal affecting both TX and RX, so it is no special need to
If you have biased symmetrically and correctly, in-spite I should say, If it is biased symmetrically, you will get equal gain. This is ac gain and there is no +/- sign for AC input signal.
You are correct but if the output is fed straight back to the inverting input (and assuming it is unity gain stable) you do not get any again amplification, you have created a signal buffer. The virtual ground is a point where the incoming signal and the feedback are equal but opposite polarities so they effectively cancel out. You will (...)
Dear all, I have designed fully differential symmetric SC-amplifier. When i load differential output with equal capacitive load (say, C1), then differential output is, say, y. But when i load differential output with unequal capacitive load (C1,C2), then differential output is y?Δ for the same input signal. Does unequal (...)
If you are allowed to use as much as 3 opamps I would recommend the first of the circuits as shown in your document. Both integrating stages may (and should) have the same time constant RoCo. This defines the frequency wo=1/RoCo. The inverting opamp has a gain of -1 (two equal valued resistors in the kohm range). Don't use the diodes as shown in th
Here you have a example of a headphone amplifier that that I designed few years ago. The supply voltage was equal to 3V.
Hi, Medium wave and short wave bands will have intermediate frequency stages scince due large bandwidth/frequecny range. the RF amplifier gain is not equal for all receiving radio chanels at different frequencies due to passive components. so what ever the desired frequency we are receiving we first convert it into IF frequency (intermediate freque
MCP601 is wired as a unity gain buffer, the output voltage will be equal to the input voltage of the positive input. When PIN4 of the jtag plug is not connected then the positive input of the opamp will have 3.3v from the output of the TSP77030 so the output of the opamp will be 3v3. When PIN4 is connected to the voltage of the target (it seems to