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99 Threads found on edaboard.com: Feedback Effect
flicker noise is a relatively low frequency effect that "upconverts" to the RF region. So if you can lower the flicker noise of the device itself, you will have less clock jitter. Some sort of audio frequency feedback loop to bias your oscialator transistor, keeping fets outside of their voltage range extremes, applying analog feedback (...)
The book DAFX - Digital Audio effects by Udo Zölzer has a good chapter about reverberation. Simple reverberation effects can be made by superimposing multiple delays and adding some feedback.
I've been studying CMRR for voltage feedback amplifiers and I came across this document from analog devices figure.3(attached) in the document it shows a circuit to measure CMRR, which is
Whenever the voltage input of the 2nd opamp gets near ground then the diode turns off and the two resistors connected to the capacitor slowly discharge it. The negative feedback resistor R5 has its value changed to change the gain then the discharge time is also changed. The value of R4 can also be changed to change the gain.
Hi guys, currently, I'm working in the design of a small robot for a nice personal project. I would like to put a feedback illumination signal, and I like a lot the effect I saw some time ago in a Disney park (please see video attached). As you can see in this small video, it's basicaly a light spinning continously, I can't see any discrete l
Yes - such a capacitor between C and E in common-B configuration provides positive feedback and, thus, increases the input impedance for rising frequencies. This can be also described by the term "invers (negative) MILLER effect". However, at the same time, the gain decreases (C parallel to Rc). Why do you expect an RHP pole? [SIZE
Higher OSR results lower quatization noise level at feedback. So effect due to ELD can be alleviated. See pp.213-215 of the following. BTW, it seems that 2nd edition of this book is published this year.
1. You have been asking about delay not gain. 2. Delay in a feedback loop affects the phase margin and in so far the closed loop behaviour
Early effect both in simulation and oscilloscope, s. the following image: 117930 For BJTs, instead of Id vs. Vds plot Ic vs. Vce, with Vbe as parameter. Va is the Early voltage. Miller effect depends on circuit design, feedback and signal amplitude, so I think an analysis method (simulation) is more suitable.
The LTC6101 model is apparently instabble with larger Rin values due insufficient compensation of the feedback amplifier. No idea if the effect occurs with the real IC.
Generally, commom mode feedback (CMFB) improves the CMRR of (any) opAmp by keeping the CM output well within the CM output range for a wide range of CM input.
It's a somewhat odd circuit. There's 2.52V on the plus input to offset the output by that amount. The op amp acts as an inverting amplifier with three feedback circuits. The series output resistor may be to make the circuit stable with high capacitance output loads, but that's not entirely clear to me.
Use "simultaneously conjugate matching" technique if you play in a narrowband.You can find some information about it in textbooks. Bu I guess you design a wideband LNA and this technique is not valid in that case.Instead, check your feedback circuits ( series and parallel ) to stabilize inout and output impedances.Since S12==0 , you will see simil
The capacitance of a transistor and its wiring is a filter and produces negative feedback from the collector to its base due to the Miller effect which reduces the hfe AC small signal current gain. The hFE DC small signal current gain is not affected so it is the highest. Fairchild's datasheet for the 2N3904 transistor shows a graph of the hfe AC
The influence of a small amount of positive feedback - as long as the system remains stable - can be found using the same formula as for negative feedback, however using a POSIVE sign for the loop gain. (Stability limit for unity loop gain). Example: Voltage controlled feedback (as for operational amplifiers): Zout=Zol/(1-A,loop) with (...)
If you compute the output resistance, you can see the effect of feedback (resistive feedback R1-R2): Rout=Rota||(R1+R2)/(1+gm*R1) gm: OTA transconductance. You can see that the OTA output resistance is reduced - however, only by the factor gm*R1.
(1) the high impedance node ("this" means the latter one). Voltage feedback should always occur from a low impedance to a high impedance node. (2) The compensation capacitor has no effect on AC swing. It serves as Miller compensation cap to separate the poles of the previous and the following stage in order to achieve one domi
If the stage is then driven by a small signal voltage which has zero output impedance, the feedback through Rg will be shorted out by the source, and Rg will have no effect on the gain. If the driving source has an output impedance greater than zero, then Rg will have an effect on the gain. To determine just what the (...)
Not S21. Maybe you meant S12? Think also the "Miller effect". How the equivalent input capacitance of an inverting amplifier gets whatever little feedback capacitance existing between the input and output terminals amplified by (1+Gain). For an oscillator, there has to be some way a portion of the output gets to wiggle the input gate r
What type of circuit allows resonant frequency and blocks non-resonant frequency? and why? A circuit that exhibits an effect called "resonance" can also be realized using an amplifier with 2nd order RC feedback. However, independent on the realization (passive or active) you should know that such a bandpass circuit a
It depends upon the design. Some RC oscillators operate as multivibrator or relaxation oscillators which generally output a square-wave. Other RC designs can output a sine wave using feedback networks such as Wein Bridge or Twin-T types. Here is a description.
That effect is called "squegging" and is normally associated with oscillators that have too much feedback, which means the very powerful RF oscillations upset the DC bias. It could be interesting to alter the value of the .1MF cap to see if it changes the squegging time constant. Frank
Real op amps are occasionally known to go into parasitic high-frequency oscillations (as Ebcir points out in post #2). It is not always easy to figure out the root cause, or to find the direct cure. Sometimes we have to try various cures, in one or more places. This design has two RC networks which can reduce AC oscillations. One is at the inpu
Hello guys, I am trying to design an IGBT based welding system and i have some questions : For current control, we need to a feedback and i know that usually Shunt resistor used for these systems. 1) would be better that we use hall effect instead of shunt resistor (at output) or not? 2) How about CT transformer? any suggestion would be h
JEFFREY< If you drive a current amplifier with a voltage source say 0.7V , you get huge variations in base current due to the Vbe vs temperature Shockley effect. and thus the drive current changes over time due to input bias problem if that is what you did. -5mV/deg C as I recall. ... feedback and self bias resistors do "work wonders" to stab
For low current, high ? ferrite feedthru is used to dampen ringing. At high current, the Q drops due to low R, but ferrite sleeve has less effect at high current due to ferrite saturation. If not in feedback path low ESR cap to ZL of inductance attenuates directly unless as you say high Q LC path and step from high load to no load. So as load
Since OA's open loop gain > 1 million, the amount of negative feedback actually the ratio of resistors determines the -ve input gain. if you had many input resistors of different value then they each have different ratios with the common feedback. The same can be done with transistors with a current gain much larger than desired voltage gain, s
An op amp tries to make the -ve terminal voltage equal to the +ve terminal voltage. To be correct: It is not the opamp alone that "tries" to do this but it is the result of the feedback applied. To understand this effect one should try to become familiar with the concept of negative feedback.
Hi all... Anyone know how to calculate the gain of the attached circuit? It is a piece of schematic from a servo controller. In reality there are some transistor between the mosfet and the opamp. I have drawn the mosfet directly for simplification. As a student, i am interested in understanding the opamp circuit. A friend
What means low output impedance for you? 1 ohm, 10 ohm 100 ohm? Do you want feedback action at 150 MHz apart from the involved effect on output impedance?
In reverse bias, there's simple junction breakdown, but no internal feedback and thus no latching effect. The SCR will behave similar to a diode in reverse breakdown.
By lagging the feedback FET, you will add peaking / sharpness to the rest of the mirror rack. This can of course be overdone.
Perhaps you meant how does negative feedback affect both gain of input noise and bandwidth of output?
Hi Palmeiras, to 1): No, it is not correct; the phase shift will be frequency dependent to 2): Yes, correct to 3): No, in will not oscillate due to 360 deg phase shift for dc. As a consequence, there is a so called "latch-up" effect. You can observe a similar behaviour for each opamp with a positive resistive feedback that exceeds negative feedbac
I mean how effect feedback on the Rout
You can think a diode connection (gate connected to drain) as a local feedback loop. Imagine Vgs as the input and Vds (= Id * Rload) as the output: raising Vgs causes Vds to decrease. When you close the loop (Vgs=Vds) the net effect is to "lock" Vds to Vth (at first order). This behaviour is probably more evident for BJTs.
The basic problem is, that the blended feedback signal is shorted to the input. A series resistor (e.g. 1 to 10 k) isolating the input from the effect send node should help. Even inserting a buffer amplifier doesn't solve the problem without a series resistor.
I also tried with resistors placed on the emitters to reduce the VCE voltage. The main effect of additional "degeneration" emitter resistances is feedback, among others increasing the output resistance and thus reducing the Early effect. You didn't mention the actual objective of your experiments. If they are targetting to real
Hey Dude! Hi Duty of T1 is voice amplification (as a CE amplifier) .T2 is an oscillator : C9 will provide positive feedback . L1 and C8 are frequency determiner (selector ) .when you give an AF signal to the base of transistor (T2) according to the miller effect, the value of internal capacitor of transistor will change due to amplitude of AF sig
is this an op-amp differential amp ? if so the gain, unity or otherwise is selected by choosing feedback resistor values. some have external compensation capacitor terminals. Otherwise additional to the feedback resistor, you can in parallel add an RC to effect the gain bandwidth. All within the capabilities of the device itself though - its (...)
It isn't easy to do. Several techniques are used commercially, some based on phasing and some on pole shifting. The phasing method uses two microphones, one you speak in to, the other is mounted very close but shielded from your voice. The idea is that by carefully subtracting the signals, feedback into the voice microphone can be cancelled out w
i need an example of slew rate which contain op-amp and with explanation, i have seen some examples but i dont understood ,thank you The best idea is to observe the slew rate effect via simulation. Take, for example, a simple universal amplifier (like LM741) with 100% feedback (unity gain configuration). Power sup
Following will be general topics: Transistor operating. Miller effect Bandgap circuit Noise amplifier feedback
there is no feedback mechanism in that block diagram, so yes, any change in Hfe of the darlington pair will cause a change in load current. Temperature change will be a big effect also. Many current sources put a resistor in series with the load, and measure the voltage across the resistor to learn what the load current is, and then vary the dr
Hi all, I've this doubt in mind from long way back.How does the input and ouput swing of open loop amplifier effect the closed loop(may be a unity feedback) parameters??
Once feedback is working, the opamp inputs are in virtual short, and R2 current is 1/1000th that of R1 (ie IL). R3 will have (beta+1)/beta the current of R1. vo=I(R3)*R3=((beta+1)/1000beta)*IL*R3 Try working out the effect of offset for yourself.
The circuit linked by albbg does the opposite of the original circuit, dividing the R feedback branch instead of the C branch, so it doesn't explain anything. A reason for the divider may be, that the OP isn't unity gain stable, but the effect depends on the diode capacitance and OP GBW, without this information, everything is just a guess. I s
You can do it but I wouldn't recommend it for several reasons: 1. possibly high voltage on the controls which might pose a safety risk, 2. the amplifier may become unstable if the gain at any frequency drops below 10, 3. the load impedance will have some effect on the feedback path which might cause instability or uneven frequency response, 4. you
Without the R1 for feedback you would have no control. You wouldn't get regulation. Without R1, the output voltage is simply fixed to 1.2V ...
The problem could be often that the CMD feedback transfer function of the external opamp connection have a less favorable effect to the connected loop gain. For better diagnosis a schematic is helpful. A strategy is to assign the CMD control less gain within the opamp. If the CMD regime is medium output voltage sense and control similar cu