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26 Threads found on edaboard.com: Positive Pole
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
1 st order are inherently stable but noisy 2nd order are less noisy and C2/C1 ratio affects pole position 3rd order are inherently unstable with >180 deg phase shift and negative feedback becomes positive feedback. Rule of Thumb ratios help provide LPF additional rolloff using lag-lead filters to give phase margin at unity gain and may have
As I recall the negative feedback loop may have zeros, but the transfer function of P(s)/Q(s) for the positive input is often 1/(1+H(s)) thus the overall gain never reaches zero unless H(s) =-1
Hi guys, I have a problem about the stability of positive feedback. Let's say for a opam, it has 3 poles (LHP) at DC and 2 zeros (LHP) at wz. Then at DC, its phase shift is -270 degree , and at wz, its phase shift is -180 degree. At wz, its gain is 4. Then we connect the opam in a unity gain negative feedback structure. The phase margin of
Hi everyone, I have a question about stability analysis of RF circuit by transfer function. Let's say I derive transfer function of a circuit and get: H(s)=N(s)/D(s) D(s) = a*s^2+b*s+c where a, b, c are function of 2 parameters x & y. N(s) is also quadratic function of s. I think I can neglect N(s) because I just focus on stability. I
please leave this figure it is not that clear, refer to the first one of my post with the separated results , the curve with the DC gain of 81 is the circuit with the positive feedback the second graph with the DC gain of 74 is without the positive feedback I dont know what you mean to say by my self, I am using mentor gr
You'll determine stability by analyzing the loop gain magnitude and phase characteristic. For the simple case of monotone magnitude characteristic, a positive phase margin implies stability. A considerably larger phase margin will be required for acceptable time domain behaviour, however. Strictly speaking, an OP with only two poles and real fee
If it is an oscillator it has ALWAYS some kind of positive feedback, otherwise it wouldn't oscillate.
Yes, I mean the Vcc, the + pole of the battery, call it how you want! Vss is how microchip calls it in the datasheets, that's why I used it like this. I never saw the positive supply node named Vss in a datasheet. If pin 2 is the positive supply node and pin 1 more negative tahn pin 2, you'll expect a PMOS FET like Si2323.
What is the inverse laplace transform of 1/(s+a)? What do you think happens when a is positive?
Grounded load only means the the load is connected one end to a ground .. In this case the current source sources current from the positive pole to negative pole (ground) through a load .. If you use current source a current limiting resistor is not required .. See: A Constant Current Source
That's not possible! Q factor by definition is always is positive. Check for calculation error.
Thanks, flatulent. I want to realize a rising phase vs frequency, so it need a positive pole, and negative zero. But in many paper and book, there are a negative pole and positive zero.
There no positive pole or negative pole concept in the alternative current, since it is continuously changing. It doesn't matter how you plug your AC since both poles will be having the same function. In the other hand, in DC DOES matter the polarization, since it is a constant voltage. From the + pin, you will have (...)
It means you have peaking caused by a conjugated pole pair around 105 MHz. This pair is probably unwanted and can be caused by a feedforward path or positive feedback at that frequency. If you used feedback, is your phase margin large enough? You might be close to instability at 105MHz.
poles and zeroes introduce phase shift frequency dependant. So at some frequencies the negative feedback can turn to positive feedback
what you see are more poles at f=1e8 . Each pole adds phase shift, but phase shift larger than minus 180 degrees is subtracted from plus 180 degrees (kinda fuzzy explanation). Basically what you are seeing is that the phase shift is clipping. Also the magnitude shows a positive peek as well again, this indicates some zeros at that point. Hope
I have derived a transfer function containing 3 poles. All the coefficients are positive but from matlab analysis, there is one LHP real pole and a pair of RHP complex poles. Previously, I have a misconception that a cubic equation having all +ve coeff will yield all poles in the LHP. Could some mathematics (...)
If I am not wrong, I read that the B+ comes from the Battery positive connector/pole used in old vacuum tubes/valve operated equipment (radios).
Dear All, why a positive zeros in a transfer function creates negative phase shift (just as a pole) ? Thanks for answering.
If you can, move load from negative pole to positive pole and then you can use TIP31 as a switch .. see picture below .. Regards, IanP
It is indeed an interesing opamp. One thing I noticed here is the role of P2 and N3: they indeed form a positive feedback with Q8, R2, R3, N3 and R7 loop. With proper amount of this positive feedback, very high impedance at E and F nodes can be obtained which makes this opamp very high gain and also eases the pole-splitting compensation. (...)
AT DC and lower frquency, PLL has positive gain margin. Why PLL is stable with positive gain margin.
Use differentilal amplifier in its basic configuration: It allows you to divide high voltages, and in your case you want to reduce 48V to 2.5V, so the values of resistors will be 160kΩ and 8.2kΩ. Connect 160kΩ resistor between positive pole of 48V supply and (+) of the opamp. Connect 8.2kΩ resistor between (+) of opamp and
hi yc, r u sure the feedback is positive for ur application ?
Quite often depth transducers (pressure transucers) are of 4-20mA type. They will accept voltages from 10-30Vd. If that's the case you will have to identify the sensors positive pole and connect a dc supply there and connect termination resistor of, say, 250Ω between sensors output and 0V. For the rannge of 4-20mA the voltage accross this


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