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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
How OPAMP ensure both input terminal are equal? Simply using negative feedback ?? Thanks
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
They don't have to be the same but it makes the math simpler. 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 fi
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.