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The best way to understand the differences between negative feedback and positive feedback is study the case when feed-forward block is dynamic system then extend the results to ideal linear block, let suppose that the transfer function of feedforward system is: The Feedforward block has a single pole at \$s=-p_1\$: $$ A(s)=\dfrac{a_0}{1+\dfrac{s}{p_1}} $$...


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In an astable configuration, is there some way to measure output frequency, compare it with some desired value, and use this to change the resistance in the config? (Basically a feedback loop) Sure, thats relatively easy, but it requires that you have something to compare it to Ie. to check whether your output signal is actually let's say 100kHz, then you ...


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The large swing slew Rate is always limited by drive current limit and load capacitance which near 0 ESR compare to load R. Thus dV/dt = Ic/C. Where Ic=V/R where for FETs, R = RdsOn or Ron and for BJT drivers , Ic is an active current limiter with the bulk Rce added to it. Where f(-3dB)=0.35/Tr , and Tr= 10~90% rise time, this is always less than the ...


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I agree with you that the Professor's statement is unclear and confusing. When used as a voltage buffer, the DC voltage at inverting and non-inverting inputs and the output are determined by the DC voltage at the non-inverting input. As you mention, the feedback loop takes care of that. Designing the output to be biased at Vdd/2 by itself (without feedback)...


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The change in voltage with current is greater on the resistor side than on the transistor side when the current is in the normal operating range for this circuit. The derivative of voltage with respect to current is R on the resistor side and the dynamic resistance of the diode on the other side. At, say, 1uA the dynamic resistance will be around 26K ohms. ...


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In addition to Justin's answer here's a mathematical way of looking at it: Using Vout = A(Vp - Vn) The incorrect assumption that I was making was that Vout == Vp == Vn which is wrong as: Vp - Vn = 0, which is a contradiction as this would make Vout = 0 Looking as Justin's answer we can see that Vout (and therefore Vn) start at 0v and rises with time until ...


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When you are considering a step response, you need to include the time delays at the different locations for the answer to be reasonable. One way to model this is to add an RC filter at the output of the op-amp. The other thing that you haven't accounted for is that the op-amp can't actually drive the output to greater than the positive supply voltage. ...


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I'm trying to get an understanding of how negative feedback works. Think of it like a control system and an error amplifier. You set the "demand" on the non-inverting input and the negative feedback causes the op-amp output to rapidly change to a voltage that makes the input error minimal. In other words, the inverting input is made to be nearly the same ...


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