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They are not completely different. They are much more similar than you think. Technically, a comparator is just an op-amp optimized to operate in the saturation region (have the output saturate since it is always either LO or HI, but be able to leave saturation quickly for fast switching). Being linear in the linear region is not a design priority. An op-...


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A comparator is often close to being a linear amplifier that is optimised to achieve rapid switching around the point where the voltage difference between two inputs are very close to zero. A comparator MAY include hysteresis which makes it less linear in action, but this is not an essential component. If you look at the block diagrams of the very old (...


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Both op-amps and comparators are types of amplifiers. The triangle symbol is used for these and many other kinds of amplifiers. You will find that power amplifiers and digital NOT gates also have very similar symbols (but only one input pin each), for example.


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From Texas Instruments "Stability Analysis of Voltage-Feedback Op Amps Including Compensation Techniques" http://www.ti.com/lit/an/sloa020a/sloa020a.pdf A is the open loop gain (the gain of the opamp itself), and \$ \beta\$ is the feedback resistors. If A\$\beta\$ is very large relative to 1, then the closed loop gain approximates as \$\frac{A}{A\beta}\$ ...


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The formula you quote is not the gain of an opamp. It is the gain of a circuit containing an opamp and several resistors. That formula only holds when the open loop gain of the opamp is much larger than that given by the formula. When that is the case, the actual value of the opamp open loop gain drops out of the equation. The derivation of the formula is ...


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Let's start with the definition of the op amp: $$e_{out}= A_{OL}(e_+ -e_-)$$ This is true for every op amp. The device is a differential amplifier, with a very high gain. Now, given that the positive terminal is grounded, $$e_+ = 0\\ e_{out}= -A_{OL}e_-$$ Next, we can apply the assumption that the input impedance is infinite, thus \$i_b=0\$. This let's ...


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This particular op amp has a note in its datasheet on page 14 about the feedback resistor and how to choose components to ensure there is no oscillation: Feedback Components Because the input currents of the LT1213/LT1214 are less than 200nA, it is possible to use high value feedback resistors to set the gain. However, care must be taken to ...


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$$\bbox[4px,border:1px solid red]{V_u=(1+\frac{R_2}{R_1})(v_+-v_-)}$$ The above assumption where you start from is actually wrong. \$(v_+-v_-)\$ is infinitesimaly small or zero ideally. The correct relation is: $$V_u = a_d.v_d + a_c.v_c$$ $$ie.,V_u = a_d.(v_+-v_-) + a_c.\frac{(v_+ +v_-)}{2}$$ $$\bbox[8px,border:1px solid black]{V_u = G_{OL}.(v_+-v_-)} $$ ...


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When two circuits - isolated from each other - form the forward path of a system with feedback, you can multiply the two transfer functions TF=TF1*TF2. As a second step, the general feedback formula (H. Black) can be applied H(s)=TF/(1-LG). LG is the loop gain (which must be negative for negative feedback). However, in your circuit, there is no overall ...


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The comparator is a mixed device with analog input and digital output; so it can be considered as a 1-bit analog-to-digital converter (just as one flip-flop can be considered as a 1-bit register). Usually, the comparator output is implemented as an open collector stage. This allows to make it compatible with various circuit inputs and to connect multiple ...


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Let's start with a few labels on the schematic: simulate this circuit – Schematic created using CircuitLab Assuming ideal opamps and a well-designed circuit operating within its limits, you know that \$V_b=V_1\$ and \$V_c=V_2\$. So it must be the case that \$i=\frac{V_1-V_2}{2\,R}\$. (I've changed the direction.) You also know that this exact same ...


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An attempt: Make use of the fact that there is negative feedback, Then V+ = V- V- = Vout - 4 V+= Vout * 30k/(20k + 30k) Then with the first equation you find Vout.


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Ok newbie, i've read your questions and comments and have a gauge of your understanding level. what classes have you taken pertaining this subject? If you're designing a DC-DC converter for stability using bode plots (poles and zeros), you received great answers already. You need to make it a goal of understanding them. This is a pretty good read: https://...


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For my opinion, in case of an inverting configuration a correct (negative) sign must be used for the open-loop gain. As a consequence, two signals must be added at the inverting input (the non-inv. input is grounded) and go into the opamp - hence the gain must be negative (-A). Of course, we still have negative feedback (because there is a minus sign in ...


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In the case of amplifiers with feedback, 'shunt' can mean to connect the input of the feedback network (e.g., a voltage divider) in parallel to the amp output... or the output of the feedback network to the amp input. I hope you know what 'in parallel' means? If the input impedance of the feedback network is high enough, it will not 'shunt' (divert current ...


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[ edited to add simple example of POLE & ZERO in a circuit ] Lets consider this circuit; the opamp in the bitmap has a RIAA vinyl compensation RCRCR network embedded in the opamp feedback element. For the schematic, examine the left most column, finding the "RIAA EQ" illustration. The feedback has two poles (at different frequencies) and 2 zeros (at ...


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Let's look at the 2 circuits he presented. The first is with Rop and C simulate this circuit – Schematic created using CircuitLab If we write out the transfer function for this, we get $$ H(s) = \cfrac{1}{1+sR_{op}C} $$. The denominator, D(s), contains a pole. A pole by itself is not bad. It depends. As the video explains, if you cross the 0 dB ...


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With a voltage step up of 3x the available current from the output is proportionately decreased by at least a factor of 3, (e.g. I/3). On top of this the overall efficiency of the step-up system also decreases the available power. If the output voltage needs to be stable the step-up voltage may even need to be higher than 36V and then regulated back to 36V,...


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It depends on what you're driving the circuit with. Your noninverting arm has infinite input impedance, and your inverting arm has relatively small input impedance. This could be problematic, but not in every situation. For example, if your input is some sort of piezo-device, you will be very unhappy. The input impedance of the inverting input is \$330R \...


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Using E24 series resistors.... Note that the op amp's positive supply needs to be at least +12V in order to achieve up to 10V output.


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