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28

There are many drawbacks to both low and high values alike. The ideal values will fall in between very large and very small for most applications. A larger resistor of same type will, for example, create more noise (by itself and through small induced noise currents) than a smaller one, though that may not always be important to you. A smaller resistor ...


25

There are downfalls with choosing very large resistors and very small resistors. These usually deal with the non-ideal behavior of components (namely Op-Amps), or other design requirements such as power and heat. Small resistors means that you need a much higher current to provide the appropriate voltage drops for the Op-amp to work. Most op amps are able ...


21

You already know that an opamp has very high open-loop amplification, typically 100 000 times. Let's look at the most simple feedback situation: The opamp will amplify the difference between \$V_+\$ and \$V_-\$: \$ V_{OUT} = 100 000 \times (V_+ - V_-) \$ Now \$V_+ = V_{IN}\$ and \$V- = V_{OUT}\$, then \$ V_{OUT} = 100 000 \times (V_{IN} - V_{...


21

You have to realize what Bandwidth actually means. Bandwidth is the frequency at which the gain starts to drop when frequency increases. So if lowering the gain (using feedback) moves that point (where the gain starts to drop) to a higher frequency then the bandwidth has increased. Let's take an example of an amplifier. It has a frequency response as shown ...


19

An ideal opamp has infinite gain. It amplifies the difference in voltage between the + and - pins. Of course in reality this gain is not infinite, but still quite large. The output of the opamp (at to some extents the input also) is constrained by the power supply, we can't get out more than the supply puts in. If we simply put signals into the opamp ...


18

Wow, it is impressive that you would ask this question, it shows admirable courage. Loop Stability Analysis in the Real World. "How does one develop a Bode-plot for circuits such as this using non-ideal op-amps that contain important poles in addition to the ones created by my passive components?" Two questions should be kept in mind while ...


18

In opamp feedback circuits, it's all about the current flowing in the feedback network, which must balance the current flowing from the input. Clearly, the transistors are intended to modify how the feedback network passes current, so the question is to figure out how they do that. The basic feedback is provided by the string of resistors in the middle, ...


17

In addition to the issues that @Asmyldof mentions, when using high resistances in the megaohms (and especially at 10M and more) environmental contamination such as dust, skin oils, soldering flux residue etc can easily reduce the effective resistance in unpredictable and time-varying ways.


15

Forget the 40kHz- this kind circuit really likes to oscillate at very high frequency- the feedback resistor is almost open (1M) at high frequencies in comparison to a few pF and the amplifier has a gain-bandwidth product of 1.75GHz. It's similar to a photodiode transimpedance amplifier in that respect. More importantly, you are measuring inputs with very ...


15

It's #2. For a "perfect" theoretical opamp, the open-loop gain is infinite, and this makes the difference at the inputs zero. When introducing opamp circuits, or when working out how things are supposed to work, people normally think about the "perfect" opamp. When thinking about the performance of a circuit, we usually have to start thinking about the ...


12

How does the opamp change its behaviour depending on the feedback? The ideal opamp behaviour itself is unchanged; it is the circuit's behaviour that is different. Isn't it something in the lines of the voltage added increases the error instead of reducing it in the case of + feedback?] That's correct as far as it goes. If we perturb (or disturb) the ...


12

"Vin is 5V, so Vout should be 50,000V." Why? The OpAmp amplifies the the difference between the + and - inputs, not just the value on the + input! OK, you might start with: the output is at 0V, and the input (connected to the + input) is 5V. What you have done is apply a 5V step to the input. Now what happens is that the OpAmp starts to rise the voltage ...


12

You are confusing 'inverting' with 'negative feedback'. Open loop simulate this circuit – Schematic created using CircuitLab Figure 1: op-amp with open-loop inverting mode. In Figure 1 the op-amp will amplify the difference between its inputs by the open loop gain. Let's say the open loop gain is 1,000,000 and we apply +1 mV at the '-' input. Since ...


12

All logic families like to use buffered inverters, because those are more reliable and use less power in digital applications. However, unbuffered inverters are useful to build crystal oscillators, so they exist in many families; search for 74xx1GU04. A 5 V-tolerant I/O has no ESD protection diode to VCC, so it tends to have less capacitance, and distorts ...


12

This op-amp boasts input noise of 0.9 nV/√Hz, which is roughly equal to the Johnson noise of a 50 Ω resistor. If you aren't putting resistors smaller than that around it, you're wasting some of this op-amp's performance, and probably should be buying something cheaper. Another useful identity is 1 kΩ ≈ 4 nV/√Hz, there being many ...


11

Most Basic Interpretation: Here is my intuitive way to understand a given op amp circuit by personification. Picture a little dude inside the op amp. The little dude has a display that indicates the difference in voltages between the + and - inputs. The little dude also has a knob. The knob adjusts the output voltage, somewhere between the voltage rails. ...


11

Okay, after a lot more research, I think I've gotten to the bottom of this. Actually I'm certain it's only approaching the bottom, as I've found this topic area quite deep, but I think I've gotten close enough to shed some light. A basic misconception A turning point in my understanding was when I realized that the equation I led off with in the OP: $$ G =...


11

Why do we get only one frequency as output in oscillators? Oscillators work at one frequency by ensuring two things: - The signal fed back to sustain oscillations is exactly in phase with the signal it is trying to sustain. Think about lightly tapping a swinging pendulum at exactly the right place and, in the right direction. The loop-gain is slightly more ...


10

The first rule of a control system is to never get information late. Unfortunately, this is exactly what low pass filters in the feedback path do. In particular, R13, C8, R12, C3, and C6 are going to cause stability problems in current limit mode. Also the fact the IC3B is being run open loop means the current shut off signal will slam back and forth. ...


10

Neither. You should use a potential divider: simulate this circuit – Schematic created using CircuitLab edit: Just to add, this will give a 0-4V wave at 24Vin, and a 0-2V wave at 12Vin. This is still fine, because the Atmega32 (supplied at 5V) will see anything above 2V as a "1".


9

Op-amp always behaves as a differential amplifier and the behavior of circuit depends on the feedback network . If negative feedback dominates, the circuit works in linear region. Else if positive feedback dominates, then in saturation region. I think the condition \$V^+ = V^-\$, the virtual short principle, is valid only when the negative feedback ...


9

There are many different BJT models, with varying degrees of usefulness in varying circumstances. (See SIDEBAR at bottom.) I'm not going to delve into any of that as it's not necessary in this case. A nice simplification is quite sufficient for your use here. Ignoring Nth order effects that aren't important here, a BJT's collector current is determined by ...


8

It's fundamental control and feedback theory. Thank you Mr Lyapunov, Mr Black, and Mr Nyquist. Consider that everyone everywhere always wants the output of their power supply to have just the right voltage, no matter what. How to manage that? The best way we know of is by using feedback. Feeding a sample of the output voltage, and current, back around to ...


8

You guys failed to recognize OP issue. If those explanation works. He won't ask about this in the first place. (methinks) Let's see the most simplest form of explanation How R3 parallel with R1? Anything wrong with that? Ditto for R2 Now you can read other answers which have good explanation why op-amp work this way. Some note on R4: Why it is there?...


8

Definition: The feedback factor beta is defined as the portion of the output voltage that is fed back to the differential opamp input (input source grounded). Therefore, regarding beta there is no difference between the non-inverting and inverting configuration. In both cases, we have $$\beta = \frac{R_{in}}{R_{in} + R_f}$$ Using the general formula for ...


8

The problem is that you mix-up two different models of the op-amp. A real, but somewhat idealized op-amp, is a differential amplifier whose output depends on the inputs as follows (neglecting saturation): $$ V_{out} = A_{Vol} \cdot (V^+ - V^-) $$ Using this simplified model (simplified because it neglects saturation, offset voltage, bias currents, ...


8

A lot of times I hear that it's useful for stability but I don't get why and how to calculate it's value. Consider that the non-inverting pin might have a parasitic capacitance of maybe 4 pF. That's the pin itself, the resistor parasites and any copper capacitance all lumped together. That 4 pF is in parallel with the 10 kohm resistor and its presence ...


7

Like you said, the fact that the two opamp inputs will be nearly equal is a simplification and depends on parameters often not explicitly stated. This is a good question in that it is essential to know the limits of any shortcuts or rules of thumb you use. As clabacchio already said, one place the assumption are violated is if the opamp output is clipped, ...


7

Fundamentally, you have way too much gain in your feedback loop, along with enough phase shift to create a very nice ~50 kHz oscillator. First, I would simplify the circuit by eliminating the MOSFET Q1; instead I would consider swapping the inputs of the LM393 and using its open-collector output to drive the Vref node directly. Secondly, I would add a ...


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