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# Tag Info

11

A transfer function of closed loop system is: $H(s) = \dfrac{F(s)}{1+F(s)}$ simulate this circuit – Schematic created using CircuitLab Wiki A transfer function of closed loop system with a feedback is: $H(s) = \dfrac{F(s)}{1+F(s)\cdot G(s)}$ simulate this circuit Now let's suppose you have a opamp buffer - the output is feedback to ...

11

The opamp in the circuit is actually a voltage comparator, it's designed to output only 2 voltage levels - high and low. The positive feedback makes it Schmitt trigger -circuit (=a comparator with hysteresis) to make the comparator a little faster and to increase noise immunity. The transistor is pulse amplifier. The capacitor makes it have low pass filter ...

10

It's a Platonic ideal of an op-amp; it's not a thing that could ever exist in the real world. So you say to yourself "I'll just assume that my system is stable" (real op-amp circuits aren't, always), and "I'll just assume that my output does whatever it needs to hold the input voltage difference at zero" (real op-amps don't, quite), and "I'll just assume my ...

9

Uin = Uout / A, so if A is big, Uin is small. That's it. Just like it took ages to "invent" zero, the concept of "infinite" is also not that easy to understand. For now, just consider that an (ideal) opamp has an amplification that is "very large", so that Uin is "very small". You may consider Uin "zero" as long as you understand that it is not really zero, ...

7

Q4, R40 and R44 form a pullup and buffer that can tolerate quite high voltages for a logic low (up to about 3.6V depending on temperature) from a relatively weak sink of perhaps < 1mA. C22 and R46 form a noise filter (-3dB at about 3.4kHz). U7 and surrounding circuitry is a comparator with significant hysteresis that also limits the output high voltage ...

6

The gain is defined as the output signal divided by the input signal. The amplifier has a voltage output. It's a virtual ground input configuration, the input will stay at 0v by feedback action. The input is therefore not a voltage signal, but a current signal. The gain therefore has dimensions volts/amps, otherwise written as ohms. This configuration is ...

5

The difference between the inputs is not exactly zero, but it is arbitrarily close to zero (i.e., infinitesimal), and the arbitrarily high gain of the opamp turns that into a finite output value. The ideal opamp is an abstraction, based on the concept of limits. Start with an amplifier that has a finite gain value. As you increase the gain to arbitrarily ...

5

It sounds like you want your Raspberry Pi to be able to control/provide a variable voltage supply for a 5V fan. I'll give you credit that you realized your RPI's pins don't have enough current to power the fan even though they have enough voltage. This next part is to educate you about the approach you are trying to do take even though it won't work (or won'...

5

"Gain" in the equation: $$GBP = Gain \times BW$$ is a linear gain, not dB. You can't plug dB directly into that equation. Furthermore, bandwidth (in most cases) is defined as the frequency where the output drops by 3dB so it's already accounted for. It's part of the definition. You don't need to account for the -3dB a second time. 20dB is the open loop ...

4

I simulated your circuit in LTspice - setting R1 to 2.222k to produce a gain of 10x, V1 to 101mV (same as your divider powered by +5V) and V2 ramped up from there by +0.25V over a period of 1 second. Instead of getting the expected 2.5V out for 0.25V in I only got 0.646V. The plot below shows why. At first the outputs of U1-A (blue line) and U1-B (red ...

4

You need significantly more capacitance from each rail to ground. Try 470uF or something like that in parallel with C3 and in parallel with C5. But if you're using a fancy high performance amplifier like that one, you're probably going to want to regulate the supply rails in reality. So that would mean a positive and a negative regulator, in addition to ...

4

Why reinvent the wheel, when this has already been done? Check out Transimpedance Amplifier Analysis by Erik Margan There are further more detailed frequency models in the paper, including those that use the DC gain/open loop gain. Source: http://www-f9.ijs.si/~margan/Articles/trans_z_amplifier.pdf

4

It is not correct to say a opamps are designed to work like comparators. They are two different components and though fundamentally similar, they are not the same. Positive feedback tends to allow op-amp behave like a comparator with hysteresis or Schmitt triggers. That said, an op-amp has so much open loop gain that it will behave like a comparator if ...

3

You've made two mistakes so far: First, an operational amplifier has lots and lots of voltage gain. When you're studying them in school, you start by assuming that they have infinite voltage gain, and for the most part you can pretend that's the case for real-world designs. So you've grounded the inverting input and put a positive voltage on the non-...

3

Your resolution is $$\huge V_\text{Resolution}=\frac{V_{DAC_\text{output range}}}{2^{DAC_\text{bit_number}}}\times GAIN$$ it is the smallest increment you can achieve for the ADJ voltage. To get a better resolution, you can: diminishing the OP AMP gain, but won't be able to have a 0-5V range for the ADJ voltage diminishing the DAC output_range (by ...

3

If your negative power supply voltage is 0V, you can't put 2V in the inputs:

3

You don't specify the length of cable with which you performed your tests so far... But in any case, if you want to carry an audio signal with reasonable signal integrity over such distances, you will probably want to use balanced audio lines so that your common mode noise gets rejected, thanks to the differential property of this transmission scheme. If ...

2

1) The input offset voltage of the LM358 is say, 1mV. The input offset voltage is multiplied by the user set closed loop gain and appears at the output as an output offset. With a gain of 2300 there's going to be a huge DC offset at the output. 2) With those input voltage values the output of U1:B will be trying to swing negative but is limited by the 0V ...

2

The circuit is equivalent to: simulate this circuit – Schematic created using CircuitLab To understand why consider pins 2 & 3 of the Op-amp. Because we have negative feedback these must be at the same potential. For this to be the case the output (pin 6) must be twice that of pin 3. The current in R2 is same as it would be if the bottom of R2 ...

2

Negative resistance happens when the circuit sources current. This doesn't happen with passive devices like resistors or capacitors. Amplifiers can sink and source current, in this circuit the amplifier will source current as the current through R1 increases. If the circuit is solved it will look like a resistor with negative resistance. R = -I/V This is ...

2

Well, with a bit of reasoning we can find: $$\text{V}_-=\text{V}_+\space\Longrightarrow\space\mathcal{H}\left(\text{s}\right):=\frac{\text{V}_\text{out}\left(\text{s}\right)}{\text{V}_\text{in}\left(\text{s}\right)}=1+\frac{\text{R}+\text{sL}}{\left(\frac{1}{\text{sC}}\right)}=\text{CLs}^2+\text{CRs}+1\tag1$$ In order to draw the bode plots, we need to ...

2

Input is signal frequency is 100Hz. I want to use the 6.5nV/sqrt(Hz) but it seems to be a spectral density, and I can't add it to a voltage This is where the bandwidth of the analog system comes in, the max bandwidth determines the noise. Let's say it's an ADC with a low pass filter at 1000Hz. The ADC would see: (1000Hz)^(1/2)*6.5nV/sqrt(Hz) = 2.05uV-rms ...

2

With the mid (possibly and Bass) controls fully CCW you have a hell of a lot of capacitance to ground hung directly on the opamp output.... Never a good look. Also, where is the decoupling on IC2? 5532s need local decoupling caps and are known to exhibit stability issues if they are absent.

2

I now want to calculate the positive voltage input. The gain of the amplifier is -(R3/R1) = -9. $V_{out}$ is calculated by $V_{out} = A(v^+ - v^-)$ In this formula, the $A$ is the open-loop gain of the op-amp, not the closed loop gain of the op-amp circuit with feedback. $A$ should be 100,000 or so, not 9. However, we wouldn't ...

2

The 100Ω resistor is likely there to improve stability. The author seems concerned about the amplifier stability because they suggest to remove capacitance from the input of the LDO and they have added the ferrite bead. The 100Ω shunts some of the feedback signal away from the inverting input and decreases the feedback factor (commonly labeled $\beta$ in ...

2

Another alternative would be to use a non-inverting adder, the output then connects to the lm317 ADJ pin. For a typical op-amp adder see SE: How to design a non inverting op amp adder for my circuit? Your DAC voltage connects at one input and you use a steady voltage reference at the other input. The voltage reference sets your lower limit and the DAC sets ...

2

You should not connect pin 2 to GND but connect pin 4 to pin 2 in order to build a voltage follower.

2

Virtual ground--the potential at the negative op amp input of your inverting amplifier--is actually at ground potential only if the gain of the op amp is infinite. If it's an ideal op amp, you can use limits (as in calculus) to work it out, but in real life the amplifier has a large but finite gain and there will be a small but extant voltage difference ...

2

The two inputs are at the same voltage because the op amp is ideal, and because there is negative feedback. If the gain is infinite and the output voltage is finite, then the difference voltage must be zero. To see how there is amplification you should follow the current from the input side to the output. Although there is no current into or out of the op ...

2

Remove R3 and it'll work as you require. Edit When input is 0V, the output is -1.25V. Perhaps missing the minus sign off was just a typo. Also assuming Vref is still +5V and that's a dash not a minus sign. That is a non-inverting amplifier, the input voltage goes into the non-inverting input. If the input is +ve with respect to Vref then the output is +ve ...

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