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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 ...


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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 ...


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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 ...


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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 ...


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This will be rather long, and does not actually answer your question, but I hope it gets you pointed in the right direction. The most obvious issue is that your readings don't make sense. First, you are not reporting your Vsource accurately. When you use 3.3k/68 ohms, you should get V1 = Vsource x (68/3368), or .113 volts for 5.6 volts, and .129 volts for ...


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The small signal response starts to drop significantly after 100kHz, flat from 500kHz to 700MHz. You should be able to see signals of a 100MHz. It could be something else. I would test the transmitter with a detector that you know that works to make sure that it's not the transmitter.


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The equation that you need is: \$ V_{out} = A_v (V_{+}-V_{-})\$ Av is the open loop gain. The resistances can be neglected if this is an ideal opamp (if it's an opamp with protection diodes, then the resistors do matter) as ideal opamps have infinite input impedance meaning zero current flows into them. If this is a real op amp, an input bias current ...


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The most simple way for explaining the negative feedback effect is to split the whole procedure in several steps: Example: Non-inverting opamp stage with two equal resistors in the negative feedback loop. Supply voltages +/- 10volts. Open-loop gain Aol. Step 1: Apply a DC input of +1V at the non-inv. terminal. Step 2: Because each amplifier has a certain ...


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Why opamp doesn't behave as datasheet "guaranteed"? It does, but only the values in the tables are guaranteed. The rest is, well, up to anyone's interpretation. How can I achieve 0V output with this opamp? Maybe by loading it with only resistor to ground. 9 mV sounds quite good to me already. If you really need to go to 0 V, use a negative supply rail! ...


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At 2kHz the impedance of C2 is 16kOhm, hence your amplifier acts like if R1 was equal to 300kOhm || 16kOhm = 15kOhm Without the capacitor, your gain is -1/3 but with the capacitor the gain becomes -6.7 and hence you saturate the op amp (1V * 6.7 = 6.7 V pick > 5V pick)


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When a parallel C(is labelled as C2) is added, the output is saturated. What is the explanation of this? At 2 kHz, the 5 nF cap has an impedance of about 16 kohm. Because it is in parallel with a resistor of value 300 kohm, the op-amp gain is increased about twenty times and it is much more likely that the op-amp output will saturate. How can i tackle ...


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