# Tag Info

7

Both capacitors are noise filters. C75 works with the Thevenin equivalent of R124 and R178 to form a single-pole lowpass filter. You want this corner frequency to be significantly below the lowest signal frequency of interest. C40 both filters noise on the power supply rail, and lowers its impedance caused by wiring and pc board trace resistances and ...

6

Why does differential op-amp require mid point biasing at its input to prevent clipping? It's not just a differential op-amp circuit, it's any linear op-amp circuit... So, you are confusing yourself by not recognizing the basic problem. The confusion arises because you have chosen an overcomplicated circuit to learn a very basic thing: - The above is not a ...

4

You are confusing open loop gain and closed loop gain. The gain in the data sheet plots is open loop gain and is very high as this is how op amps are designed. It is independent of how you are going to use the op amp (that is, inverting or non-inverting).Feedback is used in the actual circuit using the op amp to set the desired gain. The configuration of the ...

4

Is this how approximately OP284 looks like? No. Your circuit won't work close to the negative rail because the input transistors don't get enough bias voltage, but it can work close to the positive supply rail. The OP284 has rail-to-rail inputs and outputs. To handle signals close to the negative rail it has an inverted form of your circuit wired in ...

3

As pointed at by Bimpelrekkie, you are dealing with a small-signal analysis. The input voltage $V_{in}$ sets the dc operating point at which the internal $Z_{out}$ of the op-amp is obtained. Then, to determine the overall output impedance, you want to ac-modulate the output once the loop is closed. You inject an ac current $I_T$ across the output ...

3

The problem is I do not know which values of frequencies I should be testing, or at which frequency value I should know that the gain does not change. This is where you test it and find out. OK, a pro would be able to estimate where these points in the spectrum are but, given that it's an experiment, you should try it at different frequencies and discover ...

2

I am not sure if I have understood your problem in full detail. Nevertheless, perhaps the following is helpful: For any circuit, it is important that no EXTERNAL input signal (other than as injected into the output node) will contribute to the current you are going to measure through the output node. However, in case of a feedback system that means: ...

2

Why do you suppress the input when finding the output impedance of an opamp circuit? It's purely a reason of convenience. If the input is set to a value of zero then, the natural output voltage of the op-amp is also (pretty much) zero for straightforward circuits. Then, if you inject current into the output, and the voltage rises by an amount, it's simple ...

2

As shown, the body diode conducts in the MOSFET and the gate has little effect. Further, the voltage across the 0.1$\Omega$ resistor is then negative, which will destroy the LM358. It's not really clear what you are trying to do. I might guess it's an attempt at a constant current sink, but for that you will need a reference voltage to determine the ...

1

The op-amps are connected as a rail splitter to take the floating 10V supply and effectively make it +/-5V. Loud hum in old audio electronics is typically because of old dried-out electrolytic capacitors in the power supply, such as the 470uF/25V C0002 in your schematic. If you have an ESR meter check that it is within spec, or simply replace older large ...

1

No, the detector should be a current source proportional to the incident light.

1

In response to the last question of your posting I can list some very versatile filter circuits. Under B) you can find so-called "Biquadratic" or "universal filters), which - at the same time - can provide three basic filter functions: Lowpass, highpass, bandpass (bandstop). A) Two-opamp topology: GIC-Filters (Generalized Impedance Converter) -...

1

But I don't understand, why? How does the voltage follower stabilize ground? And how in the first place is the ground not stable in the configuration on the left? It's quite simple. The circuit on the left cannot sustain DC currents more than a few hundred microamps for a few seconds before a significant error is introduced in the "middle" DC ...

1

Ltspice can simulate the noise contributions. But from looking at the schematic a few things become apparent: the BJT are off at low signal level, so the noise of the feedback block is dominated by the three feedback resistors. at high signal level, the feedback impedance drops (hence rolling back gain), so that means that noise drops too. and anyway the ...

1

There are two ways to measure output impedance: Zout=dVout/dIout Make the circuit hold its voltage constant, inject an AC current I into the output, and measure output voltage variation V. Then, Zout=V/I. Input an AC voltage at the input, measure output voltage. Connect a known load, say a resistor that will draw a current from the output. Compute the ...

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