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44

Here is a $35 kit you can make, which ends up being the equivalent of a 741 op-amp using discrete 13 2N3904 and 7 2N3906 transistors. It has eight binding posts representing the eight pins of the device. Here is a link to the datasheet, which includes the schematic for the kit (shown below) and a BOM. Compare that to a "real" 741 out of the TI ...


36

The first OP-amp is actually creating the circuit ground. The 7810 creates a stable 10 volt, which is then divided by the voltage divider R2 and R3, filtered by C3 to make a stable 5 volt level relative to the most negative level. The OP-amp then buffers this, and the rest of the circuit uses its output as the reference ground. Remember that ground in a ...


27

While I agree with @pipe and in fact upvoted his answer, an additional nuanced reply is that a ground is more than "just" a reference. What I mean by this is a ground is not just a voltage, but something that can source & sink current and stay at the same potential. The ground created by that op-amp can both source and sink current and remain roughly ...


8

Your assumption "output voltage value is same either if I connect input signal into + port or if I connect input signal into - port." is wrong. Your left circuit does not work as a linear amplifier. Iuses positive Feedback. Depending on the input voltage it saturates close to one of the supply voltages (not shown in your diagramms; probably +10V and -10V; ...


7

To get the proper output for both positive and negative inputs, you need a full wave precision rectifier. Here is a workable schematic: You don't need R4 and the resistor values can be increased for relatively low frequency. This particular circuit uses an OPA2211 but other op-amp types can be substituted. The dual op-amp requires a dual supply (+/-) ...


6

"Dependent sources are useful tools to model a circuit'. But what are they really?" Regarding "dependent sources": We discriminate between four different controllable (dependent) sources: Voltage-controlled voltage source (VCVS), Current-controlled voltage source (CCVS), Voltage-controlled current source (VCCS) and current-controlled current source ...


6

The resistors at the non-inverting input are for the bias current compensation. In a non-ideal opamp the bias currents flowing into the inputs are non-zero. Parallel resistance of R7 and R8 is nearly equal to the parallel resistance of R2 and R3 (1485 Ohms). The matching in this circuit is not correct, because the output impedance (R1) of the source is not ...


5

Without a resistor in series to IN1, the current through the Z-diode is not limited. It will still protect from ESD, but any permanent voltage from a low impedance source below -0.7V or above 5.6V will destroy the diode.


4

Let's start with a statement: **The output of an op-amp is $$ V_{OUT} = A(V_{IN+} - V_{IN-}) $$ Where \$A\$ is the open loop gain, \$V_{IN+}\$ is the non-inverting input voltage and \$V_{IN-}\$ is the inverting input voltage. This is the way opamps are designed and A (the open loop gain) is a large number, typically in the order of 10\$^5\$. From this we ...


4

You've basically killed off the noise source by using a value of R2 at 100 ohms. This acts as an impedance to ground (virtual earth) and leads to only the higher end of the spectrum coming through because of the really low value of C1 (47 pF). It forms a high pass filter with a 3dB point of nearly 34 MHz. Even if you increased C1 to 47 uF (3 dB point of 34 ...


3

You are using your op amp in a single-supply configuration without providing a virtual ground. The simplest fix is to get another 14 volt supply, and connect this as a negative supply to pin 4. If you're not willing to do this, try Your output will now have an offset of about 7 volts, which may or may not be a problem, especially since any noise on the ...


3

Other answers have suggested looking at the implementation of real op-amps like a 741, but from a perspective of learning how they work, the best way to start is with a simplified system. The core of an op-amp is long-tailed pair. This can be built and operated or analysed in isolation to the rest of an op-amp, and provides the basic fundamentals of what ...


3

Are there inherent non-linear/distortion characteristics of an op-amp? Yes. Bare opamps are actually quite non-linear, or at least their linearity can't be counted on much. Or is this all contingent on implementation? No, since you're asking about opamps, not the circuits around them. The opamp is still a opamp regardless of what you put around it. ...


3

Substituting a single-supply opamp can work just fine if you make some modifications to the circuit. One of the most straightforward ways to adapt from a bipolar supply to a single supply is to reference the original GND to a "virtual GND" that is 1/2 of your supply rail. This circuit below illustrates the point. The simulation was done in LTspice, ...


3

You need to use meg instead of m for the resistor value. You have specified a 0.007\$\Omega\$ resistor, which will tend to reduce the gain to 1. Your amplifier has a GBW product of 18MHz meaning it will only have an open-loop gain of about 500 at 40kHz, not the 14,000 you appear to be expecting. However it will have extremely high gain at DC so any offset ...


3

Search for 'LM709 schematic', or 'LM 741 schematic'. Those were some of the first opamps available and have reasonably simple schematics. Modern opamps are based on similar principles, but generally have more complex circuits (because transistors are much cheaper now, and performance requirements keep increasing).


3

Without seeing your power supply turn-on and turn-off behavior, there is simply no way to be sure how it will behave on power-on. However, if you're worried about negative excursions, you have a simple way out. Connect the V- input to ground, rather than -15. You're not trying for a bipolar excursion, after all, but rather a constant 3.3 volts. Check the ...


3

The assumption that the inverting (-) and non-inverting (+) terminals are at approximately the same voltage, is only valid when the op-amp is configured for negative feedback and its output is not saturated. Negative feedback causes the output to converge such that the inverting terminal and non-inverting terminals are at very close to the same voltage. The ...


3

why don't we consider C3 and C4 for the corner(Cut-off) frequency calculation It's a simplification. Consider the following; all caps the same value and both resistors the same value. The differentially placed capacitor C2 and R2 and R3 form a filter that might have a cut-off of (say) 1 MHz. As individual wires, the cut-off frequency has to be 2 MHz ...


3

It's been more years than I can remember since using breadboard but shouldn't the op-amp sit over the trough that runs horizontally through the middle: - See also this to confirm: - Ground power also needs to connect to the common connection for Vin and Vout.


3

I read I can choose a large value such as 1M ohm for Rf Sounds like you haven't understood why/how to choose \$R_F\$, so here's a brief explanation. If you do the DC analysis of this circuit, you can find that: $$R_T = \frac{V_o}{I_p} = R_f \frac{A}{A+1}$$ where A is the gain of your amplifier (it could be an opamp, but could also be a transistor ...


3

The datasheet of the photodiode contains many useful bits of information such as: Max thermal dissipation (limits the power incident on the photodiode) Dark current (this is the signal that is present in the absence of light) Junction capacitance variation with reverse voltage Responsivity (A/W) at various wavelengths. Breakdown voltage The datasheet of ...


3

Bottom line: what are you trying to detect - that governs how much or how little of the data sheet you need to use: - Short circuit current - what it says on the tin - how much current you could expect to see under certain lighting conditions. Dark current - how much current you might expect to see in dark conditions. This will roughly double (or more) ...


2

I need the exact real amplitude of each pulse Then differentiate the signal more thoroughly like this: - The idea behind this is that you use a "strong" high pass filter that is high enough to decay the DC content within one pulse. Then measure the peak of the red signal. You can measure it at several points of course and translate each reading to a ...


2

(TL,DR: see paragraph 5) The meter module requires the voltage on its GND pin to be between its V+ and V- supply pins. It converts and displays the voltage between its IN+ pin and GND. The 7810 regulates the input to 10V between nodes labelled +5V and -5V. R2 and R3 provide a mid-point voltage with Thevenin impedance of 2.5 kOhm (= 10K // 10K) in ...


2

Your Aux input can presumably drive a speaker reasonably well. Let's say it produces .5 vrms with the speaker attached. Your amp than has a gain of 4, to produce 2 vrms. Trying to connect the speaker to the op amp output won't work - the op amp cannot usefully drive 8 ohms. When you put a 1k resistor between the op amp and the speaker, the op amp can drive ...


2

The 1 Mohm bias resistor is far too low - it will load the bass signals and attenuate them much more than mid range or the treble. This happens because the piezo has capacitance and this capacitance is in series with the signal being produced. It might only be a few hundred pF and, for instance, 1nF and an external 1 Mohm bias resistor will form a high pass ...


2

Consider the inverting op-amp circuit as driving to bring the inverting-input to the same potential as the non-inverting input. i.e. V- = 0 V. To do this I2 = I3 (the feedback current = input current). Since R2 and R3 are both 1 kΩ then the voltages across them are identical. Therefore, since Vin = 1 V then Vout = -1 V. There is no error on the output ...


2

The transistor is there to control the actual current through the load. It is essentially a voltage-controlled current limiter, in this case. The voltage/current relationship depends on a lot of things, and there are usually plenty of curves in the datasheet that shows this. The resistor at the bottom is there to convert a current to a voltage. Ohm's law. ...


2

Here is one easy way, but probably not the best way given your actual situation! Since you say you have all sorts of supply rails and references available, this way requires +/- supplies and +/- references. simulate this circuit – Schematic created using CircuitLab R2/-10V reference determines where the input meets the clamped voltage ...



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