Tag Info

Hot answers tagged

8

Let's assume that the op-amp has very high input impedance and that no current flows into the + and - pins. If current is flowing through Rs then the + input voltage will be higher than the - and the op-amp output will start to go high. This will turn on the output transistor allowing current to flow through RG1, the transistor and RL. This will cause a ...


6

Input resistance (1.5T\$\Omega\$ typical) is the change in input current for change in input voltage \$R_{in}\$ = \$\frac {\Delta V_{in}}{\Delta I_{in}}\$ It is not clear whether this figure is intended to apply to differential input voltage or to common mode voltage or both. Input current (10pA typical) is the current flowing into or out of the input ...


5

Having a capacitor directly across the op amp inputs always looks iffy to me. It can make the op amp trend to oscillate because high frequency feedback gets attenuated. If you are measuring the 1.5v "latch-up" voltage with a DC volt meter, you might simply be seeing the average voltage of an oscillation. Try placing C1 before R2 and R3 and see if that ...


5

The main amplifier is IC1A. R1+R13 and R3 set the gain, C1 and C2 are DC blocking caps, and most of the remaining mess of resistors and capacitors is for providing a low-noise DC bias at (presumably) 1/2 the supply voltage. The circuit around IC1B takes the output of the main amp and inverts it. The output of IC1B is then just the output of IC1A flipped ...


5

In your circuit you are using the tank as the load on the op-amp buffer circuit. The op-amp is trying to produce 4.5 V, but its output is short-circuited by the inductor. To generate oscillations in a linear circuit, you need to meet the Barkhausen criterion: A feedback mechanism with loop gain equal to unity Phase shift through the loop equal to a ...


4

The purpose of this circuit is to make sure the circuit has sufficient phase margin (does not oscillate). It's a particular problem with MOSFETs and not BJTs. 100nF is a very large capacitor- 1nF would work as well here (10nF is a good value), but maybe they wanted a bit of a LP filter or just wanted to be sure. The problem is that the MOSFET (with such a ...


4

You've connected your inputs backwards. A higher voltage on the non-inverting input produces a more positive output. You put the higher voltage on the inverting input, so the op amp output goes as low as it can. In this case, that's close to ground.


4

An operational amplifier will generally be designed so that when used in a suitable feedback system the inputs will always be within some specified tolerance of being equal; most op amps are designed with high-impedance input, though how high is "high" may vary. A differential amplifier will generally be designed to measure the difference in voltage between ...


4

Your output looks fine to me. There is always a period at start up of a filter where its output is off, and it converges to the correct output in a time period related to the cutoff frequency - a lower cutoff for a low pass filter means it will take longer to converge to the true value. This explains the "changing" output of your filter. Next, is the ...


3

As mentioned in another answer, the die diagram is helpful if you are buying the chip as a bare die and doing chip-on-board or hybrid assembly. The schematic is also helpful for understanding how to drive the inputs and how to load the outputs. It helps you know tings like whether pull-ups or pull-downs are needed, whether ac-coupling is needed. It might ...


3

The NIOSH states "Under dry conditions, the resistance offered by the human body may be as high as 100,000 Ohms. Wet or broken skin may drop the body's resistance to 1,000 Ohms," adding that "high-voltage electrical energy quickly breaks down human skin, reducing the human body's resistance to 500 Ohms." Let's say that body resistance is 100k ...


3

This looks like crossover distortion in the amplifier- it's really not fast enough to respond to the 30kHz input. You can try a pullup resistor of a few K on the output, but that will prevent the output from going all the way to ground, unless you have a negative supply. This kind of thing (dependence on op-amp output impedance) is one of the reasons why I ...


3

The first part of the circuit, whose output is \$v_{o1}\$, is a halfwave precision rectifier. For the positive half of the input, the diode D2 will be forward biased and D1 will be reverse biased. So it acts as an inverting amplifier with unity gain. So \$v_{o1}\$ will be inverted version of input for positive half cycle. ie, \$v_{01} = -v_{in}\$ for ...


3

Operational amplifiers are general purpose devices which can be used in many applications. If the requirements are not too stringent, they can be used to make both differential and instrumentation amplifiers. Differential amplifiers are specifically designed to amplify the difference between 2 input signals. They may include specially matched resistors to ...


3

If I understand the problem you have to invert the signal Scale the signal Remove the AC component At the very least an OPAMP is required to forfill #1 and #2 so this active component might as well be used to remove "some" of the AC component. To remove the AC component however, not just attenuate it. May I suggest a cancellation filter in the digital ...


2

If your input signal is within the constraints of +/- 2.5V then it just needs to be lifted from an average voltage of 0V to an average voltage of +2.5 volts. In doing so, the output becomes constrained to a range of 0 to +5V. If the input is generally audio/music then you can use a capacitor and voltage divider like so: - If Vref is 5V, the average dc ...


2

This circuit, a non-inverting summing amplifier, can be used for your purpose. The left op-amp is a 2.5v voltage reference (derived from your 5v supply), and the right op-amp is the summing amp. With the values shown, it takes the ±2.5v Audio input and adds it to the 2.5v offset, resulting in a 0-5v output. If R3 = R4, then Output = Gain * (Audio + Ref) ...


2

Each of the filters in the examples you posted are awkwardly drawn Multiple Feedback topology band pass filters. When designing a filter, you need to decide on: Type (bandpass, high pass, low pass) Center frequency Order (how quickly the filter rolls off) Shape (butterworth, Chebyshev, Bessel, etc) Topology (how the filter is physically implemented) If ...


2

First off, a 741 will never work. Between the feedback network and the load, the op amp needs to provide roughly 50 mA, which is way beyond a 741's capabilities. To do this the easy way, replace R1 with a 9k resistor. When the op amp is running properly, the voltage across this resistor will be 9 volts, for a total current of 1 mA. When the output is at 20 ...


2

Basically it works because the output of the Schmitt trigger inverter is either 0 or 1 (low or high). Imagine you look at the circuit at some random time. By its nature, the Schmitt trigger inverter output is either 0V or 5V (or transitioning between them but we can ignore that). If the output is 0V then the capacitor output is higher than the Schmitt ...


2

Chip topography is useful only if you can get the chip as a bare die. You need to know the precise locations of the bond pads in order to wirebond it to the board. Schematics are not exceptionally common, generally you will get block diagrams and simplified schematics of particular components, especially I/O circuitry. Schematics can be useful to ...


2

No, depending on what you are trying to do you might have to use a more complex op-amp model with finite gain, input offset voltage, input bias current, etc. If it is a real problem you can apply experience and judgment to decide what can be left out. For set problems the rule usually is that you should use all the information given. If nobody mentions ...


2

It's an output common-mode voltage, not input. So, the output is differential around whatever voltage you apply to the common-mode pin (assuming you have enough headroom for it to still act like an opamp).


2

Ideal op amps have infinite input resistance, and the voltage source is connected to only the input through Rs.


2

When an ideal op amp is connected with negative feedback, it obeys two rules: The voltages at the two input pins are equal. No current flows into either pin. In your first circuit, \$V_S\$ is only connected to the non-inverting input. By rule #2, no current flows into that input. This lets us calculate the equivalent input resistance: $$I_S = 0\ \mathrm ...


2

Possible assembly errors aside, at timecode 0:36 the guy tries to calculate the circuits gain, and comes up with 500 (which would be ~54dBV). Given that a dynamic mics voltage range is round about 40mV (Possible peaks to 80mV or more) this should ring some alarm bells: \$500 \cdot \pm 40mV = \pm20V\$. Not possible with the present supply voltages from 0 ...


2

If you are dealing with a single ended input signal you could do it with a fully differential op amp. The simpler ones have a fully diff output, i.e. the output is not referenced to ground but is the difference between two terminals. You build the usual non inverting amp with the inputs and the positive output, and feed \$-4.25\$V to the negative output. ...


2

This is one of those cases where an exact circuit is needed. Showing the circuits like you have make no sense because the input source is not specified. It is also unclear what the power rails are. Attaching a block called "noise source" to 0V is also unhelpful and meaningless. Also, 0V is 0V - that is the baseline - you cannot say 0V is noisy. However you ...


2

The opamp is in current to voltage configuration. All the current provided at the negative input node flows through R15 and becomes a voltage, the output voltage is then: $$ V_{out} = R_{15}I^- $$ Where \$I^-\$ is the current flowing into the negative input node. Please note that no current flows into the amplifier negative input, but some current flows into ...


2

The BP104 has a light sensitive area of 7.5 sq mm. If you look in the data sheet, an irradiance of 1mW per sq cm produces a current of about 33uA. The light power hitting the 7.5 sq mm can be calculated as simply: - Irradiance = 1mW per 100 sq mm therefore power hitting sensitive area (7.5 sq mm) is 75uW. So now we have 75uW producing 33uA into the TIA ...



Only top voted, non community-wiki answers of a minimum length are eligible