To design a simple integrator with an op amp, place a resistor and capacitor in series in the feedback loop, between output and inverting input. Place another resistor from circuit input to the inverting input. Ground the non-inverting input. The current through the input resistor will be balanced with the current through the feedback resistor. Since there is a capacitor also, the voltage slope at the output will be proportional to the current. If you want the capacitor to discharge faster in one direction, you can place a diode (and optional resistor) across the feedback resistor. This works because the capacitor resists a change in voltage, proportional to current, and inversely proportional to capacitance. The equation is dv/dt = i/c. This means that dv/dt is linear with constant i and c. In this configuration, a constant current input will be balanced with a linear voltage ramp on the output, limited only by the range of the op amp. Constrast this with a simple RC circuit - with constant voltage, the RC circuit will exhibit a logarithmic output. If, for instance, you were to drive this circuit with a square wave, the output would be triangular. With the diode, the output would be sawtooth.
capacitor acts as resistor because it has some resistace alos.
A capacitor is passive like a resistor is passive it performs no active amplification.
the capacitor and its associated resistor set the time constant.
The series input resistor and the feedback resistor.
To design a simple integrator with an op amp, place a resistor and capacitor in series in the feedback loop, between output and inverting input. Place another resistor from circuit input to the inverting input. Ground the non-inverting input. The current through the input resistor will be balanced with the current through the feedback resistor. Since there is a capacitor also, the voltage slope at the output will be proportional to the current. If you want the capacitor to discharge faster in one direction, you can place a diode (and optional resistor) across the feedback resistor. This works because the capacitor resists a change in voltage, proportional to current, and inversely proportional to capacitance. The equation is dv/dt = i/c. This means that dv/dt is linear with constant i and c. In this configuration, a constant current input will be balanced with a linear voltage ramp on the output, limited only by the range of the op amp. Constrast this with a simple RC circuit - with constant voltage, the RC circuit will exhibit a logarithmic output. If, for instance, you were to drive this circuit with a square wave, the output would be triangular. With the diode, the output would be sawtooth.
A: It not an element but rather a components like a resistor or capacitor and/or a combinations of both.
capacitor acts as resistor because it has some resistace alos.
Operational amplifiers are used in many applications where a well defined transfer function is needed. In former times they were widely used in analogue computers and simulators. The ideal op-amp is a phase-inverting voltage amplifier with infinite input impedance, zero output impedance and a very high voltage gain. Practical op-amps like the 741 achieve this very well. The simplest op-amp circuit is a voltage amplifier with a resistor between the input and output terminals, and a resistor in series with the signal input. The defined voltage gain is the ratio of the two added resistors, feedback resistor divided by the input resistor. The effect of the feedback resistor and the high voltage gain is to make the input terminal of the op-amp have a very low signal voltage (i.e. the output voltage divided by the intrinsic gain) and that point is called a 'virtual earth'. This allows the circuit to be considered as one in which the current flowing through the input resistor is transferred to the feedback resistor. Another simple example is to replace the feedback resistor by a capacitor. The current in the input resistor is transferred to the feedback capacitor so that the output voltage is proportional to the charge in the capacitor. Because charge is the time-integral of current, this circuit is called an 'integrator'. A loop made by two integrators and a phase inverter forms an oscillator. More interestingly it simulates a resonant system in which the damping factor can be controlled by a resistor placed across one of the capacitors. This allows the circuit to simulate the effects of a simple control system, and in this way analogue computers were widely used in the study of stability in control systems.
To add a capacitor and resistor in parallel, simply connect one terminal of the capacitor to one terminal of the resistor, and then connect the other terminal of the capacitor to the other terminal of the resistor. This creates a parallel circuit where both components share the same voltage.
resistor
An op-amp is a phase-inverting voltage amplifier (high input-impedance, low output-impedance) with a large voltage gain. Shunt feedback is connected through an impedance between the output and input terminals. In the simplest configuration there is a shunt feedback resistor - call that R2. If another resistor R1 is conncted in series with the input signal, the op-amp then produces a voltage gain of R2/R1. The new voltage gain is considerably less than the gain without any feedback, and the practical result of this is that the input terminal of the op-amp, which is also the junction of the two resistors, always has an extremely low signal voltage on it and it is termed a 'virtual earth'. The operation of the circuit can be considered by having the signal voltage inducing a signal current equal to Vin/R1 through R1, which then flows through R2 to produce a signal voltage of Vin R2/R1. If the feedback resistor R2 is replaced by a capacitor C, then the gain of the amplifer is the ratio of the capacitor's impedance and the input resistor, R, or Vout/Vin = 1/jwCR. This is know as an 'integrator' because the output voltage is the time integral of the input voltage: it has a phase-lag of 90 degrees and an amplitude that decreases at the rate of 6 dB per octave. If a constant current of i amps is applied to the input terminal, the output voltage rises on a ramp with a slope of i/CR volts per second; and if a variable current is applied, the output voltage is 1/(CR) integral i.dt. This is why an op-ap with a feedback capacitor and a series resistor i known as an analogue integrator.
I wanna use resistor , capacitor and amplifier 7173 for switch alarm circuit. How can i choice resistor and capacitor value because i wanna use 24V DC.
fully charged.
The reason why resistor voltage decreases while a capacitor discharges is because the resistor acts like a source of electrical energy. As the capacitor discharges, it draws energy from the resistor, which causes the voltage across the resistor to decrease. This is because the capacitor is acting like a drain, and is taking energy out of the resistor, thus causing the voltage across the resistor to decrease. The resistor and capacitor work together in order to create a discharge circuit. This is done by connecting the capacitor to the resistor, and then to a voltage source. The voltage source supplies the energy to the resistor, and then the resistor transfers this energy to the capacitor. As the capacitor discharges, it takes energy from the resistor, which causes the voltage across the resistor to decrease. In order to understand this process better, it is important to understand the basics of Ohm's Law. Ohm's Law states that the voltage across a resistor is equal to the current through the resistor multiplied by the resistance. As the capacitor discharges, it takes energy from the resistor, which means that the current through the resistor decreases, and therefore the voltage across the resistor will also decrease.
A capacitor is passive like a resistor is passive it performs no active amplification.
the capacitor and its associated resistor set the time constant.