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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.
is a device that smoothen your half-wave rectification into a full-wave rectification after using a 4 diode and 1 resistor , after adding a capacitor , there will be a almost steady output , it charges the capacitor when is forward biased which is the first half wave , and discharge when is reverse biased to stablelize the wave into a almost same potential difference compare to a.c
Triangular wave
first i will tell you that if we are taking output across resistor only:- if load is taken across resistor then both dc and ac currents will pass through the resistor.as we know current contains both dc and ac parts in many cases so we will get some other output... now if we add a capacitor across the resistor then all the ac component of current passes through capacitor and only dc part of current passes through the resisitor so we will get different output ..i will tell you why it happens ?? we know dc has zero frequency.. and we also know that Xc(reactance of capacitor)=1/(2*pi*frequency*c) now putting frequency =0 in above equation we get Xc =infinity.and that means capacitor's reactance is infinity so open circuited .thus no dc current passes through capacitor....... ........mrityunjay pandey (kiit university ,b.tech 2nd year)
Any system you design will have an input and an output. The output will connect to the input of another system which will load it, so when you are designing any system you have to consider how loading it will effect the circuit performance.
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.
Because the capacitor is in series with the output. Vice versa for the integrator.
Where you are measuring. A simple filter will be two elements - a capacitor or inductor and a resistor. A capacitor will tend to "trap" low frequencies. In the case of a lowpass filter made of a capacitor and resistor, the output voltage will be measured across the capacitor. Inductors are the opposite, so the output would be across the resistor.
Without a bypass capacitor it is just equal to Rc
Place a capacitor across the output or load and you have your filter. This assumes that your load is a resistor. The capacitor you use depends on the frequency of the ripple. Lower frequencies require larger capacitors. A resistor may be required to make sure you do not damage any of the electronics either in series with the capacitor or immediately after the power source.
The emitter resistor is there to provide DC bias to the base. If it is not bypassed, then the AC output signal is also dropped across this resistor, effectively lowering the output swing. When a bypass capacitor is added, the DC bias still flows through the resistor, but the capacitor acts as a short circuit for the AC signal, so that the AC signal is not reduced. The capacitor selected must be large enough so it appears as a very low resistance at the lowest frequency the amplifier will pass.
is a device that smoothen your half-wave rectification into a full-wave rectification after using a 4 diode and 1 resistor , after adding a capacitor , there will be a almost steady output , it charges the capacitor when is forward biased which is the first half wave , and discharge when is reverse biased to stablelize the wave into a almost same potential difference compare to a.c
The larger the cap the smaller the ripple at the power supplies output. It smooths the rectifiers output waveform.
A; An integrator will integrate or slowly change as a rapid input is applied. Differentiate will have just the opposite effect
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.
Triangular wave
first i will tell you that if we are taking output across resistor only:- if load is taken across resistor then both dc and ac currents will pass through the resistor.as we know current contains both dc and ac parts in many cases so we will get some other output... now if we add a capacitor across the resistor then all the ac component of current passes through capacitor and only dc part of current passes through the resisitor so we will get different output ..i will tell you why it happens ?? we know dc has zero frequency.. and we also know that Xc(reactance of capacitor)=1/(2*pi*frequency*c) now putting frequency =0 in above equation we get Xc =infinity.and that means capacitor's reactance is infinity so open circuited .thus no dc current passes through capacitor....... ........mrityunjay pandey (kiit university ,b.tech 2nd year)