Showing posts with label non-inverting amplifier. Show all posts
Showing posts with label non-inverting amplifier. Show all posts

Saturday, 23 December 2017

How to use The Integrator Amplifier Electronics. The best and right thing to understand in detail

Learning Electronics Made Easy





The Op-amp Integrating Amplifier

In the previous tutorials, we have seen circuits which show how an operational amplifier can be used as part of a positive or negative feedback amplifier or as an adder or subtractor type circuit using just pure resistances in both the input and the feedback loop. But what if we were to change the purely resistive (  ) feedback element of an inverting amplifier to that of a frequency dependent impedance, ( Z ) type complex element, such as a Capacitor, C. What would be the effect on the op-amps output voltage over its frequency range.
By replacing this feedback resistance with a capacitor we now have an RC Network connected to the operational amplifiers feedback path producing another type of operational amplifier circuit commonly called an Op-amp Integrator circuit as shown below.

Op-amp Integrator Circuit

op-amp integrator amplifier circuit
As its name implies, the Op-amp Integrator is an Operational Amplifier circuit that performs the mathematical operation of Integration, that is we can cause the output to respond to changes in the input voltage over time as the op-amp integrator produces an output voltage which is proportional to the integral of the input voltage.
In other words, the magnitude of the output signal is determined by the length of time a voltage is present at its input as the current through the feedback loop charges or discharges the capacitor as the required negative feedback occurs through the capacitor.
When a step voltage, Vin is firstly applied to the input of an integrating amplifier, the uncharged capacitor C has very little resistance and acts a bit like a short circuit allowing maximum current to flow via the input resistor, Rin as a potential difference exists between the two plates. No current flows into the amplifiers input and point X is a virtual earth resulting in zero output. As the impedance of the capacitor at this point is very low, the gain ratio of Xc/Rin is also very small giving an overall voltage gain of less than one, ( voltage follower circuit ).
As the feedback capacitor, C begins to charge up due to the influence of the input voltage, its impedance Xc slowly increase in proportion to its rate of charge. The capacitor charges up at a rate determined by the RC time constant, ( τ ) of the series RC network. Negative feedback forces the op-amp to produce an output voltage that maintains a virtual earth at the op-amp’s inverting input.
Since the capacitor is connected between the op-amp’s inverting input (which is at earth potential) and the op-amp’s output (which is negative), the potential voltage, Vc developed across the capacitor slowly increases causing the charging current to decrease as the impedance of the capacitor increases. This results in the ratio of Xc/Rin increasing producing a linearly increasing ramp output voltage that continues to increase until the capacitor is fully charged.
At this point the capacitor acts as an open circuit, blocking any more flow of DC current. The ratio of feedback capacitor to input resistor ( Xc/Rin ) is now infinite resulting in an infinite gain. The result of this high gain (similar to the op-amps open-loop gain), is that the output of the amplifier goes into saturation as shown below. (Saturation occurs when the output voltage of the amplifier swings heavily to one voltage supply rail or the other with little or no control in between).
op-amp integrator output
The rate at which the output voltage increases (the rate of change) is determined by the value of the resistor and the capacitor, “RC time constant“. By changing this RC time constant value, either by changing the value of the Capacitor, C or the Resistor, R, the time in which it takes the output voltage to reach saturation can also be changed for example.
integrator output signal
If we apply a constantly changing input signal such as a square wave to the input of an Integrator Amplifier then the capacitor will charge and discharge in response to changes in the input signal. This results in the output signal being that of a sawtooth waveform whose output is affected by theRC time constant of the resistor/capacitor combination because at higher frequencies, the capacitor has less time to fully charge. This type of circuit is also known as a Ramp Generator and the transfer function is given below.

Op-amp Integrator Ramp Generator

sawtooth waveform
We know from first principals that the voltage on the plates of a capacitor is equal to the charge on the capacitor divided by its capacitance giving Q/C. Then the voltage across the capacitor is outputVout therefore: -Vout = Q/C. If the capacitor is charging and discharging, the rate of charge of voltage across the capacitor is given as:
voltage across a capacitor
But dQ/dt is electric current and since the node voltage of the integrating op-amp at its inverting input terminal is zero, X = 0, the input current I(in) flowing through the input resistor, Rin is given as:
op-amp integrator resistor current
The current flowing through the feedback capacitor C is given as:
integrator capacitor current
Assuming that the input impedance of the op-amp is infinite (ideal op-amp), no current flows into the op-amp terminal. Therefore, the nodal equation at the inverting input terminal is given as:
integrator equation
From which we derive an ideal voltage output for the Op-amp Integrator as:
op-amp integrator equation
To simplify the math’s a little, this can also be re-written as:
simplified integrator equation
Where ω = 2πƒ and the output voltage Vout is a constant 1/RC times the integral of the input voltage Vin with respect to time. The minus sign (  ) indicates a 180o phase shift because the input signal is connected directly to the inverting input terminal of the op-amp.

The AC or Continuous Op-amp Integrator

If we changed the above square wave input signal to that of a sine wave of varying frequency theOp-amp Integrator performs less like an integrator and begins to behave more like an active “Low Pass Filter”, passing low frequency signals while attenuating the high frequencies.
At 0Hz or DC, the capacitor acts like an open circuit blocking any feedback voltage resulting in very little negative feedback from the output back to the input of the amplifier. Then with just the feedback capacitor, C, the amplifier effectively is connected as a normal open-loop amplifier which has very high open-loop gain resulting in the output voltage saturating.
This circuit connects a high value resistance in parallel with a continuously charging and discharging capacitor. The addition of this feedback resistor, R2 across the capacitor, C gives the circuit the characteristics of an inverting amplifier with finite closed-loop gain of R2/R1. The result is at very low frequencies the circuit acts as an standard integrator, while at higher frequencies the capacitor shorts out the feedback resistor, R2 due to the effects of capacitive reactance reducing the amplifiers gain.

The AC Op-amp Integrator with DC Gain Control

AC Integrator amplifier
Unlike the DC integrator amplifier above whose output voltage at any instant will be the integral of a waveform so that when the input is a square wave, the output waveform will be triangular. For an AC integrator, a sinusoidal input waveform will produce another sine wave as its output which will be 90o out-of-phase with the input producing a cosine wave.
Further more, when the input is triangular, the output waveform is also sinusoidal. This then forms the basis of a Active Low Pass Filter as seen before in the filters section tutorials with a corner frequency given as.
op-amp integrator gain
In the next tutorial about Operational Amplifiers, we will look at another type of operational amplifier circuit which is the opposite or complement of the Op-amp Integrator circuit above called the Differentiator Amplifier. As its name implies, the differentiator amplifier produces an output signal which is the mathematical operation of differentiation, that is it produces a voltage output which is proportional to the input voltage’s rate-of-change and the current flowing through the input capacitor.

Hiow to use The Summing Amplifier. step by step, simple and best guide

Learning Electronics Made Easy



The Summing Amplifier

The Summing Amplifier is a very flexible circuit based upon the standard Inverting Operational Amplifier configuration. As its name suggests, the “summing amplifier” can be used for combining the voltage present on multiple inputs into a single output voltage.
We saw previously in the Inverting Operational Amplifier that the inverting amplifier has a single input voltage, ( Vin ) applied to the inverting input terminal. If we add more input resistors to the input, each equal in value to the original input resistor, Rin we end up with another operational amplifier circuit called a Summing Amplifier, “summing inverter” or even a “voltage adder” circuit as shown below.

Summing Amplifier Circuit

summing amplifier
The output voltage, ( Vout ) now becomes proportional to the sum of the input voltages, V1V2V3etc. Then we can modify the original equation for the inverting amplifier to take account of these new inputs thus:
summing amplifier formula
However, if all the input impedances, ( Rin ) are equal in value, we can simplify the above equation to give an output voltage of:

Summing Amplifier Equation

summing amplifier equation
We now have an operational amplifier circuit that will amplify each individual input voltage and produce an output voltage signal that is proportional to the algebraic “SUM” of the three individual input voltages V1V2 and V3. We can also add more inputs if required as each individual input “see’s” their respective resistance, Rin as the only input impedance.
This is because the input signals are effectively isolated from each other by the “virtual earth” node at the inverting input of the op-amp. A direct voltage addition can also be obtained when all the resistances are of equal value and  is equal to Rin.
Scaling Summing Amplifier can be made if the individual input resistors are “NOT” equal. Then the equation would have to be modified to:
scaling summing amplifier equation
To make the math’s a little easier, we can rearrange the above formula to make the feedback resistor RF the subject of the equation giving the output voltage as:
summing amplifier feedback equation
This allows the output voltage to be easily calculated if more input resistors are connected to the amplifiers inverting input terminal. The input impedance of each individual channel is the value of their respective input resistors, ie, R1, R2, R3 … etc.
Sometimes we need a summing circuit to just add together two or more voltage signals without any amplification. By putting all of the resistances of the circuit above to the same value R, the op-amp will have a voltage gain of unity and an output voltage equal to the direct sum of all the input voltages as shown:
unity gain summing amplifier
The Summing Amplifier is a very flexible circuit indeed, enabling us to effectively “Add” or “Sum” (hence its name) together several individual input signals. If the inputs resistors, R1R2R3 etc, are all equal a “unity gain inverting adder” will be made. However, if the input resistors are of different values a “scaling summing amplifier” is produced which will output a weighted sum of the input signals.

Summing Amplifier Example No1

Find the output voltage of the following Summing Amplifier circuit.

Summing Amplifier

summing op-amp circuit
Using the previously found formula for the gain of the circuit
inverting op-amp gain
We can now substitute the values of the resistors in the circuit as follows,
summing amplifier input gain
We know that the output voltage is the sum of the two amplified input signals and is calculated as:
summing amplifier output voltage
Then the output voltage of the Summing Amplifier circuit above is given as -45 mV and is negative as its an inverting amplifier.

Summing Amplifier Applications

So what can we use summing amplifiers for?. If the input resistances of a summing amplifier are connected to potentiometers the individual input signals can be mixed together by varying amounts. For example, measuring temperature, you could add a negative offset voltage to make the output voltage or display read “0” at the freezing point or produce an audio mixer for adding or mixing together individual waveforms (sounds) from different source channels (vocals, instruments, etc) before sending them combined to an audio amplifier.

Summing Amplifier Audio Mixer

summing amplifier audio mixer circuit
Another useful application of a Summing Amplifier is as a weighted sum digital-to-analogue converter. If the input resistors, Rin of the summing amplifier double in value for each input, for example, 1kΩ, 2kΩ, 4kΩ, 8kΩ, 16kΩ, etc, then a digital logical voltage, either a logic level “0” or a logic level “1” on these inputs will produce an output which is the weighted sum of the digital inputs. Consider the circuit below.

Digital to Analogue Converter

digital to analogue converter
Of course this is a simple example. In this DAC summing amplifier circuit, the number of individual bits that make up the input data word, and in this example 4-bits, will ultimately determine the output step voltage as a percentage of the full-scale analogue output voltage.
Also, the accuracy of this full-scale analogue output depends on voltage levels of the input bits being consistently 0V for “0” and consistently 5V for “1” as well as the accuracy of the resistance values used for the input resistors, Rin.
Fortunately to overcome these errors, at least on our part, commercially available Digital-to Analogue and Analogue-to Digital devices are readily available with highly accurate resistor ladder networks already built-in.
In the next tutorial about Operational Amplifiers, we will examine the effect of the output voltage,Vout when a signal voltage is connected to the inverting input and the non-inverting input at the same time to produce another common type of operational amplifier circuit called a Differential Amplifier which can be used to “subtract” the voltages present on its inputs.

Friday, 22 December 2017

How to use Inverting Operational Amplifier. Step by step, quick and easy way to learn this article.

Inverting Operational Amplifier


The Inverting Operational Amplifier

We saw in the last tutorial that the Open Loop Gain, ( Avo ) of an operational amplifier can be very high, as much as 1,000,000 (120dB) or more. However, this very high gain is of no real use to us as it makes the amplifier both unstable and hard to control as the smallest of input signals, just a few micro-volts, (μV) would be enough to cause the output voltage to saturate and swing towards one or the other of the voltage supply rails losing complete control of the output.
As the open loop, DC gain of an Operational Amplifiers is extremely high we can, therefore, afford to lose some of this high gain by connecting a suitable resistor across the amplifier from the output terminal back to the inverting input terminal to both reduce and control the overall gain of the amplifier. This then produces an effect known commonly as Negative Feedback, and thus produces a very stable Operational Amplifier based system.
Negative Feedback is the process of “feeding back” a fraction of the output signal back to the input, but to make the feedback negative, we must feed it back to the negative or “inverting input” terminal of the op-amp using an external Feedback Resistor called . This feedback connection between the output and the inverting input terminal forces the differential input voltage towards zero.
This effect produces a closed loop circuit to the amplifier resulting in the gain of the amplifier now being called its Closed-loop Gain. Then a closed-loop inverting amplifier uses negative feedback to accurately control the overall gain of the amplifier, but at a cost in the reduction of the amplifier's gain.
This negative feedback results in the inverting input terminal having a different signal on it than the actual input voltage as it will be the sum of the input voltage plus the negative feedback voltage giving it the label or term of a Summing Point. We must, therefore, separate the real input signal from the inverting input by using an Input ResistorRin.
As we are not using the positive non-inverting input this is connected to a common ground or zero voltage terminal as shown below, but the effect of this closed loop feedback circuit results in the voltage potential at the inverting input being equal to that at the non-inverting input producing a virtual Earth summing point because it will be at the same potential as the grounded reference input. In other words, the op-amp becomes a “differential amplifier”.



Inverting Operational Amplifier Configuration

inverting operational amplifier


In this Inverting Amplifier circuit, the operational amplifier is connected with feedback to produce a closed loop operation. When dealing with operational amplifiers there are two very important rules to remember about inverting amplifiers, these are: “No current flows into the input terminal” and that “V1 always equals V2”. However, in real-world op-amp circuits, both of these rules are slightly broken.
This is because the junction of the input and feedback signal ( X ) is at the same potential as the positive ( + ) input which is at zero volts or ground then, the junction is a “Virtual Earth”. Because of this virtual earth node, the input resistance of the amplifier is equal to the value of the input resistor, Rin and the closed loop gain of the inverting amplifier can be set by the ratio of the two external resistors.
We said above that there are two very important rules to remember about Inverting Amplifiers or any operational amplifier for that matter and these are.
  • 1.  No Current Flows into the Input Terminals
  • 2.  The Differential Input Voltage is Zero as V1 = V2 = 0 (Virtual Earth)
Then by using these two rules, we can derive the equation for calculating the closed-loop gain of an inverting amplifier, using first principles.
Current ( i ) flows through the resistor network as shown.



resistor feedback circuit

inverting op-amp gain formula

Then, the Closed-Loop Voltage Gain of an Inverting Amplifier is given as.

inverting operational amplifier gain equation

and this can be transposed to give Vout as:

inverting operational amplifier gain

The negative sign in the equation indicates an inversion of the output signal with respect to the input as it is 180o out of phase. This is due to the feedback being negative in value.
The equation for the output voltage Vout also shows that the circuit is linear in nature for a fixed amplifier gain as Vout = Vin x Gain. This property can be very useful for converting a smaller sensor signal to a much larger voltage.
Another useful application of an inverting amplifier is that of a “transresistance amplifier” circuit. ATransresistance Amplifier also known as a “transimpedance amplifier”, is basically a current-to-voltage converter (Current “in” and Voltage “out”). They can be used in low-power applications to convert a very small current generated by a photo-diode or photo-detecting device etc, into a usable output voltage which is proportional to the input current as shown.
op-amp linear output

Transresistance Amplifier Circuit


trans-resistance operational amplifier

The simple light-activated circuit above, converts a current generated by the photo-diode into a voltage. The feedback resistor  sets the operating voltage point at the inverting input and controls the amount of output. The output voltage is given as Vout = Is x Rƒ. Therefore, the output voltage is proportional to the amount of input current generated by the photo-diode.

Inverting Op-amp Example No1


Find the closed loop gain of the following inverting amplifier circuit.

inverting op-amp circuit

Using the previously found formula for the gain of the circuit

inverting op-amp gain
we can now substitute the values of the resistors in the circuit as follows,
Rin = 10kΩ  and  Rƒ = 100kΩ.
and the gain of the circuit is calculated as    -Rƒ/Rin = 100k/10k = -10.
therefore, the closed loop gain of the inverting amplifier circuit above is given -10 or 20dB(20log(10)).


Inverting Op-amp Example No2

The gain of the original circuit is to be increased to 40 (32dB), find the new values of the resistors required.
Assume that the input resistor is to remain at the same value of 10KΩ, then by re-arranging the closed loop voltage gain formula we can find the new value required for the feedback resistor .
   Gain = -Rƒ/Rin
therefore,   Rƒ = Gain x Rin
  Rƒ = 40 x 10,000
  Rƒ = 400,000 or 400KΩ
The new values of resistors required for the circuit to have a gain of 40 would be,
 Rin = 10KΩ  and  Rƒ = 400KΩ.
The formula could also be rearranged to give a new value of Rin, keeping the same value of .
One final point to note about the Inverting Amplifier configuration for an operational amplifier, if the two resistors are of equal value, Rin = Rƒ  then the gain of the amplifier will be -1 producing a complementary form of the input voltage at its output as Vout = -Vin. This type of inverting amplifier configuration is generally called a Unity Gain Inverter of simply an Inverting Buffer.
In the next tutorial about Operational Amplifiers, we will analyse the complement of the Inverting Amplifier operational amplifier circuit called the Non-inverting Amplifier that produces an output signal which is “in-phase” with the input.

Charactor_At_Position

public class Char_At_Position { public static void main(String[] args) { String str = "Wankhede"; String rev=""...