As well as resistors and capacitors, Operational Amplifiers, or Op-amps as they are more commonly called, are one of the basic building blocks of Analogue Electronic Circuits.
Operational amplifiers are linear devices that have all the properties required for nearly ideal DC amplification and are therefore used extensively in signal conditioning, filtering or to perform mathematical operations such as add, subtract, integration and differentiation.
An Operational Amplifier, or op-amp for short, is fundamentally a voltage amplifying device designed to be used with external feedback components such as resistors and capacitors between its output and input terminals. These feedback components determine the resulting function or “operation” of the amplifier and by virtue of the different feedback configurations whether resistive, capacitive or both, the amplifier can perform a variety of different operations, giving rise to its name of “Operational Amplifier”.
An Operational Amplifier is basically a three-terminal device which consists of two high impedance inputs, one called the Inverting Input, marked with a negative or “minus” sign, ( – ) and the other one called the Non-inverting Input, marked with a positive or “plus” sign ( + ).
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The third terminal represents the operational amplifiers output port which can both sink and source either a voltage or a current. In a linear operational amplifier, the output signal is the amplification factor, known as the amplifiers gain ( A ) multiplied by the value of the input signal and depending on the nature of these input and output signals, there can be four different classifications of operational amplifier gain.
Since most of the circuits dealing with operational amplifiers are voltage amplifiers, we will limit the tutorials in this section to voltage amplifiers only, (Vin and Vout).
The output voltage signal from an Operational Amplifier is the difference between the signals being applied to its two individual inputs. In other words, an op-amps output signal is the difference between the two input signals as the input stage of an Operational Amplifier is in fact a differential amplifier as shown below.
The circuit below shows a generalized form of a differential amplifier with two inputs marked V1 and V2. The two identical transistors TR1 and TR2 are both biased at the same operating point with their emitters connected together and returned to the common rail, -Vee by way of resistor Re.
The circuit operates from a dual supply +Vcc and -Vee which ensures a constant supply. The voltage that appears at the output, Vout of the amplifier is the difference between the two input signals as the two base inputs are in anti-phase with each other.
So as the forward bias of transistor, TR1 is increased, the forward bias of transistor TR2 is reduced and vice versa. Then if the two transistors are perfectly matched, the current flowing through the common emitter resistor, Re will remain constant.
Like the input signal, the output signal is also balanced and since the collector voltages either swing in opposite directions (anti-phase) or in the same direction (in-phase) the output voltage signal, taken from between the two collectors is, assuming a perfectly balanced circuit the zero difference between the two collector voltages.
This is known as the Common Mode of Operation with the common mode gain of the amplifier being the output gain when the input is zero.
Operational Amplifiers also have one output (although there are ones with an additional differential output) of low impedance that is referenced to a common ground terminal and it should ignore any common mode signals that is, if an identical signal is applied to both the inverting and non-inverting inputs there should no change to the output.
However, in real amplifiers there is always some variation and the ratio of the change to the output voltage with regards to the change in the common mode input voltage is called the Common Mode Rejection Ratio or CMRR.
Operational Amplifiers on their own have a very high open loop DC gain and by applying some form of Negative Feedback we can produce an operational amplifier circuit that has a very precise gain characteristic that is dependant only on the feedback used. Note that the term “open loop” means that there are no feedback components used around the amplifier so the feedback path or loop is open.
An operational amplifier only responds to the difference between the voltages on its two input terminals, known commonly as the “Differential Input Voltage” and not to their common potential. Then if the same voltage potential is applied to both terminals the resultant output will be zero. An Operational Amplifiers gain is commonly known as the Open Loop Differential Gain, and is given the symbol (A_{o}).
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From these “idealized” characteristics above, we can see that the input resistance is infinite, so no current flows into either input terminal (the “current rule”) and that the differential input offset voltage is zero (the “voltage rule”). It is important to remember these two properties as they will help us understand the workings of the Operational Amplifier with regards to the analysis and design of op-amp circuits.
However, real Operational Amplifiers such as the commonly available uA741, for example do not have infinite gain or bandwidth but have a typical “Open Loop Gain” which is defined as the amplifiers output amplification without any external feedback signals connected to it and for a typical operational amplifier is about 100dB at DC (zero Hz). This output gain decreases linearly with frequency down to “Unity Gain” or 1, at about 1MHz and this is shown in the following open loop gain response curve.
From this frequency response curve we can see that the product of the gain against frequency is constant at any point along the curve. Also that the unity gain (0dB) frequency also determines the gain of the amplifier at any point along the curve. This constant is generally known as the Gain Bandwidth Product or GBP. Therefore:
GBP = Gain x Bandwidth = A x BW
For example, from the graph above the gain of the amplifier at 100kHz is given as 20dB or 10, then the gain bandwidth product is calculated as:
GBP = A x BW = 10 x 100,000Hz = 1,000,000.
Similarly, the operational amplifiers gain at 1kHz = 60dB or 1000, therefore the GBP is given as:
GBP = A x BW = 1,000 x 1,000Hz = 1,000,000. The same!.
The Voltage Gain (A_{V}) of the operational amplifier can be found using the following formula:
and in Decibels or (dB) is given as:
The operational amplifiers bandwidth is the frequency range over which the voltage gain of the amplifier is above 70.7% or -3dB (where 0dB is the maximum) of its maximum output value as shown below.
Here we have used the 40dB line as an example. The -3dB or 70.7% of Vmax down point from the frequency response curve is given as 37dB. Taking a line across until it intersects with the main GBP curve gives us a frequency point just above the 10kHz line at about 12 to 15kHz. We can now calculate this more accurately as we already know the GBP of the amplifier, in this particular case 1MHz.
Using the formula 20 log (A), we can calculate the bandwidth of the amplifier as:
37 = 20 log A therefore, A = anti-log (37 ÷ 20) = 70.8
GBP ÷ A = Bandwidth, therefore, 1,000,000 ÷ 70.8 = 14,124Hz, or 14kHz
Then the bandwidth of the amplifier at a gain of 40dB is given as 14kHz as previously predicted from the graph.
If the gain of the operational amplifier was reduced by half to say 20dB in the above frequency response curve, the -3dB point would now be at 17dB. This would then give the operational amplifier an overall gain of 7.08, therefore A = 7.08.
If we use the same formula as above, this new gain would give us a bandwidth of approximately 141.2kHz, ten times more than the frequency given at the 40dB point. It can therefore be seen that by reducing the overall “open loop gain” of an operational amplifier its bandwidth is increased and visa versa.
In other words, an operational amplifiers bandwidth is inversely proportional to its gain, ( A 1/∝ BW ). Also, this -3dB corner frequency point is generally known as the “half power point”, as the output power of the amplifier is at half its maximum value as shown:
We know now that an Operational amplifiers is a very high gain DC differential amplifier that uses one or more external feedback networks to control its response and characteristics. We can connect external resistors or capacitors to the op-amp in a number of different ways to form basic “building Block” circuits such as, Inverting, Non-Inverting, Voltage Follower, Summing, Differential, Integrator and Differentiator type amplifiers.
An “ideal” or perfect operational amplifier is a device with certain special characteristics such as infinite open-loop gain Ao, infinite input resistance Rin, zero output resistance Rout, infinite bandwidth 0 to ∞ and zero offset (the output is exactly zero when the input is zero).
There are a very large number of operational amplifier IC’s available to suit every possible application from standard bipolar, precision, high-speed, low-noise, high-voltage, etc, in either standard configuration or with internal Junction FET transistors.
Operational amplifiers are available in IC packages of either single, dual or quad op-amps within one single device. The most commonly available and used of all operational amplifiers in basic electronic kits and projects is the industry standard μA-741.
In the next tutorial about Operational Amplifiers, we will use negative feedback connected around the op-amp to produce a standard closed-loop amplifier circuit called an Inverting Amplifier circuit that produces an output signal which is 180^{o} “out-of-phase” with the input.
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Thank you for the wonderful topic presented and well rllaborated
we need some example for op amp
basic theory is good, but theory along with numeric problems would help to enhance the concepts 🙂
Ya,the note is good enough still it needs some modification like it needs to explain the different terminals and their use.
Thank you
it good for the student who start to learn…
Inverting amplifier inverses input signal i.e if input signal is 5V ac signal then output will be amplified but with phase difference of 180. But what happens when input signal is pure DC then how output can be inverted as DC is constant not having positive or negative peak?
Well simplified practical details