Topic 9 – Operational Amplifiers

In the exam you are expected to:

·        Know the characteristics of an ideal op-amp;

·        Be aware of open loop gain and variation of gain with frequency;

·        Describe the inverting and non-inverting inputs;

·        Describe output saturation;

·        Describe the op-amp as a voltage comparator and its use in bridge circuits;

·        Describe its use as an inverting amplifier.

·        Use

·        Describe the summing amplifier;

·        Use

·        Describe the non-inverting amplifier;

·        Use .

The operational amplifier was originally devised in the 1960’s for use in analogue computers.  These are nowadays almost completely obsolete, although a few have some very special applications.  However the operational amplifier still has many uses in control and instrumentation electronics.  Although the original circuits used discrete components and were very expensive, miniaturisation has enabled op-amps to be made as integrated circuits, available for a few pence.

Operational amplifiers require a dual power supply, which means having a central 0 volts rail, and a + 15 V rail and a – 15 V rail.  The full circuit diagram is shown below, but generally we will ignore the power supply.

Notice that the op-amp has two inputs and one output.  It is a difference amplifier and amplifies the difference between the inverting input and non-inverting input.  Be careful not to confuse the symbol with a non-inverting gate.  We need to be aware of some definitions to do with op-amps:

·        Open-loop voltage gain – ratio of the input to the output voltage with no feedback applied.  It is the D.C. gain of the amplifier or the gain at a frequency of 1 Hz.

·        Closed loop voltage gain – the voltage gain with feedback.

·        Bandwidth is the frequency range in which the output does not fall by more than 3 dB from its maximum value.

The ideal op-amp should have the following characteristics:

·        Infinite open loop gain

·        Infinite input impedance so that no current is drawn

·        Zero output impedance so that maximum current can be transferred to the load.

·        Very wide bandwidth.

In practice the maximum open loop gain is 200 000.  Beyond that limit the amplifier goes into saturation which means that the voltage cannot go any higher.  The voltage is limited, of course, by the supply voltages.  In practice the limits are rather lower than this, about 1.5 to 2 V below the value of the supply.

Question 1  What is the maximum voltage output of an op-amp connected to a ±15 V supply?

ANSWER

Question 2   Using a gain of 200 000, what change of input voltage will cause saturation?

ANSWER

Therefore the input voltage can only swing through 135 mV in total to go from a negative saturation to a positive saturation.

Notice the following:

·        The op-amp amplifies the difference between the two input voltages

·        When the difference exceeds the limits X and Y, the output is saturated.

·        In between the limits the graph is linear, so there is little distortion in these regions.

The characteristic of real op-amps makes them unsuitable for use as amplifiers in open loop form, as clipping will occur and this will distort the signal.  Therefore some of the output is returned to the amplifier by a feedback loop.  This reduces the gain and makes the amplifier more stable.  The amplifier can be used in open loop form as a voltage comparator.

The open loop frequency response of an op-amp is not very good.

We can see that the gain starts to fall away quite dramatically above a frequency of only 5 Hz, which is not very high.  It would be quite useless as an audio amplifier.  However the gain can be improved by reducing the gain with the use of negative feedback.

Question 3 What does this graph tell you about the bandwidth and feedback?

ANSWER 

Using the Op-Amp as a Voltage Comparator

The op-amp being a differential amplifier is very useful as a voltage comparator.  It will give a high positive or negative voltage, depending on which voltage is higher:

• If the non-inverting input is higher, the output voltage will be positive

• If the inverting input is higher, the output voltage will be negative.

• The voltage difference must be greater than 135 mV if saturation is to occur.  This is not difficult to achieve.

The idea of the voltage comparator is to give a digital output from an analogue input.

The diagram shows the voltage comparator used as a light operated switch.  This is how it works:

• The reference voltage is adjusted with the variable resistor so that it is equal to the voltage V₂.

• When light falls on the LDR, V₂ goes up because the resistance of the LDR goes down.

• V₂ will be bigger than the reference voltage, so that the op-amp goes into saturation.

• The LED lights up, showing a high output.

• The reverse-biased diode protects the LED for when there is a large negative voltage.

The op-amp takes a definite time to change from one state to another.  This is called the slew rate and for a typical op-amp it is 13.5 V/ms.  This means that it would take about 2ms for the op-amp to change from –13.5 V to +13.5 V.

The diagram below shows a voltage comparator used in a bridge circuit:

In this circuit, the op-amp is wired in place of the voltmeter.  The slightest voltage imbalance will ensure that the op-amp is saturated either way.  It is much more sensitive to small changes than a voltmeter.

Question 4 Use this circuit diagram to show how the op-amp will detect an imbalance in the circuit and state what outputs it would give.     ANSWER

Voltage comparators are used in a similar way for temperature monitoring, which it does by comparing the voltage from a thermistor.  Arrays of voltage comparators are used in the flashing LED’s found in signal level meters in audio systems.

Negative Feedback in Op-amps

Negative feedback is achieved by bringing a fraction of the output signal to the inverting input of the op-amp. 

Negative feedback also has the following effects:

·        The gain is reduced

·        The bandwidth (frequency response) is increased

·        The stability is improved.

This allows the op-amp to be used in many applications including audio.

The circuit below shows a typical negative feedback arrangement:

In this arrangement the non-inverting input is connected to the 0V line.  This arrangement is called a virtual earth.  Since there is very little voltage difference between the non-inverting and the inverting input, we can also say that the voltage at the inverting input is almost at 0V as well.

Let us suppose there is a current I₁ flows through R_in and a current I₂ flows through R_f. Ohm’s Law allows us to say:

·        Current through R_in, I₁ = V_in/R_in

·        Current through R_f, I₁ = V_out/R_f

Since the resistance at the inverting terminal is very high, no current can flow through the inverting input.  The sum of the currents at point X (according to Kirchhoff I) is 0, which means that the current in is the same as the current out.  Therefore the current is the same through R_in and R_f.

·        Kirchhoff I: I₁ + I₂ = 0

·        Substituting:           

·        Rearranging:          

Þ

Question 5 If  R_in = 1000 W and R_f = 100000 W, work out the gain of the inverting amplifier.

ANSWER

  The output in all inverting amplifiers relates to the input as shown below:

Question 6  What can you say about the phase relationship in this graph?   ANSWER

Summing Amplifier

This kind of amplifier is used in digital to analogue conversion, or as a mixer in an audio system.

·        This circuit uses negative feedback into the inverting input, but instead of one input, there are three inputs.

·        The point X is a virtual earth.

·        Kirchhoff I tells us that I_tot = I₁ + I₂ + I₃

So we can use Ohm’s law to rewrite this in terms of voltage and resistance.  Notice that we use the minus sign because X is at a virtual earth so we have to climb the potential difference to get to V_out.  This is Kirchhoff II.

If the values of all the resistors is the same:

            V_out = -(V₁ + V₂ + V₃)

The output is the sum of all the input signals but is of opposite polarity.

Question 7  An op-amp has 2 inputs, one having a resistance of 1000 W and the other having an input resistance of 5000 W.  The two inputs have a voltage of + 4 V and + 5 V respectively.  The feedback resistance is 2000 W.  What is the output voltage?

ANSWER

The Non-Inverting Amplifier

In this circuit the input voltage is applied to the non-inverting input.

Notice that:

·        This amplifier uses negative feedback to the inverting input. 

·        There is no difference in the voltage between the inverting and non-inverting inputs, so we can say that the voltage at the non-inverting and the inverting input is the same.  (Remember that the op-amp is a difference amplifier.  If there were only a small difference in the input voltages, the output would be saturated.) So we can say that the voltage at P is V_in.

·        Since no current is drawn by the inverting input, the current in R_a is the same as the current in R_f.  So we can treat the two resistors as a potential divider and apply the potential divider equation.

We can therefore write:

Rearranging gives:

The term Vout/Vin is the gain.  The term can be rewritten   

Therefore:

Question 8  What is the voltage gain of a non-inverting op-amp in which the feedback resistor has a value of 5000 ohms, and the input resistor of 2000 ohms?  ANSWER 

There are many other uses for op-amps:

·        Difference amplifier

·        Schmitt Trigger

·        Ramp generator

·        Oscillator.

·        Buffer amplifier.

In the voltage follower, which is a special use of the inverting amplifier, we dispense with the input and feedback resistors altogether and put the input into the non-inverting input:

Although the gain of this circuit is 1, it is remarkably useful.  It is called a voltage follower.  It matches a high impedance input source with a low impedance load.  It often used as a buffer amplifier.

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