C3 / F5 Electronics
F5.1 OPERATIONAL AMPLIFIERS
An operational amplifier (op-amp) is a voltage amplifier which amplifies the difference between the voltages on its two input terminals. Op-amps often require a dual balanced d.c. power supply, e.g. ± 15V. However, they will work over a wide range of supply voltagesfrom ±5V to ±15V. The power supply connections are often omitted from circuit diagrams for simplicity. The diagram below shows the typical connections for an op-amp.
The + input terminal is known as the non-inverting input and the – input terminal is known as the inverting input terminal.
The output voltage is given by
whereA is the voltage gain.
An op-amp is assumed to behave ideally and the general properties of an ideal op-amp include:-
- the voltage gain is very large (typically, 106 at low frequencies),
- the maximum output voltage is equal to the powersupply voltage, (in practice it is about 2V less),
- it has infinite input resistance so no current passes into the input terminals, (typically 109, so there is an input current of a few nano-amps),
- the output impedance is zero so it can supply any required current, (in practice many op-ampsare designed to limit the current to approximately 15mA),
- the output voltage is zero when the two inputs are equal, (in practice there is a small offset voltage which needs a variable resistor to balance out).
General Amplifier Definitions
An amplifier is designed to produce an output voltage or current which is a magnified copy of the input voltage or current. When power amplification occurs the extra power is provided by the external power supply. The peak power output is limited by the power supply voltage. The gain is calculated by the ratio of the output quantity to the input quantity.
F5.2 VOLTAGE AMPLIFIERS
This section shows how operational amplifiers can be used as voltage amplifiers.
The Inverting Amplifier
For most amplifiers, the voltage gain of an op-amp is too large to be of practical use. It is not possible to adjust the open loop voltage gain of an op-amp to enable them to be used in normal amplifier circuits soNegative Feedback is used to reduce the overall gain of the circuit. With negative feedback a proportion of the output signal is 'fed-back' to the input so that it cancels out some of the input signal and reduces the apparent voltage gain of the whole circuit. With an op-amp, negative feedback is the only way of controlling the voltage gain of the circuit. The voltage gain of the op-amp itself is unchanged but the overall voltage gain of the circuit is reduced significantly. The simplest example of this is the Inverting Amplifier.
The circuit diagram of an inverting amplifier is shown below. The power supply connections are not shown, but it is assumed that the circuit is operating from a dual voltage power-supply
e.g. +15V. Rf joins the output to the inverting input and provides the negative feedback.
With an inverting amplifier, as the input voltage increases, the output voltage decreases and vice versa. This is represented in the diagram of an oscilloscope display shown below.
In this diagram, both the input and output traces are to the same scale and so it can be seen that the output voltage is four times the size of the input voltage. Therefore the amplifier giving these traces has a voltage gain of 4 but the output voltage is inverted compared to the input voltage.
The Non-inverting Amplifier
The circuit for a non-inverting voltage amplifier using an op-amp is very similar to the circuit for the inverting amplifier. The only difference is that the two inputs are reversed i.e. the signal is connected to the non-inverting input of the op-amp and the free end of resistor R1 is connected to 0V. This is shown in the circuit diagram below. Again, the power supply connections are not shown, but it is assumed that the circuit is operating from a dual voltage power-supply
e.g. +15V.
Rf joins the output to the inverting input and provides the negative feedback for the circuit, so reducing the overall voltage gain for the circuit.
With a non-inverting amplifier, as the input voltage increases, the output voltage increases and vice versa. This is represented in the diagram of an oscilloscope display shown below.
Again the output voltage is four times the size of the input voltage. Therefore the amplifier giving these traces has a voltage gain of 4 but the output voltage is not inverted compared to the input voltage.
Since the input voltage is applied directly to the op-amp in this circuit, the input resistance of this amplifier will be very large (109). This should be compared with the input impedance of the inverting amplifier where the input resistance is equal to R1.
F5.3 VOLTAGE GAIN CALCULATIONS
Calculating the voltage gain of amplifiers made with op-amps is relatively straight forward.
For an op-amp, the output voltage, Vout, is given by
Vout cannot exceed the voltage of the power supply. So if A is 106 and the power supply voltage is ±10V, then,if the output voltage is to be less than the power supply voltage, the difference in voltage between the two input terminals will be
The Inverting Amplifier
For the inverting voltage amplifier the non-inverting input terminal of the op-amp is connected to 0V. This means that the inverting op-amp input terminal will be within 10µV of 0V and so will be virtually at 0V, i.e. it is a Virtual Earth Point.
This means that the input voltage, Vin, appears across the input resistor R1.
So using Ohm's Law,
Since one end of Rf is also connected to the virtual earth point, the output voltage Vout appears across the feedback resistor Rf.
So using Ohm's Law,
Assuming that the input resistance of the op-amp is so large that no current passes into its input terminals,then the current passing through R1must passthrough Rf..
The negative sign comes from the current passing from the virtual earth point to the output.
Rearranging gives
Therefore the voltage gain of the inverting amplifier is determined solely by the two external resistors and the negative sign indicates that the amplifier is inverting.
It should be noted that the voltage gain of the op-amp itself (open loop gain) has not been altered; it is still very large at low frequencies. But the voltage gain of the overall circuit (closedloop gain) has been reduced significantly. Also it is important to realise that the equation is only valid when the open-loop gain is significantly greater than the voltage gain of the whole circuit. At high frequencies the voltage gain of the circuit will decrease, since most op-amps are frequency compensated to ensure that they are stable and do not oscillate.
The Non-inverting Amplifier
When considering the voltage gain of the non-inverting voltage amplifier, it is important to remember that the output voltage of an op-amp is equal to the differential input voltage multiplied by the open loop voltage gain.
Since the open loop gain of the op-amp is very large at low frequencies, the voltage at the positive input terminal and the voltage at the negative input terminal of the op-amp will be almost identical, so long as the output has not saturated at the power supply voltage. Therefore the voltage at the negative input terminal of the op-amp will be the same as that of the junction of the two resistors Rf and R1. These two resistors form a voltage divider as shown below.
Vout causes a current to pass through Rf and R1. This current can be found from Ohm's Law.
This current causes a voltage to appear across R1, which is the voltage at the inverting input of the op-amp. Using Ohm's Law,
Substituting for I
But this will also be equal to the input voltage Vin since the two op-amp input terminals must have almost the same voltage, if the op-amp is not to saturate.
The non-inverting amplifier has the same limitations at high frequencies as the inverting amplifier owing to the frequency compensation of the op-amp, but it does have the advantage of having a very large input impedance.
F5.4 THE OP-AMP AS A COMPARATOR
The output voltage of an op-amp is given by
where A is the open loop voltage gain of the op-amp (106) and (V+-V-) is the voltage difference between the two inputs of the op-amp.
Since the op-amp has a very large open loop gain, A, only a very small difference between V+ and V- is needed for the output voltage, Vout, to equal the supply voltages. Once this has occurred, increasing the voltage difference between V+ and V- will not cause any further increase in Voutand the op-amp is said to be saturated.
Thetransfer characteristic for an op-amp is shown below, Vs is the power supply voltage.
This characteristic enables the op-amp to compare the two voltages on its input terminals.
If V+ is greater than V- the output saturates at the positive supply voltage. If V- is greater than V+ the output saturates at the negative supply voltage. This principle can be used to compare two voltages, a reference voltage and a varying input voltage.
An op-amp can be used as a comparator either with a dual power supply or with a single power supply. Both are considered below, as each have their own specific problems when used with real op-amps.
As an example of the use of a comparator circuit, consider an electronic fire alarm system.
A thermistor is a resistor made from semiconductor materials whose resistance changes with temperature. The symbol for a thermistor is shown below
The most common type of thermistor has a negative temperature coefficient (ntc) which means that as the temperature increases, the resistance of the thermistor decreases.
The circuit diagram below shows how the thermistor can be used with an op-amp comparator to sound an alarm if the temperature goes above a set temperature.
The circuit operates from a dual power supply,+Vs and -Vs. Resistors R2 and R3 set the voltage at the op-amp inverting input terminal, which gives the reference voltage for the comparison with the non-inverting input terminal. If R2 is equal to R3, then the reference voltage will be 0V, half way between +Vsand -Vs.
Resistor R1 forms a voltage divider with the thermistor. As the temperature of the thermistor increases, its resistance decreases, and so the voltage at the non-inverting input of the op-amp will increase. If V+V-, then the output of the op-amp will be at –Vs. The diode prevents the buzzer being damaged by being powered the wrong way round.
When V+>V-, the output of the op-amp will switch to +Vs and the buzzer will switch on sounding the alarm.
The temperature at which the alarm sounds can be altered by changing the value of any of the resistors, but the most convenient arrangement is to make R2 equal to R3, (e.g. 10k) and then use a variable resistor for R1.
A similar circuit could be used to provide warning lights for a freezer. The buzzer would be replaced by two LEDs (Light Emitting Diodes), one red and the other green.
When the temperature of the thermistor is high, the output of the op-amp will be positive and the red LED will light. When the temperature of the thermistor is low, the output of the op-amp will be negative and the green LED will light. The 1k resistor limits the current passing through the LEDs to a safe level.
If op-amps were ideal, then both of these circuits could be easily converted to operate from a single power supply. Unfortunately, many real op-amps suffer from the problem that their outputs do not saturate at the power supply voltages but usually approximately two volts less. This can be a real nuisance when the output of the op-amp is connected to a transistor or even a red LED, since both will be switched on permanently, irrespective of the output state of the op-amp, unless precautions are taken to prevent this from happening.
The circuit diagram below shows the fire alarm circuit modified to operate from a single power supply.
When the temperature is low, the output of a real op-amp will also be approximately 2V. This is sufficient to make some buzzers sound, and so a red LED can be connected in series with the buzzer to prevent this problem. A red LED has a forward voltage of approximately 2V, and so when the output of the op-amp is low, there will now be almost no voltage across the buzzer.
The modified circuit diagram for the freezer alarm is shown below.
The two diodes provide an additional voltage drop of 0.7V, which will ensure that each LED has insufficient voltage available to light when it is not required.
F5.5 The Schmitt trigger
When any signal is transmitted from one place to another it will suffer attenuation, (become weaker), dispersion (spread out) and it will gather noise and interference.
Consider the diagram below which shows a digital signal as transmitted and the same signal as received after passing through a transmission medium.
If the received signal was just amplified, as shown below, the original waveform would not be restored and it would be unsuitable for use as a digital signal.
If the received signal was passed through a comparator, as shown below, the noise would cause stray pulses which would again make it unsuitable as a digital signal.
In order to restore the digital signal it is necessary to use a circuit called a Schmitt trigger. This circuit uses positive feedback to create two voltage switching levels instead of the single voltage switching level of a comparator. This produces hysteresis, the voltage level needed to make the output go high is different from the voltage level needed to make it go low. As a result the circuit can be set so that it will ignore the majority of the noise on the signal and so not produce stray pulses.
The circuit diagram for an inverting Schmitt trigger operating from a single power supply is shown below.
In this circuit the op-amp functions as a comparator but the reference voltage, produced by the two 10k resistors connected across the power supply, changes with the value of Vout. Assuming an ideal op-amp, if Vout is 0V, then the feedback resistor, R, will effectively be in parallel with the lower 10k resistor as shown in the left diagram below.
The two parallel resistors form a 5k resistor as shown in the diagram above on the right. Using the voltage divider formula
In order for the output to go high, Vin must go below this voltage. When it does, V– will be less than V+ and so Vout will become 12V. The feedback resistor, R, is now effectively in parallel with the upper 10k resistor and so, using the technique above, the reference voltage at V+ will be 8V. In order for the output to go low, Vin must go above this voltage. This circuit therefore has two switching levels (4V and 8V) which are separated by 4V and is less susceptible to the effects of noise than a comparator circuit.
F5.6 Op-amp problems.
Please see the Student Work Book for suitable tasks and problems together with appropriate responses.