(OpAmps)=
So far only use has been made of discrete components like resistors, capacitors, coils and diodes. In many electronic circuits use is made of integrated circuits (ICs). ICs are small electronic blocks with specific functionality, which contain many (sometimes over a thousand) components like resistors, capacitors, coils, diodes and transistors integrated on a single device.
An example of an IC is the operational amplifier (OpAmp). This chapter covers basic concepts of the OpAmp and several exercises to become familiar with common OpAmp circuits.
The following topics will be covered:
- Differential amplifiers
- Operational amplifiers
- Feedback
- Amplifier circuits
- Differentiators and integrators
- Active filters and oscillators
For amplifiers covered in {ref}amplification a potential difference is amplified with respect to a reference potential difference (usually 0 V or ground). A situation which often occurs is that one needs to measure the difference between two nearly equal quantities. This for example is the case when both measured signals contain the same signal component (e.g.through pick up of noise). To address these sort of situations a differential amplifier has been designed. This amplifier has two inputs and one output with respect to ground, as shown in {numref}fig:oa01.
---
width: 400px
name: fig:oa01
---
Differential amplifier
An ideal differential amplifier only amplifies the signal difference
$$ CMRR = \frac{A_{\rm DM}}{A_{\rm CM}}. $$ (eq:ch6-1)
:class: note
One wants to amplify two input signals each with a common DC-component of 10 V ( =$V_{CM_{in}}$). The difference between the two input signals is in the order of 0.01 V ( =$V_{DM_{in}}$). The output signal of the amplifier is $V_{_{out}}=V_{DM_{out}}+V_{CM_{out}}$
- What is the minimal CMRR value under the condition that the unwanted part in the output signal ($V_{DM_{out}}$), due to the common signal ($V_{CM_{out}}$), is less then 1 \%?
::::{admonition} Answer :class: dropdown
Given that
$$ {\rm CMRR} = \frac{A_{\rm DM}}{A_{\rm CM}} = \frac{\frac{V_{\rm out, ; DM}}{V_{\rm in, ; DM}}}{\frac{V_{\rm out, ; CM}}{V_{\rm in, ; CM}}} = \frac{V_{\rm out, ; DM}}{V_{\rm out, ; CM}} \frac{V_{\rm in, ; CM}}{V_{\rm in, ; DM}} = \frac{V_{\rm out, ; DM}}{0.01 \cdot V_{\rm out, ; DM}} \frac{10}{0.01} = 10^5. $$ (eq:ch6-2) ::::
(OA)=
The most common form of a difference amplifier is the operational amplifier (OpAmp). {numref}fig:oa02 shows the symbols used for an OpAmp.
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width: 250px
name: fig:oa02
---
Symbol for OpAmp
The
Depending on its use an ideal amplifier should meet certain specifications. To get an insight into this the following sections will cover a number of characteristics of OpAmps.
OpAmps in most applications are used as voltage amplifiers. To amplify voltages correctly the settings for ideal OpAmps will be specified. For the 741-type of OpAmp, which will be used in the lab, settings will also be given within boundaries of proper operation.
An ideal OpAmp has an amplification factor of feedback.
The 741 has an amplification factor of 106 dB. This corresponds to a voltage amplification of
There are OpAmps available with amplification factors of over 150 dB (a voltage amplification of over
A complication is that the amplification factor is often temperature dependent, even between OpAmps of the same type. {ref}feedback covers how the stability of amplification can be improved through feedback.
Based on matching ({ref}matching) an ideal OpAmp should have an input impedance of infinity.
The 741 has an input impedance of 2 M$\Omega$.
There are OpAmps with an input impedance of
Again, based on matching ({ref}matching) an ideal OpAmp should have an output impedance of zero.
The 741 has an output impedance of 75
There are OpAmps with an output impedance of a few Ohm only.
As important as the low output impedance is the maximum output current an OpAmp can deliver.
For the 741 the maximum output current is 20 mA.
For high power applications OpAmps are available that can deliver a maximum output current of a few amperes.
An OpAmp usually has one or two input supplies. The exact values of the voltage supplies is not so important, however, the stability is.
The 741 can be operated at voltage supplies between
The output voltage of an OpAmp is limited by the voltage supply chosen. At a voltage supply of
So due to the large amplification factors, only for very small input potential differences feedback we will show that using external components larger input signals can be amplified.
An ideal OpAmp used as a difference amplifier should have a CMRR of infinity.
The 741 has a CMRR of 90 dB.
The best available OpAmps have a CMRR over 160 dB.
When the potential difference of the input signals of an OpAmp is zero for an ideal OpAmp also the currents of both inputs will be zero. For real OpAmps this is not the case (see {numref}fig:oa03). This is due to the way OpAmps internally operate. The size of these currents, which are called bias currents, depend on whether the input circuit of the OpAmp uses transistors or FETS.
The 741, based on transistors, has a bias current of about 80 nA.
For OpAmps based on FETS smaller bias currents below
The bias currents result in potential differences across resistors connected to the inputs of the OpAmp. For small input signals this may lead to errors in the output signal.
---
width: 400px
name: fig:oa03
---
Input bias currents
For an ideal OpAmp when the input potential difference is zero the output potential difference is also zero. Again, due to the internal components used in an OpAmp this is not the case for real OpAmps.
The input offset potential difference is defined as the potential difference required at the inputs to obtain a zero output potential difference.
For a 741 the offset potential difference is about 2 mV.
For better OpAmps the offset potential difference is less than 1 mV.
Most OpAmps offer to tune the offset potential difference to zero using an external voltage divider. In practice this does not function perfectly as the offset potential difference is temperature dependent.
For the 741 the temperature dependence is about
For better OpAmps this value is less than
(fOpAmp)=
An ideal OpAmp has an amplification factor which is independent of frequency. In practice this is not the case. A real OpAmp behaves like an ideal OpAmp followed by a low pass filter.
{numref}fig:oa04 shows the frequency characteristic of a 741.
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width: 500px
name: fig:oa04
---
Frequency characteristic of a 741 OpAmp
The very low cutoff frequency, and hence small bandwidth of only a few Hz, seems to make the 741 useless for nearly any application. In {ref}feedback it will be shown that feedback can enhance the bandwidth of the amplifier considerably.
(feedback)=
Not only in electronics, but also for control systems feedback plays an important role.
The importance of feedback will be illustrated by two examples of how to control the room temperature.
The first way to control the room temperature is to turn the knob of the radiator a certain amount and hope that this will result in the desired room temperature. In general this will of course result in a room temperature that is either too high or too low. Even if the person that turns the knob based on experience sets it nearly correct the desired temperature may still differ due to external factors, such as a change in the temperature outside or e.g. the opening of a window. Clearly, this example of an 'open loop' system does not provide an optimal solution.
A much better solution is to install a thermostat. The user sets the thermostat to the desired temperature. The thermostat measures the actual room temperature, compares it to the temperature set and based on the difference determines whether the heater should be turned on more or less. The knob on the radiator does not need to be adjusted (although it should be set such that water can flow through the radiator obviously). At a change of the temperature outside the thermostat will keep the room temperature at the desired setting. Such a system that includes feedback is called a 'closed loop system.
Systems including feedback use information provided by the output signal to regulate the signal at the input in such a way that the output signal reaches it desired value.
{numref}fig:oa05 illustrates schematically the difference between an open and closed loop system.
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width: 500px
name: fig:oa05
---
Open and closed loop system
The user has an objective and hopes that the result will match it. The 'forward path' converts the input signal into an output signal. In the open loop system the position of the knob on the radiator could represent the input signal and the amount of heat per second given by the radiator the output signal. In fact the final output signal is the temperature reached, which is determined by the heat given by the radiator, and the heat lost through the walls, windows, etc.
In the closed loop system apart from the 'forward path' a 'feedback path' is present. Through the latter the output signal, in the example the actual temperature, is fed to the input for comparison with the user set temperature through the 'comparator'. The difference ('error') signal is used as an input for the forward path (the heater).
Feedback systems are distinguished by positive and negative feedback.
At negative feedback the feedback signal tries to reduce the input signal, positive feedback tries to increase the input signal.
For a quantitative treatment of feedback {numref}fig:oa06 is useful.
---
width: 400px
name: fig:oa06
---
Feedback system
From {numref}fig:oa06 it follows that
$$ \frac{Y}{A} = X - BY. $$ (eq:ch6-3)
This may be re-expressed as
$$ \begin{split} G &= \frac{Y}{X} = \frac{A}{1 + AB} \ \ G &= \frac{{\rm Forward , Gain}}{1 + {\rm Loop , Gain}}. \end{split} $$ (eq:G)
:::{note}
Eqn. {eq}eq:G follows from {numref}fig:oa06 assuming the feedback signal is subtracted from the input signal. Some textbooks add both signals. In that case eqn. {eq}eq:G becomes
$$ G = \frac{A}{1 - AB}. $$ (eq:ch6-4) :::
The term 'loop gain' refers to the amplification through the forward path and back through the feedback path.
The quantity
Since the 'forward gain'
The behavior of the total system depends on the magnitudes and signs of both
-
No feedback:
$B$ = 0
Eqn. {eq}eq:Gturns into:$G = A$ .
The overall gain without feedback is equal to the forward or open loop gain. -
For the loop gain holds:
$-2 < AB < 0$
In this case the absolute value of the term$1 + AB$ is less than 1. From eqn. {eq}eq:Git then follows that$|G| > |A|$ . Effectively the feedback causes an increase of the input signal: we are dealing with positive feedback. -
The loop gain
$AB$ is equal to -1
This is a special case of feedback. Eqn.{eq}eq:Ggives$G = \infty$ .
This is a remarkable result. Any input signal is amplified infinitely. In practice this means that any input signal, even close to zero, will push the output of the system to a maximum positive or negative output signal. A system with such an extreme form of feedback turns out to be an essential part of an oscillator, as we will see later. -
The loop gain
$AB$ is positive
In this case the term$1 + AB$ is larger than 1. Eqn.{eq}eq:Ggives$G < A$ . Effectively the feedback results in a reduction of the input signal: we are dealing with negative feedback.
When the loop gain$AB >> 1$ eqn. {eq}eq:Gcan be simplified to eqn. {eq}eq:G2(check this!).
This also is a remarkable result. The 'overall gain'$G$ in this case is determined by the characteristics of the 'feedback path' B and not by the characteristics of the original system without feedback. B is also called the feedback coupling factor.
$$ G \approx \frac{1}{B}. $$ (eq:G2)
:class: tip
Assume that in the circuit shown in {numref}`fig:oa06` system $A$ is an OpAmp with an amplification factor of $A = 100000$. System $B$ is designed such that the feedback signal is 1 \% of the output signal. So $B = 1 / 100$. \
Substituting $A$ and $B$ into eqn. {eq}`eq:G` gives $G = 99.90$. \
Using eqn. {eq}`eq:G2` results in $G = 100$. The approximation in this case leads to an error of 0.1 \%. \
When $A$ is increased by a factor of 2 eqn.{eq}`eq:G` gives $G = 99.95$. So a change in $A$ by a factor of 2 causes a change in the 'overall gain' of 0.05 \% only!
The frequency characteristic of an OpAmp drops already after a few Hz as was shown in {ref}fOpAmp. In this section we will show that this frequency characteristic can be improved using negative feedback.
The characteristic of {numref}fig:oa04 strongly resembles that of a low pass filter. Therefore is seems logical to replace the 'open loop gain'
$$ A(\omega) = \frac{A_0}{1 + j \omega \tau}. $$ (eq:Alp)
eq:Alp into eqn. {eq}eq:G gives:
$$ G(\omega) = \frac{\frac{A_0}{1 + j \omega \tau}}{1 + \frac{A_0}{1 + j \omega \tau} \cdot B}. $$ (eq:Glp)
Eqn. {eq}eq:Glp can be re-written as:
$$ G(\omega) = \frac{A_0}{1 + A_0 \cdot B} \cdot \frac{1}{1 + j \omega \frac{\tau}{ {1 + A_0 B}}}. $$ (eq:Glp2)
Based on eqn. {eq}eq:Glp2 the following can be concluded:
For DC-signals the feedback results in a decrease of the amplification factor of
For AC-signals the feedback results in a decrease of the amplification factor and also of the time constant of
amplification factor x bandwidth = constant
In relation to the above the quantity 'unity-gain-bandwidth' is often used. This refers to the bandwidth where the amplification is reduced to one. So:
amplification factor x bandwidth = unity-gain-bandwidth
{numref}fig:oa07 shows the frequency characteristic of the 741 OpAmp with feedback, illustrating the effects discussed above.
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width: 500px
name: fig:oa07
---
Frequency characteristic of the 741 OpAmp with feedback
{numref}fig:oa07 shows that the unity-gain-bandwidth of the 741 OpAmp is about 1 MHz.
:class: note
- A particular OpAmp had a unity-gain-bandwidth of $10^7$ Hz and a steady state amplification of $A > 10^5$. One builds with this OpAmp an amplifier with a feedback coupling factor of 1/50. Determine the gain factor and the bandwidth of this amplifier.
- One wants to build an amplifier with a wide frequency bandwidth and a large amplification factor $D$. \
Explain why it is better to use two amplifier stages, each with a gain equal to $\sqrt{D}$, than to use one amplifier with an amplification of $D$.
::::{admonition} Answer :class: dropdown
We consider an OpAmp with a
The gain factor of the amplifier is
$$ G = \frac{A}{1 + AB} \approx \frac{1}{B} = \frac{1}{0.05} = 50. $$ (eq:ch6-6)
Since
$$ BW = \frac{UGB}{G} = \frac{10^7 ; {\rm Hz}}{50} = 2 \cdot 10^5 > 10^5 ; {\rm Hz}. $$ (eq:ch6-7)
Using two amplifiers with lower
{numref}fig:oa08 shows an amplifier with an 'open loop gain'
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width: 400px
name: fig:oa08
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OpAmp with negative feedback
The behavior of the feedback on input will be considered where the feedback signal is subtracted from the input signal:
$$ \begin{split} V_{\rm in}^{'} = V_{\rm in} - B \cdot V_{\rm out} = V_{\rm in} - AB \cdot V_{\rm in}^{'} \ \to V_{\rm in}^{'} \left( 1 + AB \right) = V_{\rm in}. \end{split} $$ (eq:Vin)
From eqn. {eq}eq:Vin and
$$ Z_{\rm in, , eff} = \frac{V_{\rm in}}{I_{\rm in}} = Z_{\rm in} \left( 1 + AB \right). $$ (eq:ch6-8)
So the effective input impedance
{numref}fig:oa09 shows the same amplifier with negative feedback as {numref}fig:oa08 but with a focus on the output signal.
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width: 400px
name: fig:oa09
---
OpAmp with negative feedback
The amplifier can be considered as a voltage source with an output voltage equal to
$$ V_{\rm out} = A \left( V_{\rm in } - B V_{\rm out} \right) + I_{\rm out}Z_{\rm out}, $$ (eq:Vout)
where it is assumed that the current through
The effective output impedance is equal to the ratio of eq:Vout gives:
$$ Z_{\rm out, , eff} = \frac{Z_{\rm out}}{1 + AB}. $$ (eq:ch6-9)
So the effective output impedance
Concluding, negative feedback by increasing the input impedance and reducing the output impedance, improves the behavior of the voltage amplifier.
An OpAmp amplifies the difference signal of the two inputs. When including feedback {numref}fig:oa06 can be replaced by {numref}fig:oa10.
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width: 300px
name: fig:oa10
---
OpAmp with feedback
Below a number of OpAmp ciruits with feedback will be discussed, assuming ideal OpAmps. In other words, we assume that:
- the 'open loop gain' is infinite
- the input impedance of the OpAmp is infinite
- the output impedance of the OpAmp is zero
{numref}fig:oa11 shows a schematic of a non-inverting amplifier.
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width: 300px
name: fig:oa11
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Non-inverting amplifier
Comparing {numref}fig:oa10 and {numref}fig:oa11 gives:
$$ B = \frac{R_1}{R_1 + R_2}. $$ (eq:ch6-10)
According to eqn. {eq}eq:G2:
$$ G = \frac{1}{B} = \frac{R_1 + R_2}{R_1}. $$ (eq:G3)
This result can also be derived as follows:
Since
$$ V_- = V_+ = V_{\rm in}. $$ (eq:Vin2)
:::{note}
This argument holds for all OpAmp circuits with negative feedback, so eqn. {eq}eq:Vin2 can be applied to all of them.
:::
Due to the infinite high input impedance no current runs to the inputs of the OpAmp. Therefore:
$$ V_- = \frac{R_1}{R_1 + R_2} V_{\rm out}. $$ (eq:Vmin)
Combining eqns. {eq}eq:Vin2 and {eq}eq:Vmin gives:
$$ G = \frac{V_{\rm out}}{V_{\rm in}} = \frac{1}{B} = \frac{R_1 + R_2}{R_1}, $$ (eq:G4)
which is the same as eqn. {eq}eq:G3.
:class: note
```{figure} /Fig-ch1/multisim.png
---
width: 50px
name: fig-multisim
align: right
---
```
- Design in Multisim a non-inverting amplifier with a gain of 10.
::::{admonition} Answer :class: dropdown
For the amplifier with a gain of 10 we used
$$ G = \frac{R_1 + R_2}{R_1} = 10. $$ (eq:ch6-11)
The Multisim circuit and result is shown in {numref}fig:63.
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width: 600px
name: fig:63
---
Amplifier with a gain of 10
::::
:::{admonition} Exercise 6.4 :class: note :name: ex64
- Give for the circuit shown in {numref}
fig:oa12an expression for$V_{\rm out}$ as a function of$V_1$ and$V_2$ . - What is the name of this circuit?
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name: fig:oa12
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A non-inverting amplifier
:::
::::{admonition} Answer :class: dropdown
The circuit under investigation is shown in {numref}fig:64.
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name: fig:64
---
Non-inverting sum amplifier
The gain is given by
$$ G = \frac{R_1 + R_2}{R_1} = \frac{5{\rm k} + 10{\rm k}}{5{\rm k}} = 3. $$ (eq:ch6-12)
The Thévenin voltages are
$$ \begin{split} V_{\rm Th}^1 &= \frac{(10{\rm k} || 10{\rm k})}{10{\rm k} + (10{\rm k} || 10{\rm k})} V_1 = \frac{5{\rm k}}{10{\rm k} + 5{\rm k}} V_1 = \frac{1}{3} V_1, \ V_{\rm Th}^2 &= \frac{1}{3} V_2, \ V_{\rm Th} &= \frac{1}{3} \left( V_{\rm Th}^1 + V_{\rm Th}^2 \right). \end{split} $$ (eq:ch6-13)
So
$$ V_{\rm out} = 3 \cdot \frac{1}{3} \left( V_{\rm Th}^1 + V_{\rm Th}^2 \right) = V_1 + V_2. $$ (eq:ch6-14)
This amplifier is called a non-inverting sum amplifier. ::::
(buffer_amplifier)=
A special case of a non-inverting amplifier is a buffer amplifier as shown in {numref}fig:oa13.
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name: fig:oa13
---
A buffer amplifier
Comparing {numref}fig:oa11 and {numref}fig:oa13 tels us eq:G4 gives:
$$ G = \frac{0 + \infty}{\infty} = 1. $$ (eq:ch6-15)
The amplification factor is 1. This is not a shocking result. However, do realize that the input signal is connected to a very high input impedance of the OpAmp. So the input source hardly experiences a load. Later we will show that the buffer amplifier has a very low output impedance, and therefore can have a substantial load.
{numref}fig:oa14 shows the schematic of an inverting amplifier.
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width: 550px
name: fig:oa14
---
Inverting amplifier
The fig:oa14 is grounded. So eq:Vin2
Because the inputs of the OpAmp draw no current, any current running throufg
$$ I = \frac{V_{\rm in } - 0}{R_1} = \frac{0 - V_{\rm out}}{R_2}. $$ (eq:I)
From eqn. {eq}eq:I it follows that:
$$ G = \frac{V_{\rm out}}{V_{\rm in}} = -\frac{R_2}{R_1}. $$ (eq:Gia)
:class: note
{numref}`fig:oa15` shows a special case of an inverting amplifier, a *current to voltage converter*.
- Give an expression for the output voltage $V_{\rm out}$ as a function of the input current $I$ for the circuit shown in {numref}`fig:oa15`.
```{figure} /Fig-ch6/OA15.png
---
width: 550px
name: fig:oa15
---
Current to voltage converter
```
If we use a resistor instead of this circuit, we can also convert current into voltage.
- What is the major advantage of the circuit in {numref}`fig:oa15` compared to a resistance.
::::{admonition} Answer :class: dropdown
- Using {numref}
fig:oa15we get:
$$ \begin{split} &V_- - V_{\rm out} = IR, \ &V_+ = V_- = 0V, \ &V_{\rm out} = - IR. \end{split} $$ (eq:ch6-16)
- Using an OpAmp is better than using a resistor as
$Z_{\rm in} = \infty$ and$Z_{\rm out} = 0$ . ::::
:class: note
{numref}`fig:oa16` also shows a special case of an inverting amplifier, using a Zener diode instead of a resistor in the feedback.
```{figure} /Fig-ch6/OA16.png
---
width: 400px
name: fig:oa16
---
Inverting amplifier with Zener diode in feedback
```
Given $V_{\rm in}$ as a sinusoidal signal with an
amplitude of 10 V and $Z_{\rm d}$ as a Zener diode with
a zener voltage of 4 V and a turn on voltage
0.7 V:
- Sketch $V_{\rm in}$ and $V_{\rm out}$ as a function of time. Explain the sketches.
```{figure} /Fig-ch1/multisim.png
---
width: 50px
name: fig-multisim
align: right
---
```
- Verify your result using Multisim.
::::{admonition} Answer :class: dropdown
The circuit concerned is show in {numref}fig:66.
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width: 600px
name: fig:66
---
Inverting amplifier with Zener diode in feedback
The signal is inverted and clipped at +4 V and -0.6 V.
When
When
The amplifier circuit shown in {numref}fig:oa17 is a combination of a non-inverting and inverting amplifier.
---
width: 400px
name: fig:oa17
---
Differential amplifier
To determine the relation between
Effect of $V_1$ on $V_{\rm out}$ ($V_2$ is grounded)
This situation resembles an inverting amplifier:
$$ V_{\rm out, , 1} = - \frac{R_2}{R_1} \cdot V_1 . $$ (eq:ch6-17)
Effect of $V_2$ on $V_{\rm out}$ ($V_1$ is grounded)
This situation resembles a non-inverting amplifier.
$$
V_+ = \frac{R_2}{R_1 + R_2} \cdot V_2.
$$ (eq:Vplus)
Combining eqns. {eq}eq:G4 and {eq}eq:Vplus results in:
$$ V_{\rm out, , 2} = \frac{R_2}{R_1} \cdot V_2. $$ (eq:ch6-18)
The total output voltage
$$ V_{\rm out} = V_{\rm out, , 1} + V_{\rm out, , 2} = \frac{R_2}{R_1} \cdot (V_2 - V_1). $$ (eq:Vdiff)
The output voltage is proportional to the potential difference between both inputs.
The differential amplifier of {numref}fig:oa17 has the disadvantage of a relatively low input impedance. For the
Another disadvantage is the relative low CMRR as a result of the tolerances of the resistors used. For the differential amplifier the output voltage is given eqn. {eq}eq:Vdiff. Only the difference between eq:Vdiff it is assumed that the two resistors
$$ {\rm CMRR} \geq \frac{1 + R_2/R_1}{4 \varepsilon}. $$ (eq:CMMR)
:class: note
```{figure} /Fig-ch1/multisim.png
---
width: 50px
name: fig-multisim
align: right
---
```
- First design in Multisim a differential amplifier ({numref}`fig:oa17`) with a gain factor of 10.
- Calculate with eqn. {eq}`eq:CMMR` the lower limit of the CMRR. The tolerance of resistor $R_1$ and $R_2$ are both 5\%.
```{figure} /Fig-ch1/multisim.png
---
width: 50px
name: fig-multisim
align: right
---
```
- Verify this result of CMRR with the Multisim design by increasing a resistance $R_1$ with 5\% and lowering the corresponding resistance of $R_2$ with 5\%, and also lowering the other $R_1$ with 5\% and increasing the corresponding resistor $R_2$ with 5\%.
::::{admonition} Answer :class: dropdown
The circuit concerned and relevant values for each component is shown in {numref}fig:67.
---
width: 600px
name: fig:67
---
Differential amplifier with a gain of 10
The lowest CMRR value is found with
$$ {\rm CMRR} \geq \frac{1 + R_2 / R_1}{4 \epsilon} \geq \frac{1 + 9500/1050}{0.2} \geq 50.24. $$ (eq:ch6-19)
From the measurement it follows that
$$ \begin{split} &V_2 = V_1 = 0.5 ; {\rm V} \rightarrow A_{\rm CM} = \frac{V_{\rm out, ; CM}}{V_{\rm in, ; CM}} = \frac{0.083173}{0.5} = 0.166346 \ &V_3 = 0.4 ; {\rm V}, ; V_4 = 0.6 ; {\rm V} \rightarrow A_{\rm DM} = \frac{V_{\rm out, ; DM}}{V_{\rm in, ; DM}} = \frac{1.909}{0.2} = 9.545 \ &{\rm CMRR} = \frac{A_{DM}}{A_{CM}} =\frac{9.545}{0.166346} = 57.4 > 50.24. \end{split} $$ (eq:ch6-20) ::::
Both disadvantages are considerably reduced for the circuit shown in {numref}fig:oa18.
---
width: 500px
name: fig:oa18
---
Instrumentation amplifier
The first part (A) of this circuit has two functions: (1) increasing the input impedances, and (2) increasing the CMRR. The second part (B) is an ordinary differential amplifier.
First, consider part A. The OpAmps 1 and 2, both with feedback
The resistor
Next consider the purely differential signal:
The input voltages of the differential amplifier are given by:
$$ \begin{split} V_1^{'} &= V_1 + (R_{\rm b}/R_{\rm a}) V_{\rm d} \ V_2^{'} &= V_2 - (R_{\rm b}/R_{\rm a}) V_{\rm d} \ \Delta V &= V_1^{'} - V_2^{'} = V_{\rm d} ( 1 + 2 R_{\rm b}/R_{\rm a}). \end{split} $$ (eq:deltaV)
The differential amplification is therefore increased by a factor
:class: note
```{figure} /Fig-ch1/multisim.png
---
width: 50px
name: fig-multisim
align: right
---
```
- Design in Multisim an instrumentation amplifier with a differential amplification factor of 30. Make use of your design of exercise 6.7.
- Check whether the CMRR is really increased by the factor $1 + 2 R_{\rm b} /R_{\rm a}$.
::::{admonition} Answer :class: dropdown
The circuit concerned and relevant values for each component is shown in {numref}fig:68.
---
width: 600px
name: fig:68
---
Instrumentation amplifier with a gain of 30
The same input voltages were used as in exercise 6.7:
The overall gain of the instrumentation amplifier is
$$ G = \frac{5.561}{0.2} = 27.8. $$ (eq:ch6-21)
This is nearly the same as
$$ {\rm CMRR} = \frac{5.561 / 0.2}{0.083173 / 0.5} = 167. $$ (eq:ch6-22)
This is nearly the same as
The integrator ({numref}fig:oa19) strongly resembles the inverting amplifier ({numref}fig:oa14), where
---
width: 400px
name: fig:oa19
---
Integrator
So:
$$ I_{\rm C} = \frac{V_{\rm in}}{R} =I_{\rm R}. $$ (eq:IC)
Since
$$ V_{\rm out} = - V_{\rm C}, $$ (eq:Vout2)
where
$$ I_{\rm C} = \frac{\partial Q_{\rm C}}{\partial t} = C \frac{\partial V_{\rm C}}{\partial t}. $$ (eq:IC2)
Combining eqns.{eq}eq:IC, {eq}eq:Vout2 and {eq}eq:IC2, the solution of the differential equation is given by:
$$ V_{\rm out} = - \frac{1}{RC} \int \limits_0^t{V_{\rm in} {\rm d}t} + V_{\rm out}(t = 0). $$ (eq:Vout3)
So the output voltage is proportional to the integrated input voltage. The proportionality constant is the time constant
A problem for the integrator is that even at an input signal of 0 after some time due to the offset voltage and the bias current of the OpAmp the capacitor fig:oa20).
---
width: 400px
name: fig:oa20
---
Integrator with a bypass resistor
Through the bypass resistor
To determine the value of fig:oa20 should be investigated. Applying eqn. {eq}eq:Gia for the inverted amplifier with
$$ A = - \frac{R_{\rm C}}{R} \cdot \frac{1}{1 + j \omega R_{\rm C} C}. $$ (eq:Abp)
Without the bypass resistor
$$ A = - \frac{1}{j \omega RC}. $$ (eq:Abp2)
From eqn. {eq}eq:Abp2 it can be seen directly that slowly varying signals, such as offset voltages and bias currents result in a large transfer function. Eqn. {eq}eq:Abp limits the transfer function for slowly varying signals to
When eq:Abp and {eq}eq:Abp2 are nearly the same, and the circuit shown in {numref}fig:oa20 therefore behaves as an integrator.
:class: note
```{figure} /Fig-ch1/multisim.png
---
width: 50px
name: fig-multisim
align: right
---
```
- Design in Multisim an integrator (with a bypass resistor $R_{\rm C}$) having a time constant of 0.01 s and check eqn. {eq}`eq:Vout3`.
- Study the frequency behavior and give the results in a Bode plot (both phase and amplitude) and check whether it corresponds with eqn. {eq}`eq:Abp`.
::::{admonition} Answer :class: dropdown
The circuit and components chosen are shown in {numref}fig:69-1.
---
width: 600px
name: fig:69-1
---
Integrator circuit
The output voltage is given by
$$ V_{\rm out} = - \frac{1}{RC} \int_0^t V_{\rm in} {\rm d}t + V_{\rm out} (t=0). $$ (eq:ch6-23)
The amplification is given by
$$ A = - \frac{R_C}{R} \cdot \frac{1}{1+ j \omega R_C C}. $$ (eq:ch6-24)
For slowly varying signals
The modulus
$$ \begin{split} &|A ( \omega) | = \frac{R_C}{\sqrt{R^2 + (\omega R R_C C)^2}} = \frac{R_C}{\sqrt{R^2 + (2 \pi f R R_C C)^2}}, \ &\Phi = \tan^{-1} \left( \frac{0}{R_C} \right) - \tan^{-1} \left( \omega R_C C \right) = - \tan^{-1} \left( 2 \pi f R_C C \right). \end{split} $$ (eq:ch6-25)
{numref}fig:69-2 to {numref}fig:69-4 show the results of the simulation and a comparison between the simulation and the theoretical calculations.
---
width: 600px
name: fig:69-2
---
Multisim simulation of the integrator
Note that in the horizontal part the gain is
---
width: 400px
name: fig:69-3
---
Comparison of modulus between Multisim and calculation of the integrator
---
width: 400px
name: fig:69-4
---
Comparison of phase between Multisim and calculation of the integrator
::::
:class: note
{numref}`fig:oa21` shows a circuit of a differentiator.
```{figure} /Fig-ch6/OA21.png
---
width: 350px
name: fig:oa21
---
Differentiator
```
- Derive using {numref}`fig:oa21` that the output voltage is given by:
$$
V_{\rm out} = - RC \frac{{\rm d}V_{\rm in}}{{\rm d}t}.
$$ (eq:Vout4)
::::{admonition} Answer :class: dropdown
The circuit of the differentiator is given in {numref}fig:oa21.
Analysis of the circuit:
$$ \begin{split} &I_R R = V_- - V_{\rm out}, ; {\rm and} ; I_C = I_R \rightarrow V_{\rm out} = - I_C R, ; {\rm or} ; I_C = \frac{- V_{\rm out}}{R}, \ &I_C = \frac{\partial Q_C}{\partial t} = C \frac{{\rm d} V_{\rm in}}{{\rm d} t} \rightarrow \frac{- V_{\rm out}}{R} = C \frac{{\rm d}V_{\rm in}}{{\rm d} t} \rightarrow V_{\rm out} = - R C \frac{{\rm d}V_{\rm in}}{{\rm d} t}. \end{split} $$ (eq:ch6-26) ::::
The proportionality constant is the time constant
:class: note
```{figure} /Fig-ch1/multisim.png
---
width: 50px
name: fig-multisim
align: right
---
```
- Design in Multisim a differentiator with a time constant of 1 ms and verify that eqn. {eq}`eq:Vout4` holds.
::::{admonition} Answer :class: dropdown
The Multisim circuit for a differentiator with a time constant of 1 ms is shown in {numref}fig:611-1.
---
width: 600px
name: fig:611-1
---
Differentiator with time constant of 1 ms
Since
$$ \begin{split} &A = - \frac{R}{1 / j \omega C} = - j \omega R C, \ &|A| = \omega R C, \ &\Phi = - \tan^{-1} \left( \frac{\omega R C}{0} \right) =- \frac{\pi}{2} \pm {\rm k} \pi. \end{split} $$ (eq:ch6-27)
The bode plot is shown in {numref}fig:611-2, the sine/cosine traces in {numref}fig:611-3.
So
---
width: 600px
name: fig:611-2
---
Modulus and phase of differentiator with time constant of 1 ms
You can also look at the sine/cosine traces, which are 90 degrees out of phase ({numref}fig:611-3).
---
width: 500px
name: fig:611-3
---
The sine/cosine traces, 90 degrees out of phase
$$ \frac{{\rm slope}}{2 \pi} = \frac{\frac{62.43594 - 0.062832}{10000 - 10}}{2 \pi} = 0.99 ; {\rm ms}. $$ (eq:ch6-28) ::::
In this section a number of active filters that make use of OpAmps are presented. These filters are called active because use is made of an active element, in this case an OpAmp.
In chapter 4 it was shown for a low pass filter made of a resistor and a capacitor that the amplitude characteristic in the bandwidth range clearly starts to decrease. This effect becomes more stronger when two of more of these types of filters are used consecutively. OpAmps in combination with well chosen values for resistors and capacitors offers the construction of filters where the amplitude characteristic starts to drop much closer toward the end of a pass band. Usually these active filters do exhibit small oscillations (ripples) in the pass band. In general these oscillations are stronger the faster the cutoff in the frequency characteristic.
For some applications the shape phase characteristic is more important than the amplitude characteristic. For example, at the transfer of pulses it is more important that all frequencies that make up the pulse experience the same shift over time such that the pulse width remains conserved. In that case the phase characteristic within the frequency band width should be linear with frequency.
There are many different filter designs. They are chosen depending on the application. In literature three basic types with characteristic features are found:
- The Butterworth filter is optimized to facilitate a flat frequency band width. As a result it has a slow cutoff near the end of the frequency band width and the phase transfer is less perfect.
- The Chebyshev filter produces a sharp transition from the pass band to the stop band. The pass band exhibits some ripples and the phase characteristic is poor. Pulse shapes are clearly deformed.
- The Bessel filter has an optimized linear phase transfer and is therefore best suited for pulse shaped signals. The amplitude characteristics is less sharp compared to the Butterworth filter and much less sharp compared to the Chebyshev filter.
The next section discusses the Butterworth filter as an example.
The amplitude transfer function of the
$$ | A | = \frac{1}{\sqrt{\left( 1 + \left( \omega / \omega_{3 , {\rm dB}} \right)^{2n} \right)}}. $$ (eq:ch6-29)
{numref}fig:oa22 shows the schematic of an active second order low pass filter, called a Sallen and Key filter.
---
width: 500px
name: fig:oa22
---
Active second order low pass filter
The transfer function that may be derived is given by:
$$ A(\omega) = \frac{1}{\left( 1 + j \omega \left( R_{\rm A} + R_{\rm B} \right) C_{\rm B} - \omega^2 \tau_{\rm A} \tau_{\rm B} \right) }, $$ (eq:ch6-30)
where
$$ | A(\omega) | = \frac{1}{\sqrt{\left( 1 - \omega^2 \tau_{\rm A} \tau_{\rm B} \right)^2 + \omega^2 \left( R_{\rm A} + R_{\rm B} \right)^2 C_{\rm B}^2 }}. $$ (eq:ch6-31)
:class: note
- Assuming $R_{\rm A} = R_{\rm B} = R$, what is the relation between $C_{\rm A}$ and $C_{\rm B}$ for the active second order low pass filter shown in {numref}`fig:oa22` to be a Butterworth filter?
::::{admonition} Answer :class: dropdown
The analysis, with
$$ | A ( \omega ) | = \frac{1}{\sqrt{\left( 1 - \omega^2 \tau_A \tau_B \right)^2 + \omega^2 \left( 2 R \right)^2 C_B^2}} = \frac{1}{\sqrt{1 + \left( \omega / \omega_{3 ; {\rm dB}} \right)^{2 n}}}. $$ (eq:ch6-32)
So
$$ \begin{split} \left( 1 - \omega_{3 ; {\rm dB}}^2 \tau_A \tau_B \right)^2 + \omega^2 \left( 2 R \right)^2 C_B^2 &= 1 + \left( \omega / \omega_{3 ; {\rm dB}} \right)^{2 n}, \ 1 - 2 \omega^2 \tau_a \tau_B + \omega^4 \tau_A^2 \tau_B^2 + 4 \omega^2 R^2 C_B^2 &= 1 + \left( \omega / \omega_{3 ; {\rm dB}} \right)^{2 n}, \ \omega^4 \tau_A^2 \tau_B^2 + 2 \omega^2 \left( - \tau_A \tau_B + 2 R^2 C_B^2 \right) &= \left( \omega / \omega_{3 ; {\rm dB}} \right)^{2 n}. \end{split} $$ (eq:ch6-33)
The second term on the left-hand side must be zero and
$$ \begin{split} 2 R^2C_B^2 &= \tau_a \tau_B, \ 2 R^2C_B^2 &= R C_A R C_B, \ C_A &= 2 C_B. \end{split} $$ (eq:ch6-34)
Then
$$ \omega_{3 ; {\rm dB}}^4 = \frac{1}{\tau_A^2 \tau_B^2} \rightarrow \omega_{3 ; {\rm dB}} = \frac{1}{\sqrt{\tau_A \tau_B}} = \frac{1}{\sqrt{2} R C_B} = \frac{1}{2} \sqrt{2} \frac{1}{R C_B}. $$ (eq:ch6-35) ::::
As mentioned in {ref}OA in the lab exercises we will use the 741 OpAmp. {numref}fig:oa23 shows the pin layout of the 741 OpAmp.
---
width: 300px
name: fig:oa23
---
Pin layout of the 741 OpAmp
{numref}fig:oa23 shows a top view so the pins are located at the bottom of the IC.
- Pin 1 and 5 are connectors to compensate the offset at the inputs (see {numref}
fig:oa24). - Pin 2 is the inverting input (-).
- Pin 3 is the non-inverting input (+).
- Pin 4 is the connection for the negative voltage supply (
$V^{-}$ ), in our case -15 V. - Pin 6 is the output.
- Pin 7 is the connection for the positive voltage supply (
$V^{+}$ ), in our case +15 V. - Pin 8 is not connected.
Particularly for amplifying DC signals reliably the input offset should be compensated. {numref}fig:oa24 shows how this is done.
---
width: 250px
name: fig:oa24
---
Offset compensation
- Both the inverting and non-inverting inputs are grounded.
- A resistor is placed between the offset pins.
- The centre terminal of the pot is connected to the negative voltage supply.
- The pot is tuned such that the output signal is 0 V.
Note that this will not be required for the lab exercises 6.13 to 6.15.
:class: note
```{figure} /Fig-ch3/Lexercise.png
---
width: 50px
align: right
name: fig-Lexercise
---
```
```{figure} /Fig-ch1/multisim.png
---
width: 50px
name: fig-multisim
align: right
---
```
- Design in Multisim an instrumentation amplifier with an adjustable differential amplification factor of 2 to 10. (Use an amplification factor of 1 for part B.)
- Build and test parts A and B of an instrumentation amplifier separately ({numref}`fig:oa18`).
- Connect parts A and B together and test the instrumentation amplifier.
- Determine the CMRR experimentally and compare this with the theoretical expected increase described below eqn. {eq}`eq:deltaV`. Investigate this for different amplification factors.
:class: note
```{figure} /Fig-ch3/Lexercise.png
---
width: 50px
align: right
name: fig-Lexercise
---
```
- Design and build an integrator with a time constant of 0.01 s.
- Verify experimentally whether the integration constant corresponds with the theoretically expected value by using $v_{out}$ and $V_{in}$ displayed as oscilloscope traces and rewriting eqn. {eq}`eq:Vout3`.
- Investigate and explain the drift of the integrator output. Why is it useful to use a square wave input signal in this investigation?
- Reduce the drift by application of a bypass resistor $R_{\rm C}$ and justify your choice of the resistance value.
- Examine the frequency behavior of the circuit ($A ( \omega )$, $| A ( \omega ) |$, asymtotic behavior) and enter the results of Multisim and NI Elvis in a Bode plot for 3 different $R_C$ values ($100 \Omega, 500 \Omega, 1500 \Omega$).
:class: note
```{figure} /Fig-ch3/Lexercise.png
---
width: 50px
align: right
name: fig-Lexercise
---
```
- Design and build a Sallen and Key filter (see {numref}`fig:oa22`) with a 3 dB point of 1 kHz, which complies with the Butterworth characteristic (see also exercise 6.12).
- Examine the behavior of the circuit and give the result in a Bode plot.
- Compare the results with exercise 4.4 (use a multiplot graph).