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Lecture Notes for Analog Electronics

Raymond E
...
uoregon
...
In electronics,
including the text, the term “voltage” refers to the physical quantity of either potential or
EMF
...

As usual, the sign convention for current I = dq/dt is that I is positive in the direction
which positive electrical charge moves
...
e
...
We will soon see, however, that these generalize to AC
...
1

Ohm’s Law

For a resistor R, as in the Fig
...


I
a

b
R

Figure 1: Voltage drop across a resistor
...
g
...

The power dissipated by the resistor is P = V I = I 2 R = V 2 /R
...
2

Kirchoff’s Laws

Consider an electrical circuit, that is a closed conductive path (for example a battery connected to a resistor via conductive wire), or a network of interconnected paths
...
For any node of the circuit in I = out I
...
So for the example of
Fig
...

2
...
g
...

Three simple, but important, applications of these “laws” follow
...


1
...
1

Resistors in series

Two resistors, R1 and R2 , connected in series have voltage drop V = I(R1 + R2 )
...
2
...


Resistors in parallel

Two resistors, R1 and R2 , connected in parallel have voltage drop V = IRp , where
Rp = [(1/R1 ) + (1/R2 )]−1
This generalizes for n parallel resistors to
n

1/Rp =
1
...
3

i=1

1/Ri

Voltage Divider

The circuit of Fig
...
It is one of the most useful and important
circuit elements we will encounter
...
3

Voltage and Current Sources

A voltage source delivers a constant voltage regardless of the current it produces
...
For example a battery can be thought of as a voltage source in series with a
small resistor (the “internal resistance” of the battery)
...

A current source delivers a constant current regardless of the output voltage
...


2

R1
Vin

R2

Vout

Figure 3: A voltage divider
...
4

Thevenin’s Theorem

Thevenin’s theorem states that any circuit consisting of resistors and EMFs has an equivalent
circuit consisting of a single voltage source VTH in series with a single resistor RTH
...
Consider a partial circuit with two output
points held at potential difference Vout which are not connected to anything
...
The
resistor RL is often said to be the “load” for the circuit
...
4
The prescription for finding the Thevenin equivalent quantities VTH and RTH is as follows:
• For an “open circuit” (RL → ∞), then VTH = Vout
...

An example of this using the voltage divider circuit follows
...


R1
Vin

R2

Vout

Figure 4: Adding a “load” resistor RL
...
5
...


To get VTH we are supposed to evaluate Vout when RL is not connected
...
Solving for Ishort and combining
with the VTH result gives
R1 R2
RTH = VTH /Ishort =
R1 + R2
Note that this is the equivalent parallel resistance of R1 and R2
...


4

Class Notes 2

1
...
)

Recall that the Thevenin Theorem states that any collection of resistors and EMFs is equivalent to a circuit of the form shown within the box labelled “Circuit A” in the figure below
...
The Thevenin idea,
however, is most useful when one considers two circuits or circuit elements, with the first
circuit’s output providing the input for the second circuit
...
6, the output of the
first circuit (A), consisitng of VTH and RTH , is fed to the second circuit element (B), which
consists simply of a load resistance (RL ) to ground
...

Circuit A
Circuit B
R TH

Vout

V TH

R

L

Figure 6: Two interacting circuits
...
5
...
In the limit that RTH → 0 the output voltage delivered to the load
RL remains at constant voltage
...

Therefore, RTH determines to what extent the output of the first circuit behaves as an
ideal voltage source
...
Since our
combined equivalent circuit (A + B) forms a simple voltage divider, we can easily see what
the requirement for RTH can be found from the following:
Vout = VTH

VTH
RL
=
RTH + RL
1 + (RTH /RL )
5

Thus, we should try to keep the ratio RTH /RL small in order to approximate ideal behavior
and avoid “loading the circuit”
...

A good power supply will have a very small RTH , typically much less than an ohm
...
The dimming of one’s car headlights
when the starter is engaged is a measure of the internal resistance of the car battery
...
5
...
(Shortly, we will generalize our discussion and substitute the term “impedance”
for resistance
...
)
• The output impedance of circuit A is simply its Thevenin equivalent resistance RTH
...

• The input impedance of circuit B is its resistance to ground from the circuit input
...

It is generally possible to reduce two complicated circuits, which are connected to each
other as an input/output pair, to an equivalent circuit like our example
...


2
2
...
3

RC Circuits in Time Domain
Capacitors

Capacitors typically consist of two electrodes separated by a non-conducting gap
...
For example, for two planar parallel electrodes
each of area A and separated by a vacuum gap d, the capacitance is (ignoring fringe fields)
C = 0 A/d, where 0 is the permittivity of vacuum
...
The SI unit of capacitance is the Farad
...

For DC voltages, no current passes through a capacitor
...
When a time
varying potential is applied, we can differentiate our defining expression above to get
I =C
for the current passing through the capacitor
...


2
...
4

A Basic RC Circuit

Consider the basic RC circuit in Fig
...
We will start by assuming that Vin is a DC voltage
source (e
...
a battery) and the time variation is introduced by the closing of a switch at
time t = 0
...

Applying Ohm’s Law across R gives Vin − Vout = IR
...
Substituting and rearranging gives (let V ≡ VC =
Vout ):
1
1
dV
+
V =
Vin
(2)
dt
RC
RC
The homogeneous solution is V = Ae−t/RC , where A is a constant, and a particular solution
is V = Vin
...

Similarly, a capacitor with a voltage Vi across it which is discharged through a resistor
to ground starting at t = 0 (for example by closing a switch) can in similar fashion be found
to obey
V (t) = Vi e−t/RC
2
...
5

The “RC Time”

In both cases above, the rate of charge/discharge is determined by the product RC which
has the dimensions of time
...
So in our charge-up
example, Equation 3, this would correspond to the time required for Vout to rise from zero
to 63% of Vin
...
0
...
Also note from our solution Eqn
...
Within this approximation, we
that the limit Vout
see clearly from Eqn
...

2
...
7

RC Differentiator

Let’s rearrange our RC circuit as shown in Fig
...


C

Vin

I

R

Vout

Figure 8: RC circuit — differentiator
...

By Ohm’s Law, VR = IR, where I = C(dVC /dt) by Eqn
...
Putting this together gives
Vout = RC
In the limit Vin

d
(Vin − Vout )
dt

Vout , we have a differentiator:
Vout = RC

dVin
dt

By a similar analysis to that of Section 2
...
6, we would see the limit of validity is the opposite
of the integrator, i
...
t
RC
...
Using the technique of the complex impedance,
we will be able to analyze time-dependent circuits algebraically, rather than by solving differential equations
...
It will probably not be particularly useful to use the text for this discussion,
and it could lead to more confusion
...


3
...
Then we can write
˜
˜
˜
V = (V ) + ı (V ) = V eıθ = V [cos θ + ı sin θ]
where ı =

√
−1
...
The operation of determining the amplitude of a
complex quantity is called taking the modulus
...
Then V = 5 + 3ı = 34eı tan (3/5)
...
It is
˜ |
...

often written |V
Since the amplitude will in general be frequency dependent, it will also be written as V (ω)
...


3
...
Therefore, we will analyze our circuits using a single Fourier
frequency component, ω = 2πf
...
Let our complex
˜
˜
Fourier components of voltage and current be written as V = V eı(ωt+φ1 ) and I = Ieı(ωt+φ2 )
...
Let’s see if this can work
...
What about capacitors and inductors?
Our expression for the current through a capacitor, I = C(dV /dt) becomes
d
˜
˜
I = C V eı(ωt+φ1 ) = ıωC V
dt
9

˜
˜˜
˜
Thus, we have an expression of the form V = I ZC for the impedance of a capacitor, ZC , if
˜C = 1/(ıωC)
...
(Note that the voltage drop has the opposite sign of the induced
EMF, which is usually how Lenz’s Law is expressed
...

˜
˜˜
To summarize our results, Ohm’s Law in the complex form V = I Z can be used to
analyze circuits which include resistors, capacitors, and inductors if we use the following:
˜
• resistor of resistance R: ZR = R
˜
• capacitor of capacitance C: ZC = 1/(ıωC) = −ı/(ωC)
˜
• inductor of self-inductance L: ZL = ıωL
3
...
1

Combining Impedances

It is significant to point out that because the algebraic form of Ohm’s Law is preserved,
impedances follow the same rules for combination in series and parallel as we obtained for
resistors previously
...
Using our definition ZC = −ı/ωC, we then recover
the familiar expression Cp = C1 + C2
...
9) to
˜
˜
Vout = Vin

˜
Z2
˜
˜
Z1 + Z2

(5)

Now we are ready to apply this technique to some examples
...
3

A High-Pass RC Filter

The configuration we wish to analyze is shown in Fig
...
Note that it is the same as Fig
...
However, this time we apply a voltage which is sinusoidal: Vin (t) = Vin eı(ωt+φ)
...
Note also that the input and
output voltages are represented in the figure only by their amplitudes Vin and Vout , which
also is common
...

10

Z

1

~
Vin
Z

2

~
V out

Figure 9: The voltage divider generalized
...


We see that we have a generalized voltage divider of the form discussed in the previous
˜
˜
section
...
5 we can write down the result if we substitute Z1 = ZC =
˜
˜
−ı/(ωC) and Z2 = ZR = R:
R
˜
˜
Vout = Vin
R − ı/(ωC)
At this point our result is general, and includes both amplitude and phase information
...
We can divide by Vin on both sides and find
the amplitude of this ratio (by multiplying by the complex conjugate then taking the square
root)
...

˜
|Vout |
Vout
ωRC
T (ω) ≡
=
=
(6)
˜
Vin
|Vin |
[1 + (ωRC)2 ]1/2
Examine the behavior of this function
...
You should convince yourself that this circuit attenuates low frequencies and “passes”
(transmits with little attenuation) high frequencies, hence the term high-pass filter
...
This is approximately equal to an attenuation of 3
decibels, which is a description often used in engineering (see below)
...
6 we see
√
that T = 1/ 2 occurs at a frequency
2πf3db = ω3db = 1/(RC)
11

(7)

The decibel scale works as follows: db= 20 log10 (A1 /A2 ), where A1 and A2 represent any
real quantity, but usually are amplitudes
...


3
...
11
...


You should find the following result for the transfer function:
T (ω) ≡

˜
Vout
1
|Vout |
=
=
˜in |
Vin
|V
[1 + (ωRC)2 ]1/2

(8)

You should verify that this indeed exhibits “low pass” behavior
...
However, the concept of high-pass and low-pass filters is much
more general, as it does not rely on an approximation
...
One can compare our results for the RC circuit using the complex impedance
technique with what one would obtain by starting with the differential equation (in time) for
an RC circuit we obtained in Section 2, taking the Fourier transform of that equation, then
solving (algebraically) for the transform of Vout
...
After all, that is what the impedance technique is doing:
transforming our time-domain formuation to one in frequency domain, which, because of
the possibility of analysis using a single Fourier frequency component, is particularly simple
...


12

Class Notes 4

3
...
)

Before we look at some more examples using our technique of complex impedance, let’s look
at some related general concepts
...
5
...
The term reactance is often used
in place of impedance for capacitors and inductors
...
2 we define the reactance of a capacitor XC to just be equal to its impedance:
˜
˜
XC ≡ −ı/(ωC)
...
This is the notation used in the text
...

This is done simply by dropping the ı from the definitions above
...
So, for example a circuit with R, L, and C in series would
have total impedance
1
˜
)
Z = R + ıX = R + ı(XL + XC ) = R + ı(ωL −
ωC
A circuit which is “reactive” is one for which X is non-negligible compared with R
...
5
...
You may wonder how our results generalize to other frequencies and
to input waveforms other than pure sine waves
...
We can formalize this within the context of the Fourier
transform, whch will also allow us to see how our time-domain differential equation became
transformed to an algebraic equation in frequency domain
...
7
...
2
...
Define the Fourier transform of V (t) as
1
v(ω) ≡ F {V (t)} = √
2π

+∞
−∞

dte−ıωt V (t)

(10)

Recall that F {dV /dt} = ıωF {V }
...
e
...
12 and 13
...
Note also that the procedure carried out to give Eqn
...


3
...
To do this, we proceed as previously, for example like the high-pass filter, but this time we preserve the phase information
˜
˜
by not taking the modulus of Vout
...
We are usually only interested in the phase difference
and the output V
φ2 − φ1 between input and output, so, for convenience, we can choose φ1 = 0 and set the
phase shift to be φ2 ≡ φ
...
Conventionally, the real part is used
...

We rewrite the expression from Section 3
...
6 of Section 3
...
But now
we also have included the phase information
...
15
...
7

Power in Reactive Circuits

Recall that for DC voltages and currents the power associated with a circuit element carrying
current I with voltage change V is just P = V I
...
We could still define an instantaneous power as the product
V (t)I(t)
...

3
...
1

General Result for AC

Since we are considering Fourier components, we will average the results over one period
T = 1/f = 2π/ω
...
Let the voltage and current be out of phase
by an arbitrary phase angle φ
...

We can plug these into the expression for < P > and simplify using the following: cos(ωt +
φ) = cos(ωt) cos(φ) − sin(ωt) sin(φ); 0T sin(ωt) cos(ωt)dt = 0; and (1/T ) 0T sin2 (ωt)dt =
(1/T ) 0T cos2 (ωt)dt = 1/2
...
Using
voltage as an example, the RMS and standard amplitudes are related by
VRMS
3
...
2

1
≡
T

T

1/2
2

V (t)dt

0

1
=
T

T
0

1/2

V02

2

cos (ωt)dt

√
= V0 / 2

(17)

Power Using Complex Quantities

˜
˜
Our results above can be simply expressed in terms of V and I
...
By noting that
start with V (t) = V0 e and I(t) = I0 e
˜ ˜
(V ∗ I) =

(V0 I0 (cos φ + ı sin φ)) = V0 I0 cos φ

we identify an expression for average power which is equivalent to Eqn
...
Several filter types are possible depending upon how Vout is
chosen
...


3
...
The RLC circuit of Fig
...

We can again calculate the output using our generalized voltage divider result of Eqn
...

˜
˜
In this case, the Z1 consists of the inductor and capacitor in series, and Z2 is simply R
...
We then obtain for the
transfer function:
T (ω) ≡

˜
R
ωγ
|Vout |
=
=
1/2
˜
˜
2
|Vin|
|R + Z1 |
[ω 2γ 2 + (ω 2 − ω0 )2 ]

where γ ≡ R/L is the “R-L frequency”
...
The quality factor Q, defined as the ratio
ω0
...
For example, a high-Q circuit, where Vin (t) is the signal on an antenna,
can be used as a receiver
...
Rather than a
resonant circuit, choosing Vout = VC yields a low-pass filter of the form
T (ω) =

2
| − ı/(ωC)|
ω0
=
1/2
˜
2
|R + Z1 |
[ω 2γ 2 + (ω 2 − ω0 )2 ]

The cutoff frequency is ω0 , and for ω
ω0 then T ∼ ω −2 (“12 db per octave”), which more
closely approaches ideal step function-like behavior than the RC low pass filter, for which
T ∼ ω −1 for ω
ω0 (“6 db per octave”)
...

a high-pass filter with cutoff at ω0 and T ∼ ω −2 for ω

16

3
...
9
...
An example is the homework problem (6) of page 59 of the text, where a high-pass and a low-pass
filter are combined to form a “band-pass” filter
...
5, it
is important to design a “stiff” circuit, in which the next circuit element does not load the
previous one, by requiring that the output impedance of the first be much smaller than the
input impedance of the second
...

3
...
2

More Powerful Filters

This technique of cascading filter elements to produce a better filter is discussed in detail in
Chapter 5 of the text
...

3
...
3

Active Filters

Filters involving LC circuits are very good, better than the simple RC filters, as discussed
above
...
In addition, filters made entirely from passive elements tend to have
a lot of attenuation
...
These typically use operational amplifiers (which we will discuss later),
which can be configured to behave like inductors, and can have provide arbitrary voltage
gain
...
When we discuss op amps later,
we will look at some examples of very simple active filters
...


4

Diode Circuits

The figure below is from Lab 2, which gives the circuit symbol for a diode and a drawing of
a diode from the lab
...
One can think of a diode
as a device which allows current to flow in only one direction
...


IF

Figure 13: Symbol and drawing for diodes
...
Semi-conductors such as silicon or germanium
can be “doped” with small concentrations of specific impurities to yield a material which
conducts electricity via electron transport (n-type) or via holes (p-type)
...
14
...
This gap is ∆V ≈ 0
...
3 V for germanium
...


When a diode is now connected to an external voltage, this can effectively increase or
decrease the potential gap
...
15
...
Conversely,
a “forward biased” configuration decreases the gap, approaching zero for an external voltage
equal to the gap, and current can flow easily
...

Thus, when reverse biased, the diode behaves much like an open switch; and when forward
biased, for currents of about 10 mA or greater, the diode gives a nearly constant voltage
drop of ≈ 0
...


18

I
Forward Biased
+
10 mA

-100 V
V
0
...


19

Class Notes 5

5

Transistors and Transistor Circuits

Although I will not follow the text in detail for the discussion of transistors, I will follow
the text’s philosophy
...
This is difficult, and the descriptions which
one gets by getting into the intrinsic properties are not particularly satisfying
...

In practice, one usually confronts transistors as components of pre-packaged circuits, for
example in the operational amplifier circuits which we will study later
...
The ability to analyze the circuit of an instrument or device is
quite valuable
...
There are other
common technologies used, particularly FET’s, which we will discuss later
...
Also, we will assume npn type
transistors, except where it is necessary to discuss pnp
...


5
...

As indicated in the figure, and as you determined in lab, the base-emitter and base-collector
pairs behave somewhat like diodes
...
In particular, for the basecollector pair this description is far off the mark
...


Figure 16: Bipolar transistor connections
...
1
...

Suppose we have the following:
1
...
1 V
...
VB > VE
3
...
We do not exceed maximum ratings for voltage differences or currents
...
When these conditions are met, then current can flow into the collector (and out
the emitter) in proportion to the current flowing into the base:
IC = hFE IB = βIB

(19)

where hFE = β is the current gain
...
) The value
of the current gain varies from transistor type to type, and within each type, too
...
Unless otherwise specified, we will assume β = 100 when we need a
number
...
This will not always
be possible, in which case the transistor will still be on, but IC < βIB
...


IC

IB

IE

Figure 17: Transistor currents
...
The simplest such control is in the form of
a switch
...
e
...
In fact, the base-emitter pair does
behave much like a diode
...
6 V
...
6Volts
21

(21)

5
...
2

Transistor Switch and Saturation

From the preceding discussion, the most straightforward way to turn the transistor “on” or
“off” is by controlling VBE
...
We will follow the lab steps again here
...

First, let R = 10 kΩ
...
When the switch
is closed, then VBE becomes positive and VB = VE + 0
...
6 V
...
6)/104 = 0
...
Hence, IC = βIB = 44 mA
...
044 = 3
...
So, VCE > 0 and VCB > 0
...

Substituting R = 1 kΩ gives IB = 4
...
Setting this equal to
IC would give VC = −9
...
This is not possible
...
2
V
...
) Hence, IC is
limited to a maximum value of IC = (5 − 0
...
So, effectively, the current gain
has been reduced to β = IC /IB = 150/4
...
In this mode of operation, the transistor is
said to be saturated
...
02 of the text
...
2

Notation

We will now look at some other typical transistor configurations, including the emitter
follower, the current source, and the common-emitter amplifier
...
We will often be considering voltages or currents which consist of a time
varying signal superposed with a constant DC value
...
Hence,
∆V = v ;
22

∆I = i

Typically, we can consider v or i to be sinusoidal functions, e
...
v(t) = vo cos(ωt + φ), and
their amplitudes vo and io (sometimes also written as v or i when their is no confusion) are
small compared with V0 or I0 , respectively
...
3

Emitter Follower

The basic emitter follower configuration is shown below in Figure 19
...
The collector is held (by a voltage source) to a constant DC voltage, VCC
...
As we shall see, the most useful
characteristic of this circuit is a large input impedance and a small output impedance
...

For an operating transistor we have Vout = VE = VB − 0
...
Hence, vout = vE = vB
...
20, IE = (β + 1)IB ⇒ iE = (β + 1)iB
...
Assuming the output connection draws
negligible current, we have by Ohm’s Law iE = vE /R
...
Now we can define the input
impedance of the follower:
Zin = vin /iin = vB /iB = R(β + 1)

(23)

By applying the Thevenin definition for equivalent impedance, we can also determine the
output impedance of the follower:
vin
Zsource
(24)
=
Zout = vin /iE =
(β + 1)iB
β+1
where Zsource is the source (i
...
output) impedance of the circuit which gave rise to vin
...
We will return to this point next
time
...
We will then return to the issue of input
and output impedance so that we can build realistic circuits using these configurations
...
4

Transistor Current Source

Figure 20 illustrates the basic configuration for a single-transistor current source
...
Hence, the base voltage VB is also a
constant, with VB = VCC R2 /(R1 + R2 )
...

Vcc

RC

R1

I

C

IB
VB

R2

RE

IE

Figure 20: Basic transistor current source
...
In addition, we have VE = VB −0
...
Solving for IB in this last equation gives IB = VE /((β + 1)RE )
...
6
VB − 0
...

Of course, there are limitations to the range of RL for which the current source behavior
is reasonable
...
2 V
...
So, for example, if we have VCC = 15 V and VB = 5 V in our circuit above,
then VE = 5 − 0
...
4 V, and the range of compliance for the collector voltage VC will be
approximately 4
...


24

5
...
To determine
the output for this circuit, we assume at this point that the input is a sum of a DC offset
voltage V0 and a time-varying signal vin , as discussed last time
...
) V0 provides the transistor “bias”, so that VB > VE , and the
signal of interest is vin
...

The incoming signal shows up on the emitter: vin = ∆(VE + 0
...
And by
Ohm’s Law, iE = vE /RE = vB /RE
...
Now, the
voltage at the output is Vout = VC = VCC − IC RC
...

Putting all of this together, vout = −iC RC ≈ −iE RC = −(vB /RE ) RC , giving the voltage
gain G:
G ≡ vout /vin = −RC /RE
(26)

5
...
In Fig
...
The same considerations we apply
here apply equally to the input of an emitter follower
...
We need to figure out what
design criteria should be applied to this design
...
We
will start by designing the DC component of the input network, that is choosing R1 and
R2
...
23
...
23
...

So our design procedure can be as follows:
25

Vcc

RC

R1

Vout

C
Vin

R2

RE

Figure 22: Common-emitter amplifier with input network
...


1
...


2
...

3
...

4
...


26

Class Notes 7
5
...
They consist of two inputs and one output,
as indicated by the generic symbol in Fig
...
The output is proportional to the difference
between the two inputs, where the proportionality constant is the gain
...
Operational amplifiers (“op amps”), which we will soon
study, are fancy differential amplifiers, and are represented by the same symbol as that of
Fig
...

in1

+
out

in2

-

Figure 24: Symbol for a differential amplifier or op amp
...
For example, a signal which
is to be transmitted and subject to noise pickup can first be replicated and inverted
...
Any
noise pickup will be approximately equal for the two inputs, and hence will not appear in
the output of the differential amplifer
...
This is often
quantified by the common-mode rejection ratio (CMRR) which is the ratio of differential
gain to common-mode gain
...

5
...
1

A Simple Design

The circuit shown in Fig
...
It looks like two
common-emitter amplifiers whose emitters are tied together at point A
...
It is simplest to analyze its output if one writes each input as the
sum of two terms, a sum and a difference
...
In general, we
can rewrite these as v1 =< v > +∆v/2 and v2 =< v > −∆v/2, where < v >= (v1 + v2 )/2 is
the average and ∆v = v1 − v2 is the difference
...

Let’s analyze the difference signal first
...
The signals at the emitters then follow the inputs, as usual, so that at point A
we have vA = vE1 + vE2 = v1 + v2 = 0
...
Hence, vout1 = −(RC /RE )v1
and vout2 = −(RC /RE )v2
...
So
Gdiff1 ≡ vout1 /∆v = −(RC /RE )v1 /(2v1 ) = −RC /(2RE )
27

VCC

RC

RC

Out2

Out1

In

In
1

RE

A

RE

2

R EE
V
EE

Figure 25: Differential amplifier design
...
Referring back to Fig
...
Keeping in
mind these results for the relative signs, it is usual to write the differential gain as a positive
quantity:
RC
(27)
Gdiff =
2RE
where the sign depends upon which is used
...
We have the
following relations:
iEE = iE1 + iE2 = 2iE ;
VA = VEE + IEE REE ⇒ vA = iEE REE = 2iE REE ;
vE − vA
vin − 2iE REE
iE =
=
RE
RE
Solving for iE in the last equation gives:
iE = vin

1
RE + 2REE

Again following the derivation for the the common-emitter amplifie, we have vout = −iC RC ≈
−iE RC
...

Building on what we learned, we can easily improve our differential amplifier design by
adding an emitter-follower stage to the output and a replacing the resistor REE with a
current source
...

5
...
2

Adding a Follower

The output impedance of the common-emitter configuration, as used in the differential amplifer, Zout ≈ RC , is not always as small as one would like
...
Hence, the input of the follower would be
connected to the output of the differential amplifier
...

Hence, in this case, we would have Zout ≈ RC /β
...

βRE
5
...
3

Adding a Current Source

In our expression for CMRR above, we see that a large REE improves performance
...

Hence, REE is limited in practice
...
In other words, point “A” in Fig
...
20
...
By limiting
variations in iEE , as provided by a current source, one effectively achieves a large dynamic
impedance
...
The latter is given by the
quiescent voltage at the inputs of the differential amplifier
...
6 V, depending upon any voltage
drop across RE
...
7
...
In this case one replaces RE in
the expressions above with the intrinsic emitter resistance discussed in Section 5
...
1 below:
RE → re = 25mV/IC
To be exact, one should replace RE in our equations with the series resistance of RE and re :
re
...
However, in most practical situations RE

29

Class Notes 8
5
...
8
...
26
...
So
in most of our examples so far in which an emitter resistor RE is present, one can simply
replace RE by the series sum RE + re
...
For example, IC = 1 mA gives re = 25 Ω, whereas RE might be typically ∼ 1 kΩ
...
In some cases an external emitter resistor RE is omitted, in which case RE → re in our
previous expressions
...


5
...
2

Input and Output Impedance of the Common-Emitter Amplifer

For convenience, the basic common-emitter amplifier is reproduced below
...
3
...
The output impedance is quite different
from that of the emitter follower, however
...
26
...

The short current is just iC , and since iC = (β/(β + 1))iE ≈ iE = vin /RE , then we have our
result
Zin = (β + 1)RE ;
Zout ≈ RC
(30)
Note that these results apply equally well to the differential amplifier configuration, which
is, as we said before, essentially two coupled common-emitter amplifiers
...


5
...
3

DC Connections and Signals

We already discussed in class the fact that for a configuration like that of the input network
of Fig
...
In other words, R1 and R2 appear, for a time-varying signal, to be both
connected to ground
...


5
...
To understand
many of the most important aspects of transistor circuits, this approach is reasonable
...
In reality, this is not the case
...
We start with the Ebers-Moll equation, which gives a foundation for
understanding one class of complications
...
9
...
We can see from the plots of Appendix K (cf pg
...
A more precise description is via the Ebers-Moll equation:
IC = IS eVBE /VT − 1 ≈ IS eVBE /VT

(31)

where VT ≡ kT /e = (25
...
Since typically VBE ≈ 600mV
is much larger than 1, and IS
IC
...

31

We see that IC is not intrinsically a function of IB , but rather is controlled by VBE
...
This means that it takes a voltage input and converts it to a current output
...
We then recover voltage gain by multiplying gm by the resistor
at the output which converts the output current to a voltage
...

The base-emitter “diode” implies a relationship between IB and VBE of the form VBE =
V0 ln(IB /I0 ), where V0 ≈ 0
...
If this form for VBE is plugged into Eqn
...

Another consequence of Ebers-Moll equation is that we see where the intrinsic emitter
resistance re , which we introduced last time, comes from
...

dVBE

From Eqn
...
So we have
re = VT /IC

(32)

where VT is again as above
...
9
...
Understanding its principle of operation requires the Ebers-Moll equation
...
(The base
currents should be negligibly small
...
The collector-base connection transfers this well-defined base
voltage to the collector, thus maintaining the voltage drop across the programming resistor
...
That is, the pair were manufactured together to have nearly identical properties
...
Thus the load current
becomes IL = IP
...

5
...
3

Other Non-ideal Effects

The following represent some of the important departures from ideal transistor behavior:
• VBE = VBE (T )
...
This can be linearized, as given in the text, to yield
approximately
∆VBE
≈ −2
...


• Early Effect
...
This affects high-frequency response
...
Just as emitter resistance
is effectively multiplied by β + 1 for input signals, so too this CCB , which is ususally
a few pF, appears to input signals as a capacitance (1 + G)CCB to ground, where G
is the voltage gain of the transistor configuration
...
The usual solution for mitigating the Miller effect is to reduce the source impedance
...
The cascode
configuration, discussed in the text, uses this
...
The β may be quite different from transistor to transistor, even of
the same model
...

To illustrate this last point, consider our earlier one transistor current source
...
Therefore, the variation in IL
induced by variation in β is
1 dIL
∆IL
∆β =
=
IL
IL dβ
33

1
β+1

∆β
β

Hence, variations in β are attenuated by the factor β + 1
...

The variation in the output of this current source resulting from the Early effect can be
evaluated similarly:
1 dIL
∆VBE
1 × 10−4
∆IL
=
∆VBE = −
=
∆VCE
IL
IL dVBE
VB − VBE
VB − VBE
which can be evaluated using the compliance range for ∆VCE
...
Using our current source, again,
to exemplify this point, we see that temperature dependence can show up both in VBE and
β
...
1 mV/◦ C
dIL
=
≈
dT
dVBE dT
RE
Therefore, we see that temperature dependence is ∝ 1/RE
...
In the case where the external resistor is omitted, then the
typically small re values can induce a large temperature dependence (cf problem 7 at the
end of Chapter 2 of the text)
...


34

Class Notes 9
6

Op-Amp Basics

The operational amplifier is one of the most useful and important components of analog electronics
...
Their primary limitation is that they
are not especially fast: The typical performance degrades rapidly for frequencies greater than
about 1 MHz, although some models are designed specifically to handle higher frequencies
...
Feedback represents a vast and interesting topic in itself
...
However, it is possible to get a feeling for the
two primary types of amplifier circuits, inverting and non-inverting, by simply postulating
a few simple rules (the “golden rules”)
...


6
...
7 with
the refinements we discussed (current source load, follower output stage), plus more, all
nicely debugged, characterized, and packaged for use
...
These two differ most significantly in that the 411 uses JFET transistors
at the inputs in order to achieve a very large input impedance (Zin ∼ 109 Ω), whereas the
741 is an all-bipolar design (Zin ∼ 106 Ω)
...
This is the
gain that would be measured from a configuration like Fig
...
A typical open-loop voltage gain is ∼ 104 –105
...


in1

+
out

in2

-

Figure 29: Operational amplifier
...
They are applicable whenever op-amps are configured
with negative feedback, as in the two amplifier circuits discussed below
...
The voltage difference between the inputs, V+ − V− , is zero
...
)

35

2
...

( This is true in the approximation that the Zin of the op-amp is much larger than any
other current path available to the inputs
...
We now use these rules to analyze the two most common op-amp configurations
...
2

Inverting Amplifier

The inverting amplifier configuration is shown in Fig
...
It is “inverting” because our
signal input comes to the “−” input, and therefore has the opposite sign to the output
...

R2

R1

-

VIN

VOUT

+

Figure 30: Inverting amplifier configuration
...
Since input + is connected to ground, then
by rule 1, input − is also at ground
...
Therefore, the voltage drop across R1 is vin − v− = vin , and the voltage drop across
R2 is vout − v− = vout
...
Since
the − input is at (virtual) ground, then the input impedance is simply R1 :
Zin = R1

(35)

The output impedance is very small (< 1 Ω), and we will discuss this again soon
...
3

Non-inverting Amplifier

This configuration is given in Fig
...
Again, its basic properties are easy to analyze in
terms of the golden rules
...
Therefore, rearranging gives
G = vout /vin =

R1 + R2
R2
=1+
R1
R1

(36)

The input impedance in this case is given by the intrinsic op-amp input impedance
...


6
...
On the other hand, they generally
describe most, if not all, observed op-amp behavior
...

• Offset voltage, VOS
...
If the
two transistors are not perfectly matched, an offset will show up as a non-zero DC
offset at the output
...
This offset
adjustment amounts to changing the ratio of currents coming from the emitters of the
two input transistors
...
The transistor inputs actually do draw some current, regardless
of golden rule 2
...
g
...
g
...
The bias current is defined to be
the average of the currents of the two inputs
...
This is the difference between the input bias currents
...
Therefore, an offset of the two currents will show up as a voltage
offset at the output
...
For example, IOS can be a
problem for bi-polar designs, in which case choosing a design with FET inputs will usually
solve the problem
...
Figure 32 shows how this might be accomplished
...
The modified design in
the figure gives a DC path from ground to the op-amp inputs which are aproximately equal
in resistance (10 kΩ), while maintaining the same gain
...

Similarly, the inverting amplifier configuration can be modified to mitigate offset currents
...
30)
...
Otherwise, charge will build up on the
effective capacitance of the inputs and the large gain will convert this voltage (= Q/C) into
a large and uncontrolled output voltage offset
...
For the non-inverting design, we have turned the very large input impedance
into a not very spectacular 10 kΩ
...
One way around this, if one is concerned only with AC signals, is
to place a capacitor in the feedback loop
...
Therefore, as stated before, it is best, where important,
to simply choose better op-amps!

6
...
It is
also possible to modify non-inverting configurations in a similar way
...
Again, one
would simply modify our derivations of the basic inverting and non-inverting gain formulae
by the replacements R → Z, as necessary
...
5
...
33, we have
vin
dvout
d(vout − v− )
vin − v−
=
= iin = iout = −C
= −C
R
R
dt
dt
So, solving for the output gives
1
vin dt
(38)
vout = −
RC
And for a single Fourier component ω, this gives for the gain
1
(39)
G(ω) = −
ωRC
Therefore, to the extent that the golden rules hold, this circuit represents an ideal integrator and a low-pass filter
...
In practice, one may need to supply a resistor in parallel with the capacitor to
give a DC path for the feedback
...


6
...
2

Differentiator

The circuit of Fig
...
We find the following:
dvin
(40)
vout = −RC
dt
G(ω) = −ωRC
(41)
So this ideally represents a perfect differentiator and an active high-pass filter
...
(The gain cannot
really increase with frequency indefinitely!)

6
...
We illustrated this with the two basic negative feedback configurations:
the inverting and the non-inverting configurations
...

39

R
C

-

IN

OUT

+

Figure 34: Op-amp differentiator or high-pass filter
...
6
...

35
...

The op-amp device itself has intrinsic gain A
...
The quantity B is
the fraction of the output which is fed back to the input
...
The gain
of the device is, as usual, G = vout /vin
...
To complete
the terminology, the product AB is called the loop gain
...

As a result of the negative feedback, the voltage at the point labelled “a” in the figure is
va = vin − Bvout
The amplifier then applies its open-loop gain to this voltage to produce vout :
vout = Ava = Avin − ABvout
Now we can solve for the closed-loop gain:
vout /vin ≡ G =

A
1 + AB

(42)

Note that there is nothing in our derivation which precludes having B (or A) be a function
of frequency
...
6
...
The figure below is meant to clearly show the relationship between
the definitions of input and output impedances and the other quantities of the circuit
...
Similarly, Ro represents the
Thevenin source (output) impedance of the open-loop device
...

We start the calculation of Zin with the definition Zin = vin /iin
...
42 gives
iin =

1
A
vin − Bvin
Ri
1 + AB

Rearanging allows one to obtain
Zin = vin /iin = Ri [1 + AB]

(43)

A similar procedure allows the calculation of Zout ≡ vopen /ishort
...
This means
that all of the current from the amplifier goes into the load, leaving none for the feedback
loop
...
So for a typical op-amp like a
741 with A = 103 , Ri = 1 MΩ, and Ro = 100 Ω, then if we have a loop with B = 0
...

6
...
3

Examples of Negative Feedback Benefits

We just demonstrated that the input and output impedance of a device employing negative
feedback are both improved by a factor 1 + AB ≈ AB, the device loop gain
...
42 in action
...
Let Amax = 104 and Amin = 103, and let
B = 0
...
We then calculate for the corresponding closed-loop gain extremes:
Gmax =

104
≈ 10(1 − 10−3 )
1 + 103

103
≈ 10(1 − 10−2 )
2
1 + 10
Hence, the factor of 10 open-loop gain variation has been reduced to a 1% variation
...
It attenuates errors which appear within the feedback loop,
either internal or external to the op-amp proper
...
For most
op-amps, A is very large, starting at > 105 for f < 100 Hz
...
An extreme example of the latter choice is the “op-amp follower” circuit,
consisting of a non-inverting amplifier (see Fig
...
In this
case, B = 1, giving G = A/(1 + A) ≈ 1
...
The
qualitative statement is that any signal irregularity which is put into the feedback loop will,
in the limit B → 1, be taken out of the output
...
Imagine a small,
steady signal vs which is added within the feedback loop
...
In the limit B = 1 the output
and feedback are identical (G = 1) and the cancellation of vs is complete
...
(See text Section 2
...
) The push-pull circuits, while boosting current, also
exhibit “cross-over distortion”, as we discussed in class and in the text
...

Gmin =

6
...
If one cascades these
filters, the phase shifts can accumulate, producing at some frequency ωπ the possibility of
a phase shift of ±π
...
This in turn tends to

42

compound circuit instabilities and can lead to oscillating circuits (as we do on purpose for
the RC relaxation oscillator)
...
However, at high frequencies (f ∼ 1 MHz or more), unintended stray capacitances can become
significant
...
Most manufacturers of op-amps confront this issue by intentionally reducing the
open-loop gain at high frequency
...
It is carried out by bypassing
one of the internal amplifier stages with a high-pass filter
...
37
...
The goal is to achieve A < 1 at ωπ , which is typically at
frequencies of 5 to 10 MHz
...

Compensation accounts for why op-amps are not very fast devices: The contribution of
the higher frequency Fourier terms are intentionally attenuated
...
Hence, their speed is typically much
greater
...


43

Class Notes 10

7
7
...
This basic idea is shown in Fig
...
The comparator has large open-loop gain A
...
We
have in the limit of very large A
vout = A(v+ − v− ) =

+Vmax v+ > v−
−|Vmin | v+ < v−

where Vmax and Vmin are aprroximately the power supply voltages
...
Hence, this can be
thought of as a 1-bit analog to digital converter (A/D or ADC)
...
In fact, as we
shall see below, comparator circuits often employ positive feedback to ensure that nothing
intermediate between the two extreme output states is utilized
...
Also, the amplifier can be optimized for speed at the expense of
linearity
...
Table 9
...


+

R
A
v out

-

Figure 38: Comparator model
...
38 the output stage consisting of a transistor with
collector connected to the comparator output
...
It is used in the 311 comparators we use in lab
...
The transistor emitter is also available as
an external connection
...
This is chosen to be ground in the figure
...
When that
44

is low it will, after passing through an inverter, turn the transistor on
...
This current produces a voltage drop
across R which pulls the output voltage (very close) to the emitter voltage (ground in our
example)
...
When the comparator inputs are in the complementary
inequality, the transistor is switched off and the output voltage goes to the voltage held by
R, which is +5 V in our example
...


7
...
39
...
Positive feedback is used to help reinforce the chosen output
state
...
The way this works is illustrated in Fig
...
Vh and
Vl refer to threshold voltages which are set up at the comparator + input by the resistor
R4 , the output states will still be determined by the pull-up
divider chain
...
For the circuit in the figure, these states are 0 and +5 V
...
Whether the
connection to +V1 and R1 is required or not depends upon whether a positive threshold is
required when Vout = 0
...


Referring to Fig
...
The output is in the +5 V state
...
When the input crosses the threshold, the output changes to the
other state, 0 V
...
Having two thresholds provides
comparator stability and noise immunity
...

45

v

in

Vh
Vl
t
Vout
+5

t

Figure 40: Examples of Schmitt trigger signals versus time
...
Bottom: vout
...
Also, a negative threshold could be
set in two ways
...


7
...
41 uses both positive and negative feedback
...
Note that the positive feedback is a Schmitt configuration
...
The output voltages are set up to be either +5 V (pull up) or −5 V (emitter
connection)
...
When the output is +5 V, the capacitor C is charged up through the
resistor R
...
42
...
Hence, the charge
up curve will eventually cross the +1 V threshold, forcing the comparator to the −5 V state,
and thereby starting a ramp-down of the capacitor voltage given by
Vc (t) = −V0 +

3V0 (t2 −t)/RC
e
2

where t2 is the time at which the output switched to −5 V
...
The output will
be a square wave, whereas Vc resembles a triangle wave
...


46

+5
100K
-

1K

10nF
v out
+
-5
80K

20K

Figure 41: RC relaxation oscillator
...


47

Class Notes 11
8

Radio Basics

In this section we will discuss some basic concepts concerning signal modulation, generation,
receiving, and demodulation
...
However, the most familiar perhaps is that of broadcast radio generation
and receiving, hence the title of the section
...
From this, we
can see how to carry over many of the concepts to other forms of signal modulation and
reception of signals
...
1

The Case for Modulation

Consider the familiar example of radio signals which carry audio information
...
Even if it were possible to broadcast
signals of such low frequency in the electromagnetic spectrum, there would be a multitude
of confusion resulting from the interference between competing broadcasts
...
An audio signal
which modulates an RF “carrier” of, say, 20 Mhz, uses only the range 20
...
02 MHz
...
10 ± 0
...
Hence, with a carrier at much higher
frequency than the signal, many channels can co-exist with little or no interference
...
It is important to remember that the signals do not have to be audio, that is only a
familiar example
...
Another familiar example is the modulation of computer-generated
signals for transmission over telephone lines
...
2

Amplitude Modulation

Figure 43 gives the general scheme
...
We see
that the carrier amplitude A cos(ωc t) is modulated by the factor 1+m cos(ωm t), where m = 0
represents the limit of no modulation and m = 1 is a miximally modulated signal
...


we can do a “poor man’s” Fourier transform of Vs :
1
Vs (t) = A cos ωc t + Am [cos ((ωc + ωm )t) + cos ((ωc − ωm )t)]
2

(46)

So we have a central carrier frequency plus two side-bands at fc ± fm
...
g
...


8
...
3
...
44
...
The filter is a resonant circuit with a sharp peak at the carrier frequency of the
√
broadcast ωc = 1/ LC
...
With this addition, and without the amplified output, the passive “crystal”
radio receiver looks like this
...


The resistor R and capacitor clearly form a low-pass filter
...
However, without the diode, the effect would be to
throw away all of the information, too, since as we saw from Eqn
...
Without the
diode, the system is linear, and no signal will be present at the output
...
In order to illustrate how this works, we
assume a specific form for the response of a forward-biased diode as I = bV 2 , where b is a
constant
...
44) in order to convert
this diode current to a voltage to be presented to the low-pass filter
...
This then gives rise to an output
current
I = bV12 cos2 ω1 t + bV22 cos2 ω2 t + 2bV1 V2 cos ω1 t cos ω2 t
Again using trigonometric identities to form the poor man’s Fourier transform, this becomes
2I/b = V12 + V22 + V12 cos 2ω1 t + V22 cos 2ω2 t + 2V1 V2 [cos((ω1 + ω2 )t) + cos((ω1 − ω2 )t)]
Therefore, from the original two frequencies, the diode has produced harmonics (twice the
original), as well as the sum and difference
...
46, where three frequencies
are originally present (ωc and ωc ± ωm ), the effect of the diode is easily generalized from the
steps above using the substitutions ω1 = ωc and ω2 = ωc + ωm or ω2 = ωc − ωm
...
Of particular interest for our receiver
is the difference frequency between the carrier and the modulated carrier
...
This can then be separated from the higher frequencies using the low-pass filter
and amplifier
...

As an aside, we note that with our example I = bV 2 , we have squared the input
...

8
...
2

Harmonic Distortion

Note that we intentionally introduced a non-linear element (the diode) to our system
...
In particular, our diode with the I = bV 2 behavior introduced first harmonics of the original frequencies at twice the original
...
For example, a nonzero b3 will generate a 2nd harmonic of the original ω at 3ω
...
Since the pattern of harmonics is what
distinguishes musical instrument types to the ear, the introduction of non-linearities should
be avoided in high-fidelity audio amplifiers
...
3
...
It uses a phase-locked
loop (PLL) circuit at the input of the receiver
...
Since one can
consider the modulation of the carrier to be a phase shift (by amount ωm t), the output
of the PLL can then produce a voltage signal proportional to these phase shifts, which in
turn is used to provide active rectification of the input at the frequency of the modulation
...
This type of PLL circuit is actually more relevant to FM detection,
which is discussed below
...
3
...
It is essentially a fancy version of our
simple heterodyne detector above
...
An example is given in Figure 13
...
Consider an input carrier of frequency 10 MHz which has amplitude modulated at
some much lower frequency
...
In the example of the text, the local oscillator has frequency tuned
to be fLO = 10
...
As with our earlier diode
example, the mixed signal includes the difference frequency, in this case 455 kHz, which in
turn has nearby sideband frequencies which differ from 455 kHz by the audio modulation
frequencies
...
One advantage here is that a relatively high-frequency carrier, which in general
will be difficult to condition using conventional electronics is effectively reduced to a more
manageable frequency, in the example from 10 MHz to 455 kHz
...

So the tuning is accomplished by adjusting the oscillator, rather than the filter
...
4

Other Modulation Schemes

Recall from Eqn
...

However, to preserve the information, the generation and receipt of the amplitude must be
linear
...
On the other hand, phase and frequency modulation (FM) do not suffer
from these complications
...
Radio broadcast by FM also has the additional advantage, by dint of historical
accident, of occupying a higher frequency band, thus allowing easy accomodation of a full
audio bandwidth
...

8
...
1

Phase Modulation

A carrier of frequency ωc is phase modulated if the resulting signal has the form
V (t) = V0 cos(ωc t + Ap cos ωm t)

(47)

where V0 and Ap are constants and ωm is the modulating frequency, as before
...

8
...
2

Frequency Modulation

The phase of a sinusoidal function, when frequency is a function of time, can in general be
expressed as
φ = ωdt
Now suppose the frequency is modulated by a frequency ω about some central carrier frequency
ω = ωc + Af cos ωm t
where Af is a constant
...

Carrying out steps analogous to those for Eqn
...


8
...
For FM detection we need
to replace the diode with something which can provide a voltage output proportional to the
input frequency modulated signal
...
The PLL scheme is reproduced in Fig
...
(For this application,
the counter is omitted
...
This is exactly what we need to detect the phase shift introduced by
FM
...

An apparent practical limitation of this technique for FM radio reception is that PLLs do
not operate at these high frequencies (∼ 100 MHz)
...
The input is mixed using a
local oscillator and the resulting lower frequency (455 kHz in our example) modulated signal
is then input to the PLL
...


52

f

Phase
Detector

Low-pass
Filter

Counter

VCO

(divide by n)

in

(V->f)

f

replica

f

out

V
out

Figure 45: PLL schematic
...


53



---