Electronic Concepts Matter and Electricity Charge, Fields, Voltage, Current, and Beams: Review main points: This should be a review for those who have taken High school Physics or have read a book on physics or electricity. 1. Conventional current flows in the direction of E field. That is it flows from positive to negative potential. From here on "current flow" means conventional current flow. 2. Current normally flows in the direction of arrows on transistors and diode symbols. 3. Current in wire and resistors is due solely to electron flow. 4. Current in transistors and diodes is due to hole flow (positively charged molecules) and electron flow. The holes and the electrons move towards each other. 5. Positive charged particles flow in the direction of current and E fields. 6. Negative charged particles flow in opposite direction of current or E fields. 7. The EMF of a battery forces positive ions to move to the positive plate of a battery. 8. The EMF of a battery forces negative ions to move to the negative plate of a battery. 9. Holes (positively charged molecules) in silicon crystal lattice carry charge in semiconductors and move in direction of current. Current is in the direction of E field. Current direction is indicated by direction of arrow on transistor or diode. About direction of current: Case A: At this very moment a beam of electrons is traveling from the gun in the back or neck of the CRT directly towards your face (pretend you are looking at
Tube [CRT] if you have a flat screen monitor). Luckily for you the beam strikes the
phosphorous-coated surface of the CRT prior to striking you. Now make a fist with your
left
hand directly in front of your face. Now point your thumb directly in the direction ofyour face. Your thumb is now pointing in the same direction as the electron beam. This
electron beam produces a magnetic flux that curls about the beam in the direction of the
fingers on your left hand.
Case B: Next, I want you to imagine that your CRT was actually an ion propulsion
engine, and instead of electrons coming straight at you there were positive ions coming
straight at you. Now make a fist with your
right hand directly in front of your face. Nowpoint your thumb directly in the direction of your face. Your thumb is now pointing in the
same direction as the positive ion beam. This ion beam produces a magnetic flux that
curls about beam in the direction of the fingers on your right hand. Next, point your
thumbs towards each other. Note that your fingers curl in the same direction. When a
negative ion (or electron) and a positive ion move towards each other, the curl of the
magnetic flux around the negative or positive ion curl in the same direction. The positive
ion moves in the same direction as the electrostatic field vector,
E. Conventional currentflow is in the same direction as the field vector
E, and hence is in the same direction asthe positive ion or charge carrier. Electrons or negative ion flow is in the opposite
direction of the
E vector.The Right Hand Rule states that when you point your thumb in the direction of the
electrostatic field (E) that is driving the beam, the curl of your fingers indicates the
direction of the Magnetic H field. The remainder of my lectures will focus on teaching
you how to troubleshoot, and actually allow you to self test your ability by using my
simulations.
Chapter 2
Kirchhoff's and Ohm's Laws
Today Kirchhoff's laws appear fairly intuitive. If water in a pipe branched into two lines,
then we would expect that the sum of the rate water flows out of the two pipes equals the
rate of flow in the single pipe. However, in Mr. Ohm and Kirchhoff's time no one new of
electrons, current was observed strictly in terms of the associated magnetic field about a
wire. Thus, Kirchhoff's Laws revealed the nature of electric current. Likewise the linear
relationship expressed by Ohm's law could not be implied from fluid mechanics. In fluid
mechanics or hydraulics, pressure drop is not linear with respect to flow rate.
On a more practical note consider what it means if Kirchhoff's law does not appear to be
being obeyed. What if you had three wires attached to a terminal, and measurement
indicated that the current going into the terminal from one wire did not equal the current
leaving the terminal in the remaining two wires. This would indicate that some current
was going to ground via the terminal and traveling a path not shown on your circuit
schematic. An obvious solder splash or a less obvious contaminant on surface of printed
circuit board could cause this. In the case of multi-layer boards, a breakdown of
insulation between layers is also a possible path of current flow.
Ohm's law states that current (in amperes) is equal to voltage (in volts) divided by the
resistance (in ohms). This form of Ohm's law is most useful for technicians, because it is
easy to measure voltage with a single probe of a grounded voltmeter. Connect the
negative lead of your voltmeter to ground, circuit common, or power source return line,
and you can measure the voltage on as many component junctions (commonly called
circuit nodes) as are accessible to your meter probe. You can determine the current
through a resistor by measuring the voltage on both sides of the resistor, calculating the
difference in voltage, and than dividing the difference voltage (delta V) by the resistance
of the resistor. This resistance is written on the resistor in letters or frequently indicated
by a color code. Resistor value can also be read from the circuit schematic. Usually one
current calculation and a few voltage measurements tell you all you need to know about
transistor, diode, operational amplifier, or vacuum tube circuits. Thus, you must be able
to use Ohm's law to analyze what is going on not only in purely resistive circuits but also
in the more complex circuits that you will study in future lessons.
When the power to a circuit is off, resistance of circuit component can be measured with
an ohmmeter. Sometimes when you measure resistance across any component,
measurement may also include resistance of parallel circuit branches. You can prevent
this by lifting one end of the resistor from the circuit board. This e-book does not teach
troubleshooting using an ohmmeter. It teaches fault isolation using only voltage, current,
and frequency data.
Ohm's Law Animated Lesson:
There are two quantities that technicians keep track of in order to
monitor circuit performance, for the purpose of fault isolation. They are current and
voltage. The top circuit on this page shows current as a function of voltage. Or otherwise
stated, it animates the equation current = voltage divided by resistance. This relationship
is known as Ohm's law. Voltage is usually the easiest quantity to measure, because it can
be read simply by probing a single point. You can record the voltage on both ends of a
resistor, and then read the resistor value from the circuit schematic. With this data Ohm's
law can be used to calculate the current flowing through the resistor. The Animation
shows current increasing as the voltage increases.
This circuit shows that when a constant current goes through a variable
resistance, the voltage increases in proportion to the resistance. This is in accordance with
Ohm's law in the form of Voltage = Current times Resistance. Power dissipation in a
resistor is proportional to the product of current and voltage. In this case current is
constant; therefore, power dissipation is directly proportional to the voltage. This
relationship is clearly demonstrated by the animation, when voltage and power are
observed. Note that the power meter is connected in series with and across resistor, so
that meter monitors both current and voltage.
This animation shows that the current through a fixed resistor is
proportional to the applied voltage. Power is proportional to the product of voltage and
current which are both increasing in step. From the equation, Power = Voltage times
current, it follows that Power = voltage squared divided by resistance. If you care to do
the algebra, simply substitute voltage divided by resistance into the equation. Remember
to convert milli-amps, and milli-volts, to volts when you calculate power in units of
watts. It is easy to do this using scientific notation.
P = I
* VI = V/R
P = V
2/RChapter 3
Series Circuits
Calculating Series Resistance:
In a series resistive circuit the same current flows in each resistor. That
is the same current flows through R1, R2, R3, R4 and Rs (Rs is the internal resistance of
the power source.) Note that all the resistors in the power circuit are equal to one ohm.
Therefore, the voltage drop across each resistor is the same. Click TP1 button and note
the voltage across the whole circuit is 8 volts. Click the other buttons and note the
voltage at each test point. Calculate the voltage across each resistor. The results should
be 2 volts.
The current in the circuit can be calculated using ohm's law. I = 2 volts/1 ohm = 2
amperes, or 8 volts/4 ohms = 2 amperes.
Note that Rs, the internal resistance of the power source, and Vs, the electromotive force
of the power source (EMF) have not been measured. To measure these quantities you
must follow the following procedure:
1. Break the circuit. Removing a resistor would break circuit.
2. Measure voltage at TP1. The result would be 10 volts. (This step not
simulated)
3. Vs = EMF = 10 volts. In the closed circuit, 2 volts was dropped across Rs of
the power source. In the case of a battery Rs is distributed throughout the
battery.
4. Thus Rs = V/I = 2 volts/ 2 amperes = 1 ohm.
The total resistance of a series circuit is the sum of resistors in the circuit. In this case Rt
= R1 + R2 + R3 + R4 = 4 ohms. I = Vtp1/ 4 ohms = 8 volts/ 4 ohms = 2 Amps. Where:
Vtp1 is the voltage at TP1.
General procedure for testing resistors in series:
To test a two-resistor series circuit, measure the voltage drop across each resistor. Use
Ohm's law to calculate current through each resistor. If they are not equal (with in the
limits of tolerance specified on the resistors), than one of the resistors is defective. If you
have three or more resistors in series, calculate the current through each of the resistors.
If they are not equal (with in the limits of tolerance specified on the resistors), than one of
the resistors is defective. The resistor with a significantly different calculated current is
the defective one. The two resistors with the same calculated current are good.
Chapter 4
Parallel Circuits
Calculating Parallel Resistance:
In parallel circuits the same voltage is present across each resistor in
the circuit. The current through R
1is: I1= V/ R1and the current through R2is : I2=V/R
2 . For any number branch elements we can say: IN=V/ RN. The total current in aparallel circuit is simply the sum of the currents in each branch:
I
T = I1+ I2 … + INI
T = V/ R1+ V/ R2 … + V/ RNI
T = V/ RTI
T = V/ R1+ V/ R2 … + V/ RNV/ R
T = V/ R1+ V/ R2 … + V/ RN1/ R
T= 1/ R1+ 1/ R2 … + 1/ RNYou can use above equation to calculate value
RT. Substitute resistor values into rightside of equations, and calculate the decimal value of each fraction on right side of
equation. Then add the fractions and take the reciprocal. I teach how to program this
calculation on my JavaScript site.
Estimating and Measuring Parallel Resistance:
In the parallel circuits the applied voltage is the same across each parallel branch, and the
current in each branch equals the applied voltage divided by the resistance of the branch.
The total current in a parallel circuit is always more than the current in any one branch
even if one branch carries most of the current. This is true because the total current is
simply the sum of the branch currents. This is usually obvious to most students. What this
implies that is not always obvious to many students is the relationship of the total
resistance to the branch resistance. The combined resistance of resistors in parallel is
always less than the smallest branch resistance. For example, if you had 100,000, 10,000
and 1000 ohm resistor in parallel, the combined resistance would be less than 1000-ohms.
If you connected an ohmmeter across the 1000-ohm resistor you should measure slightly
less than 1000-ohms. If you measure across the 10,000 or 100,000 ohm resistor you
should also measure slightly less than 1000-ohm. If you measure across the 1000-ohm
resistor and read more than 1000-ohms, than you can be certain the 1000-ohm resistor is
out of tolerance or open. If you don't measure the same resistance across all three
resistors than they are not really connected in parallel anymore, and you should start
looking for an open wire or land or a cold solder joint at one of the resistor nodes.
Of course you can also make ohmmeter measurements when circuit or system power is
off. You may not realize that it is also better to make resistance measurement with the
circuit board out of the electronic system. For example, if you had 1000-ohm resistor in
series with a 2000-ohm resistor and one resistor went to ground and the other resistor
went to 5 volts. Then, when you measured across both resistors you would get a reading
of 3000 ohms if the board were out of the system. However, a measurement made on the
same circuit board when circuit board is in system might be much lower. In above
example, even though the power is turned off you might still read much less than 3000
ohms. That is because the power source provides a parallel pathway for meter current to
flow from the 5-volt bus to ground. That is the power supply is in parallel with the 1000-
ohm and 2000-ohm series combination.
Parallel Circuits Animated Lesson:
In a parallel circuit the voltage across each resistor is the same. Thus,
current is strictly a function of resistance. In the animation shown on this page, resistors
are switched in and out of a parallel circuit, causing total current to the circuit to vary.
You are asked to deduce from the total current variations whether the circuit is defective.
If it is defective, you should determine which resistor is suspect. Previously I stated that
current is not easy to measure. However, total current readings are frequently available.
Sometimes they can be read directly off a laboratory power supply. Sometimes it is easy
to connect a current probe around a wire between the power source and the circuit board.
Chapter 5
Series and Parallel Combination
We analyze series-parallel circuits in stages. We replace several resistor values with a
single one. For example, if two series resistors (R1 and R2) were in series with a parallel
pair of resistors (R3 + R4) we would write.
RsumA = R1 + R2, Where two resistors R1 and R2 are in series.
1/RsumB = 1/R3 + 1/R3, Where two resistors R3 and R4 are in parallel.
Rtotal = RsumA + RsumB
Recommended outside reading:
Most electronic circuits are a combination of series and parallel
components. When we analyze parallel, series circuits we reduce the circuit to an
equivalent purely series circuit or equivalent purely parallel circuit. For example, when
the two-milli-ampere range is selected, resistor one, two, and three can be combined into
a single resistor, that we can call "branch one total". Now we have a parallel circuit
consisting of "branch one total" resistance and the meter resistance. We can now
calculate the current that should be flowing through the three resistors and the resistance
of the meter. Troubleshooting is often much easier. If a circuit has a fault, we can simply
measure the voltage drop across each resistor of the voltage divider. Then using Ohm’s
law calculate the current in each resistor. If for example, we find that resistor number
two’s calculated current is different than the other two calculated currents, we know that
"resistor number two" is the defective component. Our troubleshooting is complete. If we
find that the voltage divider is not defective, then the meter must be the problem.
Observe that the
Unit Under Test (UUT) meter reading is not intended tosimulate a digital meter, but indicates the position of a needle on an analog meter
movement. The Meter reads pegged when the reading is more than 105% of full scale.
You can study normal operation of the meter or insert a fault. You must step the voltage
up or down in order to activate the simulation. Then you will have total control of the
power supply, and meter switch position. You can also select which point you wish to
probe with your virtual voltmeter.
Here we see a GIF circuit animation consisting of a parallel circuit in
series with a resistor Rs (R series). Recall that in the previous chapters, I stated that the
total resistance of resistors in parallel was always less than the smallest parallel resistor.
In this case the smallest resistor is 1k. Thus, I would estimate that the resistance of the 1k
and 10k resistor was .9k. The exact calculated value of the 1k and 10k resistor in parallel
is 909 ohms (Rp = 909 ohms). The reason I emphasize estimating techniques is that it is
hard to use a calculator when you have a voltmeter probe in your hands. Thus we have
reduced this complex circuit to a simple series circuit with the equation: Rtotal = Rs +
Rp.
Chapter 6
Alternating Current
Power Generation:
Rotating a coil in a magnetic field creates alternating current, and is the method that is
used to create most of the world's electricity. Even DC generators create a sinusoidal
voltage that is rectified by a commutator. AC electricity is better suited for long distance
power transmission than DC electricity. Alternating current voltage can be stepped up to
over 100,000 volts, making it possible to transmit power for hundreds of miles over High
Voltage transmission lines. The voltage can than be reduced to 120 volts (US single
phase) or 220 volts (European single phase) for safe home use. The transformers that you
see on poles or behind chain link fences are used to step down high voltage to a safer
level.
Power Generation is the province of electrical engineering not electronic engineering.
One aspect of electronics deals with AC signal generation, filtering, and amplification. It
encompasses audio signals, radio signals, radar signals and the like. Of course most
electronic system do have power supplies that convert AC to the DC voltages required by
vacuum tube, transistor, and integrated circuits.
Electronic AC Signal Generation:
This is one of the simplest ways to generate a sinusoidal signal. The capacitor is charged by DC supply
voltage. Potential energy is stored in the electrostaticE field between the plates of the
capacitor. The capacitor discharges through the coil. When the capacitor discharges to
zero volts the current in coil reaches a maximum. At this point the potential energy of the
E
field is reduced to zero and the current is at a maximum. All the potential energy ofcapacitor is now kinetic energy stored in the coil magnetic
H field. As the H fieldcollapses, the change in magnetic flux sustains the current flow through it. This current
causes the capacitor to charge to a level opposite in polarity with respect to original
voltage. This transfers the energy back to the capacitor
E field. This cycle repeatsindefinitely unless damped. This oscillating circuit is frequently referred to as a tank
circuit. An ideal tank circuit has no resistance and will oscillate indefinitely. Of course
the wire of the inductors has some resistance, which will course some damping of the
oscillations. Heat can also be generated in the electrolyte of the capacitor. This will also
contribute to oscillation damping. An electromagnetic oscillator can also radiate energy
in the form of electromagnetic radiation. Electromagnetic radiation consists of an
oscillating E and H fields in space. This radiation propagates at the speed of light.
AC Signals are usually sinusoidal:
The equation below defines a sinusoidal waveform. Electromagnetic radiation in space
propagates as a sinusoidal wave moving at the speed of light.
V = A sin
wt where w = 2 *Pi*f or angular velocity.Ohm's and Kirchhoff's Laws:
In purely capacitive and purely inductive circuits, Ohm's and Kirchhoff's Laws still
apply. Perform AC calculation in the same manner as you performed DC calculations.
Simply substitute Vac for Vdc. In purely inductive or purely capacitive circuits, you
must first calculate the inductive or capacitive reactance before applying Ohm's law.
Inductive and capacitive reactance are a function of frequency. Reactance is expressed in
ohms.
Chapter 7
Capacitive Reactance
Capacitive Reactance:
Capacitors block direct current and pass high frequencies. Capacitance reactance is high
for low frequencies and low for high frequencies.
X
C = 1/wC where w = 2*Pi*f or w = angular velocity.Pi = 3.141… and f = frequency.
Capacitive Reactance in Series Circuit:
Once you calculate impedance for a strictly capacitive circuit, use Ohm's law directly.
For example let us consider a voltage divider consisting of three capacitors C1, C2, and
C3 in series. Total impedance (
Z total ) of the series circuit is as follows:Z total = X
C1+ XC2+ XC3, I = V/ Z totalV = I(Ztotal) where V is the voltage across the three capacitors.
The voltage across each component is given by:
V1 = I
*XC1,V2 = I *XC2 , V3 = I *XC3The general form of the equation is:
Z total = X
C1+ XC2+ XC3 … + XCNCapacitive Reactance in a Parallel Circuit:
1/Z total = 1/ X
C1+ 1/ XC2+ 1/ XC3… + 1/ XCNCapacitor and Resistor in Series:
The analysis of
RCL circuit usually involves vectors or complex algebra. RC circuit'svector analysis only requires a straightforward application of the Pythagorean theorem to
derive the following formulae.
Z total = (X
C+ R2)1/2, I = V/ Z total,where V is the total voltage across RC circuit.The above equation is the result of vector addition using the Pythagorean theorem. Z
total is the hypotenuse, and vectors
XCand R form the two sides of the triangle.Capacitors in Series:
Capacitors in series result in a total capacitance less than the smallest capacitor in the
series circuit. Two 4-microfarad capacitors in series produce a total capacitance of 2
microfarad.
1/C
total= 1/C1+1/C2+1/C3… +1/CNCapacitors in Parallel:
Capacitors in parallel result in a total capacitance equal to the sum of the capacitor
values.
C
total= C1+C2+C3… CN18
Chapter 8
Inductive Reactance
Inductive Reactance:
Inductors block high frequencies and pass direct current. Inductive reactance is high for
high frequencies and low for low frequencies.
X
L = w*L,where w = 2*Pi*f or w = angular velocity.Inductive Reactance in Series Circuit:
Once you calculate impedance for a strictly capacitive circuit you can apply Ohm's law.
For example, let us consider a voltage divider consisting of three inductor L1, L2, and L3
in series. Total impedance (
Z total) of the series circuit is as follows:Z total = X
L1+ XL2+ XL3, I = V/ Z totalV = I(Z total) where V is the voltage across the three capacitors.
The voltage across each component is given by:
V1 = I X
L1,V2 = I XL2 , V3 = I XL3The general form of the equation is:
Z total = X
L1+ XL2+ XL3 … + XLNInductive Reactance in a Parallel Circuit:
1/Z total = 1/ X
L1+ 1/ XL2+ 1/ XL3… + 1/ XLNInductor and Resistor in Series:
The analysis of
RCL circuit usually involves vectors or complex algebra. RL circuit'svector analysis only requires a straightforward application of the Pythagorean to derive
the following formulae.
Z total = (X
L + R2)1/2, I = V/ Z total,where V is the total voltage across RLcircuit.
The above equation is the result of vector addition using the Pythagorean theorem. Z
total is the hypotenuse, and vectors
XLand R form the two sides of the triangle.Inductors in Parallel:
Inductors in parallel result in a total inductance less than the smallest inductor in the
series circuit. Two 4 milli-henry inductors in parallel produce an inductance of 2 millihenries.
1/L
total= 1/L1+1/L2+1/L3… +1/LNInductors in Series:
Inductors in series result in a total inductance equal to the sum of the inductor values.
L
total= L1+ L2+L3… LNInductor Physics:
Self-inductance is the ability of wire to produce an induced voltage in opposition to the
applied voltage. The inductance unit of measure is the Henry (symbol H). Typical values
are usually in the milli-henry (symbol mH) or micro-henry (
uH ) range. The longer awire the greater the inductance of the wire. Inductance effects power transmission lines,
communication lines, and antennas. Electronics circuits use inductors consisting of coils
of wire. For air core inductors inductance is proportional to the area of loop and the
number loops. Iron cores increase an inductor's inductance greatly.
Chapter 9
Resonance
The inductance reactance vector and the capacitive reactance vector point in the opposite
directions. Thus, when an inductor and capacitor are placed in series, their reactances
subtract from one and other. At high frequencies the series circuit will appear as an
inductive reactance. At low frequencies the inductor and capacitor series circuit will
appear as a capacitive reactance. At some in between frequency the inductance and the
capacitance will be equal. The sum of these vector quantities will be zero. This
frequency is called the resonant frequency.
w =
1/(LC)1/2where w = 2*Pif or simply w = angular velocity.Pi = 3.141
…,and f = frequency.Note that at frequency of about 700 kHz, the voltage at TP1 dips to 2.23 V. At 700 kHz the series
circuit formed by C1 and L1 exhibits resonance. The impedance of a resonant circuit is
low compared to Rs, and therefore a dip in signal amplitude occurs at TP1. At this
resonant frequency the current in the series circuit Rs/C1/L1 is maximum. Thus, the
voltage across L1 peaks to 16.29 volts. This is characteristic of resonance.
If you calculated the resonant frequency for L1 and C1 you would get 877 kHz, not the
700 kHz we observed in the bottom animation. That is because the second stage of this
filter effects the resonant frequency of the first stage. In the top animation the first stage
of the filter has been isolated. Note that the voltage at TP1 dips to 43 milli-volts at 877
kHz. The voltage at TP1 would dip even lower if L1 were an ideal inductor. L1 has a
resistance of .1 ohms. This prevents the impedance of L1/C1 series circuit from going to
zero at the resonant frequency.
21
Chapter 10
Filters
An electronic filter separates signals of different frequencies from one another. For
example, an audio signal could also be carrying a radio frequency (RF) signal on it. A
low-pass filter could be used to pass audio frequencies below 20 kHz and stop RF
signals, which are well above 20 kHz. A low-pass filter is also used to separate DC
power from AC ripple passed by the power source. A simple RC circuit can be used as a
filter. For a simple RC circuit the cutoff frequency can be calculated as 1/2*Pi*R*C (Pi =3.141…). Thus, we can relate cutoff frequency directly to the time constant (R*C) of a circuit. The product RC is
referred to as the time constant. A time constant is the time it takes a capacitor to charge
to 63% of DC voltage applied through a resistor. You can see the effect of a time
constant on a square wave in oscilloscope animation (center of Web page).
The cutoff frequency is the point that the amplitude of the output signal is half the value
of the input signal. Comparing input to output is the standard method of defining the
cutoff frequency of a filter. A second method is to simply compare the amplitude of the
signal output in the pass band to the signal output above the pass band. This second
method allows you to find the cutoff frequency with only a single meter or scope
connection. Simply connect probe to output of filter and adjust frequency roughly to
where you get highest output. This puts you in the band pass range of the filter. When
frequency is well in the pass band, you can vary the frequency up and down without
causing a change in the output amplitude. Next, increase or decrease the frequencies
rapidly until the output signal amplitude starts to drop of rapidly. Now adjust the
frequency carefully until the output voltage amplitude is halved. The power fall off rate is far greater than the 6 decibels per octave power fall off rate of a simple RC filter. This filter is designed for a fifty-ohm
signal source and a fifty-ohm load. Note that at high frequencies the filter acts almost like
a short. If you think of the filter as a short, the circuit reduces to a voltage divider
consisting of two fifty-ohm resistors. Thus, at all frequencies in the band pass the
resistors divide the voltage by a factor of two, which is a three decibel reduction in
voltage, and a six-decibel reduction in power. Note that at some frequencies the output is
greater than the input at test point one. This makes sense if you consider that the
capacitors and inductors form series resonant circuits. However, the coils pass low
frequencies, while the capacitor short circuit very high frequencies to ground. In the High
Pass filter the capacitors passed high frequencies and the coils shorted low frequencies to
ground. The Radio Frequency and Microwave filters that you are likely to encounter in
industry will be inside of metal containers. This type of construction prevents radiation
output and external radiation pickup. It also provides a defined ground plane for the filter
components. The metal container does not eliminate stray capacitance between
components and ground, but provides the filter designer with a well-defined ground plane
on which filter components can be mounted. A design engineer can consider the ground
plane of the metal container as constant rather variable that is subject to the placement of
other system components or wires. Encapsulation also ensures that once the filter is built,
touching any component cannot effect the tuning. Encapsulation also simplifies fault
isolation, because repair is accomplished simply by replacing the entire encapsulated
filter.
Chapter 11
Diodes
A simple pn-junction forms a diode. Current can flow through diode only in one
direction. If you think of the diode symbol as an arrowhead touching a vertical line, then
the arrowhead points in the direction of conventional current. I hope you recall that
conventional current flow is in the opposite direction of electrons. The arrowhead
represents the anode of the diode, and the vertical line represents the cathode of the
diode. The anode is composed of the p-type material and the cathode is made of the ntype
material. When the diode is reverse biased no current will flow, and the diode
appears open. When a diode is forward biased current flows freely through the diode.
The ideal forward biased diode acts as a short. Silicon diodes require .6V and a
germanium diode requires only .3V to be forward biased. An ideal diode can be viewed
as a switch that closes when forward biased and opens when reverse biased. This simple
view of a diode is adequate for qualitative analysis of circuits and is often all that is
required to troubleshoot a system. For the purpose of discussion we will assume a diode
is silicon unless otherwise specified.
PN-Junction Physics:
Pure silicon is an insulator. The silicon is doped with impurities to produce n-type and ptype
silicon. The doped silicon is a semiconductor. The n-type silicon is doped with a
metal such as Aluminum or Gallium. The n-type silicon has an excess of electrons in its
outer shell. Free electrons carry the current in n-type material. The p-type material is
doped with a none-metal such as phosphorous or arsenic. The p-type material is said to
have a shortage of electrons in its outer shell. In the p-type material current is carried by
positively charged holes. Note that having excess or a shortage of electrons describes the
outer band of an element and does not imply that the elements carry a static charge. For
those of you who have had High school chemistry recall the chapter on valance. You do
not need to understand valance or chemistry to understand diode or transistor electronics.
Just remember that we call the positive charge carriers in p-type silicon
holes and theexcess charge carriers in the n-type material free electrons. Both holes and free electrons
make the silicon more conductive than an insulator and less conductive than a metal.
What occurs at the junction of these semiconductors makes solid state diodes and
transistors possible.
Current in a diode consists of a combination of hole and free electron flow. In the n-type
material electrons move toward the PN junction. In the p-type material positively
charged holes move toward the junction. The holes move in the direction of the
arrowhead on the diode symbol. When the forward bias becomes strong enough to
overcome the barrier potential of the junction (referred to as depletion layer), then the
electrons from the valance band of the n-type materials can cross the depletion layer and
fill the holes in the p-type material resulting in current. Note that current consists of
electrons and positive holes moving towards each other. Note that the arrow on a diode
symbol points in direction of positive hole flow and conventional current flow. The
electrons move in the opposite direction of conventional current. You can learn more
about the physics of semiconductors from sites on our Reading List . Semiconductor physics is not required to troubleshoot or even to design andbuild a circuit; therefore, I will not spend much time on the solid state physics or quantum mechanics.
Types of Diodes:
The diode is the simplest of the semiconductor components. There are many diode types
with specific applications. The term " diode" by itself refers to a basic PN junction. The
basic PN junction diode is most frequently used for rectification. All other types of
diodes must be specified by type, such as Zener diode, Light Emitting Diode (LED), and
many others. The n-type material of the PN junction is called the cathode, and the p-type
material is called the anode. When the Anode is positive with respect to the cathode
current will flow. The rules for diode conduction can be simply stated as:
(1) The arrow must point toward the more negative terminal. In other words the
anode must be positive.
(2) The voltage difference across diode must exceed 0.3V for germanium and
0.7V for silicon. When forward biased, restriction of current by the depletion
area is eliminated and resistance of diode is reduced to its bulk resistance
(typically 25 Ohms).
(3) When the diode is reverse biased the depletion area is enlarged and the current
is reduced to near zero.
Diode Rectifier:
The transformer provides avoltage changing in polarity to the anodes of both diodes. Current can only flow in the
direction of diode arrows. The diode output current varies in amplitude, but it can only
flow in direction of arrows. Thus, the output of the circuit is called direct current or
rectified AC. The capacitor reduces the amplitude of the output variations. If the
capacitor is made large enough, the output will appear to be a constant DC voltage.
Observe the oscilloscope display of the rectifier output. The sine wave is the voltage at
TP1. The second trace displays the output of the rectifier. Note that the output exhibits a
peak voltage at both the maximum and minimum of the sine wave. The peak at the
minimum is the result of current through CR2, while the in phase peak is the result of
current flowing through CR1. This process is called rectification. The reduction of the
output fluctuation amplitude by the capacitor is called filtration. Those fluctuations not
filtered ride on the DC voltage and are called " ripple voltage". Note that the ripple
voltage displayed on scope is twice the frequency of the AC input.
Zener Diode:
When diodes are reverse biased, they ideally draw no current. Real diodes draw very
little current when reverse biased, because the depletion layer size increases as reverse
bias increases. This only holds true as long as the reverse bias is maintained below the
peak reverse voltage
(VRRM). When this voltage is exceeded, the diode will break downand conduct hard preventing voltage from going much above the V
RRM. This suddensurge in current is called the avalanche current. Avalanche current results from free
electrons bumping more electrons free of their molecular bonds, which in turn bump
more electrons free. This chain reaction or avalanche will destroy the PN junction in the
same manner that arcing would destroy an insulator in a cable or the dielectric of a
capacitor. Zener diodes are specially designed to withstand this avalanche current; thus,
their peak reverse voltage makes a good voltage regulator.
Thevoltage at TP2 is regulated at 7.4 V by the Zener diode. The current draw through the
220-ohm is resistor is:
VR2 =15V - 7.4V = 7.6V
IR2 = 7.6 V/ R2 = 7.6 V/ 220-ohms = 34.5- milli amps.
As long as less than 34 milli-amps is drawn through R2, the voltage at TP2 could be used
directly as a regulated voltage source. Adding the transistor results in a regulated current
output many times higher than 34 milli-amps. Note that the PN (Base Emitter) junction of
the transistor drops the voltage regulator output voltage by.7V to 6.77V. At first glance
you might think that the transistor has only added another diode to the Regulated Power
Supply. In one respect it has, some current does flow from TP2 through the transistor
base emitter junction to the Rload. The difference between a transistor and a pair of
diodes is power amplification. When the base emitter PN junction is forward biased and
conducts a small amount of current, a large current flows from TP1 to Rload via the
collector to emitter current flow. This process of controlling a lot of power with a little
power is called Amplification.
Varactor Diode:
A Varactor diode is designed to exhibit considerable capacitance when reversed biased.
In this state it like a polarized capacitor in that it does not conduct direct current. The ptype
material constitutes the negative plate of the capacitor, and n-type material
constitutes the positive plate of the capacitor. The depletion region of the diode is
depleted of charge carriers; therefore it acts as an insulator or the dielectric of the
capacitor. As negative bias is increased, the depletion region becomes greater and the
distance between the two plates is effectively increased. Capacitance decreases as
distance between plates increase. Thus, the capacitance decreases as reverse bias
increases. This provides a way of varying capacitance with voltage. This means a circuit
designer can use a Varactor to vary capacitance electronically. When this capacitor is part
of a tuned radio frequency oscillator, the frequency of oscillator can be controlled
electronically. If the inverse bias voltage were varied with an audio signal, the radio
frequency oscillator frequency would vary with the amplitude of the audio frequency.
This is called frequency modulation or simply FM.
26
Chapter 12
Transistors
A transistor can be viewed as two diodes back to back. The base collector junction
forms one diode, and the base emitter junction forms the other. In a circuit the transistor
base collector PN junction should always be reverse biased. Ideally no current will flow
between the base and collector. When the base emitter is reverse biased, no current will
flow from the base to emitter, and the transistor will be off (no conduction). When the
base emitter is forward biased, current will flow from the base to emitter, and the
transistor will be on (conductive). Current will not only flow between base and emitter,
but a much greater current will flow from collector to emitter. Current is amplified, and
voltage is not attenuated; therefore, power is also amplified.
The n-type collector should be positive with respect to both the n-type emitter and the p-type base. Ideally no current will flow from the base to collector, because the collector/base pn-junction is reverse biased.
Observe animation! Note that the n-type collector of the transistor is connected to + 15V.
The base is connected to ground. This arrangement assures that the collector/base pnjunction
is reverse biased. The base/emitter pn-junction controls transistor operation. A
negative voltage is applied to emitter to forward bias the base/emitter pn-junction. The
current meters indicate a rapid increase of base current as the bias goes from .6 V to .8 V.
The meters also indicate that the collector current also increases and contributes 99% of
the emitter current. A small base current is controlling a large collector current. The
current gain is well over a hundred. This clearly indicates that forward biasing of
base/emitter junction results in collector to emitter current flow. Note, that this collector
to emitter is n-type to n-type current flow. This current flow cannot be explained with the
simple concept of back to back diodes. This current amplification is what makes the
transistor different than back to back diodes. A simple explanation of this is that the
forward bias reduces the depletion area, and therefore current can flow. The actual
physics is not simple.Physics is not needed to troubleshoot a circuit or design a transistor circuit. Any
explanation I would give would be to complex for those who have not had physics and
chemistry, or it would appear lame to those who know solid state physics and quantum
mechanics. I will teach transistor circuit operation and troubleshooting using
simulations. We will study a transistor voltage regulator in chapter 14 and a transistor
amplifier in chapter 15.
Chapter 13
Transistor Regulated Power Supply
A power supply consists of an unregulated power source and a voltage regulator. The unregulated voltage must be
higher than the desired regulated voltage to maintain proper bias for the series regulator
transistor. Simply stated a series regulator reduces input voltage to the desired regulated
output voltage. In this simulation, I used a battery for the unregulated input. Since a
battery outputs only DC, I varied the load to generate changing current. This changing
current resulted in a changing voltage drop across the internal resistance of battery, Rs.
This enables simulation to show how regulator adjusts to accommodate a varying load.
Feedback is the key to regulator operation. Assume output voltage goes lower for any
reason. Then, the lower voltage is picked off by the potentiometer, and this voltage is
compared to the Zener reference via the base/emitter junction of the transistor. The lower
voltage causes transistor to conduct less; thus, the voltage at the transistor collector goes
higher. This higher voltage causes transistor T2 and T1 to conduct harder, and raises
output voltage.
Observe baseline operation of regulated power supply below. Then troubleshoot fault number one
to fault number four below. This voltage regulator simulation uses a battery rather than a
rectifier as a power source. Since batteries do not generate ripple voltage, I provided a
dynamic load to demonstrate that voltage regulators not only adjust for slow changes in
current but also can attenuate ripple voltages. They also adjust for rapid changes in load
current demands. The frequencies I used were 50 hertz, 60, 400, and 1200 hertz
corresponding to European, American, single-phase 400 hertz and three phase 400 hertz
ripple frequencies respectively. Thus, it illustrates the effect of the regulator on ripple
voltage or rapid load changes.
28
Chapter 14
Transistor Amplifiers
The first two frames of animation show DC levels. The lather frames show small signal amplification over a range of
frequencies. The amplifier consists of three transistors with a common 12V (Vcc =12V)
power source. A small signal input of 1 milli-volt is amplified to as much as 1 volt at 100
hz. Most of the current through Q1 and Q2 flows from collector to emitter. Only a very
small amount of the current passes from the base to the emitter. Therefore, the same
current flows through both the emitter and collector resistor. Likewise, a change in
transistor conduction results in a small change in both the collector and emitter current,
which we will refer to as delta I. The change in collector voltage is: deltaVc = delta I *
Rc. The change in emitter voltage is: deltaVE= delta I * RE.It follows that the ratio of
deltaVc/ deltaV
E= Rc/ RE. For small signal variations the emitter voltage tracks thebase voltage,
delta VB = delta VE . As derived below Voltage Gain = Rc/ RE .delta Vc/ delta V
E= Rc/ REdelta V
B= delta VEdelta Vc/ delta V
B= Rc/ REdelta Vc/ delta V
B= Voltage Gain by definition.Voltage Gain = Rc/ R
EIn the first amplifier stage, the ratio of collector resistance to emitter resistance is 110
[22,000/200 = 110]. The animation shows us that at 100 Hz the voltage gain is about 90.
The difference can be explained by other factors such as loading by the next stage. The
transistor parameters will effect actual voltage output. The ratio
RC/ RE is actually themaximum DC or low frequency gain that can be achieved by this transistor circuit. It is a
handy indicator of how the circuit is intended to operate. When troubleshooting this
calculation can be performed in ones head; thus, you don't have to waste time putting
down a meter or scope probe in order to pick up a pencil or a calculator.
I set the operating point of the Q1 circuit at 2.7V by selecting a base resistance of 4.4-M
(4,400,000.00-ohm) for the resistor in the base circuit of Q1. The Q1 operating point is
not centered. Since only small signals are being amplified by the first stage, centering of
operating point is not critical. There is little chance that the 1 milli-volt-input signal will
drive the transistor into saturation or cutoff.
In the Amplifier second stage, Q2's operating point is better centered. The ratio
RC/REis5.2 [12000/2300 = 5.217]. Observing the animation at 100 Hz, we see that the voltage
gain is 5.38 [481.68/89.5 = 5.38]
.Thus, the gain at low frequency is determined by theratio of the collector resistance to the emitter resistance
.The slightly higher actual gain isthe result of some AC signal being bypassed by the 100
uemitter capacitor.At higher frequencies, the emitter impedance of Q2 is reduced to near zero, and the gain
of transistor Q2 increases until limited by the transistor's band pass. The emitter bypass
capacitor in the first and second stages of the amplifier increases the band pass of the
amplifier. The voltage across the emitter resistors decreases the base emitter forward bias
of the transistor. Thus, the voltage across the emitter resistor is a form of negative
feedback. The more current driven through the emitter resistors, the greater the negative
feedback. Looking at the animation we see that the emitter voltages (feedback) decrease
as frequency increases. This results in the gain of amplifier increasing as frequency
increases through the 10 kHz and 100kHz range. At 1 MHz and above transistor gain
drops off drastically due to internal transistor capacitances. The emitter bypass capacitors
still extend the band pass of the amplifier well past 1 MHz.
The third stage of the amplifier is called an emitter follower. The voltage gain of the Q3
circuit is unity and current gain is very large. This third stage gives the three-stage
amplifier low output impedance and high current gain. The first two stages give the
amplifier voltage gain that exceeds 100. Thus, the amplifier exhibits both high voltage
and current gain. Or simply overall high power gain is P = I*V *(cosine A), where A is
the phase angle between current and voltage. At low frequency
cosine A goes to one,and power formulae reduces to P = I*V. Note the circuit does not generate power.
Signal output power increases by several orders of magnitude with respect to signal input
power, but the signal output derives its energy from the DC supply voltage. Only a
fraction of the DC power is transmitted to next stage of system. The remainder of power
is converted to heat.
30
Transistor Circuit Troubleshooting:
To troubleshoot transistor circuits all that is required is a full understanding of Ohm's law
and a little understanding of semiconductors. First measure the voltage drop across
RCand R
E,and than calculate the current across each transistor using Ohm's law. If they areclose to equal you can conclude
RCand REare good. Thus, the fault must lie in thetransistor or base-biasing resistor. This method only works when the transistor is
conducting, because when the transistor is not conducting there is zero voltage drop
across
RCand RE.Next, consider the case where the transistor is in cutoff (V
CE= VCC). In this case thatsimply means the collector voltage reads 12V. With no current flowing through
RC,theratio of voltage drops across
RCand REare meaningless; however, we can analyze thebase/emitter circuit. In the absence of collector current, we have a series circuit
consisting of
RB,a base/emitter pn-junction, and RE. The same current flows through RBand R
E. If the voltage drops across RBand REare proportional to their specified values,than
RBand REare good, else one of the resistors is defective. The transistor pn-junctionshould exhibit a voltage drop of about .7V. If it exhibited 0 volts that would indicate that
the pn-junction was shorted. If the voltage across the pn-junction were far above .7V,
that would indicate an open junction. Either way the transistor would be bad.
If current is flowing through
RBand REvia the base emitter junction, than the transistorshould be conducting. If base/emitter current does not induce collector current, then the
transistor is defective.
Troubleshooting Frequency Response Problems:
If this amplifier worked well at low frequency but exhibited to little gain at high
frequency, one would expect that a capacitor or a transistor was at fault. At a frequency
of 10 kHz you expect that all bypass and coupling capacitors would exhibit very little
voltage drop. The bypass capacitors would exhibit near zero AC signals. The coupling
capacitors (capacitors between stages) should exhibit nearly the same voltage at both
ends of capacitor. That is to say that no significant attenuation should be observed across
any of the coupling capacitors at frequencies above the lower band pass limit. A bad
capacitor will exhibit far greater voltage drop across it than a good capacitor would. If
frequency response problem does not localize to a capacitor, than a transistor probably
has poor frequency response.
The point that I am trying make here is that simple network analysis is often all that is
required to troubleshoot transistor circuits. My Web sites include many troubleshooting
exercises. Use them to practice fault isolation. You will see that Ohm's law and simple
network analysis leads to fault isolation. This is not only true for transistor circuits but
also applies to integrated circuits such as operational amplifiers.
Chapter 15
Differential Amplifiers
Differential Amplifiers:
A differential amplifier has two ungrounded inputs. Its output is proportional to the
difference between the two inputs. This means that when both inputs are zero volts the
output should be zero, and if both inputs are 6V the output should still be zero. In an
ideal differential amplifier no output is produced when the input signals are identical;
only the difference in input signals are amplified. Signals common to both inputs are
known as common mode signals. Sometimes common mode signals could be a DC level
from a bridge type sensor, or they could be radio frequency noise riding on both input
lines. The ability of a differential amplifier to prevent conversion of a common-mode
signal to an output signal is referred to as the common mode rejection ratio (CMRR).
CMRR is the ratio of normal mode gain to common mode gain. If differential amplifier
inputs were 0.0V and 0.1V and an output was 1.40 V, its normal mode gain would be 14.
If both inputs changed from 0.00V to 0.10V and output changed from .000V to .001V,
then the common mode gain would be
.10. The CMRR would be 140 [CMRR = 14/ .1 =140].
Click number 18!
The schematic shows a Whetstone Bridge connected to a differentialamplifier. The technician balances the bridge so that the differential output of the bridge
is zero. When the bridge is balanced, both outputs of the bridge are 2.50 volts. When the
Whetstone Bridge is balanced the value of Rx can be calculated from the value that Rs is
set to. Rx/Rs = 1000/200 reduces to Rx = 5 Rs. In the past scientists would test the
bridge for balance with a sensitive Galvanometer. When the bridge was balanced the
galvanometer would not deflect because no current would flow through galvanometer.
Since the galvanometer draws no current, it does not effect precision of the bridge. Here,
the differential amplifier replaces the galvanometer as the bridge interface.
The differential amplifier also allows us to measure only the difference voltage output of
the bridge. The common mode rejection of the amplifier makes the bridge operation
independent of the supply voltage. Differential amplifiers consist of a constant current
source feeding the emitters of a matched pair of transistors. If the voltage at the bases of
matched transistor pair increase an equal amount, the current through each transistor does
not increase. The current through both transistors cannot decrease or increase in unison,
because both transistors are fed by a single constant current source. Since the current
does not change, there is no voltage change at the collector output of the first stage
differential amplifier. This gives it low (ideally zero) common mode gain. If the voltage
increases on one of the differential inputs, the current will increase through that
transistor. Since the current to the transistor pair is constant, the current to the other
transistor decreases. This results in high normal mode gain.
Chapter 16
Operational Amplifiers
Operational amplifiers are very high gain, DC differential amplifiers. Their circuits are
extremely complex and have many transistors, resistors and other discrete components.
Fortunately, the development of integrated circuits simplified electronic system design
and troubleshooting. The triangular symbols represent operational amplifiers. Integrated circuits (IC's) can contain
thousands of discrete components in a single package. One integrated circuit can contain
several operational amplifiers. Frequently the schematics of large systems show only
signal input and output pins. Other inputs such as power and ground are listed on separate
tables.
Open loop gain (gain with no feedback) of most operational amplifiers exceeds 200,000.
Thus, if there was as much as one milli-volt difference in the input voltages, the output
tries to slew to over 200 volts. Since typical supply voltages for operational amplifiers
are in the +15 to -15 volt range, less than one milli-volt would drive the amplifiers into
saturation. Therefore, these high gain amplifiers require negative feedback to stabilize
their operation. This negative feedback determines the closed loop gain of amplifier.
The feedback resistor current forces voltage between IC inputs below one milli-volt. For
an ideal operational amplifier the gain approaches infinity. The input impedance also
approaches infinity, which implies that the ideal operational draws no current. Real
operational amplifiers have gains far above 200,000 and input impedance well above four
million ohms. These parameters result in operational amplifier gain being a function of
the feedback resistors. Only Ohm's law is required to calculate their gain.
Since one input is connected to ground, the other input is forced to zero volts via feedback resistor. When
the input is 2 mV, the circuit input current is [2mV/1k-ohm = 2 uA]. The operational
amplifier input draws no current, thus the only path for the 2uA current flow is the 100kohm
feedback resistor. Applying Ohm's law to calculate the output voltage indicates the
output should be negative 400mV [2uA*100k-ohm = 400mV]. That is to say that the
output must be -400mV, in order to maintain the voltage at the inverting input at zero. If
the voltage at the output drifts a little bit negative of 400mV, than in accordance with
Ohm's law the inverting input node would also go negative. This negative voltage at the
inverting node would drive the output of operational amplifier positive, back towards
negative 400mV.
Remember that the inputs of operational amplifiers need not be maintained at zero volts,
but the difference between inputs must be maintained at zero volts. In the lower
animation both operational amplifier inputs are forced to level of the signal input. The
non-inverting input is connected to signal. Feedback forces the inverting input to same
level as the non-inverting input. The output can than be calculated with Ohm's Law. Try
it and compare your results with values displayed by the animation.
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