Introduction to Engineering
Walnut Hills High School
Pre-lab Assignment for Camera Lab 4
What You Have To Do Before Lab 4
1. Read the material for Camera Lab 4. If you like, you can even visit a library
or use the web to find out more about these things.
2. Perform the pre-lab assignment described in the Camera Lab 4 material.
The pre-lab tasks are to be performed and turned in by your team. You will
NOT have time to do it at the beginning of Lab 4, but will be given some
class time to work on the assignments.
Note: The pre-lab assignment is not a lab report. It does not have to follow the
lab report format and may be handwritten.
What You Have To Do After Lab 4
1. Analyze the results of the measurements you performed in Lab 4.
2. Prepare a team report on Lab 4. You will be given a set of instructions that
are specific to the Lab 4 report. Follow these instructions when writing your
report.
OSU College of Engineering
ENG 181 - Introduction to Engineering I
Camera Lab 4 - The Flash Circuit
Table of Contents
Table of Contents................................................................................................ 2
Camera Lab 4 - The Flash Circuit ...................................................................... 3
1
Introduction ................................................................................................ 3
1.1 Objectives of Lab 4 ................................................................................. 3
2
Background ................................................................................................ 4
2.1 The Core and Auxiliary Functions of the Flash Circuit ............................ 4
2.1.1 The Battery and the Flash Tube ...................................................... 4
2.1.2 The Core Functions ......................................................................... 5
2.1.3 Auxiliary Functions .......................................................................... 6
2.2 A Few Basic Concepts for Electronic Circuits ......................................... 7
2.2.1 Objective ......................................................................................... 7
2.2.2 Electric Current ................................................................................ 7
2.2.3 Voltage ............................................................................................ 8
2.2.4 Resistors ......................................................................................... 8
2.2.5 Capacitors ..................................................................................... 10
2.2.6 Series and Parallel Combinations of Circuit Elements................... 12
2.2.7 Other Circuit Components ............................................................. 17
2.3 The Circuit Board and Circuit Schematic .............................................. 19
2.4 What's an Oscilloscope? ...................................................................... 22
3
Pre-lab 4 Assignments ............................................................................ 24
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Introduction to Engineering
Walnut Hills High School
Camera Lab 4 - The Flash Circuit
1 Introduction
This handout contains:
 A description of the core and auxiliary functions performed by the flash circuit
in the Kodak Max Flash camera
 A discussion of the circuit board and circuit schematic
 A discussion of some basic circuits concepts including a description of the
components of the flash circuit on which you will be performing
measurements.
 A discussion of how the flash circuit performs the core functions
 The Pre-lab Assignment you are to perform before coming to Lab 4
 An outline of the measurements you will perform in Lab 4
You should read this entire document and complete the pre-lab assignments
before coming to Lab 4 as part of your preparation for lab.
1.1 Objectives of Lab 4
The Kodak Max Flash camera contains a circuit for operating the flash.
We do not expect you to become an instant expert on the operation of this circuit.
We do expect you to:
 Develop an understanding of the distinction between the core functions and
auxiliary functions of the flash circuit and acquire a "big picture" view of what
the circuit must do to perform the core functions.
 Acquire experience with the relationship between the actual circuit board in
the camera and the "schematic" representation of the circuit.
 Acquire experience in using a voltmeter and an oscilloscope to measure
voltages in the circuit.
 Be able to calculate currents flowing in certain parts of the circuit using the
measured voltages together with the values of some of the components.
 Develop an understanding of how and why the speeds of various parts of the
flash circuit are related to the speeds of other parts of the camera.
The components and operation of some parts of the circuit are more complex
than others are. Understanding the operation of the more complex sections
requires more background than we will be able to give you in this quick touch on
electronic circuits. These more complex parts of the circuit will receive only a
cursory treatment in this laboratory.
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2 Background
2.1
The Core and Auxiliary Functions of the Flash Circuit
2.1.1 The Battery and the Flash Tube
The overall objective that the flash circuit must meet is to provide a short
flash of light that is synchronized with the shutter of the camera. This is basically
a problem in synchronized energy conversion that engineers solved with the
flash circuit. Before we can look at the core functions of the camera circuit we
must take a brief look at the battery at the input end of the system and the flash
lamp at the output.
The energy for the flash of light comes from the AA battery in the camera.
The battery is basically an energy storage and conversion device. It stores
energy in chemical form and converts it to electrical form.
In general, a battery makes its electrical energy available when the user
"completes a circuit" between the battery's positive and negative terminals. In
this case to "complete the circuit" means to provide a path for electrical current to
flow out of the positive terminal of the battery, through the device that the battery
is powering, and back to the negative terminal of the battery. The term "electrical
current" refers to the motion of electrons through the wires and components of
the circuit. More will be said about current in the Basic Concepts section of this
document. Note that the current eventually comes back to where it started. This
is the origin of the term "circuit." In general, current will not flow unless such a
closed path exists.
The AA battery is a 1.5 volt battery. It has a voltage difference of 1.5 volts
between its positive and negative terminals. The battery supplies current to the
circuit at an electric potential, or voltage, of 1.5 volts. (For a discussion and
definitions of voltage, electric potential and current see the Basic Concepts
section.) The current that returns to the battery comes back at an electric
potential that is 1.5 volts lower than when it left the battery. This difference is
related to the energy delivered to the components that the current passed
through in the circuit.
The flash lamp in the camera is also an energy conversion device. It
converts electrical energy into optical energy. The flash lamp is very different
than a typical incandescent light bulb since flash photography requires a very fast
and bright light pulse.
The flash lamp has the form of a sealed glass tube with an electrode at
each end. The flash tube requires a voltage difference of several hundred volts
across the two electrodes for it to operate. However, it is not sufficient to merely
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apply this voltage. Unlike an incandescent light bulb, the flash tube does not
have a wire inside it between the two electrodes. Rather, the tube is filled with a
gas. Under normal conditions the gas does not provide a path for current flow
from one electrode to the other. That is, under normal conditions the gas is an
"open circuit" and has a resistance to current flow that is so large we can
consider it to be infinite for most practical purposes. It is not our purpose to study
the physics of what happens inside the flash tube during a flash. It will suffice to
say that under the right conditions the gas in the tube can be made to break
down and form a glow discharge. In the glow discharge some of the atoms have
one electron removed. The atoms that are short one electron have a positive
charge and are known as ions. The voltage applied between the electrodes of
the lamp cause the negative electrons and positive ions to move (in opposite
directions), and moving charge is current. As a result the glow discharge has a
relatively small resistance to current flow. Light is emitted as the electrons relax
back to their normal locations on the atoms.
The right condition to trigger the flash is a voltage pulse applied to a third
electrode (a trigger electrode) placed very close to the side of the flash tube. The
metal mirror, or reflector, behind the flash tube has double duty. In addition to
reflecting the light that comes out of the back of the lamp toward the front of the
camera, it serves as the trigger electrode. The required trigger voltage pulse is
more than a thousand volts and causes the gas in the flash tube to break down.
Once the flash tube is triggered the glow discharge in the tube provides a low
resistance path for current to flow between the electrodes at the two ends of the
tube.
The trigger voltage is only needed to start the discharge. Once the glow
discharge is started the voltage applied across the tube (the capacitor voltage)
will maintain the glow discharge for a time. The current flow will persist as long
as the voltage across the two electrodes is above some minimum value (typically
a few tens of volts).
2.1.2 The Core Functions
It should now be apparent that there is a mismatch between the battery
and the flash tube. The battery provides a voltage of 1.5 volts, but the flash tube
needs much larger voltages. The flash circuit provides the interface between the
battery and the flash tube. It must take what the battery supplies and convert it
into something that is useable by the flash tube.
While it may be possible in principle to design a circuit that will directly
match the battery to the flash tube as soon as the shutter is triggered, such a
circuit would definitely be expensive compared to the low cost of the camera. It
would probably also be too large to fit inside the existing camera box. A circuit
that can accomplish the required goals can be made economical and small
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enough to fit in the camera by breaking the overall process into two parts. These
two parts are the core functions of the flash circuit.
The first core function (charging) starts when the user presses the
charging button on the front of the camera. As a result, chemical potential
energy stored in the battery is delivered to the circuit as a relatively large current
at only 1.5 volts. The circuit steps the voltage up to a much larger value, but the
process of increasing the voltage necessarily decreases the current. The small
current at larger voltage is delivered to a large capacitor in the circuit. As a result
of the small current the capacitor slowly "charges up" to close to 350 volts. This
takes several seconds, much longer than the time the shutter is opened, so it
must be done before the picture is actually taken (In Lab 4 you will measure how
long this takes). The capacitor is used to temporarily store the electrical energy
delivered from the battery through the circuit, but now the energy is at a large
enough voltage to be useful for the flash lamp. The flash tube is connected "in
parallel" (see section 2.2.6 for an explanation of the usage of parallel here) with
the capacitor, so the voltage on the capacitor also appears across the flash tube.
However, since the flash tube normally has very high resistance, it does not flash
yet.
The second core function (trigger) starts when the shutter is opened in the
process of taking a picture. The circuit uses a small fraction of the stored
electrical energy to generate a very quick flash trigger pulse to initiate the glow
discharge in the flash tube. Once the glow discharge is started the flash tube
temporarily has a very low resistance, allowing the energy stored in the capacitor
to quickly discharge through the flash tube, where much of it is converted from
electrical energy into optical energy. (The remainder is converted into waste
heat.) This happens very quickly, while the shutter is open (In Lab 4 you will
measure just how fast these processes are).
2.1.3 Auxiliary Functions
The circuit performs a few convenience functions in addition to the two
core functions. One is to illuminate a pilot light to let the user know when the
circuit is charged and ready for flash photography. The charge stored on the
capacitor slowly leaks off. If the amount of stored charge becomes too low, the
flash will be too weak or may not happen at all. The neon light goes out if the
voltage across the flash lamp becomes too small. This lets the user know they
need to restart the charging circuit before taking a flash picture.
The two core functions and the pilot light were the only functions
performed by the flash circuit in the first generation of the Max Flash camera.
The user of the first generation camera had to press and hold the charging button
until the pilot light turned on. A flash photograph could then be taken within a few
minutes (before the pilot light went out). In order to take another flash
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photograph the user had to press and hold the charging button until the pilot light
came on again.
The flash circuit in the current generation of the MAX Flash camera has a
few additional functions. First, the user only has to momentarily push the charge
button to start the charging cycle. The circuit latches into an "on state" and
keeps charging the capacitor until the flash capacitor is fully charged. Since the
charging circuit is latched on, additional modifications had to be added to
automatically unlatch it after the capacitor is fully charged. Otherwise the battery
would be quickly drained.
Another convenience feature that was added causes the flash circuit to
automatically recharge after a flash photograph has been taken. In this way the
user doesn't have to remember to press the charge switch between taking flash
photographs.
2.2
A Few Basic Concepts for Electronic Circuits
2.2.1 Objective
The objective of this section is to provide you with an introduction to some
aspects of electronic circuits enabling you to perform and understand
measurements characterizing the camera flash circuit. Some terms are defined,
the electrical characteristics of some components of the flash circuit are
discussed and an overview of some aspects of circuits is presented.
2.2.2 Electric Current
You may have an intuitive grasp that when we speak of an electric current
we are referring to the motion of electronic charges. In this discussion we will
confine ourselves to currents flowing through wires or electronic components.
Charge is measured in units of coulombs. The current is the amount of charge
that moves past a location on the wire per unit time (coulomb/second = ampere).
(Or the charge that moves into or out of a lead of a component per unit time.)
We can conceive of moving positive or negative charges. Historically, the
existence of electronic current was known before it was known what physical
object was actually moving. A convention was established in which positive
current flowed from point of higher electric potential to points of lower electric
potential. However, they got the convention backward! It was eventually
discovered that the particle moving in metal wires was the electron and that it
moved in a direction opposite to the established convention. The convention was
so well established that it wasn’t changed. Therefore a positive current flowing in
a wire actually corresponds to the motion of negatively charged electrons moving
in the opposite direction.
By the way, an electron has a charge of
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-1.6  10-19 coulomb. A current of 1 Amp = 1 coulomb/sec therefore corresponds
to a very large number (~ 6,250,000,000,000,000,000) electrons moving past a
location on the wire each second.
2.2.3 Voltage
Voltage is related to the force that causes current to flow. (Important: pay
attention to the wording in the previous sentence, in particular to "related to."
While voltage is related to the force that causes charges to move, it is not the
force itself.) An analogy that you may find useful is water flowing in a garden
hose. (It's not a perfect analogy but ignore that for now.) The current of water is
analogous to the electric current in a wire and the pressure that causes the water
to flow is analogous to the voltage.
When someone quotes a number for pressure in a hose or air pressure in
a tire, that pressure is always stated relative to some reference. The pressure
quoted is actually the pressure difference between the thing being measured and
the reference. A common reference is the atmospheric pressure of the
surrounding air, in which case the pressure is known as gage pressure (in units
of psig - pounds per square inch gage - for example). Another reference that is
sometimes used is vacuum, in which case the pressure is known as the absolute
pressure (psia).
In a similar fashion, when a number for a voltage is stated, it is actually the
difference in voltage (or difference in electric potential) between two different
points. Voltages are always stated relative to some reference. A common
reference that is often used is earth ground. That is the reference used by
electrical utilities for electric power generation and distribution to homes and
industry. The 120 V supplied to your residence by the electric company is
measured relative to earth ground. Some circuits are isolated from earth ground,
and in those cases a reference point in that circuit may be picked. This is often
also called the "ground" for that circuit, although it may not actually be connected
to earth ground in any way.
Another way in which voltage is used is to describe the difference in
electric potential across an electronic component. When we speak of the voltage
across a component we are measuring the voltage on one lead (wire) of the
component while using another lead of the same component as the reference
point. (In the water hose analogy we could talk about the pressure difference
across a device, such as a filter, through which the water is flowing.)
2.2.4 Resistors
A resistor is an electronic device that resists the flow of current and obeys Ohm's
law. Ohm's law states that the voltage across a resistance is proportional to the
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current through it. (Note the emphasis on the words "across" for voltage and
"through" for current.)
Ohm's law can be stated mathematically as
V  IR
(Ohm's law)
(Equation 1)
or in words:
(voltage across resistor) = (current through resistor)  (value of resistance)
where V stands for the voltage across the resistor, R stands for the resistance of
the resistor, and I stands for the current through the resistor. The unit of
resistance is the ohm (from
Table 1, ohm = volt/ampere, which agrees with the equation). The symbol for a
resistor is shown in Figure 1. Also shown is the polarity of the voltage drop
across the resistor if the current is flowing in the direction indicated by the arrow.
+
V
-
I
Figure 1: Symbol of the
resistor.
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EXAMPLE: A 10 k resistor that has 10 volts
across it will have a current of I =
10 V/10 k = 1.0 mA flowing through it.
EXAMPLE: If the same resistor had 0.5 mA
flowing through it the voltage across it would
be V = (0.5 mA)(10 k) = 5 volts.
Camera Lab 4 - Page 9
2.2.5 Capacitors
A capacitor is an electronic device that stores a charge and can be
discharged very quickly when the electricity is needed; for example, to set off a
flash for a camera. The more charge you want to store, the bigger the capacitor
you need. Capacitors come in standard sizes, and when you buy one, you
specify the size you want by giving the capacitance. The unit of capacitance is
farads (F) or microfarads (F). (µ=10-6) See tables 1 and 2 for details about
units.
A capacitor consists of two metal plates separated from each other by a
thin insulator. Since an insulator separates the two plates a steady state DC
current cannot flow through a capacitor. On the other hand, if a positive current
is applied to one plate of the capacitor a positive charge will build up on that
plate. Since like charges repel each other and opposite charges attract, this
positive charge will force an equal amount of positive charge off of the other
plate, inducing a net negative charge of equal magnitude on it. There will be an
electric field between the positive and negative charges, and a voltage difference
between the two plates, and hence across the capacitor. The voltage difference
across the capacitor is proportional to the charge on the capacitor.
In equation form:
V 
Q
C
or
Q  CV
(Equation 2)
where V is the voltage (volts) across the capacitor, Q is the charge (coulombs)
on the capacitor, and C is the capacitance (farads) of the capacitor. Restating
this with words
(voltage across capacitor) = (charge on capacitor) / (value of capacitor)
or
(charge on capacitor) = (value of capacitor)  (voltage across capacitor).
Capacitance is measured in farads (farad = coulomb/volt). The symbol for a
capacitor is shown in Figure 2.
EXAMPLE: In order to charge the 120 F capacitor in the flash circuit up to
300 V we would have to add a charge of 0.036 coulombs [= (120 F)  (300 V)]
to the positive plate of the capacitor.
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+
V
-
If a current (I) is flowing into a capacitor the
charge on the capacitor has to be changing.
Mathematically
I
I
dQ CdV

.
dt
dt
(Equation 3)
Note: dt means “change in t.” It is similar to
t.
Figure 2: Symbol of the
capacitor.
Since Q  CV where C is a constant, dQ=CdV.
CdV
I dV

So I 
or
dt
C dt
In words:
(current entering capacitor)
= (time rate of change of charge on capacitor)
= (value of capacitance)  (time rate of change of voltage across
capacitor)
dQ
dV
is the time rate of change of the charge.
is the time rate of change of
dt
dt
voltage.
For example, if you force a constant current of 1 mA onto the discharged
120 F capacitor the voltage across the capacitor will change at a rate of
dV I
1 mA
10 3 coulomb/se c
 

 8.33 volt/sec
dt C 120 F 120  10 6 coulomb/vo lt
The current flowing into a capacitor is proportional to the time rate of
change of the voltage across the capacitor. If the voltage is not changing, the
rate of change (derivative) with respect to time will be zero, and the current will
therefore be zero. Conversely, if a current is flowing in or out from the capacitor,
the voltage across it must be changing.
A charged capacitor stores electrical potential energy via the electric field
inside it. The amount of electrical potential energy stored in a charged capacitor
can be expressed as
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U
1 Q2 1
 CV 2 .
2 C
2
(Equation 4)
In words:
(electric potential energy)
= (1/2)  (charge on capacitor)2 / (value of capacitor)
= (1/2)  (value of capacitor)  (voltage across capacitor)2
The stored electrical energy can be released by allowing the capacitor to
discharge through a circuit attached across it. The energy stored on a 120 F
capacitor charged to 300 V is (1/2)  (120 F)  (300 V)2 = 5.4 Joules. In Lab 2
you measured the speed of the flash. You probably saw that it was on the order
of a few milliseconds long. If we could completely discharge that capacitor in 1
ms, the average power leaving the capacitor during that 1 ms would be 5.4 J/ 1
ms = 5,400 watts! The reason I stated this is the average power is that the
energy does not leave the capacitor at a uniform rate during the flash discharge.
The energy leaves the capacitor faster at the beginning of the discharge so the
peak power is even larger! The flash capacitor delivers power on the order of
kilowatts to the flash lamp (But fortunately only for a very short time, otherwise
components of the flash circuit would probably melt!).
2.2.6 Series and Parallel Combinations of Circuit Elements
Circuits are formed by combining electronic devices in such a way that
current can flow from one terminal of an electrical energy source, through the
devices, and back to the other terminal of the energy source. The emphasis here
is on the "circular" nature of the current flow, in that the current must be able to
return to the energy source. This is the origin of the term "circuit" and the
phrases "completing the circuit" or "closed circuit." If a path does not exist for the
current to return to the energy source there is an "open circuit" and current will
not flow out of the energy source.
.2.2.6.1
Series Circuit
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


Figure 3: A simple series circuit
example.
We will begin by looking at the
example of three resistors with a power
supply connected as shown in Figure 3.
The positive direction of current flow will be
out of the positive terminal of the 5 V
supply and then through the 10 K
2.2 K and 4.7 K (K=x103) resistors and
finally back to the negative terminal (not
labeled) of the 5 V supply. Since the same
current must flow through all of them, the
components are said to be connected in
series, and this is a series circuit.
IMPORTANT: Although this example only
has resistors in it the concept of series
current flow applies to other components
as well. If two components are connected
in series with each other then the current
flowing through them must be equal.
Any point in a circuit can only have a single voltage at any instant of time.
That necessarily means that the voltage at any point in the circuit is independent
of the path to that point from the reference point. For the circuit in this example
let’s take the “bottom” of the circuit as the reference point (we define V = 0 there).
If a clockwise path from the reference is followed the path goes through the 5 volt
power supply. That means the voltage at the node between the 5 volt supply and
the 10K resistor must be 5 volts.
Note: Most people think of a “node” as a point. In this document, the
word “node,” refers to the portion of the circuit between two components in
the circuit. The voltage at any point between those two components will
be the same and you could connect one probe of your voltmeter at any
convenient point between those two components to make a measurement.
If we follow a counterclockwise path from the reference the path goes
through all three resistors to the top of the 5 volt power supply. Therefore the
voltage across the three resistors must exactly add up to 5 volts. Since the
voltage across each resistor is equal to the current through the resistor multiplied
by the value of the resistance, we can write
5 V  I (10K)  I 2.2K  I 4.7K  I 10K  2.2K  4.7K  I 16.9K .
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5V
 0.296 mA . From Figure 3, we also see
16.9 K
that resistors that are connected purely in series add together. The three
resistors in series are equivalent to one resistor with a value of 16.9 K
From this we can find I 
To continue the example we can find the voltage at every node in the
circuit. You might want to mark the voltages on Figure 3 as we work through this
example. This circuit has four nodes, one between each pair of components.
We will use the node at the negative terminal of the voltage supply as our
reference and say it has a potential of zero volts. (Mark this on Figure 3.) The
other voltages will be referenced to it.
The voltage across the 4.7 K resistor is (0.296 mA)(4.7K) = 1.39 volts.
Since the node at the bottom of this resistor is our reference, the node between
this resistor and the 2.2K resistor will be at 1.39 volts (relative to our reference mark it on Figure 3).
The voltage across the 2.2 K resistor is (0.296 mA)(2.2K) = 0.65 volts.
Since the node at the bottom of this resistor is at 1.39 volts, the node between
this resistor and the 10 K resistor will be at 1.39 + 0.65 = 2.04 volts (relative to
our reference - mark it on Figure 3).
Finally, the voltage across the 10 K resistor is (0.296 mA)(10K) = 2.96
volts. Since the node at the bottom of this resistor is at 2.04 volts, the node
between this resistor and the positive terminal of the supply will be at 2.04 + 2.96
= 5 volts (relative to our reference), just as we expected.
Before moving on to the next example it is worth making one more
observation that we will use in Lab 4. The circuit above is a form of a voltage
divider. In deriving the current we used
I
5V
.
10K  2.2K  4.7 K
Then, in finding the voltage across each of the resistors we multiplied the current
by the resistance, or in general
VR  IR  5 V
R
10K  2.2K  4.7 K
where VR is the voltage across resistor R and R has a value of 10K, 2.2K or 4.7K.
This shows that the voltage across any of the resistors is a fraction of the voltage
across all of them, and the fraction is given by the ratio of the resistor in question
to the sum of the resistances. This is known as voltage division, and a series
connection of resistors can be called a voltage divider.
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One very important feature of series connected components should
be repeated for emphasis. Components connected purely in series have
the same current flowing through them.
.2.2.6.2
Parallel Circuits
Our next example will be
circuit
components
connected as shown in

F

Figure 4. This circuit has
only two nodes, one at
the top of the figure and
the other at the bottom.
Figure 4: A simple parallel circuit.
As a result, both of the
resistors
and
the
capacitor each have identical voltage across them, 5 volts. When components
are connected as shown in Figure 4, the circuit is called parallel. Using Ohm's
law, the current through the 4.7 K resistor is (5 V)/(4.7K) = 1.064 mA. The
current through the 10 K resistor is (5 V)/(10K) = 0.500 mA. Since this is a DC
circuit, the current through the capacitor is zero. However, since there is a
voltage across the capacitor the charge on the capacitor is (5 V)(10 F) =
5  10-5 coulombs (Equation 2) and the energy stored in it is (0.5)(10 F)(5 V)2 =
125 Joule (Equation 4). When the voltage supply in this circuit was first turned
on there would have been a very large current flow to the capacitor for a very
short time to charge it to 5  10-5 coulombs and 5 volts.
The current flowing into the two resistors is flowing out of the positive
terminal of the 5 volt supply. The algebraic sum of all currents entering a
node must be equal to the sum of all currents leaving a node. For this circuit
the current flowing out of the 5 volt supply can be found by adding the current
leaving the node at the top of the circuit. It is 1.064 mA + 0.500 mA = 1.564 mA.
Note that if we use Ohm's law and divide the voltage by the total current we
obtain a resistance of 3.20 K. This is the "parallel" equivalent resistance of the
two resistors. We did this using numbers, but a general formula could be derived
following a similar procedure using variables.
When two or more resistors are combined purely in parallel they have an
equivalent resistance given by
 1

1
1
R parallel   

 ... 
 R1 R2 R3

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1
(Equation 5)
Camera Lab 4 - Page 15
where one term is included for each resistor in the parallel combination.
Reiterating an important point for you to remember from this
example, circuit elements that are connected in parallel have the same
voltage across them.
.2.2.6.3
Combination Circuit
We will look at one more
example, which has both parallel and
series circuitry, before moving on to
the specifics of the flash circuit. The
circuit is shown in Figure 5.
4.7K
In this example the capacitor
is
starting
out charged to a voltage of
5V
5 volts. While the switch is open
there can be no current flow. The
4.7K
10K
voltage across each resistor is zero
and the current through them is also
zero. As soon as the switch is
closed the capacitor voltage will be
applied across the resistors, and a
current will flow. That means that
Figure 5: A circuit with both parallel
charge will flow out of the capacitor,
and series connections.
and the voltage will therefore
decrease, which will mean the
current will decrease too. Our objective will be to only find the current and
voltage of each resistor and current leaving the capacitor immediately after the
switch is closed (t = 0+).
10F
First we note that there is a 4.7 K resistor in parallel with a 10 K
resistor. We can use Equation 5 to find the equivalent resistance of this parallel
combination to be
1 
 1
Req  


 4.7 K 10 K 
1
 3.20 K .
We also see that this equivalent resistance is in series with the upper
4.7 K resistor, so the total resistance connected across the capacitor is 4.7 K
+ 3.2 K = 7.9 K
As stated above, right after the switch closes a total of 5 volts will appear
across this combination of resistors. We can thus use Ohm's law to find that the
initial total current flowing into the resistors will be (5 volts/7.9 K = 0.633 mA.
This current is flowing out of the capacitor and into the upper 4.7 K resistor.
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The voltage drop across the upper 4.7 K resistor is therefore
(4.7 K0.633 A) = 2.975 V (from Ohm's law again). That leaves 5 V - 2.975 V
= 2.025 V across the parallel combination of the lower resistors.
We will use Ohm's law two more times to find the current through the
lower two resistors, which each have 2.025 V across them. The lower 4.7 K
resistor will have a current of (2.025 V)/( 4.7 K = 0.431 mA flowing through it.
The 10 K resistor will have a current of (2.025 V)/( 10 K = 0.202 mA. Note
that these two currents add up to 0.633 mA, as they should, since they must
equal the current leaving the upper resistor.
As stated above, since there is current leaving the capacitor the total
voltage will immediately begin to fall below 5 volts. We can use the capacitor
equations to calculate the initial rate of decrease at t = 0. From Equation 3
dV
dt

t 0
I (t  0  ) 0.633 mA

 63.3 volt/sec .
C
10 F
The voltage will drop quickly as seen in Figure 6.
Capacitor Voltage (V)
5
4
3
2
1
0
0
100
200
300
400
500
Time (ms)
Figure 6: Voltage vs. time for the circuit of Figure 5.
2.2.7 Other Circuit Components
Another circuit element that is important for operation of the flash circuit is
the transformer. We will briefly describe the operation of an ideal transformer. A
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transformer basically consists of two coils of wire wound so that the magnetic
field generated by a current passing through one of the coils will be intercepted
by the other coil. A changing magnetic field, created by a changing current in
one coil, will induce a changing current in the other coil. This only works for a
changing current (and voltage). A transformer does not work for steady DC
current. The relative values of the currents (and of the voltages across each of
the coils) will depend on the relative number of turns in the two coils. One of the
coils in the transformer is usually referred to as the primary, and the other is
referred to as the secondary. In an ideal transformer if the secondary has n
times as many turns as the primary, the voltage across the secondary will be n
times larger than across the primary and the current through the secondary will
be 1/n times the current through the primary. The actual operation of real
transformers is a bit more complicated due to non-ideal properties, but this is
enough to understand the basics of how the transformers are used in the flash
circuit.
A final circuit element that we need to discuss it the diode rectifier. It is
labeled D1 in and Figure. As mentioned earlier, the diode allows easy current
flow in the "forward direction", but almost no current flow in the "reverse
direction." The diode symbol is arrow-like, and the "arrow" points in the direction
of easy current flow. Large currents can flow with very small voltages in the
forward direction. Forward voltages will typically be between 0.5 and 1 volt for all
forward currents that a diode is rated for. In the reverse direction only very small
currents can flow as long as the breakdown voltage of the diode is not exceeded.
In the flash circuit, the diode allows charging currents to flow between the
capacitor and the iron core transformer, but blocks currents in the opposite
direction that would discharge the capacitor through the iron core transformer.
This allows a large voltage to build up on the capacitor as the charging circuit
delivers current to it.
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2.3
The Circuit Board and Circuit Schematic
The flash circuit for the camera is made on a printed circuit board. The
body of the board is made of an electrically insulating material. Much of the
"wiring" in a printed circuit board takes the form of thin metal traces printed on
the board, rather than actual wires. That's why it's called a "printed" circuit board.
Holes are drilled through the board in appropriate locations. When the
circuit is being manufactured components are connected to the metal traces on
the board by inserting the wire leads of the components into the holes and joining
them to the metal traces with solder. In the original versions of the flash circuit
board all of the components were attached to the board this way. In some more
recent versions of the flash circuit board some of the components are attached
by a technique known as "surface mount." Components that are attached by
surface mount do not have wire leads designed to go through holes. Rather,
they have metal tabs that are directly soldered to the metal traces on the board,
without holes. Since the cameras are recycled, some of you will have boards
with surface mount components and some of you will not.
A circuit schematic is a symbolic representation of an electronic circuit.
Each symbol on the schematic represents a component in the circuit. The lines
in the schematic represent the wires or metal traces on the circuit board used to
connect the components together.
A partial schematic of the 5C3406 generation of flash circuit is shown in
Figure 7. In both circuits, the more complex circuitry that we will not study in this
lab is not explicitly shown. It is represented by the box labeled "More complex
circuitry for converting DC to AC." Most of the differences between the two
generations of camera occur inside this box. There are a few other differences
that are important for this lab, and they will be discussed below.
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2K
More Complex
Circuitry for
Converting DC to AC
G
B
C
2M
A
Figure 7: Schematic of the 5C3406 generation of flash circuit.
A few general features about some conventions used in these schematics
should be understood.
Wires connecting the various components are
represented by lines. A connection between two devices is known as a "node."
Wire intersections that are marked with a dot represent a connection between the
wires. Wires that are connected together will have the same voltage. They form
a single node and will have the same "node voltage." The term "node" does not
mean the dot at the intersections; all of the wires that are connected make up the
node and have the same voltage. Wires that cross without a dot are not
connected together - one wire bridges over the other. They are not the same
node.
Note: There is an alternate convention for drawing wires in
schematics. In this second convention any wires that intersect as
straight lines are connected (no dots). To draw wires that cross
without a connection one uses a small "loop," typically drawn as a
half-circle, in one of the wires to show it "bridging" over the other.
This second convention is not used in these instructions.
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Table 1: International System of Units
Quantity
Unit
Symbol
Dimensions
Length
Meter
m
Mass
Kilogram
kg
Time
Second
s
Temperature
Kelvin
K
Current
Ampere (Amp)*
A
Frequency
Hertz
Hz
1/s
Force
Newton
N
kg-m/s2
Pressure
Pascal
Pa
N/m2
Energy
Joule
J
N-m
Power
Watt
W
J/s
Electric charge
Coulomb
C
A-s
Potential
Volt
V
J/C
Conductance
Seimens
S
A/V
Resistance
Ohm
V/A

Capacitance
Farad
F
C/V
Magnetic flux
Weber
Wb
V-s
Magnetic
Tesla
T
Wb/m2
induction
Inductance
Henry
H
Wb/A
* "Amp" is a commonly used abbreviation for "Ampere"
The layout of the components in the schematic does not reflect the physical
position of the components on the circuit board. Rather, it shows the electrical
connections between the components. In the schematic the components are
frequently grouped according to the function of various parts of the circuit. On
the circuit board the layout of the components might be chosen for a variety of
different reasons. Some of these might be ease of manufacture, minimization of
distance between components for speed of operation, or fitting the circuit board
inside it’s housing. Can you think of others?
Table 2: Unit Prefixes
Multiple
Prefix
Symbol
Multiple
Prefix
Symbol
18
-1
10
exa
E
10
deci
d
15
-2
10
peta
P
10
centi
c
1012
tera
T
10-3
milli
m
9
-6
10
giga
G
10
micro
 (u)**
6
-9
10
mega
M
10
nano
n
103
kilo
K
10-12
pico
p
2
-15
10
hecto
H
10
femto
f
-18
10
deka
Da
10
atto
a
** The Greek letter mu is the formally correct symbol for micro. In cases when 
is not available in the typeset being used to prepare a schematic the letter "u" is
substituted.
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2.4
What's an Oscilloscope?
An Oscilloscope is an instrument that is typically used to measure voltage
as a function of time. The oscilloscope you will be using is a digital oscilloscope
that has two "channels." With two channels you will be able to display the
voltage at two different points in your circuit as a function of time. You will use
both channels in Task 1 of this lab, but will only use one channel for Tasks 2 and
3. A picture of the front panel of the oscilloscope is shown in Figure 9. The front
panel of the oscilloscope is divided into sections by function. These sections are
described below.
Horizontal
Run Control
Trigger
File Control
Display
Vertical
Input
Connections
Soft Keys
ON/OFF Switch
Floppy Disk Drive
Figure 9: Front panel of Oscilloscope
Display Screen: The largest section is the display. The results of the
measurements will be plotted on a graph on the display. The horizontal axis will
be time, and the vertical axis will be voltage. There are several soft keys at the
bottom of the display. What they do will depend on the last button you used on
other parts of the front panel. The function of these buttons at any particular time
will be displayed at the bottom of the screen. In addition, the values of
parameters you set with the controls (for example, the vertical scale) will be
displayed on the screen (usually near the top of the screen), and the values of
measurements you make can be displayed near the bottom of the graph.
Vertical: The controls marked vertical are used for controlling how the measured
voltage is displayed on the screen. There are two channels, 1 and 2. The push
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buttons in this section are used to select the channels. The position of the zero
for the vertical axis of the display can be set with the small knobs ("Position").
Separate "zeros" (reference value) can be set for each channel. The scale of the
vertical axis (in Volts/division) can be set with the large knobs. Different scales
can be set for each channel.
Horizontal: The controls marked horizontal are used for controlling how time is
displayed on the horizontal axis of the screen. For this lab you will use the small
knob to control the position of t = 0 on the screen. The large knob is used to set
the scale of the horizontal axis in time units per display division.
Trigger: While the oscilloscope can be set to continuously display the voltages
being measured, that will usually result in a jumble on the screen. We usually
want to trigger the display at some point in time, so that the signal we want to
observe gets displayed. You will use the controls in this section to define when
the oscilloscope will start to display the voltages (the t = 0 point of the
measurements). You will do this be setting a trigger level voltage, the trigger
mode, and the trigger slope. With the trigger level you will tell the oscilloscope to
define t = 0 at the point the voltage crosses some value you set with the trigger
level knob. With trigger mode you tell the oscilloscope to trigger in an “AUTO,”
“AUTOLEVEL,” or “NORMAL” way. We will always use “NORMAL.” With the
trigger slope you tell the oscilloscope to trigger when the voltage crosses the
trigger level with either a positive slope or a negative slope (and not to trigger if
the voltage crosses the trigger level with the wrong slope.)
Run Control: One of the buttons in this section is the "RUN/STOP" button. You
can use this button to tell the oscilloscope when you are done adjusting the other
controls and are ready to measure. The “single” button will be used to get only a
single time history trace on the display and then stop.
Measure: You will use the controls in this section to move cursors on the screen.
The voltages or times corresponding to the location of the cursors will be
displayed at the bottom of the graph. This will allow you to record the values of
different points on the voltage vs. time graphs on the display.
Floppy Disk Drive: You will save your oscilloscope data to this disk drive by
using file control buttons.
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3 Pre-lab 4 Assignments
This section describes tasks that you should perform before coming to Lab
4. The Pre-lab Tasks should be done together with your teammates. Include
your Team number and at the names of all members of the team at the top of the
first page.
Guidance on performing these assignments can be found in various
sections of the reading. In most cases you will need information from different
sections to perform each of the assignments, but to help you out the most
important sections for each assignment are listed after the assignment number.
Pre-lab Assignment 1: (See section 2.2)
Find the currents and voltages in the adjacent
circuit by the following steps.
Refer to sections 2.2.4, 2.2.5, and 2.2.6 for help
with this.
Purpose: To get you familiar with using Ohm's
law to analyze current flow in a simple circuit.
Team Assignment to be Turned In
a) What is the current flowing into the capacitor in this DC circuit?
b) The 3.9 M and 10 M resistors are in parallel. Find their equivalent
resistance?
c) If we neglect the capacitor, the 1M resistor is in series with the equivalent
resistance calculated in part b). What is the total resistance?
d) Why is it OK to neglect the capacitor for part c)? [Think about part a).]
e) Use the resistance calculated in part c) to calculate the total current flowing
from the voltage source into the 1 M resistor. (Use Ohm's law)
f) What is the voltage across the 1 M resistor? (Use Ohm's law)
g) The capacitor, 3.9 M resistor and 10 M resistor all have the same voltage
across them. Calculate this voltage.
h) Calculate the currents flowing through the 3.9 M and 10 Mresistors. (Use
Ohm's law twice)
i) Add the currents through the capacitor, and the 3.9 M and 10 M resistors
together. Compare the result to the current through the 1 M resistor.
j) Calculate the charge on the capacitor.
Pre-lab Assignment 2
Assume you are required to measure the following voltage signal with an
oscilloscope.
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V = 5 volts + (15 volts)  sin[ (2 sec-1) 
]
The results will be displayed on the following grid.
Team Assignment to be Turned In
Answer the following questions.
a) At which horizontal gridline would you position V = 0 (pick from this list: the
bottom, 3rd from the bottom, 3rd from the top, 2nd from the top) and which of
the following voltage scales would you use for the vertical axis to make the
display as large as possible without going over the edge of the screen? (1
V/div, 2 V/div, 5 V/div or 10 V/div). You must consider both choices together
to get the correct answer.
b) Which of the following time scales would you use for the horizontal axis to
show one period of the sine wave on the screen? (50 ms/div, 100 ms/div,
200 ms/div, 500 ms/div, 1 s/div).
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