# Resistance Characteristic

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```					     Resistance Characteristic
Remember what happened when we plotted the current flowing
through a resistor versus the voltage across the resistor, while
varying the voltage? We got a straight line, passing through the
origin (If VR = 0, then IR = 0), with slope = 1/R. This is called the
characteristic curve of the resistor.

IIR
slope  1
R
IR
+
+
E               R    VR
-
VR                          -

VR  E
Resistance Characteristic
The slope-intercept form of the equation for a line is y = mx + b,
where y is the vertical coordinate, x is the horizontal coordinate, m
is the slope and b is the y-intercept. In the case of the
characteristic curve of a resistor R, m = 1/R, b = 0 (because the
line passes through the origin), y = IR, and x = VR

The resulting equation is:
IIR
The I-V characteristic                                   1
maps a particular       slope  1
R                 I R  VR  0
voltage to a particular
current.
R
or,
VR
IR 
VR                R
where VR and IR are in Volts and
Amperes, respectively, and R is
the resistance in ohms.
Resistance Characteristic
Intuitively, this makes sense: Increasing the battery Voltage E
increases VR (according to KVL), and is like increasing the
pressure at the inlet to a pipe. Increasing pressure increases the
flow rate, which is analogous to current IR. Reducing the
resistance is analogous to increasing the pipe’s diameter, which
also increases the flow rate.

IIR

slope  1
R
Increasing Voltage                                           IR
increases current                                                     +
+
E                   R    VR
-
-

VR  E
VR
IR 
R
Ohm’s Law
VR
The equation    IR 
R
is called Ohm’s Law, and it’s true for any resistance: The
resistance of an ideal resistor, the dynamic resistance (the slope of
the characteristic curve at a particular point) of an element with a
nonlinear characteristic curve (e.g., a diode or transistor), or the
resistance of a wire (which we’d rather
IIR                was zero). if we know R and IR, but not
slope  1
VR, we can solve the equation
R                                  VR
IR 
R
for VR:
VR  I R R
VR
if we know VR and IR, but not R, we
can solve for R:
VR
R
IR
Ohm’s Law
Let’s take a simple example: A battery (Voltage = E) in series with
a resistor R1. KVL tells us that E = V1, and KCL tells us that the
same current, I, flows through both the battery and the resistor.
Ohms law tells us
V
V1  IR1 or I  1
R

+
+
E                      R1   V1
-      I
-
Ohm’s Law
Let’s add another resistor in series, as shown. Now,
E  V2  V1
E  IR1  IR2
E  I R1  R2                E
So,   I
R2                                   R1  R2
and
+ V    -                                        E 
2
 R  R  R1
V1  IR1          
+                     +                             1    2 
E                        R1       V1
-                                                 R1
-                 V1  E
R1  R2
Ohm’s Law
Turning our attention to R2,
 E 
 R  R  R2
V2  IR2          
 1    2 

R2
V1  E
R2              R1  R2

+ V    -
2                      Ohm’s law, KVL and KCL
+                    +        allowed us to fully analyze this
E                      R1       V1   circuit. We were able to find
-                    -        the current flowing around the
loop, and the Voltages across
each of the resistors.
Nonlinear Resistance
The characteristic of an ideal resistor is a straight line. In other words, a
resistor has a linear I-V characteristic.
Some devices, such as transistors and diodes, have nonlinear I-V
characteristics. This means plotting current versus voltage does not
result in a straight line.
diode              II             resistor
characteristic                    characteristic
reverse bias            Forward Bias

The diode’s characteristic can
V       be divided into three regions,
reverse                  each of which is approximately
breakdown                linear: reverse breakdown,
reverse bias, and forward bias.
Nonlinear Resistance
Pick a voltage in the forward-bias region. Now, if the voltage is
increased by a small amount, the current increases by a large amount. If
the Voltage is reduced slightly (but stays in the forward-bias region) the
current diminishes by a small amount. Within the forward-bias region,
the characteristic can be approximated by a line with a large slope.

II

Recall that the slope of the
characteristic is 1/R, so a large
(steep) slope means small
V   resistance. We can say that a
linear              forward-biased diode has a
approximation to    small (ideally, zero) dynamic
diode in forward-   resistance. In other words, the
bias region         forward-biased diode conducts
current easily.
Nonlinear Resistance
Next, consider a point on the diode characteristic in the reverse-bias
region. A large increase or decrease in voltage (but not so large as to
leave the reverse-bias region) results in a very small increas or
decrease in current. The line approximating the characteristic in this
region has very small slope, so the dynamic resistance of a reverse-
biased diode is very large –
II                     ideally infinite, like an open
circuit.

V
linear
approximation to
diode in reverse-
bias region
Nonlinear Resistance
Finally, consider a point on the diode characteristic in the reverse-
breakdown region. A small increase or decrease in voltage (staying
within the reverse-breakdown region ) results in a large increase or
decrease in current. The line approximating the characteristic in this
region has large slope, so the dynamic resistance of a diode in reverse-
breakdown is very small.
II

V
linear approximation to
diode in reverse-
breakdown region
Nonlinear Resistance
Finally, consider a point on the diode characteristic in the reverse-
breakdown region. A small increase or decrease in voltage (staying
within the reverse-breakdown region ) results in a large increase or
decrease in current. The line approximating the characteristic in this
region has large slope, so the dynamic resistance of a diode in reverse-
breakdown is very small.
II

V
linear approximation to
diode in reverse-
breakdown region
Nonlinear Resistance
Here’s an example. We’d like to light an LED. We can increase the
LED’s brightness by increasing the current flowing through it, until the
current becomes so large it destroys the LED. We need to make sure
the current is great enough to make the LED as bright as we want, but
not so great that the LED destroys itself. W do this with a resistor in
series with the LED.
II
VR
+        -

R
I
12V      +                    +
VLED
VLED                                  -
-
2.5 V
Nonlinear Resistance
The LED’s I-V characteristic shows that its forward-bias voltage is about
2.5 V. The LED must be forward biased to light the LED. Only a very
large (large enough to destroy the LED) increase in current could
produce an increase beyond 2.6 V or so, so it’s a reasonable
approximation to say
VLED  2.5 V
II
VR
+        -

R
I
12V    +                    +
VLED
VLED          -                    -

2.5 V
Nonlinear Resistance
Examining the manufacturer’s data sheet, we find that an LED current
of 20 mA should make the LED sufficiently bright. Applying KVL around
the loop,
12 V  VR  VLED
12 V  VR  2.5 V
9.5 V  VR
II
VR
+        -

R
I
12V        +                   +
VLED
VLED              -                   -

2.5 V
Nonlinear Resistance
Applying Ohm’s law,    9.5 V  VR  IR                 475 is not a standard
VR    9. 5 V               resistor value, but 470W
R                475 
I   20 mA                 and 510 are.

II
VR
+        -

R
I
12V       +                    +
VLED
VLED              -                    -

2.5 V
Power Rule
We’ve already seen that the power consumed by a circuit element (e.g.,
a resistor) is given
P  IV
Recall that power is the rate at which energy is converted from one form
to another. For example, a 1 Hp motor converts electrical energy to
kinetic energy at a rate of 746 J/sec., or 746 Watts (1 Hp = 746 W).
In other words, if K represents energy and t = time,
K
P
I
+
t
+                        Current is the rate of flow of charge, so if
E                  R    VR         Q = charge and I = current,
-                                              Q
-
I
t unit charge:
and Voltage is energy per
K
V
Q
Power Rule
So we can prove P = IV, the power rule:
Q K K
P  IV     P
t Q T
Resistive elements consume electrical energy by converting it to heat.
This is called dissipation, because as heat is produced it is radiated
away to the surrounding. The faster heat is consumed (that is, the faster
electrical energy is converted to heat, which means greater power),
the faster it must be radiated away. The only
way (aside from increasing the surface area of
I            the component) to make it radiate heat faster is
+
+                 for the temperature to rise, but if the temperature
E                  R   VR   rises too much, damage can result. All
-
-    components have a maximum power dissipation
rating, which must not be exceeded.
Power Rule
Consider our simple LED example. We found previously that
VLED  2.5 V
I  20 mA.
PLED  20 mA. 2.5 V   50 mW
So

The LED’s power dissipation rating had better be at least 50 mW.
We also found that VR = 9.5 V, so the
VR
+        -             resistor dissipates
R
I
PR  20 mA. 9.5 V   190 mW
12V     +                   +      And the power delivered by the battery is
VLED
-
-                         Pbattery  190 mW  50 mW  240 mW
which can also be found like this:
Pbattery  12 V 20 mA   240 mW
Power Rule
We’ve already seen that the power consumed by a circuit element (e.g.,
a resistor) is given
P  IV
In other words, the power consumed by an element is equal to the
product of the voltage drop across it and the current flowing through it.
For the resistor shown below,
PR  IVR
Ohm’s law tells us that
I                          VR  IR
+
+                       Substituting,
E
PR  I IR   I 2 R
R   VR
-
-
Now we have another way of calculating the
power consumed by a resistor: It’s the
product of the resistance and the square of
the current through the resistor.
Power Rule
Another form of the power rule is found as follows: Start with the basic
power rule:              P  IV
R     R

If we know VR and R, we can solve Ohm’s law for I:
V
I R
R
Substituting,
VR2
PR  VR  
VR
I                       R         R
+
+
E                R            We now have yet another way of calculating
VR
-                     the power consumed by a resistor: It’s the
-
quotient of the square of the voltage across
the resistor divided by the resistance.
Power Ratings
Through-hole-mount resistors, the kind we use in the lab, are available
with power ratings of 1/8 W, ¼ W, ½ W, 1 W, and 2 W. Suppose we
have a 470  resistor (a standard value) rated at ¼ W. What is the
maximum voltage it can drop?
V2
P       V 2  Pm axR
R
Vm ax  PR  0.25 W 470  

I
Vm ax  10.84 V
+
+
E                 R    VR
-
-
A BJT Circuit
Ohm’s law tells us                                     Let’s look at the
3V                           +                  example in the book,
RC         300                                     with a few
10 mA                    3V       RC
-                  differences. We see
+         10 mA    that IC = 10 mA, and
2V       LED
IB = 430 mA. KCL
-                  gives us
430 mA                                        I E  I C  I B  10 .47 mA
VRB                           +
+            -            +
The definition of
4V    9V
+                             b gives us
RB                               -
+                     0.7 V -      -
10K                                  I C  bI B
5V
I C 10 mA
-                                                 b              23.3
I B 430 mA
A BJT Circuit
Let’s let b increase
+                  to 40. The base-
emitter loop is
6V       RC             unchanged, so IB is
-
+         20 mA    still 430 mA.
2V       LED             I C  bI B  40430 mA 
-
430 mA
 20 mA.
VRB                           +     The voltage across
+          -            +
4V    9V         RC is now
+
VRC  I C RC  20 mA.300  
RB                             -
+                   0.7 V -      -
10K
5V                                                   6V
-
A BJT Circuit
a. The base current
+                   is unchanged, so the
power generated by
6V       RC              the 5 V supply is still
-
+         20 mA     2.15 mW
2V       LED
-                     b. and c. Similarly,
430 mA                                         the power dissipated
VRB                           +      by RB is still 1.85
+          -            +
4V    9V          mW, and the power
+                             dissipated by the
RB                             -
+                   0.7 V -      -
base-emitter junction
10K
5V                                                      is still 301 mW.
-
d. The power delivered
by th 9 V supply is
doubled, to 180 mW
A BJT Circuit
e. The power dissipated
by RC is quadrupled,
+                     to 120 mW.
6V       RC
-                  f. power dissipated by
+         20 mA       the LED is doubled, to
2V       LED               40 mW.
-
430 mA
VRB                           +
+          -            +
4V    9V
+
RB                             -
+                   0.7 V -      -
10K
5V
-
A BJT Circuit
g. KVL gives us the
+                   collector-emitter
voltage, VCE
6V       RC
-                    9 V  VRC  2 V  VCE
+         20 mA
VCE  9 V  VRC  2 V
2V       LED
-                     9 V  6 V  2 V 1V
430 mA                                          The power
VRB                           +       dissipated by the
+          -            +
+
4V    9V           collector-emitter
RB                             -       output is now
+                   0.7 V -      -
10K                                VBJTce  I CVCE  20 mA.1V 
5V
 20 mW
-
A MOSFET Circuit
Here’s the circuit used by Herrick in his example. Let’s work that
example, except that the load voltage VLoad is doubled to 6 V.

If the load voltage is doubled, the source current must also be doubled,
to 2 A. Examination of the MOSFET characteristic reveals that
VGS  4.2 A
Applying KVL
around the gate-
+                  +
source loop yields
VDS         12 V
+
-   VDD   VIN  VGS  VLoad
+             VGS -     -
+                             4.2 V  6 V
3          Vload                     10.2 V
-                      -
A MOSFET Circuit
The load power is now given by

6 V 2
3
so the load power is more than doubled. The supply power is doubled,
to 18 W. The MOSFET output power is       PMOSFET  PSUPPLY  PLoad
 18 W  8 W  10 W

+                          +
VDS                 12 V
+                                        VDD
-
+           VGS -     -
+
-                    -
An Opamp Circuit
Here’s Herrick’s opamp example. Suppose we change VIN from 1 V to
2 V:
The ideal op-amp assumptions say that Va = VIN, because the Voltage
difference between the inverting and noninverting inputs must be zero.

+Esupply               Having found Va, we
Ini                                          see that the VRi = Va, so
VIN                                +             Iout                     Va   2V
2V                                                VOUT     I Ri              2 mA.
-                                      Ri 1000 
Iinv                                        KCL tells us that IRf = IRi,
-Esupply               so Ohm’s law says
IRf
2V                           Rf=2.7K                     VRf  I Rf R f  I Ri R f
a   Ri          -             +          Rload
+                           VRf              1K           Va         Rf
VRi
1K
 R f  Vin
-           IRi                                              Ri         Ri
An Opamp Circuit
KVL tells us that
VOut  VRf  VRi
Rf
 Vin          Vin
Ri
+Esupply
   R 
Ini                                                      1  f 
 Vin 
VIN                                +             Iout                          Ri 

2V                                                VOUT
-                                         2700 
Iinv                                               2V 1         7.4 V
-Esupply                            1000 
IRf                                     so
2V                           Rf=2.7K
VOUT   7.4 V
a                                                  I Load                 7.4 mA.
Ri          -
VRf
+       1K                                  1K
VRi
-           IRi
Energy
Our lab exercises all require the use of a power supply. What does a
power supply do? It supplies power!
What does that mean? What’s Power?
Power is the rate at which energy is converted from one form to another.
If K represents the energy converted in a time interval Dt, the rate of
conversion is K/Dt, and the power which must be supplied for
conversion is
K
P
Dt
An electric motor converts electrical energy to mechanical (kinetic)
energy. A motor which is 100% efficient (that is, it converts all of the
electrical energy supplied to it to useful work, none is wasted as heat or
other forms) and is rated at 1 horsepower (1 hp = 746 Watts) converts
energy at the rate of                 746 Joules
P  746 Watts 
1 second.
Energy
So a 1 hp. motor can convert 746 Joules of electrical energy to 746
Joules of work every second. Where does the electrical energy come
from? A power supply!
120 V
I
+                      1 horsepower
M        DC motor
-

Here the power supply is a 120 Volt battery. The motor converts 746
J/sec. of energy, so the battery must supply 746 J/sec., or 746 W. of
power. remember that P = IV, so the current supplied must be
P 746 W
I           6.22 A
V 120 V
Energy
Notice that in this example the current is flowing out of the battery’s +
terminal, and into the more positive terminal of the motor. There is a
voltage rise across the battery terminals, and a drop across the motor.
120 V
I    +
+                     1 horsepower
VM   M    DC motor
-            -

Since current flows out of the positive battery terminal, the battery is
supplying energy to the motor.
Energy
Some batteries, like the one in your car, are rechargeable. That is, they
can convert electrical energy to chemical energy for storage, then
reverse the process to supply electrical energy.
120 V
I    +
+                    DC generator
VM   M
-           -

Here’s the same battery, connected to a DC generator (which may be
identical to the DC motor, except that its shaft is driven by a gasoline or
diesel engine). The generator converts mechanical energy to electrical
energy. Current flows out of the generator’s positive terminal and into
the battery’s positive terminal, so now the generator is supplying energy
to the battery.
Energy
The battery now receives energy from the generator at the rate of
Pch arg e  120 V I Ch arg e Joules per sec., or Watts

120 V
ICharge +
+                       DC generator
VM      M
-          -

So the battery stores energy at that rate. If the generator supplies
charging power to the battery at 746 Watts, then the battery must
charge 1 minute for every minute the motor is to run at full power.
Energy
Of course, the wires connecting the battery to the generator have some
resistance. Let’s say it’s 1 Ohm. The generator voltage must be slightly
greater than the battery voltage, to force current to flow from the
generator to the battery. To
1                        charge at a 746 Watt rate, the
current must be 6.22 A, so
VM  6.22 A 1   120 V  126 V
+
+     ICharge
120 V             VM        M   DC generator
-               -

Similarly, the voltage across the motor when it’s receiving power from
the battery must be lower than the battery voltage, so current flows out
of the battery. to run at full power,
VM  120 V  6.22 A 1   120 V  113 .78 V
Energy
A DC motor does produce a voltage across its terminals while its
running, called the back emf, or counter emf. That’s where VM comes
from.
1

+    ICharge +
120 V            VM      M   DC Motor
-            -

```
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