Transcript Phys 345 Electronics for Scientists

Lecture 4 Overview
• More circuit analysis
– Thevenin’s Theorem
– Norton’s Theorem
Announcements
• Assignment 0 due Thursday
• Lab reports due today and tomorrow
• Physics Colloquium 4pm tommorrow
Method 3: Thevenin and Norton
Equivalent Circuits
Léon Charles Thévenin
1857-1926
vTH= open circuit voltage at
terminal (a.k.a. port)
RTH= Resistance of the
network as seen from port
(Vm’s, In’s set to zero)
Method 3: Thevenin and Norton
Equivalent Circuits
Any network of sources
and resistors will appear
to the circuit connected
to it as a voltage source
and a series resistance
vTH= open circuit voltage at
terminal (a.k.a. port)
RTH= Resistance of the
network as seen from port
(Vm’s, In’s set to zero)
Norton Equivalent Circuit
Any network of sources
and resistors will appear
to the circuit connected
to it as a current source
and a parallel resistance
Ed Norton – Bell Labs, 1898-1983
Calculation of RT and RN
• RT=RN ; same calculation (voltage and current sources set to zero)
• Remove the load.
• Set all sources to zero (‘kill’ the sources)
– Short voltage sources (replace with a wire)
– Open current sources (replace with a break)
Calculation of RT and RN continued
• Calculate equivalent resistance seen by the load
Calculation of VT
• Remove the load and calculate the open circuit voltage
VOC  VR 2 
R2
VS
R1  R2
(Voltage Divider)
Example
• Use Thevenin’s theorem to calculate the current
through Resistor R6.
– (solution RTH=6.67Ω, VTH=12V, I=0.95A)
Exercise: Draw the Thevenin Equivalent
• To find RTH remove the load, kill the sources (short
voltage sources, break current sources) and find the
equivalent resistance.
• To find VTH Remove the load and calculate the open
circuit voltage
Exercise: Draw the Thevenin Equivalent
• To find RTH kill the sources (short voltage sources, break
current sources) and find the equivalent resistance.
• To find VTH Remove the load and calculate the open
circuit voltage
Exercise: Draw the Thevenin Equivalent
• To find RTH kill the sources (short voltage sources, break
current sources) and find the equivalent resistance.
• To find VTH Remove the load and calculate the open
circuit voltage
VAB = 20 - (20Ω x 0.33amps) = 13.33V
Exercise: Draw the Thevenin Equivalent
• To find RTH kill the sources (short voltage sources, break
current sources) and find the equivalent resistance.
• To find VTH Remove the load and calculate the open
circuit voltage
VAB = 20 - (20Ω x 0.33amps) = 13.33V
Exercise: Draw the Thevenin Equivalent
• To find RTH kill the sources (short voltage sources, break
current sources) and find the equivalent resistance.
• To find VTH Remove the load and calculate the open
circuit voltage
Calculation of IN
• Short the load and calculate the short circuit current
(mesh analysis)
(R1+R2)i1 - R2iSC = vs
-R2i1 + (R2+R3)iSC = 0
(KCL at v)
RN=RTH
Source Transformation
Summary: Thevenin’s Theorem
• Any two-terminal linear circuit can be replaced with a voltage source
and a series resistor which will produce the same effects at the
terminals
• VTH is the open-circuit voltage VOC between the two terminals of the
circuit that the Thevenin generator is replacing
• RTH is the ratio of VOC to the short-circuit current ISC; In linear circuits
this is equivalent to “killing” the sources and evaluating the
resistance between the terminals. Voltage sources are killed by
shorting them, current sources are killed by opening them.
Summary: Norton’s Theorem
• Any two-terminal linear circuit can be replaced with a current source
and a parallel resistor which will produce the same effects at the
terminals
• IN is the short-circuit current ISC of the circuit that the Norton
generator is replacing
• Again, RN is the ratio of VOC to the short-circuit current ISC; In linear
circuits this is equivalent to “killing” the sources and evaluating the
resistance between the terminals. Voltage sources are killed by
shorting them, current sources are killed by opening them.
• For a given circuit, RN=RTH
Maximum Power Transfer
• Why use Thevenin and Norton equivalents?
– Very easy to calculate load related
quantities
– E.g. Maximum power transfer to the load
• It is often important to design circuits that
transfer power from a source to a load.
There are two basic types of power transfer
– Efficient power transfer: least power loss.
Power is usually large (e.g. power utility)
– Maximum power transfer (e.g.
communications circuits)
• Want to transfer an electrical signal
(data, information etc.) from the source
to a destination with the most power
reaching the destination. Power is
usually small so efficiency is not a
concern.
Maximum Power Transfer: Impedance matching
2
 vT 
 RL
p  i RL  
 RT  RL 
2
Differentiate w.r.t. RL using quotient rule:
du
dv
u
d u
 dx 2 dx
dx v
v
v
Set to zero to find maximum:
2
dp
2  ( RT  RL )  2 RL ( RT  RL ) 
  0
 vT 
4
dRL
( RT  RL )


vT2 ( RT  RL ) 2 2 RL vT2 ( RT  RL )

( RT  RL ) 4
( RT  RL ) 4
so maximum power transfer occurs when
RT  RL
http://circuitscan.homestead.com/files/ancircp/maxpower1.htm
and
vT2
pmax 
4 RL
Maximum Power Transfer: Impedance matching
2
 vT 
 RL
p  i RL  
 RT  RL 
2
Differentiate w.r.t. RL using quotient rule:
du
dv
u
d u
 dx 2 dx
dx v
v
v
Set to zero to find maximum:
2
dp
2  ( RT  RL )  2 RL ( RT  RL ) 
  0
 vT 
4
dRL
( RT  RL )


vT2 ( RT  RL ) 2 2 RL vT2 ( RT  RL )

( RT  RL ) 4
( RT  RL ) 4
so maximum power transfer occurs when
RT  RL
http://circuitscan.homestead.com/files/ancircp/maxpower1.htm
and
vT2
pmax 
4 RL