Unit 2 content

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Electric Fields
A charged object experiences
a force inside an electric field
Electric Fields
• The field does work on charged
objects
• Charges accelerate
•W=QxV
Potential Difference
• Work done to move 1 coulomb
of charge
• 1V = 1 J C-1
Internal Resistance
Internal
resistance,
r
E.m.f.
(E)
Current, I
E = I.R + I.r
Load, R
Internal Resistance
• Resistance of power supply itself
• work is done to push charges through power
supply hence ‘Lost Volts’
• E.M.F. Maximum energy to push unit
charge around circuit
• Terminal potential difference , work to push
unit charge through external circuit ( load )
Internal Resistance
• E = V t.p.d. + V lost
• V t.p.d. = E - V lost
V t.p.d = E - I x r
V t.p.d
E.M.F
open
circuit
p.d. NO
lost volts
- slope = r
I
Short circuit
current E = V lost
Wheatstone Bridge
R1
R2
R3
R3
R4
Wheatstone Bridge
• At balance point
R1
R3
For out of balance bridge
V
r
=
R2
R4
Alternating Current
Voltage ( V )
Alternating Voltage
Peak Voltage
2
1
0
-1 0
-2
V peak = 2 x Vr .m.s
5
10
Time ( ms )
15
r.m.s. voltage
Alternating Current
I peak = 2 x I r .m.s
Resistance is independent of frequency
Capacitors
• Dielectric
0V
+ 5V
Charge builds up on plates ( does not flow
through dielectric )
1 F = 1 C V-1
Q
C=
V
Capacitors
• Work is done charging up Capacitor
• Energy ( charge) is stored in Capacitor
• Ee = 0.5.Q.V
Q
EE = 0.5 C.V 2
Q2
EE = 0.5
C
V
Area under graph = Energy Stored
Slope = Capacitance
Capacitors
•
•
•
•
Capacitors block d.c. yet pass a.c
Capacitors supply time delays
Capacitors used to smooth a.c.
Capacitors used as microphones
Capacitors
• I directly proportional to frequency
I
f
Op Amps
• Voltage Amplifier
• Ideal Op Amp a) Input current = 0 A
• b) p.d. across inverting and non inverting
pin = 0V
Op Amps : Inverting Mode
Gain =
Rf
Vout
=
Rin Vin
Rf
Rin
Vin
Vout
Op Aps : Inverting Mode
• Saturation occurs at c.a. 85 % of p.supply
voltage
Vout
Vin
Op Amps : Summing Amp
2 inputs can be
added together
Op Amps :Differential Mode
• Difference between two inputs is amplified
Rf
R4
Gain = (V2 V1 )
Rin
Rf
V1
V2
R3
Vout
R
i

Rf
Rin
R3
R4
Op Amps :Differential Mode
• Can be connected to a wheatstone bridge
• Used in ECG to subtract the 50 Hz mains
hum from the heart signal
Rf
R3
=
Rin R4
Op Amps : Output
• Output current is of micro amp order
• To drive a speaker or motor a power amp
must be used I.e. an NPN transistor or a
MOSFET transistor
• This increases the current
Op Amps : Output
NPN is switched
on by output
from op amp
MOSFET
is
switched
by output
from op
amp