ElEn 236
1.
voltage
source
notes
Symbols
resistor
ground
cap
inductor
signal
source
volt
meter
current
meter
V
diode
npn bjt
pnp bjt
n ch jfet
p ch jfet
mos fet
I
op-amp
+
2.
Conventional current ( dc )
Even though current is really electrons moving in the direction of out from
the negative terminal of the source, through the circuit and back through the
positive terminal, we will follow the convention that current is flowing
from the positive terminal of the source, through the circuit and back through
the negative terminal of the source.
3.
Ohms Law
The amount of current in amperes that flows through a resistor is equal to the V
voltage in volts that is applied across the resistor divided by the resistance of
the resistance in Ohms.
I = V / R so V = IR and R = V / I
eg if the voltage drop across a 200 Ω resistor is 12 V then the current through
it will be 0.06 A or 60 mA
4.
Resistors in Series ;
I
R
Kirchoffs voltage law
R1
The applied voltage must equal the sum of the voltage drops. That is, the sum
of the voltage drops across R1 and R2 must equal the applied voltage E .
Therefore the total resistance seen by the applied voltage is R1 + R2 .
So I = E / ( R1 + R2 )
eg if : R1 = 2 kΩ , R2 = 3 kΩ and E = 20 V then I = 4 mA
5.
I
E
R2
Voltage Dividers
In the case above, the voltage drop across R1 would be 8 V ( V = I R ) and the voltage drop
across R2 would be 12 V . VR1 = E ( R1 / ( R1 + R2 ) )
I
6.
Voltage and Current Measurement
An ammeter must be placed in series with the circuit while a voltmeter must
be placed across the device whose voltage drop is to be measured.. In order
to measure the voltage and current for R1 in the previous case, the meters
would have to be placed as shown.
R1
E
R2
V
7.
Resistors in Parallel ; Kirchoffs Current Law
I
The total current will equal the sum of the parallel currents
The parallel components will have the same voltage drop
So I = I1 + I2
E
R1
I1 R2
I2
The resistance ( RT ) seen by the source E will be :
RT = 1 / ( 1/R1 + 1/R2 )
eg
if R1 = 1 kΩ , R2 = 2 kΩ , E = 10 V then :
I1 = 10 mA , I2 = 5 mA , I = 15 ma
RT = 1 / ( 1 / 1 kΩ + 1 / 2 kΩ ) = 666.7 Ω
I = 10 V / 666.7 Ω = 15 mA
8.
Current Dividers
If the total current is known , and there are only two parallel resistors , then a short-cut way tofind
The current through one of the resistors is : I1 = I ( R2 / ( R1 + R2 ) )
eg
in the circuit above I1 = 15 mA ( 2 kΩ / 3 kΩ ) = 10 mA
9.
Series and Parallel Symbols
If R1 and R2 are in series , we write : R1 + R2
If R1 and R2 are in parallel , we write : R1 // R2
10.
Power
Power in Watts is equal to the voltage across a device in Volts multiplied by the current through the
2
2
device in Amps. i.e. P = VI . If the device is a resistor then P = V / R or P = I R
For AC voltages or currents, if they are in rms equivalents , then we can used the DC equations .
11.
Resistors in Series / Parallel Circuits
In the circuit shown , the resistance ( RT ) seen by the source ( E )
is R1 + ( R2 // R3 ) or :
RT = R1 + 1/( 1/R1 + 1/R2 )
Eg if R1 = 1 kΩ , R2 = 3 kΩ , R3 = 6 kΩ , E = 12 V , then :
I
R1
E
R2
R3
R2 // R3 = 1 / ( 1/3000 Ω + 1/6000 Ω ) = 1 / ( 333.3 μS + 166.7 μS )
= 1 / 500 μS = 2000 Ω
RT = R1 + ( R2 // R3 ) = 1 kΩ + 2 kΩ = 3 kΩ
IT = 12 V / 3 kΩ = 4 mA
VR1 = 4 mA x 1 kΩ = 4 V , V R2 // R3 = 4 mA x 2 kΩ = 8 V , or V R2 // R3 = 12 V – 4 V = 8 V
IR2 = 8 V / 3 kΩ = 2.667 mA , IR3 = 8 V / 6 kΩ = 1.333 mA
IE = 4 mA , IR1 = 4 mA , IR2 + IR3 = 4 mA
PE = 4 mA x 12 V = 48 mW
PR1 = 4 mA x 4V = 16 mW, PR2 = 2.667 mA x 8V = 21.333 mW, PR3 = 1.333 mA x 8V = 10.667 mW
PR1 + PR2 + PR3 = 48 mW
PE = PR1 + PR2 + PR3
12.
Energy Storage in Capacitors
A capacitor is simply two pieces of metal ( usually aluminum ) foil separated by an insulating film.
The two pieces of foil each have a lead exiting from the cap. When a voltage is applied across the
cap, electrons will forced onto the negative side of the cap and the same number of electrons removed
from the positive side of the cap. In other words, there will be a displacement charge on the cap.
Capacitance is measured in Farads. A one Farad cap will have one Coulomb of electrons moved from
the positive plate to the negative plate when the voltage across it equals one volt. The displacement
charge stored on the cap can be calculated from the equation Q = CV, where Q is the charge in
Coulombs, C is the capacitance in Farads, and V is the voltage across the cap in Volts. The energy
2
that is stored by the cap can be calculated from the equation :
W = 0.5 C V .
13.
Transient Response of RC Circuits
S1
In the circuit shown, assume that the switch is open and the voltage
R1
across the capacitor is zero volts. Then at time zero, the switch closes.
I
Current will start to flow and the voltage across the cap will become
E
more and more positive ( top relative to bottom ) as time progresses
C1
until the voltage across the cap eventually equals the applied voltage.
The cap will charge quickly at first because when the switch closes,
The voltage across the cap is initially zero so all of the applied voltage
E is across the resistor, but as the capacitor charges up, there is less and less voltage for the resistor
and the current decreases and the cap charges more slowly. Initially the current would be E / R
and since the equation relating charge, capacitance and voltage is Q = CV and Q = It ( current
times time ) then i t = CV or V/t = i/C or the initial rate of change in voltage across the cap
would be the initial current divide by the capacitance or V/t = ( E/R ) / C or E/RC. If the cap
continued to charge at this rate then the time that it would take the cap to get up to E volts would be
E / ( E/RC ) or RC this is called the time constant ( Tau ) of the circuit. Ohms times Farads is
equal to seconds. So the time constant for an RC circuit is simply T = RC .
The equation for the voltage across the cap ( top relative to bottom ) with respect to time is :
Vc = E ( 1 – e ^ ( – t/T ) ) = E (1 – e ^ ( – t/RC ) )
Once the cap is charged up to E volts ( we say that this will by the time that 5 Tau have passed
even though it is not quite true ) then if the switch is opened, the cap will retain the displacement
charge. In fact, caps do have a small amount of leakage so eventually the cap would discharge to
zero volts but this would take quite some time. Once the cap is charged up to E volts and we
remove it from the original circuit and discharge it through a resistor R then the equation for the
voltage across the cap will be Vc = E ( e ^ ( – t/T ) ) = E ( e ^ ( – t/RC ) ) .
It is not wise to charge or discharge a cap with a short circuit as this will cause a very short but
very large current that could damage the cap.