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KVL Example Resistor Voltage Divider
- Consider a series of resistors and a voltage source
- Then using KVL
V − V 1 − V 2 = 0
V 1 = I 1 R 1 V 2 = I 1 R 2
V − I 1 R 1 − I 1 R 2 = V − I 1 ( R 1 + R 2 ) = 0
mA R R
V
I 1
1 2
- i.e. get the resistors in series formula
Rtotal = R 1 + R 2 = 5 K Ω
KVL Example Resistor Voltage Divider Continued
- What is the voltage across each resistor
- Now we can relate V 1 and V 2 to the applied V
- With the substitution
1 2
1
R R
V
I
V
R R
VR
V I R 2
1 2
1 1 1 1 =
V
R R
VR
V I R 3
1 2
2 1 1 2 =
Usefulness of Resistor Voltage Divider
- A voltage divider can generate several voltages from a fixed source
- Common circuits (eg IC’s) have one supply voltage
- Use voltage dividers to create other values at low cost/complexity
- Eg. Need different supply voltages for many transistors
- Eg. Common computer outputs 5V (called TTL)
- But modern chips (CMOS) are lower voltage (eg. 2.5 or 1.8V)
- Quick interface – use a voltage divider on computer output
- Gives desired input to the chip
Variable Voltage and Resistor Voltage Divider
- If have one fixed and one variable resistor (rheostat)
- Changing variable resistor controls out Voltage across rheostat
- Simple power supplies use this
- Warning: ideally no additional loads can be applied.
- Loads are current drawing devices
- In practice the load resistance >> the divider output resistor
- Best if Rload >100Rk
Current Divider Continued
- To get the currents through R 1 and R 2
2
1 2 1
1 1
R
V
I
R
V
I = KK =
- First get the voltage from the KCL equation
1 1
1 2
1
− −
R total
I
R R
V I
1 2
1
1
1 1 1 1
R R
R
I
R
V
I
- Similarly solving for I 2
1 2
2
2
1 2 1 1
R R
R
I
R
V
I
Example of Current Divider
- Consider 4KΩ and 2KΩ in parallel to a 3 mA current source
- The by the current divider for I 1
mA
R R
R
I
R
V
I 1
1 2
1
1
1 1 =
mA
R R
R
I
R
V
I 2
1 2
2
2
1 2 =
- Note the smaller resistor = larger current
- Checking: the voltage across the resistors
V 1 = I 1 R 1 = 0. 001 × 4000 = 4 V
V 1 = I 2 R 2 = 0. 002 × 2000 = 4 V
Practical Current Divider
- Create current dividers for use with current sources
- Less common that Voltage dividers as a circuit application
- Again any load used must not be significant
- Load in this case in series with the output resistor
- Load must be very small compared to Rk
- Best if load is <<0.01 of Rk
General Current Divider using Conductance
- Often better with parallel circuits to use conductance
- Again the KCL says at the node
=
N
j 1
I Ij
- Total conductance is resistors in parallel is
= =
N
j (^1) j
N
j 1
total j
R
G G
- The general current divider equation for Ik through resistor Rk
=
N
j
j
k k
G
IG
I
1
- conductance calculations useful for parallel resistors
- conductance equation is I equivalent of voltage divider equation
- Note for resistors in series then conductance is
=
N
total j^1 Gj
G
Example Solving Circuits with equivalent Resistors
- Consider circuit with R 2 , R 3 in parallel R 1 all in series with R 4
- For the R2, R 3 side
R 2 + 3 = R 2 + R 3 = 1000 + 3000 = 4000
- Now get the parallel equivalent
1 || 2 3 1 2 3
R R R
+ = =^1333.^3 Ω
R 1 || 2 3
- Adding the series resistance
Rtotal = R 4 + R 1 || 2 + 3 = 1000 + 1333. 3 = 2333. 3 Ω
- Thus current from the source is
_6 mA
- 3_
R
V
I
total
total = = =
Example Circuits with equivalent Resistors Continued
- Voltage across R 4 and parallel section is
VR 4 = I 4 R 4 = 1000 × 0. 006 = 6 V
V 1 = V − I 4 R 4 = 14 − 1000 × 0. 006 = 8 V
- And the current in the parallel resistors
4 mA 2000
R
V
I
1
1 1 = = =
2 mA 4000
R
V
I
2 3
1 2 = = =
V R 2 = I 2 R 2 = 0. 002 × 1000 = 2 V
V R 3 = I 2 R 3 = 0. 002 × 3000 = 6 V
Measuring Small Values: - Wheatstone Bridge
- Resistor dividers are set by ratios of resistance
- Thus can compare unknown R to a known set of R
- Called a Wheatstone Bridge
- Left side know resistance R 1 and variable resistor R 3
- Right side known R 2 and unknown Rs
- Place a very sensitive meter between the middle nodes
- Best is a galvonometer
- Voltages balance and no current ig flows if
2
s
1
3
R
R
R
R
- If know the R 1 R 2 R 3 very accurately can measure Rs accurately
2 1
3 s R R
R
R =
- Must use very accurate variable resistance
Circuit Analysis with Kirchhoff's Laws Circuits (EC 4)
- Task of Circuit analysis:
- Find the current in and the voltage across every element
Four methods used:
- Resistor substitution
- Mesh analysis (KVL)
- Node analysis (KCL)
- Superposition (simple circuits)
- Computer methods use aspects of these
Linear & Nonlinear Circuit Elements
Linear devices
Response is linear for the applied Voltage or Current
eg Double voltage get twice the current
eg devices: resistors, capacitors, inductors (coils)
Non-Linear devices
Response is non-linear for applied Voltage or Current
eg may have different response for different polarity of V
eg devices Semiconductor Diodes, iron core inductors
Kirchhoff's Laws and complex circuits
- Kirchoff's laws provide all the equations for a circuit
- But if know the currents then can calculate the voltages
- If know the voltages then can calculate the currents
- Thus only need to solve for one or the other.
- Use the other laws to obtain the missing quantity
