KVL Example Resistor Voltage Divider
• Consider a series of resistors and a voltage source • Then using KVL
V −V1 −V2 = 0
• Since by Ohm’s law
V1 = I1R1 V2 = I1R2
• Then
V − I1R1 − I1R2 = V − I1()R1 + R2 = 0
• Thus
V 5 I1 = = = 1 mA R1 + R2 2000 + 3000
• i.e. get the resistors in series formula
Rtotal = R1 + R2 = 5 KΩ
KVL Example Resistor Voltage Divider Continued
• What is the voltage across each resistor • Now we can relate V1 and V2 to the applied V • With the substitution
V I 1 = R1 + R 2
• Thus V1
VR1 5(2000) V1 = I1R1 = = = 2 V R1 + R2 2000 + 3000
• Similarly for the V2
VR2 5(3000) V1 = I1R2 = = = 3V R1 + R2 2000 + 3000
General Resistor Voltage Divider
• Consider a long series of resistors and a voltage source • Then using KVL or series resistance get
N V V = I1∑ R j KorKI1 = N j=1 ∑ R j j=1
• The general voltage Vk across resistor Rk is
VRk Vk = I1Rk = N ∑ R j j=1
• Note important assumption: current is the same in all Rj
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: Example of KCL
• KCL equivalent of voltage divider is a current divider • Consider a current source with resistors in parallel • At node 1 the KCL laws state:
I − I1 − I2 = 0
• Define V1 as the voltage between node 1 & node 0 • Then
V1 V1 I1 = KKI2 = R1 R2
• Thus from KCL
V1 V1 1 1 I = I1 + I2 = + = V1 + R1 R2 R1 R2
• This produces the parallel resistors formula
1 1 V1 I = V1 + = R1 R2 Rtotal
Current Divider Continued
• To get the currents through R1 and R2
V1 V1 I1 = KK I 2 = R1 R2
• First get the voltage from the KCL equation
−1 −1 1 1 1 V1 = I + = I R1 R2 Rtotal
• Solving for I1
1 V1 R1 I1 = = I R1 1 1 + R1 R2
• Similarly solving for I2 1 V1 R2 I2 = = I R2 1 1 + R1 R2
Example of Current Divider
• Consider 4KΩ and 2KΩ in parallel to a 3 mA current source
• The by the current divider for I1
1 1 V R 4000 I = 1 = I 1 = 0.003 =1 mA 1 R 1 1 1 1 1 + + 4000 2000 R1 R2
• Similarly for I2
1 1 V R 2000 I = 1 = I 2 = 0.003 = 2 mA 2 R 1 1 1 1 2 + + 4000 2000 R1 R2
• Note the smaller resistor = larger current
• Checking: the voltage across the resistors
V1 = I1R1 = 0.001× 4000 = 4V
V1 = I2R2 = 0.002× 2000 = 4V
General Current Divider
• Using KCL to get the currents into the node
n I = ∑ I j j=1
• Getting the voltage from the KCL equation
−1 −1 N 1 1 V1 = I ∑ = I j=1 Rj Rtotal
• Solving for Rk with it’s the current Ik
1
Rk Ik = I N 1 ∑ j=1 R j
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 I = ∑ I j j=1
• Total conductance is resistors in parallel is N N 1 Gtotal = ∑ G j = ∑ j =1 j =1 R j
• The general current divider equation for Ik through resistor Rk
IGk Ik = N ∑G j j=1
• conductance calculations useful for parallel resistors • conductance equation is I equivalent of voltage divider equation • Note for resistors in series then conductance is 1 N 1 = ∑ Gtotal j=1 G j
Solving Circuits with Equivalent Resistors
• Series and parallel Resistor equivalents can solve some circuits • Method, make equivalent resistance to simplify • Go between series and parallel as needed • Produce one final equivalent resistance • Use voltage and current divider equations • Get I & V for each element
Example Solving Circuits with equivalent Resistors
• Consider circuit with R2, R3 in parallel R1 all in series with R4 • For the R2, R3 side
R2+3 = R2 + R3 =1000 + 3000 = 4000
• Now get the parallel equivalent
1 1 1 1 1 3 = + = + = R1||2+3 R1 R2+3 2000 4000 4000
4000 R = = 1333.3 Ω 1||2+3 3
• Adding the series resistance
Rtotal = R4 + R1||2+3 =1000 +1333.3 = 2333.3 Ω
• Thus current from the source is
V 14 Itotal = = = 6 mA Rtotal 2333.3
Example Circuits with equivalent Resistors Continued
• Voltage across R4 and parallel section is
VR4 = I4 R4 = 1000 ×0.006 = 6 V
V1 = V − I4 R4 = 14 − 1000 ×0.006 = 8 V
• And the current in the parallel resistors
V1 8 I1 = = = 4 mA R1 2000
V1 8 I2 = = = 2 mA R2+3 4000
• Solving for the voltages
VR2 = I2 R2 = 0.002 ×1000 = 2 V
VR3 = I2 R3 = 0.002 × 3000 = 6 V
Advantages & Disadvantages: equivalent Resistors Method
Advantages • Simply guided by pattern of circuit • Easy to understand
Disadvantages • Can be quite time consuming • Some circuits cannot be solved this way
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 R1 and variable resistor R3 • Right side known R2 and unknown Rs • Place a very sensitive meter between the middle nodes • Best is a galvonometer • Voltages balance and no current ig flows if
R R 3 = s R1 R2
• If know the R1 R2 R3 very accurately can measure Rs accurately
R3 Rs = R2 R1 • 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
Circuit Definitions
• Node: point where several current paths meet:
• Branch: a current path connecting only two node • Branch contains 1 or more devices eg resistors • Note: a node may have many branch connections
• If 2 nodes are connected by a wire • Then combine them into a single node
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