Lab 7
Gareth Beckham
Brief7VoltageDividersandReducingPhaseShift.pptxCapstone: Two Circuit Problems
This lab presents a pair of circuit design problems which are designed to test the knowledge you have gained in this class.
Problem 1 is a simple resistor-based voltage divider, capable of providing a voltage of any value lower than the supply voltage.
Problem 2 deals with reducing the power factor in an AC circuit – a way of reducing the current drawn in an AC circuit that has inductive circuit elements.
Without proof, if we have any number of series resistors, we can determine the voltage across resistor by using the voltage divider equation.
Voltage Dividers
Resistive Voltage Divider Application
A resistive voltage divider uses an input DC voltage, VIn . We can choose R1 and R2 to produce the output Voltage, Vout that is used to power another device. In this case, the resistance values start off as “Unknowns”. Procedure:
Find the equivalent resistance:
Find the total current using Ohm’s Law
Since the output voltage is the voltage across R2:
We choose R1 and R2 so that produces the fraction of VIn to get our desired VOut
Voltage Divider Analysis
For example: We have a device that requires 3 Volts DC to operate, but we only have a 10 Volt DC Supply.
So, VOut = 0.3 VIn . From the Previous Slide:
Then
Rearranging this equation:
Or,
Voltage Divider Analysis (2)
Let’s choose R2 to be (maybe because we happen to already own a resistor!) What value of R1 would give us our desired VOut?
Check our Work using the Voltage Divider Equation:
3 Volts
Check!
Adding the Load Resistor
Now assume that RL is connected to the divider circuit (which, after all, is the purpose of the divider). The resistance below R1 is the parallel resistance of R2 and RL. From previous labs we know that the equivalent parallel resistance of RL and R2 is
The resistance of the circuit is no longer . It is now
The addition of RL changes the total current, which will change value of VOut.
We must therefore recalculate the value of the output resistance to determine the effect on VOut.
For example, for V=10VDC, R1 = 2330Ω, and R2 = 1000Ω, assume that RL is 1000Ω, and the desired Vout = 3V. Then the parallel combination of RL and R2 is:
Using the Current Divider Formula:
The Voltage across the Load Resistance is therefore:
This is not close to our desired VOut of 3Volts!
Adding the Load Resistor (2)
Adding the Load Resistor created another current path in our circuit! It changes the resistance, which changes
We can get around this problem by choosing our resistors such that RL >> R2 :
It turns out that a simple “rule of thumb” for a voltage divider is that RL ≥ 10 R2.
If RL is big compared to R2, most of the current will still go through R2, and very little current will flow through RL , and the circuit will “behave” almost as if RL was not there, so VOut will be nearly the same
Why Didn’t it Work?
Rule of Thumb:
From before: so choose
We still require:
Using the Current Divider Formula:
The Voltage across the Load Resistance is therefore:
This is close to our desired VOut of 3Volts!
Let’s Fix It!
Real-World Voltage Dividers
A voltage divider is usually specified with a range of acceptable output voltages, VOut.
The range of RL is also normally specified. For example, say that the range of RL is 50KΩ-100KΩ, and that the desired VOut is 3V, ± 10%, with an input voltage of 10V
Using RLMin and RLMax, we need to choose our R1 and R2 such that
Rule of Thumb:
In our example, RL MIN = 50 KΩ. Then we can choose R2 = 5 KΩ.
We are once again shooting for 3 Volts, so we arrive at the same relation between R1 and R2 (refer to the “Voltage Divider Analysis” slides):
Or, R1 = 11.7KΩ.
Resistors we can buy have standard values. The nearest standard values to the values above are 5.1KΩ and 12KΩ. So, choose R2 = 5.1 KΩ, and R1 = 12KΩ.
Finding the Correct Resistors
According to our “spec,” VOut MIN = 2.7V.
The parallel resistance of RL MIN and R2 is
Then VOut MIN = Check!
The “spec,” states that VOut MAX = 3.3V.
The parallel resistance of RL MAX and R2 is RL2=(5100·100000)/(5100+100000)=4852Ω
Then 2.88V. Check!
Our voltage divider, with the resistor values as shown, satisfies the “spec.”
Finding the Correct Resistors (2)
Reducing the V-I Phase Angle
All industries use AC motors (air conditioners, assembly line motors, etc.).
Electric motors work due to large coils of wire. A coiled wire is an inductor!
These large inductors represent an inductive impedance, which consumes power
Reminder of RL AC Circuit Analysis
In an RL circuit with a sinusoidal AC voltage applied, current lags the voltage by phase angle
From Lab 5, the ω-domain current is
where Vp is peak voltage, L is the inductance in Henry’s, and
I is a complex number in the ω-domain, with the imaginary part due to the inductive impedance, jωL. If the inductive impedance in the circuit was reduced to 0, all inductive current would cease and the overall current magnitude would decrease.
Inductive Currrent is NOT imaginary
Inductive current is mathematically imaginary, but physically real; it can be observed and measured. It requires that additional current be provided by the power company, and it is carried by a transmission line into the circuit.
Removing inductive current from the circuit would reduce the amount of current that is supplied to the circuit.
Adding Capacitance
The impedance of an Inductor:
Recall that the Impedance of a capacitor is
In a sinusoidal AC circuit, inductors and capacitors produce imaginary impedances of opposite signs!
In the circuit at right, the total impedance is:
)
Hence, the imaginary impedance “goes away” when:
For a given frequency, you can solve this for Capacitance as a function of inductance. Then, given the inductance and frequency, you can solve for the capacitance that will give you the negative capacitive impedance that will cancel out the inductive Impedance. The phase angle between the current and voltage will be zero when this is accomplished.
