Lecture Week 7. Quiz 4 - KCL/KVL Capacitors RC Circuits and Phasor Analysis RC filters Workshop

Transcription

1 Lecture Week 7 Quiz 4 - KCL/KVL Capacitors RC Circuits and Phasor Analysis RC filters Workshop

2 Quiz 5 KCL/KVL Please clear desks and turn off phones and put them in back packs You need a pencil, straight edge and calculator 20 minutes Keep eyes on your own paper Follow the same format as for homework

3 Capacitors A capacitor is a passive device that stores electrical charge in the form of an electric field A simple capacitor is made of two metal plates separated by a dielectric (insulator) The charge stored in a capacitor is given by the following equation Q = CV

4 Capacitor characteristics The current flow through a segment in a wire is: i t = dq dt Since Q = CV, we can derive another equation for i(t) in terms of capacitance, i t = CdV dt = C dv dt This means that current will flow across a capacitor, as long as the voltage is changing.

5 Capacitor under DC conditions A capacitor will behave as an open circuit under DC conditions. This means that if you apply a constant voltage to a capacitor and let it fully charge, it will stay fully charged until it is discharged. Current Flowing No current or charge will pass across the capacitor once it is fully charged!

6 RC circuits An RC circuit is of special interest since it is used in many fundamental electric circuits (timing circuits, filters, sensing, power storage) In this course, our main application will be to filter a signal. Therefore, we will see how an RC circuit can be used to filter, how it works, and why it is useful.

7 RC Circuit Analysis We can derive an equation for the RC circuit by using KVL 1 Capacitance Eq: I = C dv c dt 2 KVL: V s + V R + V C = 0 3 Ohm s Law: V s = IR + V C By replacing Eq (1) into (3), we obtain 4 V s = RC dvc dt + V C, a differential equation

8 RC Circuit Analysis Continued The RC relationship is a differential equation. V s = RC dvc dt + V C You will learn how to solve differential equations in MATH You will soon find out that solving differential equations is tedious, and it usually involves complicated integration steps. Math Wizardry! Pay attention to MATH 2326 if you want to be able to do this!

9 Phasor Domain Analysis: What is it? Phasor analysis allow us to convert complicated differential equations into simple algebraic problems using complex numbers (Z) Complex Number: Z = x + iy Imaginary Axis y Z = x + iy Real Part Imaginary Part x Real Axis

10 Phasor analysis: Capacitors and Resistors The details of how to go from time domain to phasor domain will be covered in a future lecture, but for now we will assume that the Phasor Machine takes time domain functions as inputs, and will generate a frequency domain complex function as an output. The resultant function is what we call the impedance. For a capacitor C PHASOR MACHINE 1 jwc For a resistor R PHASOR MACHINE R

11 Phasor Analysis: Capacitors and Resistors It will be easier once you see how it is done, and prove that it works!

12 1. Convert every component (R and C) to its phasor domain representation 2. Find the equivalent impedance by adding individual impedances in series or in parallel 3. Calculate the unknown variable by using ohms law, voltage divider, KCL or KVL RC Circuit in Phasor Domain Once we transform the circuit into the phasor domain we can analyze it using the circuit analysis tools we have learned so far (Series, Parallel, KCL, KVL, Voltage Divider, etc.) By following these steps you can analyze any RC circuit in the phasor domain:

13 Example: RC Circuit Analysis on Phasor Domain Lets derive the Phasor domain equation of the following circuit (using R for resistors, and 1 for capacitors jωc

14 Exercise: Phasor Domain Use the Voltage Divider method and Phasor Domain Analysis to Calculate V out Hint: in Phasor Domain, V in R = R and C = 1 jωc

15 Low-pass RC filter What happens as w 0? Z c =, open circuit, V out = V in What happens as w? Z c = 0, short circuit, V out = 0 V c = V in 1 jwrc + 1

16 High-pass RC filter What happens as w 0? Z c = 0, short circuit, V out = 0 What happens as w? Z c =, open circuit, V out = V in V in + jwrc V V R R = V in jwrc + 1 -

17 Bandpass RC filter A band-pass filter is a combination of a high-pass and a low-pass filter. Do you know how to build this circuit?

18 Bode Plots Will help to visualize and understand what is happening in an RC circuit and/or filter!!!

19 Bode Plots and Cutoff frequency A bode plot tell us the magnitude of the RC filter with respect to a range of frequencies. It is a useful graphical tool to determine where the amplitude of the output is starting to drop. The equation to find the frequency at which the amplitude will start to drop is give by the following equation: f c = 1 2πRC (Cutoff frequency)

20 Low-Pass Filter: Bode Plot Vin R Vout C low pass filter

21 High-Pass Filter: Bode Plot Vin C R Vout high pass filter

22 Bandpass Filter: Bode Plot

23 10 minute Break

24 Workshop

25 P1 (a) Determine the cut-off frequency for the low-pass filter shown below. Include all units and unit conversions for full credit. (b) If a signal frequency is increased to 10 khz, will its amplitude be diminished or will it remain intact as it is processed by the filter?

26 P2 Assume that the circuit below has reached equilibrium and the capacitor is fully charged. (a) Determine the value for VA, (b) the current flowing through R1, R2, and R3, and (c) the power consumed by R3. Show all equations required to solve this problem, all units, and all unit conversions for full credit.

27 What s Next in Week 8? Will introduce LAB Module IV EKG LECTURE Quiz 6 Cut-off frequency and bode plots Op-Amp Theory and Analysis Please bring laptops to all lectures and labs.

28 Questions?