Powerpoint

Impedance and AC Analysis

Introduction

The purpose of this lab is to analyze inductors and the affect that changing frequencies can have on the impedance in a circuit.

Procedure

We set up the following circuit






The function generator was set to generate 5V at 1000 Hz

We measured each component and garnered the following values:

R_L = 8.3 Ω

R_ext = 67.2 Ω

V_in,rms = 4.93 V
I_in,rms = 65.0 mA

Note: The voltage reading differs from the function generator because of its internal resistance

We then used the measured values to complete the following calculations:

V_in/I_in = Z_L = 75.8 Ω

Z_L = sqrt((R_ext + R_L)^2 + (ωL)^2)

ω = 2πf = 6280 rad/s
L = 1.16 mH

We then found the value of a capacitor needed to cancel the impedance provided by the inductor

1/ωC = ωL

C = 1/(ω^2*L) = 2.19 *10^-5 F

We then built the following circuit:

V_pp, CH1 = 1.5V
V_pp, CH2 = 20mV


deltaT = 0.4 ms


Phase difference = (deltaT) *6240* 360 ° = 184.32 °

Analysis

Frequency (kHz)	V_in (V)	I_in (A)	|Z_in| (Ω)
5	4.6	0.073	63.0137
10	4.2	0.0719	58.41446
20	3.51	0.0691	50.79595
30	3.51	0.0653	53.75191
50	4.5	0.0558	80.64516

1. The largest current should be seen at 1 kHz, as this is when the imaginary part cancels out.

2. V_L = Z_L/Z*V_in

V_L = 2.8574 + 1.2737j

Phasor = 24.0243 °

3. The circuit is more capacitive at frequencies below 1kHz because of the fact that the impedance provided by the capacitor increases when frequency is decreased.

4. The circuit is more inductive at frequencies above 1kHz because of the fact that the impedance provided by the inductor increases when frequency is increased.