Difference between revisions of "Lab 6 RS"

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[[File:Tek00023.png | 600 px]]
 
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:Input voltage given by <math>V_{in} = V_0 \left ( 1 - e^{-t/\tau}\right ),\ V_0 = 6\ V</math>
 
:Input voltage given by <math>V_{in} = V_0 \left ( 1 - e^{-t/\tau}\right ),\ V_0 = 6\ V</math>
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:From the plot above my measurement of rise time is:
 
:From the plot above my measurement of rise time is:
  
  <math>\tau = 53\ us</math>  
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  <math>\tau = 53\ us</math>
 
 
  
 
==='''Rise time <math>\tau^{\prime}</math> measurements  '''===
 
==='''Rise time <math>\tau^{\prime}</math> measurements  '''===

Revision as of 22:26, 11 February 2011

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Lab 6 Pulses and RC Filters

Differentiator

Construct the circuit below selecting an RC combination such that RC [math]\lt 10^{-4} [/math] s

TF EIM Lab6b.png


Taking [math]R = 989\ \Omega[/math] and [math]C_{out} = 1.255\ \mu F\ \Rightarrow \tau = RC = 1.24\ ms[/math]

Adjust the pulse generator to output square pulses which at [math]\tau \approx[/math] RC/10, RC and 10 RC

1) [math]\tau \approx \frac{1.24\ ms}{10} \approx 0.124\ ms [/math]

2) [math]\tau \approx 1.24\ ms [/math]

3) [math]\tau \approx 10 \cdot 12.4\ ms \approx 124\ ms [/math]

Measure[math] V_{in}[/math] and [math]V_{out}[/math]. Sketch a picture comparing[math] V_{out}[/math] and [math]V_{in}[/math]. ( 3*5 = 15 pnts.)

Table1. Differentiator circuit for different pulse width
[math]\tau \approx 0.124\ ms [/math] [math]\tau \approx 1.24\ ms [/math] [math]\tau \approx 12.4\ ms [/math]
Tek00017.png Tek00018.png Tek00019.png


From the plots above (at point a):

[math]\tau = 0.124\ ms \rightarrow V_1 = 8.0\ V,\ V_2 = 8.0\ V[/math]
[math]\tau = 1.24\ ms \rightarrow V_1 = 8.0\ V,\ V_2 = 8.0\ V[/math]
[math]\tau = 12.4\ ms \rightarrow V_1 = 8.0\ V,\ V_2 = 8.0\ V[/math]

Questions

What happens if the amplitude of [math]V_{in}[/math] is doubled.( 5 pnts.)

The amplitude of [math]V_{out}[/math] is doubled as well.

What happens if R is doubled and C is halved?( 5 pnts.)

The times constant [math]\tau_{RC} = \frac{1}{(R/2)(2C)} = \frac{1}{RC}[/math] becomes unchangeable so nothing is really happened


Integrator

Now repeat the above experiment with the resistor and capacitor swapped to form the low pass circuit below.

TF EIM PulsedRCLowpass.png

Table1. Integrating circuit for different pulse width
[math]\tau \approx 0.124\ ms [/math] [math]\tau \approx 1.24\ ms [/math] [math]\tau \approx 12.4\ ms [/math]
Tek00020.png Tek00021.png Tek00022.png


From the plots above (at point b):

[math]\tau = 0.124\ ms \rightarrow V_1 = 8.0\ V,\ V_2 = 0.4\ V[/math]
[math]\tau = 1.24\ ms \rightarrow V_1 = 8.0\ V,\ V_2 = 4.4\ V[/math]
[math]\tau = 12.4\ ms \rightarrow V_1 = 8.0\ V,\ V_2 = 8.8\ V[/math]

Questions

What happens if the amplitude of [math]V_{in}[/math] is doubled.( 5 pnts.)

The amplitude of [math]V_{out}[/math] is doubled as well.

What happens if R is doubled and C is halved?( 5 pnts.)

The times constant [math]\tau_{RC} = \frac{1}{(R/2)(2C)} = \frac{1}{RC}[/math] becomes unchangeable so nothing is really happened

Pulse Sharpener

The goal of this section is to demonstrate how well the circuit below can sharpen an input pulse

TF EIM Lab6a.png

1.) The first step is to create an input pulse which is rounded, similar to the output of the integrator circuit when RC = 10 [math]\tau[/math]. You can do this using a capacitor shorted across the output of the pulse generator. This will essential be coupled to the input impedance of the pulse generator and form a low pass circuit.

As a result the input voltage is given as

[math]V_{in} = V_0 \left ( 1 - e^{-t/\tau}\right )[/math]

where

[math]\tau=R_{out} C_{out}[/math]
[math]R_{out}[/math] = impedance of the function generator at output which produces V_{in}
[math]C_{out}[/math] = capacitor shorting the function generator output to ground (not shown in the above picture)


Taking [math]R_{out} = 52\ \Omega[/math] and [math]C_{out} = 1.255\ \mu F\ \Rightarrow \tau = RC = 65.3\ us[/math]

2.) The output should be given by

[math]V_{out} = V_0^{\prime} \left ( 1 - e^{-t/\tau^{\prime}}\right )[/math]

where

[math]\tau^{\prime} =\left ( \frac{R_1 R}{R_1+ R}\right )C_1[/math]


Taking [math]R = 49.9\ \Omega ,[/math] [math]R_1 = 51.2\ \Omega[/math] and [math]C_1 = 1.225\ \mu F\ \Rightarrow \tau^{\prime} = \left ( \frac{R_1 R}{R_1+ R}\right )C_1 = 31.0\ us[/math]

Make measurements of the rise time [math]\tau[/math] and [math]\tau^{\prime}[/math]. The rise time is defined as the time it take the pulse to go from 10% of its max value to 90% of its max value.( 5 pnts.)

Rise time [math]\tau[/math] measurements

Tek00023.png


Input voltage given by [math]V_{in} = V_0 \left ( 1 - e^{-t/\tau}\right ),\ V_0 = 6\ V[/math]
At time [math]t = 0 \rightarrow V_{in} = 0[/math]
At time [math]t = \tau \rightarrow V_{in} = 6 \left ( 1 - e^{-1}\right ) = 3.79\ V[/math]
From the plot above my measurement of rise time is:
[math]\tau = 53\ us[/math]

Rise time [math]\tau^{\prime}[/math] measurements

Tek00024.png

Output voltage given by [math]V_{out} = _0^{\prime} \left ( 1 - e^{-t/\tau^{\prime}}\right ),\ _0^{\prime} = 3\ V[/math]
At time [math]t = 0 \rightarrow V_0^{\prime} = 0[/math]
At time [math]t = \tau^{\prime} \rightarrow V_0^{\prime} = 6 \left ( 1 - e^{-1}\right ) = 1.90\ V[/math]
From the plot above my measurement of rise time is:
[math]\tau = 22\ us[/math] 


Compare the measurement of [math]\tau^{\prime}[/math] to what you expected based on your measured values of [math]C_1[/math], [math]R_1[/math] and[math] R[/math].( 15 pnts.)

Questions

1.) Qualitatify, why is [math]\tau^{\prime} \lt \tau[/math]?( 10 pnts.)

2.) How is [math]V_{out}[/math] worse than [math]V_{in}[/math]( 10 pnts.)



Forest_Electronic_Instrumentation_and_Measurement

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