BIOEN 316 Biomedical Signals and Sensors Spring 2016
Homework 6
Due May 6 in lecture
Part 1. Bridge circuits
Figure 6 in Homework 5 is an example of a Wheatstone bridge circuit, which contains four resistive elements in a square or diamond. A voltage is applied across two opposite corners, while the output is obtained across the two other corners. It is useful because it allows us to measure a change in resistance as a change in voltage, and it allows the output to be positive or negative depending on which way the resistance changes, even though the actual voltages in the circuit are all positive when compared to ground. The following example will help explain this concept.
Thermistors are temperature-sensitive resistors. Typically their resistance decreases as temperature increases. They are good candidates for a bridge circuit, or at least a voltage divider. As an example, to build a –20° freezer one might use an Ice Cube Sensor from
Vishay. It is an NTC thermistor, meaning that it has a negative temperature coefficient: as temperature increases the resistance decreases. A thermistor’s temperature coefficient varies according to temperature, but at any given temperature it is very reliable. The Ice
Cube thermistor has a resistance of 96.4 kΩ at –20° C, and for every Kelvin increase in temperature its resistance drops by 5.8%. Note that the function R ( T ) is non-linear, but we can linearize the function around –20° C by evaluating its first derivative at that temperature.
1a. Show that you could use a simple voltage divider to achieve an output of 2.0 V when the temperature is –20°. A supply voltage of +5 V is applied across three components in series: the thermistor, a fixed resistance R
1
and a potentiometer of 10 kΩ or less.
1 Set up the circuit so the output voltage increases when the temperature increases. For R
1
, use any ¼-
W resistors available at EE stores.
2 These resistors will have some manufacturing tolerance (error), and you can adjust the potentiometer to accommodate the resistor tolerance. Determine the sensitivity of this circuit in volts/Kelvin.
1b. Show that you could get an output voltage difference of 0 V by using a Wheatstone bridge and any power supply voltages V plus
and V minus
. In figure 6, the two R
1
resistances should be equal, fixed (not adjustable), and within 10% or so of the thermistor value. R
2 should be replaced with the thermistor. R3, which is the sum of potentiometer plus resistor, should match the thermistor. Find the sensitivity of the bridge circuit in V/K.
1 A potentiometer is a variable voltage divider that is used in this case like a variable resistor. It has a screw that you can turn to change the resistance to any value from 0 to 10 kΩ. Try to find a fixed resistance R
1
2 adjust the variable resistance to get the total resistance you desire.
See the online catalog at www.ee.washington.edu/stores . I suggest starting with the “Resistors, 5%, 1/4 W” category, but note that there are other tolerance options. You may combine two resistors to get R 1.
that is
5 kΩ less than the total resistance you want. You will find something close but not exactly right, then you can
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BIOEN 316 Biomedical Signals and Sensors Spring 2016
Exercise 2.
In lecture we saw that a resistor and capacitor together act as a low-pass filter when the input voltage is applied across the two components in series and the output voltage is measured across the capacitor. To do this we used the impedance divider formula, G(jω) ≡
V
OUT
/ V
IN
= Z
C
/( Z
C
+ Z
R
), where V = V(jω) = F { v ( t )}. We then determined the magnitude and phase of G(jω) at ω=0, ω → ∞ , and the half-power frequency. The half-power frequency is also called the cutoff frequency f
C
or ω
C
.
Using the technique described above, show that a resistor and inductor together act as a low-pass filter when the input voltage is applied across the two components in series and the output voltage is measured across the resistor. a) Generate the complex formula for G(jω). If you like, you may rename L / R or R / L as a single variable to make the equations easier to write. b) Make the following three plots, in which ω is on a log scale (0.1, 1, 10, 100, etc.).
|G(jω)| vs. ω, where |G(jω)| is on a linear scale,
|G(jω)| vs. ω, where |G(jω)| is on a decibel scale, and
∠G(jω) vs. ω, where ∠G(jω) is on a linear scale from –π to π.
Notes:
⟿ You may use frequency in Hz or rad/sec, but remember the factor of 2π. The cutoff frequency that you calculate from R and L is in rad/sec.
⟿ You cannot get ω to reach zero on a logarithmic scale, so just use some small value as your minimum.
⟿ When making the decibel scale plot, find |G(jω)| as ω → 0, ω → ∞, ω
C
, ω
C
/10, and
ω
C
/100.
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BIOEN 316 Biomedical Signals and Sensors Spring 2016
Exercise 3.
The figure below shows a series RLC circuit with the output measured across the capacitor.
The diagram includes two probes ( V in a circle pointing to a wire). The left probe measures the input and the right probe measures the output voltage. Both are referenced to ground. a) Use impedances to derive the complex gain formula for this circuit, V
C
(jω)/ V
IN
(jω). b) What general type of filter does this circuit provide? c) Assume that R = 2 kΩ, L = 60 mH, and C = 25 μF. Can you find a frequency that makes
G(jω) completely imaginary?
Circuit for Exercise 3
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BIOEN 316 Biomedical Signals and Sensors Spring 2016
Vocabulary
Voltage gain, G
V
(jω), is the ratio of the output voltage to the input voltage. It is a generally a complex number, so it can be written as a magnitude and phase shift. Its complex nature implies that it is a vector, so it is sometimes written in bold.
Current gain, G
I
(jω), is the ratio of the output current to the input current.
Half-power frequency is the frequency at which the magnitude of the voltage or current gain, |G(jω)|, is 1/√2 of the maximum gain. Therefore, it is also the frequency where the power gain (square of the voltage or current gain) is ½. If the maximum gain is 1, i.e.
0 dB, then the voltage gain at the half-power frequency is 0.707 or –3 dB. RC and RL circuits usually have one half-power frequency, but RLC circuits (band-pass filters and others with a peak in the magnitude spectrum) often have two (one on the way up to the peak and one on the way down).
Cutoff frequency and break frequency are other names for the half-power frequency when used in high- or low-pass filters. The transition from pass to stop band is defined to be at this frequency.
A current divider is a pair of components in parallel, such that the incoming current is split between them. The combined resistance or impedance is smaller than the resistance or impedance of either one. The formula, using R’s, is
R
Parallel
=


1
R
1
+
1
R
2


− 1
Example: Finding the cutoff frequency in an R-L circuit. For this problem, assume the input voltage is applied across the two components in series and the output voltage is measured across the inductor.
Using an impedance divider formula, we find the complex gain.
G
RL
( j ω ) ≡
Vout (
Vin ( j ω ) j ω )
=
V
L
(
V
L j ω )
( j ω
+ V
R
)
( j ω )
=
Z
R
Z
L
+ Z
L
=
R j ω L
+ j ω L
=
R
L j ω
+ j ω
We then set the magnitude of the voltage gain to be 1/√2 of maximum and solve for ω.
Here, the maximum gain is 1 as ω  ∞.
1
2
=
R
L j ω
+ j ω
= j ω
R
L
+ j ω
=
ω
( ) 2
+ ω 2
( ) 2
+ ω 2
= 2 ω 2
ω = ± R
L
We know that frequencies always occur in +/- pairs in order to produce complex conjugates that produce real sinusoids, so we usually report only the positive ω. Note that
ω is in radians/sec, so divide by 2π to get Hertz.
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