Download Analog Path Amplification/Attenuation Resistive divider --

1. Analog Path
a. Amplification/Attenuation
i. Resistive divider ---LEAVE OUT?
ii. Op amp design
iii. Two inverted op amps
iv. Programmable gain Amplifier
b. Filter Design
i. 1st order filter
1. -20dB per decade
ii. 8th order filter
1. -160dB per decade
c. ADC in Micro controller
2. Power Management
a. 5Volt
b. Battery Charging
i. USB charging
ii. LiPO charging
3. PCB Design
Analog Sensor
Amplification/
Attenuation
8th Order Low
Pass Filter
Analog to
Digital
Converter
Figure 1: Block Diagram of the Analog Path
Figure 1 above shows the pathway that the analog signal needs to take going through this system.
The analog signal is first generated from the analog sensor. This sensor can be measuring
anything from temperature, pressure, vibration or etc. The next step is the amplification or
attenuation stage. The amplification or attenuation will be done by using a programmable gain
amplifier. The next stage is the Low Pass Filter. The filtering for this system will be done using
an 8th order low pass filter. The final stage is the Analog to Digital converter. This will be done
inside of the micro controller that was chosen.
The analog sensor will be connected to the circuit through the 2.5mm jack. From the customer
requirements, the analog sensor can input an analog voltage anywhere between 0 and 10V into
this system.
Amplification and Attenuation
The original customer requirement for this project was for a universal sensor input. This could
not be accomplished while accomplishing the other requirements as well. So the input of the
system was changed to be from 0-10V. The microcontroller that was selected only has an input
range of 0-3.3V. To accommodate for this different range in voltages, the input range from the
sensors will need to be attenuated.
There are a few different ways that the voltage can be attenuated. One way that the voltage can
be attenuated is a resistor divider circuit. This is not the best way to do the attenuation because
the voltage division would change for different load impedances. Figure 1 represents one kind of
resistor divider circuit. To find the output voltage you would use Equation. With this type of
circuit the load gain of the circuit, the output divided by the input, would change if the input
voltage changed. Another reason why this is not the best way to attenuate voltage is because if
another resistor was added in series or parallel at Vout, the gain would change. The gain of this
circuit will change depending on the load impedances.
Figure 1: Voltage Divider Circuit
𝑉𝑜𝑢𝑡 =
𝑅2
𝑅1 ∗𝑅2
∗ 𝑉𝑖𝑛
Eq. 1
Figure 2: Second Voltage Divider Circuit
𝐺=
𝑉𝐿
𝑉𝑆
=[
1
][
𝑅3
)
(𝑅2 +𝑅𝐿 )(𝑅3 +𝑅𝐿 )
1+𝑅𝑠 (
1
𝑅
2+𝑅3
]
Eq. 2
𝐿
Figure 2 shows a different type of voltage divider. The gain for this circuit can be seen in
Equation 2. This circuit has the same problem as the other resistor divider circuit in that the
output changes if the load changes. For both of these circuits the load impedance will also affect
the output voltage. When the load changes the output resistance of the voltage divider circuit will
change. This change of resistance will cause the output voltage to increase or decrease depending on
the change in the load resistance.
One way to overcome this impedance matching issue is to use op amps to attenuate the voltage.
Op amps have zero output impedance so there is no need to worry about any impedance
matching. Op amps can be used to have a positive or negative gain in a circuit. Equation 1 shows
the equation needed to find the gain of a non-inverting op amp. The gain of a non-inverting op
amp can never be less than one. Equation 4 shows the equation used to find the gain of an
inverting op amp. Using two inverting op amps in series a positive gain of less than 1 can be
achieved.
𝑉𝑂𝑢𝑡
=1+
𝑉𝐼𝑛
𝑉𝑜𝑢𝑡
𝑉𝑖𝑛
=
−𝑅𝑓
𝑅𝑔
𝑅𝑓
𝑅𝑔
Eq. 3
Eq. 4
Since the voltage for this system needs to have a gain of less than 1, to inverting op amps can be
used. The circuit in Figure 3 shows a circuit that attenuates a 10V signal to a 3.3V signal. The op
amp that was chosen was the LT1001 precision op amp. This op amp was chosen because of the
high accuracy and the low noise. The feedback resistor in the first op amp is 33kΩ and the input
resistor is 100kΩ which yields a gain of -1/3. The second op amp has a gain of -1 because the
feedback resistor and the input resistor have the same value. This results in a total gain of 1/3.
Figure 3: Attenuation using Two Inverting Op Amps
Figure 4 shows the simulation result of the circuit in Figure 3. The green curve is the input signal
which is a sine wave that ranges from 0-10V. The output signal is the blue curve which is a sine
wave from 0-3.3V.
Figure 4: Simulation Results of Op Amp Attenuation Circuit
Figure 5 shows the results from building the circuit shown in Figure 3 on a bread board. The
input signal on the function generator was set to a sine wave with amplitude of 5V and a DC
offset of 5V. This input signal should have been a 10Vp-p signal but the oscilloscope was
reading a 9.68Vp-p sine wave. This could result from a problem with the function generator not
reading the exact signal that it is putting out. All of the resistor values were the same that were
used in the circuit used in the simulation in Figure 3. These resistors were not ideal and there
actual values can be seen in Table 1. The gain in the actual circuit was 0.38 instead of the 0.33
that was simulated. Using Equation 4 and the actual resistance values from Table 1 the gain was
calculated to be 0.34. The percent error in the actual gain measured and the gain that was
calculated using the resistor values is 10.5%. This percent error is high and one of the reasons for
this is that the same oscilloscope probe was not the same for both the input and output. Another
reason that this might not be the most accurate measurement is because of non-idealities and
noise that can be caused from other components that were on the breadboard.
Figure 5: Hardware Results of Op Amp Attenuation Circuit
Resistor
Ideal Value(kΩ)
Actual Value
(kΩ)
Rg
100
98.4
RF
33
32.9
R1
10
9.78
R2
10
9.87
Table 1: Actual Resistance Values used in Op Amp Attenuation
𝑃𝑒𝑟𝑐𝑒𝑛𝑡 𝑒𝑟𝑟𝑜𝑟 =
(𝐴𝑐𝑡𝑢𝑎𝑙−𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑)∗100
𝐴𝑐𝑡𝑢𝑎𝑙
=
(0.38−0.34)∗100
0.38
= 10.5%
Eq. 5
A programmable gain amplifier was used in this system. The chip that was chosen was the
MAX9939 PGA. This PGA is a SPI PGA with gains from 0.2Volts/Volt to 157 Volts/Volt. The
maximum input signal to this DAQ can be 16.5V because the input of the ADC on the
microcontroller is 3.3V and the minimum gain is 0.2V/V.
Don’t have any simulations
What Else?
Low Pass Filter Design
A low pass filter is used to pass low frequency signals through a system and filter out the
unwanted high frequency components of a signal. The cutoff frequency is the frequency point
where the energy starts to be reduced. Any signal that is at a higher frequency than the cutoff
will be attenuated. Low pass filters are used for smoothing data.
Figure 6: First Order RC Low Pass Filter Circuit
Figure 6 above shows a circuit design for a first order low pass filter. Figure 7 shows the
frequency response for a first order filter. In Figure 7 the cutoff frequency is set to 1 rad/sec. At
the cutoff frequency the magnitude of the signal going through the filter starts to decrease a 20dB per decade. Equation 6 shows how to calculate the gain in dB. So when the gain is
decreasing at -20dB per decade that means the output voltage is decreasing as the input voltage is
remaining constant. Signals with frequencies higher than the cutoff, will be attenuated.
𝑉
𝐺𝑎𝑖𝑛(𝑑𝐵) = 20 ∗ log(𝑉 𝑜 )
𝑖𝑛
Eq. 6
Figure 7: Frequency Analysis of a First Order Low Pass Filter
For this design the MAXIM MAX7403 8th order filter low pass filter (LPF) was used. An 8th
order filter attenuates the signal at -160dB per decade past the cutoff frequency. Since the 8th
order filter attenuates faster, more of the unwanted high frequency noise will not be allowed to
pass through. Figure 8 shows the frequency response of the 8th order filter. After the cutoff
frequency, 10 kHz, the slope of the curve is 8 times steeper than the first order filter shown in
Figure 7. Figure 8 shows that all signals with a frequency of 10 kHz or less will have a gain of
0dB. A gain of 0dB is the equivalent of multiplying by 1. Anything with a frequency higher than
10k Hz will have a gain of something much less than 0dB which is equivalent of multiplying by
a number that approaches 0.
Figure 8: 8th Order Filter Frequency Response Simulation Results
Figure 9 shows the actual measurements that were taking using the MAX7403 8th order LPF. As
you can see the actual measurements look very similar to the simulation shown in Figure 8. This
proves that this chip will work in our system as a very efficient low pass filter.
Figure 9: 8th Order Filter Frequency Response Hardware Results
Figure 10: Input and Output of the 8th Order Filter Simulation Results
Figure 10 shows the input and output of the filter. The yellow curve is the input to the filter and
the red curve is the output of the filter. There is some time delay between the input and output of
the filter but it is in the µs range. The Input signal was changed and for all signals with a
frequency below 10 kHz the output had the same amplitude as the input. For signals with higher
than 10 kHz frequencies, they were attenuated and the output was zero. This helps prove that the
filter is an efficient filter. The output of the filter cannot go below 0.25V. This can easily be
fixed by using the offset pin on the chip. The signal will be offset by 0.25V but the ADC on the
microcontroller can only have an input of 3.3V. This will change our maximum input signal to
the whole system from 16.5V to 15.25V. Even though we have zener diodes protecting the
inputs to the ADC, we will round down the maximum input voltage allowed from 15.25V to 15V
to be safe.
Figure 11 below shows the actual measurement that was taken using the MAX7403 chip
Figure 11: Input and Output of the 8th Order Filter Hardware Results
Power Management
This DAQ device is being powered by a 3.7V Lithium Polymer battery. This battery will be
charged using a USB. USB was chosen because USB has a 5V pin. Also the FRDM KL25Z
board is charged via a USB. The MCP73831 chip will be used to charge the LiPo battery. This
chip is a linear charge management in a small 2 mm x 3 mm package. This chip along with the
USB charging will allow for quick easy charging of this DAQ device.
None of the components in this device require 3.7V that the LiPo battery can supply. This is
resolved by using a 5V boost converter. The converter that was chosen was the TI TPS61200.
This chip was chosen for because it has high efficiency and it can operate with a varying input.
This important because the voltage of the battery might not always be the same. When the
battery is in use and not charging, the output voltage will decrease slowly over time.
The MCP73831 LiPo charging chip and the TPS61200 were chosen because these chips were
used on the power cell LiPo Charger/Booster. This charger/Booster was purchased to test these
chips and charging a LiPo battery.
PCB Design
After going through a few iterations, the final PCB design for this will be a shield board that will
go on top of the Freescale KL25Z devolvement board. The development board is used because it
is reducing a lot of risk. The development board has the KL25 microcontroller already on the
board. All of the pins are routed to the holes on the outside of the board. Pins will be put into
these holes and these holes line up directly with our shield board. Figure 12 shows the PCB
layout for the shield board.
Figure 12: Finalized PCB Layout