PowerPoint Presentation
Sensing Systems and Signal Processing
Dr Richard
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Signal Conditioning
Signal Conditioning
How big is the signal?
Does the signal currently meet the input requirements of the next stage?
Is it in the right form to be captured by an acquisition system?
Is there noise on the signal, and where does the noise come from?
Is there sufficient signal to noise to see the signal of interest?
Can we improve the signal to noise ?
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Signal Conditioning
Next Stage
Conversion
Amplification
Signal conditioning
Anything you need to do to get the signal into the right form for subsequent processing comes under signal conditioning.
Linearization – if the sensor produces a nonlinear output you may deal with this here, these days this tends to be done in software
Amplification – to get the signal big enough for the next stage
Filtering – to remove out of band noise or drift.
Other things: Electrical isolation, Power to the sensor (excitation) and monitoring, conversion (current to voltage).
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Conversion
For a simple Photodiode: we get a photocurrent that is proportional to the light falling on the device.
We generally convert the photocurrent to a voltage using a load resistor (100R) such that our output signal for the DC light level is simpy
dc out = Ip. Rl
We do this because it is much easier to deal with voltages than currents in circuits. This conversion makes life much simpler.
We will revisit this when we look in detail at the SRAS instrument.
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Amplification
The signals we often encounter are small and so we may need to amplify them to get them into a range that can be digitised.
This might be done in a single stage with just one amplifier, but if a lot of gain is required it may need to be achieved with several stages.
For example: +/-1mV signal needs to interface to an ADC with an input range of +/-1V
We need a voltage gain of 1000, which is 60dB of gain.
A single amplifier might be able to do this (depending on the bandwidth) or several might be chained together.
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Amplification
Consideration must be given to the impedance of the sensor circuit and the impedance of the amplifier circuit.
The sensing circuit will have a particular impedance (which you may have no control over). This impedance will influence the signal transfer to the amplification circuit.
Let’s look at a simple example:
Microphone and loudspeaker
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Amplification
A microphone converts sound energy into electrical energy, so we can represent this as a signal generator and a source resistance. Typical values of the source resistance are ~500 ohms.
A loud speaker is essentially an amplifier and a device that converts electrical signals into sound (coil: magnet and cone). The input resistance of the loudspeaker is typically ~10 ohms.
When we connect them together we get this equivalent circuit:
Loud Speaker Driver
Microphone
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Amplification
If Vs is 25mV and the gain of the driver A is 100, what is Vout?
We’ll use Rs = 500 ohms, Ri 10ohms.
Vin = Vs x (Ri/(Rs+Ri)) = 25mV x (10/510) = 0.49mV
Vout = A*Vin = 49mV
This voltage probably wont be big enough to driver the speaker.
We might have thought we were going to get a gain of 100, but the actual output signal is only ~ twice as big as the source.
This is because the source impedance is mismatched to our driver circuit.
Loud Speaker Driver
Microphone
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Amplification
Loud Speaker Driver
Microphone
What can we do about it?
We could increase the gain A so the speaker works – but this will also amplify noise and become unstable for high gains
We need to reduce Rs or increase Ri.
We may not be able to change Rs as this could be an inherent property of the sensor design. Ri will have limitations as this is part of the design of the driver circuit (e.g. changing it could impact the frequency response)
We need a device that has:
a high input impedance – so the voltage get drops across it’s inputs
a gain of 1 – stable operation and it doesn’t change the signal scale
a low output impedance – so it can interface to other circuits with moderate input impedance (the driver in this case)
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Amplification
A device that meets these requirements is a Buffer.
Example: Ra = 100k ohm, Rb = 1 ohm, other values as before
V across Ra = 25mV. (100k/(100k+0.5k))=24.88mV
Vin = 24.88mV.(10/(10+1))=22.61mV
Vout = A.Vin =22.61mV*100 =2.26 V
Much better!
What can we use for a buffer?
microphone
loud speaker
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Amplification
Voltage Buffers commonly used are:
Common collector amplifier:
Disadvantages
Adds delay to the signal which might be important if timing is important.
Can reduce the bandwidth of the circuit, or distort the signal due to differences in the response for different frequencies
Consume power
Common collector
Voltage gain is ~1
High input impedance
Low output impedance
Refer back to H62EPC
Very high input impedance (~Mohm)
Low output impedance (~10 ohms)
Gain is 1 if no feedback resistor used
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If the signal of interest is within a small known frequency range then the time domain SNR can be improved by filtering.
The noise in this case is white noise (shot noise) and covers the entire spectrum removing this noise improves the SNR of the signal.
Can be even more important if there are other noise sources present.
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1/f corner ~10’s Hz
50Hz mains pickup
Comm’s bands
Frequency Hz
White noise (shot)
Other interference signals could be present from power supplies, equipment etc.
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SNR – Noise Sources
Let’s look at noise for a specific case – the photodiode we looked at last week.
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Noise Sources
Minimum detectable power with a PD
Photodiodes are not ideal devices – they are subject to various sources of spurious signal (noise)
Main sources are noise are
Shot noise due to dark leakage current
Johnson (thermal) noise due to the shunt resistance of the photodiode
Dark noise, from current that flows in the absence of light
Shot noise
real reverse-biased photodiodes have a non-zero current
this is called the dark leakage current (iD)
the dark leakage current produces a shot noise current
IS = (2.e.iD.B)0.5
IS is the shot noise current (Arms)
e = 1.6 x 10-19 C is the electronic charge
iD = dark leakage current (A)
B = system bandwidth (Hz)
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Noise Sources
Johnson noise
generated by the shunt resistance of the device
IJ = (4.k.T.B / Rd)0.5
IJ = Johnson noise current (Arms)
k = Boltzmann’s constant (1.33 x 10-23 JK-1)
T = absolute temperature (K)
Rd = shunt resistance (Ω)
Noise Equivalent Power (NEP)
NEP = RMS noise current (A) / R (AW-1)
The shot noise and Johnson noise are uncorrelated, so they are added in quadrature
IN = (IS2 + IJ2)0.5
NEP = IN / R(λ)
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A photodiode has a dark leakage current of 1 nA, a shunt resistance of 500 MΩ and a responsivity of 0.5 AW-1 at the wavelength we are using, if the bandwidth B is 1 Hz, standard room temperature (25 degrees Celsius):
calculate:
shot noise current =
Johnson noise current =
total noise current =
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A photodiode has a dark leakage current of 1 nA, a shunt resistance of 500 MΩ and a responsivity of 0.5 AW-1 at the wavelength we are using, if the bandwidth B is 1 Hz:
shot noise current = 1.79 x 10-14 A
Johnson noise current = 5.63 x 10-15 A
total noise current = 1.88 x 10-14 A
NEP = 3.76 x 10-14 W
IJ = (4.k.T.B / Rd)0.5
IJ = Johnson noise current (Arms)
k = Boltzmann’s constant (1.33 x 10-23 JK-1)
T = absolute temperature (K)
Rd = shunt resistance (Ω)
IS = (2.e.iD.B)0.5
IS is the shot noise current (Arms)
e = 1.6 x 10-19 C is the electronic charge
iD = dark leakage current (A)
B = system bandwidth (Hz)
IN = (IS2 + IJ2)0.5
NEP = IN / R(λ)
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Noise Sources
A photodiode has a dark leakage current of 1 nA, a shunt resistance of 500 MΩ and a responsivity of 0.5 AW-1 at the wavelength we are using, if the bandwidth B is 1 Hz:
shot noise current = 1.79 x 10-14 A
Johnson noise current = 5.63 x 10-15 A
total noise current = 1.88 x 10-14 A
NEP = 3.75 x 10-14 W
Shot noise tends to dominate in reverse-biased photodiodes, especially for large area devices operated at high bias
If a photodiode is operated at zero bias, then Johnson noise tends to dominate
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Conversion
A photodiode is essentially a current generator or a light controlled current source
In general, instrumentation prefers to work in voltage mode, so the photo-current must be converted to a voltage
Simplest method for converting a current to a voltage is to pass it through a resistor
Is = signal current
Il = leakage current
In = noise current
Cd = diode junction capacity
Rd = diode parallel shunt resistance
Rs = diode series resistance
Rl = load resistance
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Conversion
Very simple
Very linear
(neglecting noise)
Transimpedance gain = Rl Ω
Should we just use a very large resistor?
Is = signal current
Il = leakage current
In = noise current
Cd = diode junction capacity
Rd = diode parallel shunt resistance
Rs = diode series resistance
Rl = load resistance
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Conversion
Referring to the PD equivalent circuit, the bandwidth of the PD and resistor front-end depends on the PD capacitance and the value of the load resistor
fRC = 1 / (2πRl Cd)
e.g. if Rl is 1 MΩ and Cd = 10 pF, the output rolls off at around 16 kHz, far too low for many applications
So should we just use a very small resistor?
Is = signal current
Il = leakage current
In = noise current
Cd = diode junction capacity
Rd = diode parallel shunt resistance
Rs = diode series resistance
Rl = load resistance
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Conversion
A small resistor will certainly speed the circuit up
However, a small resistor has larger Johnson noise, so the system SNR is reduced
In general, using a small load resistor is not a good idea
IJ = (4.k.T.B / R)0.5
Is = signal current
Il = leakage current
In = noise current
Cd = diode junction capacity
Rd = diode parallel shunt resistance
Rs = diode series resistance
Rl = load resistance
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Conversion
Can we increase the system bandwidth without reducing the SNR?
We can reduce Cd by applying a large reverse bias
Cd is a depletion capacitance
increasing the reverse-bias increases the depth of the depletion region (w), so capacitance goes down as Cd = εA/w
this can reduce Cd by up to a factor of 10
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Conversion
When using a load resistor, the full signal swing appears across Cd
If we make both terminals of the PD operate at constant voltage:
no swing across Cd
no capacitive current
Reducing the voltage swing implies a small load impedance (which makes Johnston noise bad)
How can we achieve this without degrading the SNR?
Is = signal current
Il = leakage current
In = noise current
Cd = diode junction capacity
Rd = diode parallel shunt resistance
Rs = diode series resistance
Rl = load resistance
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Conversion
Use an operational amplifier (op amp)
very high open-loop gain
input can be a virtual earth
The inverting input draws no current
However, feedback resistor R forces the voltage at the inverting input to be close to zero at all times
The op amp and feedback “clamps” the voltage across the PD
the swing across Cd is greatly reduced
the bandwidth is greatly improved
Vout=-Rf.Iin
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