Lecture Examples Week 6 Question on Conversion
We often need to convert a signal from one form to another, for example in the case of a photodiode we generate a current proportional to the light falling on the device. Currents are often more difficult to deal with when designing circuitry and so we need to conver this to a voltage.
What is a simple way we can do this?
We can pass the current through a known resistor to produce a voltage proportional to the current
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Question: For the system in the diagram above, if the light falling on the photodiode is 25mW and the photodiode has a responsivity of 0.4 A/W at this wavelength. Choose a suitable load resistance to produce an output voltage which can be digitised by a simple microcontroller.
We need to work out the photocurrent.
Then use V=IR to work out a sensible value for R. In this question you were not given the value for V, but in your experience you have worked with different microcontrollers and so can pick a sensible value. e.g. 5V for arduino, 3.3 V ESP32,RPi etc.
Ip = 25e-3*0.4 %Amps
Ip = 0.0100
V = 3.3; %Volts
R = V/Ip %Ohms
R = 330.0000
Here 330 Ohm resistor will map to max value of micro controller ADC range, so use something slightly smaller to have some head room.
300 Ohm is sensible here.
Can you think of any limitations where this approach may cause issues?
A limitation to this approach is where you care about the bandwidth of the device (RC time constant). if the capacitance of the device is large then you may not be able to use a larger load resistance to match to your ADC and meet your bandwidth requirements.
Question on Gain and ADC range
An ADC has an input range of 0-3.3V we wish to digitise a signal where the variation in voltage produced by the sensor is 5.5mV. How much voltage gain do we need to use the full range of the ADC? Express your answer in dB.
Gain_V = 600
Gain_dB = 20*log10(Gain_V) %Voltage Gain in dB
Gain_dB = 55.5630
If the signal had a DC offset what would we need to do differently?
Question on impedance matching and buffers
A sensor has an internal series resitance of 1k it is connected to a digitiser with an input impednce of 50 . What proportion of the sensor voltage is seen by the digitiser?
delta_V = 5.5e-3;
range = 3.3-0 ;
Gain_V = range ./delta_V
%range of signal
%range of ADC input (max – min voltage)
%gain in volts
The equivlanent circuit is a voltage divider with the sensor series resitance in series with the digitiser input resistance.
Rs = 1000;
% not given but also not really needed, if include choose a value
% digitiser 50 ohms
% sensor votlage 1000 ohms
Vi = Vs*(Rd/(Rs+Rd)) % Vi value (using Vs above)
Vi = 0.0476
Vpct = 100*Vi/Vs % as a % of Vs Vpct = 4.7619
Vpct = 4.7619
How could this be improved? Give an example device and show how this improves the situation.
Use a Buffer, for example an op-amp in voltage follower configuration, where the gain is one. This has a very high input impedance ~1M and a small output impedance ~10 .
Vbi = 99.9001
% without using Vs the ratio is just the votlage divider ratio.
Vpct = 100*(Rd/(Rs+Rd)) % just looking at the resitance ratio
Rbi = 1e6; % buffer input R
Rs = 1000; % sensor votlage 1000 ohms
% without using Vs the ratio is just the voltage divider ratio.
Vbi = 100*(Rbi/(Rs+Rbi)) % just looking at the resitance ratio
At the input to the buffer we now drop 99.9% of the signal voltage.
The output of the buffer will be dropped across the input of the digitiser. Here we use the output resistance of the buffer (~10 ) with the digitiser impedance.
Vd = 83.3333
tot = Vd*Vbi / (100)
tot = 83.2501
Here we drop 83% across the input to the digitiser.
Before we had ~4.7% from sensor to digitiser. Now we have 99.9%*83.3% = 83.25%
A much improved situation. We could have used a buffer with a lower output impedance to improve this still futher.
Question on Quantisation noise
A 16 bit ADC with and input range of +/-2.5V is used to digitise a signal. What is the expected level of quantisation noise for this device?
We need to know the change in voltage per level.
Rbo = 10; % buffer output R
Rd = 50; % Digitiser input R
% without using Vs the ratio is just the voltage divider ratio.
Vd = 100*(Rd/(Rbo+Rd)) % just looking at the resitance ratio
range = 2.5- -2.5
levels = 2^bits
levels = 65536
delta = range / levels
delta = 7.6294e-05
%Vax – vmin
% number of levels
%voltage representing a change for 1 level
Quant_noise = delta / sqrt(12) % standard deviation of the quantisation noise in volts Quant_noise = 2.2024e-05
What is the max SNR that can be obtained with this device?
We can use the simple equation here.
SNR_max = 1.76+6.02*bits %in dB SNR_max = 98.0800
Lectures Examples Week 8 Example Sheet Questions Biomedical question
Describe how signals and signal processing are used in biomedical applications. Explain the impact this has to the wider world giving examples of specific applications. Discuss the types of techniques and approaches used.
Medical ultrasound uses an array of piezoelectric transducers to produce images of soft tissues according to mechanical contrast, particularly at discontinuities in acoustic properties. Although various modes of operation are available, typically B-scans are used to produce images along the direction of propogation of ultrasound energy – where back-reflections are presented as varying contrast along each penetration vector. Applications include scanning the abdomen of pregnant women to track fetal development and morphology, or to detect injuries or malformations of soft tissues. Impacts include the diagnosis of disease, abnormalities, ultrasound- guided surgery and injection.
NDE application question
Why is the microstructure of aerospace materials important? Briefly describe a technique that can be used to measure the grain structure of materials.
Several important material properties are structure sensitive:
• yield strength,
• fracture toughness, • thermal conductivity
These are all sensitive to microstructure parameters, such as:
• Mean grain size
• Degree of randomness, (both size and orientation) • Clusters of grains all oriented in the same direction
SRAS can be used to measure the grain structure because the speed of sound of surface wave depends on the crystallographic orientation of the material. By measureing the speed of sound across a sample the grains can be image.
SRAS does this by measuring the frequency of the surface wave that is generated with a fixed wavelength. The speed of sound (c) is then determined by using c=fL with a known wavelength(L) and the measured frequency (f)
Biomedical ultrasound question
Why is ultrasound useful for imaging in the body? How is an image reconstructed in a simple linear array ultrasound machine?
Ultrasound will pass through the body and allow us to sense in places where we cannot see directly with optical techniques. Differences in the mechanical properties cause differences in reflection and absorption of the sound leading to contrast in the image.
Ultrasound is non-ionising – the energy the sound carries is sufficiently low that it does not cause damage to the tissue it is propagating in (there is an exception to this for HIFU where it is used to destroy cancerous tissue).
Ultrasound scanners provide real time, low cost, with reasonable resolution and contrast and are widely used in hospitals around the world for a wide range of applications.
You could scan the transducer around (in a line) to produce a cross sectional image or B scan. However, transducer arrays are usually used so that the scanning can be done electronically.
The transducer array emits ultrasound pulse on each element in turn. It records the trace and then moves onto the next element. Each signal is processed and forms an image of x vs z (space vs depth)
This gives an image of a slice through the object.
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