Embedded Files
Microphone Amplifier Filter
Amplifies a very small signal such as that typically supplied from a sensor coil (magnetic field detector or a microphone) with Broken Coil Detection
Amplifies a very small signal such as that typically supplied from a sensor coil (magnetic field detector or a microphone) with Broken Coil Detection
If the coil is intact, then both the plus and minus inputs (Coil+ and Coil-) will be equal to that voltage and will cancel out in the differential op-amp circuit. Thus, only the signal differential generated by the coil will be amplified in this system.
Detects if the coil or the connecting wiring breaks
Band pass (a high pass and a low pass stage)
Band pass (a high pass and a low pass stage)
Low Pass filter: 3 poles and has a gain of 2. -3dB at 2.9 KHz
Final output of the circuit: high gain with mainly low pass filtering (Stages 1 to 5)
Final output of the circuit: high gain with mainly low pass filtering (Stages 1 to 5)
High pass filter: -3dB at 3.4 Hz
Triangular/Square Signal Generator
Triangular/Square Signal Generator
Triangular/Square Signal Generator
(Only Vcc+ and Vcc- with no Input Voltage)
(Only Vcc+ and Vcc- with no Input Voltage)
Outputs in different stages to get the desired gain/amplitude/ waveform
Generates a triangular waveform with variable frequency and adjustable minimum and maximum output voltage as shown in my built circuit and schematic.
Generates a derived square waveform as a second stage.
Final Output with two waveforms
The amplitude of the triangular waveform can be set with the two potentiometers.
Square Root Operation Circuit
Using an op-amp and BJTs, I achieved the square root operation between Vout and Vin to be:
Vout=0.1 √(Vin )
In Detail description and calculation:
In Detail description and calculation:
Firstly, we have a voltage divider, getting a 4k/(36k+4k) Vin=Vin/10 at OA1, Op-amp with Q5 would act as a voltage to current converter: IC5=V+/1k=Vin/10k A, Q6 and Q7 section would be simply a current source of 100µA as many examples we did at the beginning of this course. Afterwards, looking at the BJT loop Q1, Q2, Q3, Q4 while ignoring base currents, we get the following relationship: IC1*IC2=IC3*IC4, and looking at Q1 and Q5, we know that IC1=-IC5=-Vin/10k A, while I_C2= -100µA fixed by our current source while IC3=-IC4. Therefore, Vin/10k*100µ=I_C3^2 and rearranging we get: I_C3= -√(Vin/(10k*100µ)) and simplifying we get, I_C3=-√(Vin )/10k .
Lastly, the 1kΩ resistor I added would simply act as a current to voltage converter and gives us the relationship we want: Vout= - IC3*1kΩ=√(Vin )/10k*1kΩ, and simplifying this expression:
Vout=0.1 √(Vin )
Pet Classification
Implemented code for a machine learning pipeline that determines whether a 64×64 image is a cat or dog. Several approaches were used to distinguish between images, such as: closest average, nearest neighbor, linear regression.
Example#1
Example#1
Dog average
Example#2
Example#2
Cat average
Convolutional neural network method for better accuracy
Convolutional neural network method for better accuracy
Training Accuracy: 97.8
Test Accuracy: 97.25
Save an iPhone
Designed a case for an iPhone that electro-mechanically deploys when the phone is dropped
PCB and schematic design shown above
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