Personal Projects

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

  • 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)

  • 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)

  • High pass filter: -3dB at 3.4 Hz

Triangular/Square Signal Generator

Triangular/Square Signal Generator

(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:

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

Dog average

Example#2

Cat average

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