Design Details

Through we could find the existing circuit diagram on the Internet, my partner and I still decided to design our own circuit, even it will contain any problems. After searching many materials and master the basic knowledge of a regulated power supply, we started to draw our first design.

Firstly, we determined to devise every important part of the whole circuit and combine them together to run after the simulation of every part works correctly.

Now, I will show you every step of our design.

Step 1. Simulate the part as power supply

From the right simulation figure, it can be found that a sinewave with amplitude at about 2 Volts is shifted upward by 5 Volts. It shows the combined two power supply can provide the expected input voltage. However, it disappears a problem：the output AC signal always shifts to the left, that means, the phase of the mixed power supply always changes.

Step 2. Invert the orignall signal and mix all signals

This step is to mix the original signal from source power and the inverting signal processed by the operational amplifier. If the two signals can be completely offset, then the output voltage ought to be a straight line at zero voltage, and that means, the designed circuit is reasonable to invert the original signal. From the below simulation diagram, it shows the output verify our thinking.


Step 3. Solve shifting-phase problem and filter DC signal

From the right simulation diagram, V1 is the signal voltage from the sources and V0 is the output voltage filtered by two capacitors. They just have 5 volts in difference without any phase difference.

Step 4. Connect the inverting amplifier to the filtered DC signal circuit

Step 5. Use transistor package to adjust and amplify the filtering DC signal

Step 6. Finally regulate and filter output voltage

Step 7. Combine all the parts together

Step 8. Modify the circuit

Step 9. Modify the design circuit again

Step 10. Change a little of the thinking of the design and draw the final draft

3rd Lab

We have our successfully simulated circuit shown as below. The output voltage range of this circuit is around 0 to 12.5 volts, and it can provide an appropriate current for a load resistor of 50 ohms.

Component List
D1: 6.8V Zener Diode
C1: 1mF
C2: 1uF
C3: 1uF
T1: BC107
OP1 and OP2: LM741CN
R1 and R2: 1k ohm
R3 and R4: 10k ohm
R5: 1k ohm
R6: 4k ohm
R7: 1k ohm potentiometer
R8 and R9: 1k ohm

We built the circuit according to the circuit diagram above.

• First test:

1. Progress: The output voltage is steady, it is a DC output.
2. Problem1: The output voltage cannot be adjusted when we change the resistance of the potentiometer.
3. Problem2: The output voltage is not zero, but quite low compared to the input voltage.
4. Actions taken: We used a multimeter to measure the voltage drop accross each component, to see whether it was working properly.
5. Discovery: We find out that the voltage drop across R4 is too large. It was strange for the voltage to drop from 5V  to 1V after passing R4. The voltage signal was not amplified by the operational amplifier. Aactually, the voltage signal drops from 1V to 0.2V.

The supervisor told us there exists some potential problem in our design. But he could not tell where is the mistake. To keep in process, I decided to build the circuit from google, while my partner continues with the designing part.

I finished building the circuit from google.

• Second test:

1. Progress: The output voltage is DC and the amplitude of it is much larger than that of the designed circuit.
2. Problem: The output voltage does not change with the variation of potentiometer.
3. Actions taken: I suspect there is some connection mistake in the circuit. So I check the connection line by line.
4. Discovery: I find out there is one connection error around the potentiometer. The third pin of the potentiometer is not connected to the circuit at all. It means that the whole potentiometer is connected to the circuit, it functions as a normal resistor. As a result, the output voltage will not change with the potentiometer.
5. Result: After the correction of the mistake, the output voltage can response to the change of potentiometer.

Here is the plot of the input and output voltege.

2nd Lab

We came up with our first designed circuit shown as below, and the simulation of the circuit was correct. (Detailed designing processes are in another post named "Our complete designing processes of this project".)

The first draft of our design

We tried to build the circuit on the breadboard. After we built the circuit, we tried to power the circuit and display the output voltage on the oscilloscope  But the output voltage was always zero and could not be adjusted when we tried to change the resistance of the potentiometer. The supervisor told us that there existed some problem in our design. In his opinion, the whole circuit works like a potentiometer rather than a power supply. He suggest we add a 50 ohms load to the output ports and try to power it. If we can manege to generate a current of several amperes through it, then the circuit is approximately successful.

We tried to simulate the circuit using PSpice. However, PSpice is not advanced enough and it lacks of some essential components that the circuit needs. Simulation failed.

We tried another soft ware called LTspice which was recommended by the supervisor. But there was some problem with the input transformer, the circuit cannot be simulated.

The demonstrator came and tried to fix the problem. But she cannot find out the problem in the circuit either.

We searched Internet to learn about how to use LTspice, because it is a brand new software for us who come from XJTLU. It took us the rest of the lab day.

1st Lab

We received the electronic components we ordered before Christmas holiday. There are 2 kinds of transistors. Unfortunately, they are not exactly required by the circuit. Besides, transistors only are not enough to build the circuit.

We discussed together and decided to order new components which are exactly what the circuit need. We printed a new component list, and search the Internet to find out more information about every component we need. We also went to the technicians to check whether they have the resistors and capacitors we need. Finally, we need to order 3 kinds of transistors from the website. And we handed in the final component list to the technicians, hoping the components could arrive before the next lab.

We tried to design our own circuit this afternoon.  We went to the computer room to search further about the project. What's more, we reviewed some knowledge from the first year to learn more about the important component: operational amplifier.

An operational amplifier has two differential inputs but usually only a single-ended output and should be connected to a couple of opposite DC power supply. It is a high-gain electronic voltage amplifier since the output from an op-amp is hundreds of thousands of times higher than the voltage difference between its input terminals. Most op-amp are Integrated Circuit (IC) chips. What we chose is UA741CN.

The power supply voltage Vcc and VEE power the operational amplifier and in general define the output voltage range of the amplifier.

In this project, in order to invert the original AC signal, we need to use a negative feedback to the fundamental op-amp configurations to constitute a inverting amplifier.

The feedback circuit provides a fraction of the output signal, βVo, which is subtracted from the input source signal, Vs. The resulting signal, Vi, which is also called the error signal, is the input to the amplifier which in turn produces the output signal Vo = AVi. The gain Vo/Vs of the inverting amplifier is given by

In the inverting amplifier, output voltage that is fed back into the inverting input by the feedback resistor network and the voltage gain can be calculated as 

The negative sign for the gain indicates that the polarity of Vo is opposite to the polarity of Vin.