Hi EEVBlog Community,
this is my first Post in this Forum although I am a EEVblog viewer for a long time. I first want to introduce myself a little bit to give you an overview on what my knowledge with electronics is.
I'm currently graduating in Computer Science and came to electronics about two years ago. Until now I've only done projects in the typical digital/Arduino Domain and some of them actually got featured on Hackaday:
here and
here.
I wanted to dig into some more "analog stuff" and for this I thought about reverse engineering an already available product. I've chosen a small PSU because I thought it would be perfect for the purpose and it still is rather simple. I used the PSU that I bought some weeks ago. It runs under the name QJ1502C and is quite readily available on eBay and amazon. It is a 15V/2A unit.
The internal construction of the PSU consists of 4 seperate boards:
- The main "power board"
- a voltage/current adjust board (that really only consists of 2 pots and 2 connectors)
- a output board that holds the shunt and also the 2 banana plugs
- the display board with 2 ICL7107
Other components are just a Fused IEC input jack (the Earth connection is properly secured with a shape proof washer to the case
) and the main transformer. it has 4 different windings as it seems (I relly don't have any expirience with transformers):
- primary (obvious)
- 0V/13V/22V as three fat wires
- 0V/12V with thin wires
- 0V/17V/17V with thin wires
As a reverse engineering method I used the photo and print on transparent foil that Dave used in one of his videos. I started off by adding all the components on the board to the layout and give them the correct value/name that matches the real board. then i started to add the traces and tried to group the components together while i was moving on.
The power board is only single layer so it was quite an easy task to get the schematic. The output and adjust board also was quite simple. I didn't bother to reverse engineer the display board for now but it basically consists of two identical configurations of the ICL7107 and a local 7805.
I think I understand how the voltage regulation happens in the board but I'm struggling with the current limiting and a few other stuff (I've commented on the layout where I thought it was necessary and where I've thought I know that a component/section is good for.
My questions (also commented on the schematic):
The output of IC2B goes into a diode in reverse direction. Just today I've watched Daves video about the 4XXX-Series Clock and learned that you can use such a configuration to build an OR-Gate (very interesting
), but for this you need a pull up resistor on the anode of the diode. In this configuration all diode anodes are connected together and nothing else is connected there (I've double checked this over and over...). How is the output of the OpAmp switching the transistor?
V13 and V14 are in darlington arrangement and they switch the big transistor on the back of the unit (connected on SWITCH) which is a
2N3055 which has a base-emitter voltage of 1.5V. I understand that you maybe don't want to drive the 3055 directly with the opamp, but why the second (D313) transistor (which looks quite beefy (TO220) and also has a dedicated heatsink).
How does the current limiting work? I think that it is handled by IC1A (Which is also connected to the 4 common anode arrangement) but it looks like IC1A is only connected to the voltage adjust pot. The current adjust goes directly into the non inverting input of IC2B. Am I wrong and IC1A actually is the voltage adjust and IC2B is the current adjust?
Why did they use an extra chip for the third opamp (IC1A) when they still had 2 unused opamps in IC2
how does IC2A switch the relay to the 22V rail? I noticed it switches when I set the voltage above ~7.3V but the voltage divider R24 and R35 produces a 4.6V at the inverting input of IC2A. However, if I would switch the values of R24 and R35, the math works out to ~7.3V (coincidence?).
Why isn't the negative supply of IC1 and IC2 connected directly to ground but instead with the resistor and the diode? wouldn't then the negative supply be liftet by 0.6V due to the diode drop? But why are R2 and R3 then not the same value?
I think thats all for now. I hope I'm not asking for too much.
Of course I may have made some mistakes in the schematic. If you find something that doesn't make sense at all, please let me know.
Again, I'm still a beginner, please don't be so hard to me for my missing knowledge in, what for you may be, basic electronics.
Daniel