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Circuit Design Help


I am just starting to play with the micro-framework and decided to build a power supply and volt meter for my first circuit.

The software has not been a problem but my circuit design is a bit rusty… like 30 years ago rusty. I think this is correct but can I get some feedback. I am unsure on the ground wiring and if I can multipurpose the bridge for the power supply and voltage reading.




The transformer in the drawing is a 1:1 line transformer, maybe you picked that one from a drawing program, but of coarse it needs to be a 230 : 12V

After bridge rectification the voltage is about 12 * 1.414 ± 17 Volt. Using the resistor divider you will have 4.2. volt in stead of the 3.3

I would suggest to use a transformer with a lower secondary voltage otherwise your voltage regulator can get pretty hot… It needs to dissipate (in)17 - (out)5 volt = 12 volt. using 1Amp produces 12 Watt op heat.


Thanks for the help.
Yes the transformer is a step down transformer, I am using visio as my design and it only has the simple transformer. Its actually a LP-580 110/220 : 24v (12-0-12).

I forgot about the 1.414 so I switched my transformer to an LP-430 6-0-6 and swapped out the 3k resistor for a 2k resistor. This should be 2.8 v max. I happen to have the 6-0-6 laying around as well.

So now I need to dissipate 8.5-5 = 3.5 volts using 1 amp = 3.5 watts.

Does this seem correct?

Another question, do I need to supply the mainboard with 5volts or 3.3? The pinout on the chip shows 5 but the power modules appear to provide 5.


There’s something wrong in both schematics that i didn’t notice initially. You don’t need the 2 left most diodes since you use a center tapped transformer. The center tab of the transformer remains your ground point. your 2.8 Volt and dissipation calculations are correct. The 10uF capacitor at the input of the 7805 is a to small to have a stable voltage. I would suggest to take 1000 uF

Depending on the type of 7805 u use, if i remember well some need at least 3 volt higher as the output, which mean 8V min in… In the calculation of 6 * 1.414 i did not take into account the lost in the diodes. Probably you need a transformer with a bit higher voltage 8 - 9 volt or use a 7805 with a lower dropout… Not sure if they exist.

Depending on your mainbord you use, most use 3.3Volt but some modules require 5 Volt too. You could add a Low Dropout 3.3 Volt regulator and connect it to the 5 Volt


I will remove the two lead diodes. The 10uf capacitor and min voltage are from the manufacturer. Its the CUI 7805-1000, I also have the 7803. According the doc 6.5 volt is the min. I will build it this weekend, I still need some more parts. Alternatively, I can try the 24 volt transformer and try it to see how hot it gets.


I see that you are using the switching version of the old 7805, which is great… no problems with heat !

In the attachment are some options to connect your transformer


T2 and T3 was were circuits I initially looked at, but I was not sure what to do with the terminal that is not connected. I dont like the idea of just leaving it live and not connected to something. The center tap transformers I have do not have a jumper on the center terminal.


Ok, then you need T2


Wouldn’t T1 and T2 have the same result?


Yes, they have the same result, but i understood (maybe wrong) that you have a transformer without center tap.


It has a center tap but only one wire on the center so I cannot jumper them.


I guess i am bit blond today :wink: I do not fully understand what you mean with “i can not jumper them”


@ moadib2k: Hi, you are using the bridge rectifier the connections are ok but there is no need to connect the center tap of the transfer to the ground because the you picked up the ground accurately from the rectifier and one more thing just for your knowledge that the voltage across the bridge rectifier is 24V not 12V because you are using two 12V terminals of center tap transformers, so it would affect your output voltage. Keep working on micro-frame work.