Doubt in output of 7805

Hi,

I attached two circuits here. These two boards share same A,B,C lines for mux and decoder and common ground.





I used to power these boards separately using 9V supply but these two board are interfaced always.

My doubt is when i power the board 1 only and not the board 2, the POWER ON led in board 2 is lightly glowing. When i measure the voltage i really shocked because it measures 3V at the output of 7805 and 2V at the input of 7805 in board 2.

The above case happens when i power the board 2 and not power the board 1.

My question is...
1. How the output of 7805 is greater than its input
2. Without powering the board 2,from where does the voltage come from?

Please clarify my doubts...

Thanks,
Murugesh

Board 1 is partly powered by the interface lines through CMOS substrate diodes, in this case MO-L and PIC RB0. To avoid exceeding IC maximum ratings, the interface lines between different power supply domains should have series resistors, or use buffer ICs without substrate diodes.

I dont understand how to do this. Can you please elaborate me little more.

FvM said:

the interface lines between different power supply domains should have series resistors, or use buffer ICs without substrate diodes.

Click to expand...

Also i removed the connection (MO-L and RB0). Now only have the a,b,c lines of mux and decoder, enable line for decoder, ground as common between two boards.
But i still have a output of 3V and input of 2V at the 7805 of board 2 when i power the board 1 only.

The board 1 feeds the power (high output) through the outputs, in the input many ICs (usually used as input and output) there will be a diode which is for protection of the IC which maintains the input below the Vcc that will give the more input voltage into Vcc line so that is given reversly to the 7805...

whatever the pin is if you construct a buffer in the board 2 inputs, It will eliminate the problem...

Thanks venkadesh..
How do i construct a buffer or is there any ic's readily available?

Also since you have mentioned like this..

Venkadesh_M said:

there will be a diode which is for protection of the IC which maintains the input below the Vcc that will give the more input voltage into Vcc line so that is given reversly to the 7805...

Click to expand...

Since the voltage given reversely to the 7805, will it affect or destroy 7805?

you can use ULN series with output pulled up with 7404 in series (to not affecting your existing logics) or you can use cmos non inverting buffers... whatever to a power IC you shdnt feed it reversely...

But i still have a output of 3V and input of 2V at the 7805 of board 2 when i power the board 1 only.

Click to expand...

Why do you say "but"?

In this case (opposite to the original discussed power board 2 case) lines a,b,c are feeding the reverse power.

or you can use cmos non inverting buffers...

Click to expand...

Not generally. Standard CMOS families have substrate diodes also with non-inverting buffers. I keep my suggestion of series resistors (e.g. 1 - 10 k). There might be still a certain reverse powering, but safe due to resistor current limiting.

I dont know whether i am correct or not. But i just want to know about my point. What happens if a place a diode (anode at output and cathode at input of 7805)?
Will this help?

Also instead of buffer ic if i use the transistor as a switch(output taken from collector with a pull up resistor) . Placing 7 transistors in board 2 ( 3 mux abc, 3 decoder abc, 1 decoder enable) and powering these transistor from 7805 in board 2 - To avoid reverse powering of board 2 when board 1 is only powered ON.

Use a single transistor in board 1 for MO-L pic B0 line and powering this transistor from 7805 in board 1 - To avoid reverse powering of board 1 when board 2 is only powered ON.

Please correct me if i am wrong...