Applying 5vdc after the (unpowered) regulator

wapo54001

Senior Member
I have an 08M soldered into a circuit, and I'm figuring out how to program it on the board without disturbing any other components. The 10K and 22K resistors are already in the circuit.

To do this programming, I would like to deliver the +5, ground, program in, and program out signals from a separate connector while the board is not powered up through the on-board regulator (board is physically disconnected from normal 12v supply).

That means I will be delivering the +5v after the regulator (LM2931cm), which I know is normally frowned upon. The only component connected to the regulator on the 12v supply side is a .22uF ceramic capacitor, and a 100uF capacitor on the 08M side.

I'm thinkiing that this will be safe (regulator will not be harmed), but I'm not sure. Am I right?
 

Andrew Cowan

Senior Member
I know that if you connect up a regulator and put 12V into the output of a LM7805, there is a bang and lots of flying casing. Although I don't know if 5V would be any different, I would avoid doing it.

Andrew
 
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hippy

Technical Support
Staff member
I've seen a number of circuits where a regulator is used or disconnected and power is supplied to regulator output ( The AXE210 XBee/GPS board is one, page 9 ), so while not recommended, people do it and it seems to work. Perhaps it depends on the make/model of regulator ?

If you were concerned and could stand a slight voltage drop you could diode mix the regulator output and programming voltage. You can put a diode in the common leg of the regulator ( pointy-end to 0V ) to bump the regulator output up so after the drop it's still 5V.
 

sghioto

Senior Member
You could connect the external 5 volts directly to the input of the LM2931. Specs show a 4.5 volt output with a 5 volt input. Still plenty of voltage for programming.

Steve G.
 

wapo54001

Senior Member
Actually, the board is made without accomodation for a diode, so that's not going to work.

I read and re-read the datasheet for the 2931 and found no mention of this problem or any schematic with a diode wrapping from output to input to handle a reverse voltage situation. That's what gave me the courage to go this route. Just getting cold feet and was hoping someone could put my concerns to rest.
 

papaof2

Senior Member
This is one of those cases where a current-limited power supply is ideal - set the output for 5 volts and just enough current to operate the PICAXE.

John
 

premelec

Senior Member
A diode from regulator output to input [backwards] often will keep bad things from happening as then voltage put on the output will charge the regulator input capacitor and only subject the regulator to less than 1 volt reverse polarity.
 

wapo54001

Senior Member
I looked for that trick in the application notes, and it did not appear. No reference to anything about the output being higher than the input.

Also, I wondered if disconnecting the power at the input would eliminate a path that could cause problems
 

sghioto

Senior Member
As suggested before, just connect the 5 volts to the regulator input the same place where the 12 volts is connected. No diodes needed. You stated the board is disconnected from the 12 volt supply so what's the problem.

Steve G.
 

Dippy

Moderator
The diode 'trick' is (certainly used to be) commonly used and I've seen it mentioned in Regulator data sheets now and then.

But why not just try Steve's suggestion and apply 5V to the reg. normal input?? It won't blow up. What's the problem?
 

wapo54001

Senior Member
The problem is that the board is laid out and built. I've designed in four pads in a row just in front of the 08M, one each connected to the board's +5, Ground, and the 08M's Program In and Program Out pins, with the appropriate 22K and 10K in place.

I want to be able to program the board by pressing a gizmo with four spring-loaded pins against the four pads to make connection, but I got nervous about the 5V regulator, and decided to aske for opinions.

If I have to deliver +5 to the input of the regulator (which is accessible through one pin of a separate Sub-D connector) I may as well deliver +12, let the regulator work, and abandon the idea of programming the board solely through my gizmo.

Edit: Gizmo = Mill Max Spring-loaded connectors.
 
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hippy

Technical Support
Staff member
This is where there's a chorus of "should have prototyped and tested before committing to PCB, that way there's no doubt if it will or will not work".

As it is, given that other people do it you may as well try what you have and see how it goes.
 

moxhamj

New Member
Re Premelec "A diode from regulator output to input [backwards] often will keep bad things from happening as then voltage put on the output will charge the regulator input capacitor and only subject the regulator to less than 1 volt reverse polarity."

A 4001 will be less than 0.6V and a schottky even less.

I've done this before and nothing bad happened.

No room on the board? Put the diode on the track side across the pins of the regulator. There is nothing wrong with fixes like this, especially on prototypes. Just fix it next time you rebuild the board.
 

boriz

Senior Member
Internal circuit of 7805: http://datasheet.digchip.com/321/321-4-000511-7805.pdf

At first glance, it looks to me as though you will reverse bias a few transistors, but 5v is unlikely to do any harm.

I suggest that you connect up your 5v to the 7805 output and measure the current flowing into the 7805 output. It should be almost zero. If it is, then you are probably ok. If not, then more experiment is necessary.
 

boriz

Senior Member
OOPS. Drinking and typing!. Sorry chaps.

I like the diode idea. Prolly work fine. I would still try the current measurement thing though.
 
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