Reverse Polarity Protection, Question

[LBNote]

Hi,

I've built a circuit using a 7805 regulator and a pic, I was wanting to run
it of a 6v Lantern battery. I was using a diode for reverse protection but I
since discovered the voltage drop caused the output of the regulator to
somewhere around 2.5 volts.

I was wanting to know is there a simple way to achieve reverse polarity
protection with out the consequences of the diode / voltage drop ?

Let me know if more information is required.


Lorne...

 

[Phil Allison]

Hi,

I've built a circuit using a 7805 regulator and a pic, I was wanting to run
it of a 6v Lantern battery.

** That is your first problem - a dry battery will drop to 1 volt per
cell before it is discharged.

1 x 4 = 4 .

See the problem ?

I was using a diode for reverse protection but I
since discovered the voltage drop caused the output of the regulator to
somewhere around 2.5 volts.

** There seems to be a missing word here.

I was wanting to know is there a simple way to achieve reverse polarity
protection with out the consequences of the diode / voltage drop ?

** Yep - use a fuse and a diode wired across the device input.

Wrong polarity = need for new fuse.


............. Phil

 

[LBNote]

** That is your first problem - a dry battery will drop to 1 volt per
cell before it is discharged.

1 x 4 = 4 .

See the problem ?





** There seems to be a missing word here.




** Yep - use a fuse and a diode wired across the device input.

Wrong polarity = need for new fuse.


............ Phil

Yep I see the problem, I'm going to be wasting a lot of battery life.
Regulator is going to cut out long time before the battery is exhausted.

Thanks for your help.

Lorne...

 

[Alan]

Hi,

I've built a circuit using a 7805 regulator and a pic, I was wanting to run
it of a 6v Lantern battery. I was using a diode for reverse protection but I
since discovered the voltage drop caused the output of the regulator to
somewhere around 2.5 volts.

I was wanting to know is there a simple way to achieve reverse polarity
protection with out the consequences of the diode / voltage drop ?

Let me know if more information is required.


Lorne...

First thing to remember is that the 7805 needs about 2.5v drop ACROSS
it (ie between input and output) for it to work and takes a fair few
milliamps itself to keep going.

I suggest you forget about the regulator and feed the pic from the 6v
battery via a diode. Won't the pic run of any voltage from 2.5v to
5.5v or so??? I haven't played with them myself.

If you really want to use a regulator consider using a low drop out
regulator like the LM2931 series. They are reverse protected as well
so you don't need the diode in series

HTH

Alan


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[The Real Andy]

If you really want to use a regulator consider using a low drop out
regulator like the LM2931 series. They are reverse protected as well

Good waste of a battery. I would be using a SEPIC switcher for maximum
efficency and to extract every last drop out of that battery. Linear
regulators are crap for battery powered devices.

 

[Mainlander]

Hi,

I've built a circuit using a 7805 regulator and a pic, I was wanting to run
it of a 6v Lantern battery. I was using a diode for reverse protection but I
since discovered the voltage drop caused the output of the regulator to
somewhere around 2.5 volts.

I was wanting to know is there a simple way to achieve reverse polarity
protection with out the consequences of the diode / voltage drop ?

Let me know if more information is required.

You are wrong, there is a voltage drop across the regulator, around 3
volts is quite normal for these regulators

The diode will only drop about 0.6 volts

You may have to use a zener diode for the voltage regulation if you
absoluetly must have 5V from a 6V source, a three terminal reg is not
going to do it.

 

[David L. Jones]

First thing to remember is that the 7805 needs about 2.5v drop ACROSS
it (ie between input and output) for it to work and takes a fair few
milliamps itself to keep going.

I suggest you forget about the regulator and feed the pic from the 6v
battery via a diode. Won't the pic run of any voltage from 2.5v to
5.5v or so??? I haven't played with them myself.

It does.
That is the easiest way if you don't have any voltage sensitive
devices hanging off the PIC.
If your battery voltage falls to say 4V then your PIC outputs will be
4V minus the doide drop (under no load) etc.

If you really want to use a regulator consider using a low drop out
regulator like the LM2931 series. They are reverse protected as well
so you don't need the diode in series

You still won't be fully utilising the battery though.
Best to use a DC-DC convertor in this case.
Could run your circuit from any battery voltage, even a single cell.
You can even get drop in DC-DC convertor replacements for the 7805.

Dave

 

[[email protected]]

Hi,

I've built a circuit using a 7805 regulator and a pic, I was wanting to run
it of a 6v Lantern battery. I was using a diode for reverse protection but I
since discovered the voltage drop caused the output of the regulator to
somewhere around 2.5 volts.

I was wanting to know is there a simple way to achieve reverse polarity
protection with out the consequences of the diode / voltage drop ?

Let me know if more information is required.


Lorne...

You could use a polarised connector. You wouldn't have the problem
of a wasted voltage drop across a diode. A DC-DC converter chip from
National Semiconductor (many available from RS Components, or
Farnell) can then be utilised to give you a consistant voltage even
if the battery voltage drops. Nat Semi have got a website that can
help you to design a converter. Failing that, Linear Technology
have a number of DC-DC converter chips, and have SwitcherCad,
which can also help design a solution.

www.national.com
www.linear.com

Regards
David Milne

 

[LBNote]

You could use a polarised connector. You wouldn't have the problem
of a wasted voltage drop across a diode. A DC-DC converter chip from
National Semiconductor (many available from RS Components, or
Farnell) can then be utilised to give you a consistant voltage even
if the battery voltage drops. Nat Semi have got a website that can
help you to design a converter. Failing that, Linear Technology
have a number of DC-DC converter chips, and have SwitcherCad,
which can also help design a solution.

www.national.com
www.linear.com

Regards
David Milne

Thanks Group.

After doing a bit of reading on the subject of DC-DC converters there was
mention about supply noise/ripple generated from the switching nature of the
converter. Not being a real teckie on these matters, My design uses a UHF TX
and RX Module around 433Mhz, will the converter (no specific on in mind
here ) produce to much noise/ for my UHF receiver to function properly? Or
are there counter measures I can take to get around this ?

I really like the idea of being able to power the device from 2 C cells or 2
D Cells much cheaper then a 6v lantern battery.

Regards

Lorne...

 

[David L. Jones]

After doing a bit of reading on the subject of DC-DC converters there was
mention about supply noise/ripple generated from the switching nature of the
converter. Not being a real teckie on these matters, My design uses a UHF TX
and RX Module around 433Mhz, will the converter (no specific on in mind
here ) produce to much noise/ for my UHF receiver to function properly? Or
are there counter measures I can take to get around this ?

Noise shouldn't be a problem. The usual filter caps on the output of
the converter should be enough, at worst you might need a choke on the
output too. Best to check with the UHF module data sheet though.
It's quite common to have a 7805 type linear reg on the output of a
DC-DC converter if really nice clean output is desired. For instance,
you might choose a 3V to 8V DC-DC converter and use a linear reg. Not
all that efficient, but clean power.
Some DC-DC modules are fully regulated and have really clean outputs.

I really like the idea of being able to power the device from 2 C cells or 2
D Cells much cheaper then a 6v lantern battery.

Sure is. D cells are the go.

Dave

 

[trash]

Hi,

I've built a circuit using a 7805 regulator and a pic, I was wanting to run
it of a 6v Lantern battery. I was using a diode for reverse protection but I
since discovered the voltage drop caused the output of the regulator to
somewhere around 2.5 volts.

I was wanting to know is there a simple way to achieve reverse polarity
protection with out the consequences of the diode / voltage drop ?

Let me know if more information is required.


Lorne...

Technology is getting smarter and we're getting dumber.
A bit of Occam's theorem here... the simplest solution is often the
best.

Why not just have 2 x 1N4004's in series. The voltage drop across
each will bring you down to 5V give or take a few points and the
condition of the battery.

Now you get the best of everything.
The voltage drop works in your favour
It doesn't use any extra power
It's cheaper
Simpler
provides you will double the protection. (hate to think one diode
wasn't enough)

Best of all.... you can even put a fuse in if it makes you feel safer



- trash

Yes I am an agent of Satan, but my duties are largely ceremonial.

 

[David L. Jones]

Technology is getting smarter and we're getting dumber.
A bit of Occam's theorem here... the simplest solution is often the
best.

Why not just have 2 x 1N4004's in series. The voltage drop across
each will bring you down to 5V give or take a few points and the
condition of the battery.

Now you get the best of everything.
The voltage drop works in your favour
It doesn't use any extra power

Yes it does. 0.6V x Ipic

It's cheaper

Not cheaper than 1 x 1N4004

Simpler

Not simpler than 1 x 1N4004

provides you will double the protection. (hate to think one diode
wasn't enough)

Best of all.... you can even put a fuse in if it makes you feel safer



- trash

Yes I am an agent of Satan, but my duties are largely ceremonial.

If you have voltage intolerant parts, then one 1N4004 would be enough
in this case. The PIC works to 6V, no need to drop to 5V.
If you have parts external to the PIC that need 5V then you can't use
this solution, unless you want to underutilise your battery capacity.

Occams only works if you have all details :->

Regards
Dave

 

[The Real Andy]

Technology is getting smarter and we're getting dumber.
A bit of Occam's theorem here... the simplest solution is often the
best.

Here the simple solution is not the best. Analyse the whole problem. Read
the whole thread.

 

[trash]

Here the simple solution is not the best. Analyse the whole problem. Read
the whole thread.

Ummm.. let me see. I never was one for passing an English test, but
I take it we want 5 volts from a 6 volt lantern battery ?
We also want reverse polarity protection ?

Now sometimes I don't believe everything I'm told, and in this case...
It works for me ! I have several 5V circuits that run off 6V
batteries in this fashion.





- trash

Yes I am an agent of Satan, but my duties are largely ceremonial.

 

[David L. Jones]

Ummm.. let me see. I never was one for passing an English test, but
I take it we want 5 volts from a 6 volt lantern battery ?
Yep

We also want reverse polarity protection ?

Yep

And presumably one would also want to extract the maximum life from
the battery, right?

Now sometimes I don't believe everything I'm told, and in this case...
It works for me ! I have several 5V circuits that run off 6V
batteries in this fashion.

What is the minimum operating voltage of your "5V" circuit?
What happens when the battery drops to 4V?, which would be about the
end of the life of a typical 6V battery. This gives only about 3V to
your circuit (4V - 0.5Vx2), will it still work at this voltage? If you
have a 5V circuit with a minimum operating voltage of 4.75V (very
common) then you are in trouble when your 6V battery drops to only
5.75V
Not mention that the voltage drop of a diode varies with current. A
series diode is not a good solution for a true 5V circuit.

Dave

 

[trash]

What is the minimum operating voltage of your "5V" circuit?
What happens when the battery drops to 4V?, which would be about the
end of the life of a typical 6V battery. This gives only about 3V to
your circuit (4V - 0.5Vx2), will it still work at this voltage? If you
have a 5V circuit with a minimum operating voltage of 4.75V (very
common) then you are in trouble when your 6V battery drops to only
5.75V
Not mention that the voltage drop of a diode varies with current. A
series diode is not a good solution for a true 5V circuit.

Ah yes, now I see where your going.
Yes, makes a lot of sense when you want to extend useable life.

I haven't ever had that problem. The circuits I've used have only
drawn a small amount of current compared to what the battery has to
offer and the battery was often recharged immediately after use.


- trash

Yes I am an agent of Satan, but my duties are largely ceremonial.

 

[LBNote]

It does.
That is the easiest way if you don't have any voltage sensitive
devices hanging off the PIC.
If your battery voltage falls to say 4V then your PIC outputs will be
4V minus the doide drop (under no load) etc.


You still won't be fully utilising the battery though.
Best to use a DC-DC convertor in this case.
Could run your circuit from any battery voltage, even a single cell.
You can even get drop in DC-DC convertor replacements for the 7805.

Dave

Ok Found a DC -DC converter, Got some samples of a national LM2621 on there
way, after reading the datasheet (most of it sunk in )
but I still have a simple question, what if the Voltage In is Higher then
the Voltage Out, as this seems to be a booster type setup,

What will happen in this case? should a just try it and see ? I'm only
talking about Vin being 1.5 > Vout while the batter is new.

Also where in OZ would be the best (Cheapest) place to source the inductors
for the circuit ?

Thanks

Lorne...

 

[David L. Jones]

Ok Found a DC -DC converter, Got some samples of a national LM2621 on there
way, after reading the datasheet (most of it sunk in )
but I still have a simple question, what if the Voltage In is Higher then
the Voltage Out, as this seems to be a booster type setup,
What will happen in this case? should a just try it and see ? I'm only
talking about Vin being 1.5 > Vout while the batter is new.

The datasheet should tell you the maximum input voltage. Don't go
higher than that, it's not designed for it.
If you want a lage input range that extends from below Vout to above
Vout then you'll need a combined boost/buck converter. These are
available, just don't have details on hand.

I thought you were going to use only a couple of cells?, in which case
the voltage will always be < Vout.

Also where in OZ would be the best (Cheapest) place to source the inductors
for the circuit ?

Try the usual places like Jaycar, Altronics, Oatley, Rockby etc.
Failing that, Farnell and RS I guess.

Dave

 

[LBNote]

The datasheet should tell you the maximum input voltage. Don't go
higher than that, it's not designed for it.
If you want a lage input range that extends from below Vout to above
Vout then you'll need a combined boost/buck converter. These are
available, just don't have details on hand.

I thought you were going to use only a couple of cells?, in which case
the voltage will always be < Vout.


Try the usual places like Jaycar, Altronics, Oatley, Rockby etc.
Failing that, Farnell and RS I guess.

Dave

Yeah I was only going to use a couple of cells, but I thought I 'd just ask
the question and if it was easy enough then widen the input voltage range of
the circuit.

Thanks

 

[Foley U. Matthews]

You could use a polarised connector. You wouldn't have the problem
of a wasted voltage drop across a diode. A DC-DC converter chip from
National Semiconductor (many available from RS Components, or
Farnell) can then be utilised to give you a consistant voltage even
if the battery voltage drops. Nat Semi have got a website that can
help you to design a converter. Failing that, Linear Technology
have a number of DC-DC converter chips, and have SwitcherCad,
which can also help design a solution.

www.national.com
www.linear.com

Regards
David Milne

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