Simple voltage question

[DorkyGrin]

I'm trying to get a steady 5 volts from a 6 volt battery. The 5v will
power a circuit that draws a variable amount of current (between 70
and 100 ma). I've used LM7805 in the past but I know they need more
then just one volt difference.

I do have a 1/2watt 5 volt zener I could use but I can't seem to get
it to work properly. Would it be wired in series?

Or should I just use a dropping resistor, even though the current draw
changes?

Thanks

 

[Tom Biasi]

"DorkyGrin" <DorkyGrin@yahoo.com> wrote in message
news:d8ff209d-60f8-4ee4-80fe-902dad6a07cd@r66g2000hsg.googlegroups.com...

I'm trying to get a steady 5 volts from a 6 volt battery. The 5v will
power a circuit that draws a variable amount of current (between 70
and 100 ma). I've used LM7805 in the past but I know they need more
then just one volt difference.

I do have a 1/2watt 5 volt zener I could use but I can't seem to get
it to work properly. Would it be wired in series?

Or should I just use a dropping resistor, even though the current draw
changes?

Thanks

You could use the zener. The series resistor feeds the zener and you tap
your 5 volts from the zener.
The series resistor needs to allow for all the current that you need and the
zener needs to be able to handle what you take from it.
PS: The resistor sees both currents, the zener current and the load.
Also there are low overhead VRs.
Can you look up any info on this?


Tom

 

[Tim Wescott]

On Sat, 11 Oct 2008 17:27:27 -0700, DorkyGrin wrote:

I'm trying to get a steady 5 volts from a 6 volt battery. The 5v will
power a circuit that draws a variable amount of current (between 70 and
100 ma). I've used LM7805 in the past but I know they need more then
just one volt difference.

I do have a 1/2watt 5 volt zener I could use but I can't seem to get it
to work properly. Would it be wired in series?

Or should I just use a dropping resistor, even though the current draw
changes?

Thanks

You want a "low drop out" voltage regulator.

But most batteries have quite a range of voltage outputs possible, and
that voltage output is usually less than the nominal voltage. Most cell
technologies will still have a lot of useful capacity left when a nominal
"6V" battery drops to 5V.

Examples:

A "6V" battery may be five NiCd or NiMH cells, which are 1.2V/cell
nominal, 1.25V/cell at full charge, and as low as 0.9V/cell after you've
wrung all you can get out of the cell without damage, for 4.5V.

It may be four dry cells which are 1.5V/cell nominal, 0.9V/cell if you're
really sucking it dry, or 3.6V for the battery.

It may be three lead-acid cells , which are 2.0V/cell nominal, up to 2.3V/
cell when fully charged, down to 1.8V or less/cell depending on the
battery when it's discharged, or about 5.4V -- and there are some lead-
acid technologies that are supposed to be good much below that.

So your "6V" battery may not be good for it's full advertised capacity,
even with your LDO.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" gives you just what it says.
See details at http://www.wescottdesign.com/actfes/actfes.html

 

[Allen Bong]

On Oct 12, 8:27 am, DorkyGrin <DorkyG...@yahoo.com> wrote:

I'm trying to get a steady 5 volts from a 6 volt battery. The 5v will
power a circuit that draws a variable amount of current (between 70
and 100 ma). I've used LM7805 in the past but I know they need more
then just one volt difference.

I do have a 1/2watt 5 volt zener I could use but I can't seem to get
it to work properly. Would it be wired in series?

Or should I just use a dropping resistor, even though the current draw
changes?

Thanks

Do you need exactly 5V or is 5.4V Ok for you?
The simplest way to get 5.4V is using a 1N4001 connected as below:

[Battery +]--------->|---------[Load]--------[Battery -]
1N4001

If you must use 5V zener connect it this way:

[Battery +]---------[ 6.8 ohm]---------+-----------+
| |
K 5v zener [RL]
A |
| |
[Battery -]----------------------------+-----------+

Allen

 

[DorkyGrin]

You want a "low drop out" voltage regulator.

But most batteries have quite a range of voltage outputs possible, and
that voltage output is usually less than the nominal voltage.  Most cell
technologies will still have a lot of useful capacity left when a nominal
"6V" battery drops to 5V.

Examples:

A "6V" battery may be five NiCd or NiMH cells, which are 1.2V/cell
nominal, 1.25V/cell at full charge, and as low as 0.9V/cell after you've
wrung all you can get out of the cell without damage, for 4.5V.

It may be four dry cells which are 1.5V/cell nominal, 0.9V/cell if you're
really sucking it dry, or 3.6V for the battery.

It may be three lead-acid cells , which are 2.0V/cell nominal, up to 2.3V/
cell when fully charged, down to 1.8V or less/cell depending on the
battery when it's discharged, or about 5.4V -- and there are some lead-
acid technologies that are supposed to be good much below that.

So your "6V" battery may not be good for it's full advertised capacity,
even with your LDO.

--

Tim Wescott
Wescott Design Serviceshttp://www.wescottdesign.com

I got the zener to work using the methods above but you brought up a
good point - the battery is a lead acid battery and I thought I needed
the voltage to be quite stable at 5v. As it turns out, I only my
circuit only needs 3.6 volts and I was lucky enough to find a circuit
that has a 7136A regulator in it. It's a perfect source and I think my
immediate issue is solved.

Thanks for all the great and fast replies!