converting USB 5v to 3.6v

M

Mark Lin

Guest
Hi group:
I have a little circuit that takes 3.6 battery. I dont know its
current draw, but the original battery was a Saft 3.6 volt with 50 mA
output. The battery is running out way too fast, so I have the idea
of getting power from my USB port. I know it's possible because they
also sell the USB powered version of the circuit but it wasn't
available at the time when I bought it. I know USB port can put out
maximum of 5V and 500 mA(with data port used) or 100mA(without using
data port). Can some enlight me on building the voltage converter and
maybe I need a current regulaor(?).

Regards,
Mark
 
Hey

What about using a 3,6V Zenerdiode and a resistor ??

+5V----RESISTOR------ZENERDIODE (-|<-)---- GND
|
3,6V out to GND
--
Mvh Kasper

Tjek Elektronik Chatten pĺ -> http://www.angelfire.com/alt/elektronik/
"Mark Lin" <linmark333@yahoo.com> skrev i en meddelelse
news:40121488.0307040334.312c8464@posting.google.com...
Hi group:
I have a little circuit that takes 3.6 battery. I dont know its
current draw, but the original battery was a Saft 3.6 volt with 50 mA
output. The battery is running out way too fast, so I have the idea
of getting power from my USB port. I know it's possible because they
also sell the USB powered version of the circuit but it wasn't
available at the time when I bought it. I know USB port can put out
maximum of 5V and 500 mA(with data port used) or 100mA(without using
data port). Can some enlight me on building the voltage converter and
maybe I need a current regulaor(?).

Regards,
Mark
Click to expand...
 
Kasper wrote:
Hey

What about using a 3,6V Zenerdiode and a resistor ??

+5V----RESISTOR------ZENERDIODE (-|<-)---- GND
|
3,6V out to GND
--
Mvh Kasper

Tjek Elektronik Chatten pĺ -> http://www.angelfire.com/alt/elektronik/
"Mark Lin" <linmark333@yahoo.com> skrev i en meddelelse
news:40121488.0307040334.312c8464@posting.google.com...

Hi group:
I have a little circuit that takes 3.6 battery. I dont know its
current draw, but the original battery was a Saft 3.6 volt with 50 mA
output. The battery is running out way too fast, so I have the idea
of getting power from my USB port. I know it's possible because they
also sell the USB powered version of the circuit but it wasn't
available at the time when I bought it. I know USB port can put out
maximum of 5V and 500 mA(with data port used) or 100mA(without using
data port). Can some enlight me on building the voltage converter and
maybe I need a current regulaor(?).

Regards,
Mark



DIODEs drop .7 volts, two will drop 1.4. 5v-1.4v=3.6v Will that work?
Click to expand...

5V-|>|--|>|--3.6v--LOAD---GND

Will that work?
 
Larry Hatch <lhatch@flite.net> wrote in
news:vgavihh8upft29@corp.supernews.com:

DIODEs drop .7 volts, two will drop 1.4. 5v-1.4v=3.6v Will that work?

5V-|>|--|>|--3.6v--LOAD---GND

Will that work?

It will work, and save power, but you are vulnerable to a short circuit
Click to expand...
which could take out the supply.

With a conventional if the load shorts it does not short the supply, so if
power consumtion is not the main consideration it's prolly better

Mike R
 
Larry Hatch <lhatch@flite.net> wrote in
news:vgcj7g3gd4qg03@corp.supernews.com:
With a conventional if the load shorts it does not short the supply,
so if power consumtion is not the main consideration it's prolly
better

Mike R
So 2 diodes and a current limiting resistor. Or I need to stock up on
3.6V zeners.

But if you use a series current limiting resistor the voltage across that
Click to expand...
resistor will vary with losd current and you won't have a constant voltage
source.

If your load is costant, just use the appropriate series resistor and
forget the semiconductors.

If it's not, the zener diode _in parallel_ with the load will soak up
current variations within design limits, ie how much current is available
and how much you are prepared to "waste".

Incidentally I've just looked at your numbers again and you quote .7 volts
as diode drop. IIRC (I'm long retired), that a geranium diode drop, a
silicon diode, much more bulletproof and common, drops 1.4 ish volts, so
one of them will do in series - but the vulnerability to resultant damage
from a short still applies.

You can also use a series zener of any voltage to get a required volts
drop, but the s/c condition still applies (and when I last looked zeners
didn't go down to 1.4 volts, I don't know if thats because it's impossible,
or given the standard diode drop, it would be silly)

Mike R
 
Mike,
So using a silicon diode with 1.4 volts drop would be a good
choice? Can I purchase the part in Radio Shack? So I guess the
scheme would like this:
power ---> diode ---> circuit?



Mike Ring <mike.ring@Michaelbtinternet.com> wrote in message news:<Xns93AF72726E021mikeringbtinternetco@217.32.252.50>...
Larry Hatch <lhatch@flite.net> wrote in
news:vgcj7g3gd4qg03@corp.supernews.com:

With a conventional if the load shorts it does not short the supply,
so if power consumtion is not the main consideration it's prolly
better

Mike R
So 2 diodes and a current limiting resistor. Or I need to stock up on
3.6V zeners.

But if you use a series current limiting resistor the voltage across that
resistor will vary with losd current and you won't have a constant voltage
source.

If your load is costant, just use the appropriate series resistor and
forget the semiconductors.

If it's not, the zener diode _in parallel_ with the load will soak up
current variations within design limits, ie how much current is available
and how much you are prepared to "waste".

Incidentally I've just looked at your numbers again and you quote .7 volts
as diode drop. IIRC (I'm long retired), that a geranium diode drop, a
silicon diode, much more bulletproof and common, drops 1.4 ish volts, so
one of them will do in series - but the vulnerability to resultant damage
from a short still applies.

You can also use a series zener of any voltage to get a required volts
drop, but the s/c condition still applies (and when I last looked zeners
didn't go down to 1.4 volts, I don't know if thats because it's impossible,
or given the standard diode drop, it would be silly)

Mike R
Click to expand...
 
I disagree with Mike.
As a rule of thumb, I'd point
- 0.7 V drop on a silicon diode;
- 0.5 V drop on a germanium diode;

Go for two 1N4007 (silicon diodes) in series like this


power------|>|---|>|----circuit
|
|
gnd-------------------------

Done.

Ricardo

"Mark Lin" <linmark333@yahoo.com> wrote in message
news:40121488.0307070857.7a2ab7a@posting.google.com...
Mike,
So using a silicon diode with 1.4 volts drop would be a good
choice? Can I purchase the part in Radio Shack? So I guess the
scheme would like this:
power ---> diode ---> circuit?



Mike Ring <mike.ring@Michaelbtinternet.com> wrote in message
news:<Xns93AF72726E021mikeringbtinternetco@217.32.252.50>...
Larry Hatch <lhatch@flite.net> wrote in
news:vgcj7g3gd4qg03@corp.supernews.com:

With a conventional if the load shorts it does not short the supply,
so if power consumtion is not the main consideration it's prolly
better

Mike R
So 2 diodes and a current limiting resistor. Or I need to stock up on
3.6V zeners.

But if you use a series current limiting resistor the voltage across
that
resistor will vary with losd current and you won't have a constant
voltage
source.

If your load is costant, just use the appropriate series resistor and
forget the semiconductors.

If it's not, the zener diode _in parallel_ with the load will soak up
current variations within design limits, ie how much current is
available
and how much you are prepared to "waste".

Incidentally I've just looked at your numbers again and you quote .7
volts
as diode drop. IIRC (I'm long retired), that a geranium diode drop, a
silicon diode, much more bulletproof and common, drops 1.4 ish volts, so
one of them will do in series - but the vulnerability to resultant
damage
from a short still applies.

You can also use a series zener of any voltage to get a required volts
drop, but the s/c condition still applies (and when I last looked zeners
didn't go down to 1.4 volts, I don't know if thats because it's
impossible,
or given the standard diode drop, it would be silly)

Mike R
Click to expand...
 
"Ricardo Matos Abreu" <spam_is_not_for_me_etmabreu@spam_blocker.ua.pt>
wrote in news:1057597487.220891@proxy1.ua.pt:

I disagree with Mike.
As a rule of thumb, I'd point
- 0.7 V drop on a silicon diode;
- 0.5 V drop on a germanium diode;

Go for two 1N4007 (silicon diodes) in series like this


power------|>|---|>|----circuit
|
|
gnd-------------------------

Done.

Ricardo

Many aplogies for duff gen: Horowitz and Hill (what I stole on early
Click to expand...
retirement) says that it's .6 - .7 for a geranium and .9 - 1.0v for a
silicon (.9 for the 1N4007)

I used to use the 1N4006 for everythig, steering, quenching, rectifying
being small and bulletproof, etc.

I think my 1.4 V may have been the foreard drop of a LED (good for giving a
light as well as a volts drop - IIRC the backwards drop of a LED was a
handy 2V, but I stand to be corrected on all that, it's more music theroy
and piano for me these days.

Any how, experiment!

Mike R
 
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