Using more voltage than specified

Hi,
I would like to check that I won't damage a device I am using by
connecting an external power pack to it, rather than the supplied
power pack.

The supplied power pack is 3.7v,750mah. The power pack I intend to use
is 5v, 3700mah.

Would the extra voltage do an adverse damage to the unit?, or is it
safe?

Thanks for any help.

Regards,
Sean

 

[John Fields]

On Mon, 06 Oct 2008 17:45:23 +0100, jneil wrote:

Hi,
I would like to check that I won't damage a device I am using by
connecting an external power pack to it, rather than the supplied
power pack.

The supplied power pack is 3.7v,750mah. The power pack I intend to use
is 5v, 3700mah.

Would the extra voltage do an adverse damage to the unit?, or is it
safe?

---
It wouldn't assume it to be safe if directly connected to the device.

However, putting a couple of diodes in series between the 5V power pack
and the device would drop enough voltage (about 1.4V) to make it safe:
(View in Courier)

5V 3.6V
/ /
+--[DIODE>]---[DIODE>]--+
| |
|+ |+
[5V BAT] [DEVICE]
| |
| |
+-----------------------+


That diodes need to be connected so their cathodes point toward the
load.

Also, there will be power wasted in the diodes, however, and depending
on how much current the device draws the diodes may get hot.

Do you know how much current the device draws? If you do, and can post
what it is, I can recommend suitable diodes.

JF

 

On Mon, 06 Oct 2008 13:42:34 -0500, John Fields
<jfields@austininstruments.com> wrote:

On Mon, 06 Oct 2008 17:45:23 +0100, jneil wrote:

Hi,
I would like to check that I won't damage a device I am using by
connecting an external power pack to it, rather than the supplied
power pack.

The supplied power pack is 3.7v,750mah. The power pack I intend to use
is 5v, 3700mah.

Would the extra voltage do an adverse damage to the unit?, or is it
safe?

---
It wouldn't assume it to be safe if directly connected to the device.

However, putting a couple of diodes in series between the 5V power pack
and the device would drop enough voltage (about 1.4V) to make it safe:
(View in Courier)

5V 3.6V
/ /
+--[DIODE>]---[DIODE>]--+
| |
|+ |+
[5V BAT] [DEVICE]
| |
| |
+-----------------------+


That diodes need to be connected so their cathodes point toward the
load.

Also, there will be power wasted in the diodes, however, and depending
on how much current the device draws the diodes may get hot.

Do you know how much current the device draws? If you do, and can post
what it is, I can recommend suitable diodes.

JF

There is nothing on the device that specifies the mah but I do know
that the 750mah battery lasts about 120 minutes. Thanks for the info.

Regards,
Sean

 

[John Fields]

On Mon, 06 Oct 2008 21:11:01 +0100, jneil wrote:

On Mon, 06 Oct 2008 13:42:34 -0500, John Fields
jfields@austininstruments.com> wrote:

On Mon, 06 Oct 2008 17:45:23 +0100, jneil wrote:

Hi,
I would like to check that I won't damage a device I am using by
connecting an external power pack to it, rather than the supplied
power pack.

The supplied power pack is 3.7v,750mah. The power pack I intend to use
is 5v, 3700mah.

Would the extra voltage do an adverse damage to the unit?, or is it
safe?

---
It wouldn't assume it to be safe if directly connected to the device.

However, putting a couple of diodes in series between the 5V power pack
and the device would drop enough voltage (about 1.4V) to make it safe:
(View in Courier)

5V 3.6V
/ /
+--[DIODE>]---[DIODE>]--+
| |
|+ |+
[5V BAT] [DEVICE]
| |
| |
+-----------------------+


That diodes need to be connected so their cathodes point toward the
load.

Also, there will be power wasted in the diodes, however, and depending
on how much current the device draws the diodes may get hot.

Do you know how much current the device draws? If you do, and can post
what it is, I can recommend suitable diodes.

JF

There is nothing on the device that specifies the mah but I do know
that the 750mah battery lasts about 120 minutes. Thanks for the info.

You're welcome.

The device wouldn't specify milliampere hours, it would specify
milliamperes.

The milliampere-hour specification refers to the battery's ability to
supply rated current before the battery voltage falls to an unacceptable
level.

Since your battery lasts for two hours, then the load is probably
pulling close to 375 mA.

Since currents in series-connected elements are the same, that means
that if you used the larger battery with the two series-connected diodes
between it and the load they would also have 375mA through them and the
power each would dissipate would be:

P = IE = 0.375A * 0.7V ~ 0.26W

That's just a little over a quarter of a watt, so they'd get warm but,
if you used something like a 1N4000 or a 1N4001, just gently warm and
certainly not something you'd have to worry about.

A bonus, if your 5V battery decayed at the same rate as the 3.7V one, is
that it'll last:

C1 3750mAh
T = ---- = --------- = 5 times longer
C2 750mAh

It should actually last somewhat longer because you'd be taking less
from its total capacity, per hour, than you would be from the smaller
pack, so figure on something like six hours.


JF

 

[John Fields]

On Mon, 06 Oct 2008 16:22:09 -0500, John Fields
<jfields@austininstruments.com> wrote:

A bonus, if your 5V battery decayed at the same rate as the 3.7V one, is
that it'll last:

C1 3750mAh
T = ---- = --------- = 5 times longer
C2 750mAh

It should actually last somewhat longer because you'd be taking less
from its total capacity, per hour, than you would be from the smaller
pack, so figure on something like six hours.

---
Oops...
If your small battery lasts for two hours, the large one should last at
least five times that, or ten hours.

More than likely though, eleven or so.

JF

 

certainly not something you'd have to worry about.

A bonus, if your 5V battery decayed at the same rate as the 3.7V one, is
that it'll last:

C1 3750mAh
T = ---- = --------- = 5 times longer
C2 750mAh

It should actually last somewhat longer because you'd be taking less
from its total capacity, per hour, than you would be from the smaller
pack, so figure on something like six hours.


JF

Thanks for your help, I will certainly try what you have suggested.

Regards,
Sean