# using battery to power device not intended to be

P

#### Patrick Chu

Jan 1, 1970
0
How, if at all, can I use batteries to power a device that was meant

Here are the specs:

Power Consumption: 0.4 Watts
Current: 75 mA +/- 5 mA
Input Voltage: 5V DC

Instead of having to have it plugged into a USB port, I'd like to be
able to use batteries to power it free-standing. What kind of
batteries and other electrical hardware would I need, if this is even
possible?

Thanks,
Patrick

X

#### XandMS

Jan 1, 1970
0
try to use 9V batteries, and use 78L05 to convert 9V to 5V. It's a simply
circuit

P

#### Patrick Chu

Jan 1, 1970
0
Actually, let me try, using my Google-assisted layman's knowledge, to
answer my own question, and hopefully an expert can verify or correct
me:

If I take a 9-volt battery and some kind of voltage regulator, I could
drop it down to 5 volts.

Then, if ohms = volts/amps, and the required current is something like
75 mA, or 0.075 A, then I would need a 5/0.075 ~= 65 ohm resistor
somewhere in the mix?

Using another formula to double-check, ohms = (volts^2)/watts, the
ohms comes out to 25/0.4 = 63, which is around mid-60s like the
previous calculation said.

So to recap, I need a 9V battery, a +5V voltage regulator, and a ~65
ohm resistor? Can all be found at Radio Shack?

Thanks,
Patrick

J

#### John Popelish

Jan 1, 1970
0
Patrick said:
Actually, let me try, using my Google-assisted layman's knowledge, to
answer my own question, and hopefully an expert can verify or correct
me:

If I take a 9-volt battery and some kind of voltage regulator, I could
drop it down to 5 volts.

Yes. A linear regulator wastes the extra 4 volts.
Then, if ohms = volts/amps, and the required current is something like
75 mA, or 0.075 A, then I would need a 5/0.075 ~= 65 ohm resistor
somewhere in the mix?

Actually, that is the effective resistance of a 5 volt load that draws
..075 amp. The regulator that wastes the extra 4 volts would have an
effective resistance of 4/.075~53 ohms. But if the load current
varies a bit, or the battery voltage decreases, the regulator will
have to have some other resistance to waste only the excess voltage.
Fortunately, regulators are an active resistance that varies as
needed.
Using another formula to double-check, ohms = (volts^2)/watts, the
ohms comes out to 25/0.4 = 63, which is around mid-60s like the
previous calculation said.
So to recap, I need a 9V battery, a +5V voltage regulator, and a ~65
ohm resistor? Can all be found at Radio Shack?

A 5 volt regulator and a couple capacitors are all you need. It can
handle a load current up to 100 ma. Try a 78L05. See the data sheet
for the typical capacitors recommended.

http://cache.national.com/ds/LM/LM78L05.pdf
available from Digikey for $0.40 C #### CFoley1064 Jan 1, 1970 0 From: [email protected] (Patrick Chu) Date: 2/6/2004 5:24 PM Central Standard Time Message-id: <[email protected]> How, if at all, can I use batteries to power a device that was meant to be powered by USB (universal serial bus)? Here are the specs: Power Consumption: 0.4 Watts Current: 75 mA +/- 5 mA Input Voltage: 5V DC Instead of having to have it plugged into a USB port, I'd like to be able to use batteries to power it free-standing. What kind of batteries and other electrical hardware would I need, if this is even possible? Thanks, Patrick Thanks for a good problem description. 75 mA is quite a lot to be drawing from a 9V "transistor" battery. It won't last very long. If you just need a quick fix, you might want to try one of the powered USB hubs -- that would be your quickest fix, and wouldn't be expensive in comparison with batteries, perfboard, IC regulator, caps, switch, enclosure, &c. That might be your best solution if you've got an outlet handy. If not, or if you really want to use batteries, you could try 6 "C" or "D" batteries in series to give you the 9V to start with. You could then use the information given by other respondents about the 78L05 to get a solution that will last more than a day or two. Good luck Chris J #### Jeffrey Turner Jan 1, 1970 0 John said: Yes. A linear regulator wastes the extra 4 volts. Actually, that is the effective resistance of a 5 volt load that draws .075 amp. The regulator that wastes the extra 4 volts would have an effective resistance of 4/.075~53 ohms. But if the load current varies a bit, or the battery voltage decreases, the regulator will have to have some other resistance to waste only the excess voltage. Fortunately, regulators are an active resistance that varies as needed. A 5 volt regulator and a couple capacitors are all you need. It can handle a load current up to 100 ma. Try a 78L05. See the data sheet for the typical capacitors recommended. http://cache.national.com/ds/LM/LM78L05.pdf available from Digikey for$0.40

You might want to look into a switching regulator instead. It will
save a lot of battery life. It's a slightly more complicated circuit
and I don't have any part numbers off hand, but look into it - your
battery life could easily increase by 2/3.

--Jeff

V

#### vic

Jan 1, 1970
0
Patrick said:
Actually, let me try, using my Google-assisted layman's knowledge, to
answer my own question, and hopefully an expert can verify or correct
me:

If I take a 9-volt battery and some kind of voltage regulator, I could
drop it down to 5 volts.

Then, if ohms = volts/amps, and the required current is something like
75 mA, or 0.075 A, then I would need a 5/0.075 ~= 65 ohm resistor
somewhere in the mix?

Using another formula to double-check, ohms = (volts^2)/watts, the
ohms comes out to 25/0.4 = 63, which is around mid-60s like the
previous calculation said.

So to recap, I need a 9V battery, a +5V voltage regulator, and a ~65
ohm resistor? Can all be found at Radio Shack?

Thanks,
Patrick

Well, if you use a 9V battery, the exceeding 4V will be converted to
heat by the regulator and you will lose this energy. You'd better use 4
1.5V batteries and a low drop regulator or a simple diode (1N4004) wich
has a drop voltage of 0.6V, so this is an easy way to get 5.4V or 4.8V.
(your circuit should work OK with this kind of supply voltage.

Also you do not need to insert a current limiting resistor. In fact, if
you do so, supposing the circuit draws 75mA, the supply would be :
5 - .075 * 65 = ~0V which is not what you want.

The circuit will draw itself the current it needs.

As others have suggested you may add 100uF in parallel with 100nF to
filter out parasites and transients.

Note that if you use a 9V battery the lifespan will be about :
300mAh / 75mA = 4 hours (rough estimation for an alcaline battery)

vic

G

#### George

Jan 1, 1970
0
Jeffrey Turner said:
You might want to look into a switching regulator instead. It will
save a lot of battery life. It's a slightly more complicated circuit
and I don't have any part numbers off hand, but look into it - your
battery life could easily increase by 2/3.

--Jeff

Try an LM2574N-5.0 or something like it. It's more expensive, but more
efficient. $1.86 at digikey with about 80% efficiency. You can get 90% efficiency if you are willing to pay$3.10 for an LM2674N-5.0.

http://www.national.com/appinfo/power/

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