Assume I have a 9v battery conected to the input of an lm7805. My question is, the lm5805 will produce 5v @ 1Amp... i know the voltage is constant (as long as input voltage is fine) but is the amperage variable.
I guess what I am trying to ask is, will the connected circuitry only draw as much current as it requires, or will the lm7805 produce 1A that will need to be lowered.
Thanks for looking at this silly question, Robert
The IC will draw a minimal current at idle. It should be around several milliamps. You would have to refer to the spec sheet to figure out the no load current draw.
The load will determine the mount of current it will require. Take care that a 9 volt battery will not last very long when running much of a load. These batteries are designed to run devices that require less than 20 to 50 ma.
Jerry G. ======
Robert Menotte wrote:
Maybe a little background would help,
I'm trying to make a portable battery recharge unit... mainly for a cellphone, ipod, etc. It would have a usb interface for connecting to these devices and supplying power from the 9V batt.
my question is just a very basic electronics question. Is the amount of current provided based on the need of the attached circuitry to the
7805, or will the current be 1A. (I would assume the former but I am unsure)Thank you very much for such a quick reply!
-Robert
Jerry G. wrote:
it almost always does.
The 9V battery, if it's one of those things with 6 AAA cells inside it, won't go far, but may be enough in an emergency.
yes, it will be "based on need".
-- Bye. Jasen
Thanks for the reply!
One more question. I was wondering if placing a capacitor for voltage smoothing would be a good idea on the input leg of the 7805. It is comming from a standard square 9v battery.
If so, how would I calculate the required size of the capacitor needed?
Also, how does a cap work really. A friend told be it actually blocks dc current... but I doubt that.
Sorry for the silliy questi> > > Maybe a little background would help,
The current drawn will be: the current drawn by the load (e.g. your cellphone) plus the current, required by the LM7805 to operate.
This is however, sorry to say, a lousy setup. Say, you want 50 mA to charge your cellphone. Your battery would need to supply 5 mA (for the 7805) plus 50 mA (for the cellphone). That's 55 mA total. So, from your battery you'll draw 9 volts * .055 A = 0.495 Watt.
Your cellphone would receive 5 volts * 0.05 A = 0.25 Watt. So, in the process almost half the high priced battery power just gets wasted!
(And in practice its even worse. You feed 50 mA into the cellphone. But that needs to be stored first and retreived from the internal cell later. Count 80% efficency, both times. Which leaves 0.16 Watts for 'call power' out of .495 Watts from your battery.)
-- Kind regards, Gerard Bok
The data sheet recommends a tantalum capacitor across both I/P & O/P of
3-terminal regulators to minimise the risk of the regulator bursting into oscillation. About 22 to 47uF should do.
What you will learn very quickly is what the others have told you, that using a 9V battery is inefficient, wasteful and expensive! A chemist shop near me often has a display rack of gadgets selling cheap - recently they had a mobile phone emergency charge thing that hangs on a keyring for UK£1.99p it contains 4x AAA cells and comes with an assortment of connector adapters.
probably not needed.
look at the data sheet for the 7805
it actually does, if it didn't it'd run your battery down.
for details oof how try howstuffworks.com or a physics text.
they behave a bit like a rechargable battery, by connecting them across a noisy supply they can swallow the spikes and/or provide power in times of peak load, (this is a very short term thing, miliseconds or less)
-- Bye. Jasen
My understanding of it is that you shouldn't need a "smoothing" capacitor if the supply is coming from a 9 volt battery(or any battery). Usually it's needed for transformers(AC to DC). Some are better than others for ripple.
The current from a LM7805 can be up to 1 amp but if the device needs that much you will need to provide a heatsink for the regulator. There are many circuit diagrams on the internet for these regulators, have a Googol for them.
Good luck, Boozo.
That's just one of its functions. Another function is, to provide a low impedance AC path. Your dry cell won't supply that, so you need a capacitor.
No. In his situation, the LM7805 won't carry 1 Amp. At least: not for more than a few seconds :-) (In case you wonder: the dry cell is just not capable of sourcing
1 Amp for several minutes.)In this situation, I would suggest to use an 78L05. Capable enough and with the added advantage of a 100 mA current limit.
-- Kind regards, Gerard Bok
I was talking about the capacity of the regulator not his power supply. And the 9volt battery won't give you 100mA for very long either. A "low impedance AC path"???
Boozo.
His 'power supply' is a 9 volt dry cell :-)
No :-)
As any analog circuit, a linear regulator requires a capacitor across it's supply terminals. Need to know more ? Copy the term --including the quotes-- in google and enjoy your personal teacher :-)
-- Kind regards, Gerard Bok
ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.