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Old 14th October 2004, 04:48 AM   #1
Morbid is offline Morbid  Israel
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Default 7805 to get 5V from car battery

Hi there,

i've made a car circuit that can remote control the power doors
with remote control.
but i need 5V to use this circuit
can i use the 7805 5V regulator? and it wont get burn or short the circuit after a while?

i had problems connection him a PS rated 14.4V/1.5A he didnt gave me stable supply.

i also guess that after a time it will get pretty warm.
i use 78L05 and it almost explode. (the small module)
is there any other way geting 5V from the car battery?
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Old 14th October 2004, 05:16 AM   #2
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How much current does your circuit draw at 5V? My guess is that power doors draw more current than what the typical 7805 can provide. You may need to look into using a DC-DC converter (switch-mode power supply, aka buck-regulator) that can provide a lot more current than a 7805.

-Vinnie
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Old 14th October 2004, 03:25 PM   #3
pukka is offline pukka  Spain
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A 7805 (T0220 package!!) can supply 1A max of current. Also, it can stand a maximum wattage disipation, always with the aid of a heatsink; in a car you can expect 13.4V from the battery or car generator. 13.4 - 5= 8.4V, it will have to "eat" (disspate) those 8,4V multiplied by the current you draw through it.

Maybe a DC-DC solution will be a little cumbersome for a car, specially as you only have to decrease voltage. I think a better solution, if you need in excess of 3A (use a LM340 or alike in such case, they're rated for 3A) will be using a source-output bypass transistor with your trusty 7805.

See http://homepages.which.net/~paul.hil...rSupplies.html
there are examples on how to use these also in the datasheets. You can get even more current rating by paralleling transistors.
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Old 14th October 2004, 05:42 PM   #4
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I'm guessing that the 5V is needed for the control electronics, and not for the actual power doors. Current should be small, and an LM7805 should be fine. I'm using a few of these in various circuits for car use and they work decently.

If power dissipation is a concern, then you can use a simple voltage divider before the input, but this will come at the expense of current draw when the circuit is inactive. In a car environment, this is something that I feel should be avoided, because it will drain the battery over an extended period of time. Again, if this is the control circuit that's waiting for the remote control (And possibly triggers a couple of relays) dissipation across the regulator is unlikely to be a problem, and if it IS a problem, then maybe y ou should revisit your circuit design. A quiescent draw of more than 100mA IN TOTAL in a car battery is (IMO) excessive, and should never be needed. a 100mA draw will drain 2.4AH per day which will have a signifigant impact on batery performance and life, especially if your car ever gets left for more than a weekend
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Old 14th October 2004, 06:09 PM   #5
Morbid is offline Morbid  Israel
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the current i need is less than 0.5A, max. for operation a relay..
the relay gets to drive the heavy stuff

7805 on its simple configuration will give me stable 5V?

3 - In
2 - Gnd
1 - Vout

no need for minimum load or caps?

cuz when i connected a simple IR reciever to 7805 it burned
but when i connected the IR reciever (its a module) with the circuit it worked well...
what can be wrong?

another question...when the car is started...the Vcc source is not stable as it when the car is not started...isn't it? beacuse of noise from the alternator (ac-dc convert)
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Old 14th October 2004, 10:26 PM   #6
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The National Semi datasheet for the LM7805 lists:
1 vin
2 gnd
3 vout

Reversing this could cause issues.

A bypass capacitor is a good idea. So is a zener before the voltage to prevent voltage spikes caused by load dumps while the alternator is running.

In my most recent circuit, I have the +12V rail bypassed with a 0.1uF and 220uF cap, and there are 2 regs, so there are 2 of each cap. This is probably overkill, but it was easy enough to do.

Also, a cap on the voltage output is a good idea. This is generally "as large as is practical" but I've found that 10uF is generally adequate.
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Old 15th October 2004, 12:52 AM   #7
Morbid is offline Morbid  Israel
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i thought reducing the voltage by passing the 7805 thru some diodes (every diode is 0.75V reduce?) as close to 7V-9V it will be easy to keep stable and safe voltage from the 7805
and as i describe the pinout i mention from the right to left....
_____
|====|
|=o==|
|====|
|+ + +|
| | |
| | | vout
| | gnd
| vin

the 78L05 is the same as 7805 but running low current...
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Old 15th October 2004, 01:12 AM   #8
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do everything supraguy says
also .1uf on output
USE A HEAT SINK USE A HEAT SINK USE A HEAT SINK USE A HEAT SINK
the 7805 should shut down if and not hurt itself if you do all above
the diodes cannot hurt 1N4004/6/7 etc
you could always use a little relay to drive the bigger one

some people cheat and put 7805s in parallel (same lot# plz)
never seen factory approval on this
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Old 15th October 2004, 08:59 PM   #9
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Default LM2876

Look up the simple switcher series on the National web site. Like the LM2876 or LM2592. These come in a fixed 5.0V version too. I find that most any air core coil out of a crossover network will work for the coil.
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Old 15th October 2004, 10:16 PM   #10
cpemma is offline cpemma  United Kingdom
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Quote:
Originally posted by SupraGuy
...If power dissipation is a concern, then you can use a simple voltage divider before the input, but this will come at the expense of current draw when the circuit is inactive...
If 500mA is the max current, a 10R 5W resistor before the 7805 will drop 5V, leaving 7-10V for the regulator to deal with. At lower currents it won't drop as much but the regulator will have an easier time watt-wise.
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