Serin serout problem

[lerameur]

HI,
hope you are doing good.
I read a thread on this site concerning the 7805 overheating.
My 7805 is overheating even with a heat sink and I just have a LCD a pic chip and a RF receiving module. I tried the tricks they say but none is working.
I tried conencting the backlight directly to the 12v battery using a 150 IOhm resistor, or t o the 5V output of the 7805 with a 22 ohm resistor. Whils always keeping the LCD Vdd to the 5v of the 7805.
It heats op in every case ?
any ideas ?
Lcd 16 x2: UNIQ/eVision GC-1602I1

[skimask]

HI,
hope you are doing good.
I read a thread on this site concerning the 7805 overheating.
My 7805 is overheating even with a heat sink and I just have a LCD a pic chip and a RF receiving module. I tried the tricks they say but none is working.
I tried conencting the backlight directly to the 12v battery using a 150 IOhm resistor, or t o the 5V output of the 7805 with a 22 ohm resistor. Whils always keeping the LCD Vdd to the 5v of the 7805.
It heats op in every case ?
any ideas ?
Lcd 16 x2: UNIQ/eVision GC-1602I1

What is your input voltage to the 7805? 12v?
What do you mean by 'overheating'? It's actually overheating and shutting down or just getting hot?
Put a meter across the 5v and ground lines at their input to the board. What does it read when using the resistance setting? What does it read when using the diode-check (continuity) setting?

This should help get it narrowed down...

[lerameur]

It is 12v
I am getting .22amp It is getting very hot, but not shutting down.
Should I put a bigger heat sink or maybe two Lm7805 ?

[skimask]

It is 12v
I am getting .22amp It is getting very hot, but not shutting down.
Should I put a bigger heat sink or maybe two Lm7805 ?

Here's a bit of math for ya...
12v input @ .22 amp = 2.64 watts total
5v output, drawing .22 amp = 1.1 watt.
(don't know if you're familiar with this stuff yet)

The regulator has to 'dump' 7v @ .22amp, which is 1.54 watts. That doesn't seem like a lot, but it's enough to make the 7805 hot, even if you've got a heat sink on it.

Do you have any heat sink compound between the 7805 and the heat sink face itself?

(oh...and .22amp for your configuration sounds about right)

[Archangel]

HI,
hope you are doing good.
I read a thread on this site concerning the 7805 overheating.
My 7805 is overheating even with a heat sink and I just have a LCD a pic chip and a RF receiving module. I tried the tricks they say but none is working.
I tried conencting the backlight directly to the 12v battery using a 150 IOhm resistor, or t o the 5V output of the 7805 with a 22 ohm resistor. Whils always keeping the LCD Vdd to the 5v of the 7805.
It heats op in every case ?
any ideas ?
Lcd 16 x2: UNIQ/eVision GC-1602I1

Hi Ken,
Here are some figures for you to "SMOKE OVER" pun intended.
you need to add up the current draw of all of your outputs, and other loads and make sure you are not overloading your 7805 and your PIC. OR hook your DVM in series on the amp scale.
LED @ 4 volts at 45 ma hooked to 5 volt = 22 ohm
LED @ 4 volts at 20 ma hooked to 5 volt = 50 ohm
LED @ 5 volts at 20ma hooked to 12v =350 ohm
LED @ 5 volts at 40ma hooked to 12v = 140 ohm
LED @ 4 volts at 20 ma hooked to 12 volts = 400 ohms
LED @ 4 volts at 40 ma hooked to 12 volts = 200 ohms
the formula is still Rs = Vin - V led / I led. Where V in is input voltage, V led is led voltage and I led is LED current in ma.
Hope this is helpful.
JS

[lerameur]

well because i still need all of these components. Its drawing way too much. 1.1 a out of 1amp max.

what about putting 2 Lm7805 in parallel?

[Archangel]

well because i still need all of these components. Its drawing way too much.
what about putting 2 Lm7805 in parallel?

Honestly Ken,
I do not know if that works or not, That is a linear IC and I am not too sure how they will interact. I think a better solution would be to use a bigger regulator IC, or use 2 or more 7805s as seperate supplies using a common ground. Your pic can sink current from the secondary supply, and your display could also run from it. Note: check the data sheet for maximum current the pic can handle in total, so as not to exceed.

[skimask]

well because i still need all of these components. Its drawing way too much. 1.1 a out of 1amp max.

what about putting 2 Lm7805 in parallel?

You just said you were drawing .22amp...
Which one is it? 1.1 a or .22 a ?

[lerameur]

sorry 1.1watt
These babies can draw 1amp, why cant I draw more then .22 without being very hot ?
I dont see the wattage value on the spec sheet of this chip,

[skimask]

sorry 1.1watt
These babies can draw 1amp, why cant I draw more then .22 without being very hot ?
I dont see the wattage value on the spec sheet of this chip,

There's a wattage value buried in the 7805 datasheet somewhere. And remember, your circuit is drawing 1.1 watt, but your regulator is dissipating 1.54watt, and you are drawing 2.64watts from your battery total.

The amount of temperature rise on the back of the 7805/heatsink/whatever, is directly related to the ability of that 'place' to dissipate heat.
If you've got something that just doesn't give up heat, it'll get hot...if it's very conductive, it'll stay cool.
So, with that being said, even 1 watt can feel very hot depending on what the heatsink is, if there is any heat conductive compound in between the 7805 and the heatsink, ambient temperature, etc.etc.

Think about it this way....take a 40 watt light bulb... It's hot right? All the way around it. It's also pretty big when compared to a 7805 with a heatsink. Now, take that same 40 watt bulb, shrink it 40 times, and apply 1/40th of the heat...sounds like it might be in that same ballpark. Obviously I don't know what the actual numbers would be, I'm generalizing quite a bit, I think it sounds reasonable.

And take another look at what I was saying earlier about how much power the regulator has to dissipate. Do you understand that much of it? (As in why the regulator dissipates 1.54watt when you power it with 12v and only draw .22amp from the regulator?) Remember, a linear regulator, such as a 7805, regulator by dumping extra voltage off as heat...that's the way they work (in simple terms anyways)...kinda like an infinetely variable power resistor that's always turning itself to keep the output voltage at a certain point.

[lerameur]

well it dissipates 1.54watt because it can take 1watt , so 2.54 watt total from the battery. I guess I need this part: L78S05CV for now