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[SOLVED] can a single L7805 is sufficient for powering up a PIC and an LCD?

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salman.

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Is a single L7805 is sufficient for powering up a PIC and an LCD?

Hi All, my question is "can a single L7805 is sufficient for powering up a PIC and an LCD?"
 
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It depends on the power draw of the LCD, especially the backlight. However, the 7805 should be sufficient. It's good for upto 1A current and the LCD shouldn't be drawing that much current.

Remember to mount the 7805 on a sufficient heatsink. Keep in mind that a linear regulator works by dropping voltage through heat dissipation.

Hope this helps.
Tahmid.
 
I personally feel that 7805 is plenty for LCD, i am working on 3.2TFT with some additional circuit, it only takes 0.4Amps. I hope you are not having a LCD bigger that this one.
 
Potential difference for LCD and PIC will be 5v but the current drawn by PIC and LCD is different. So you have to find out the max current drawn by PIC and LCD in your circuit. Then Check if 7805 can provide double that current. If yes, you can use 7805. You have to check for double the current because you should not draw the max current it can provide. It will heat up.
 
I have used l7805 with a Pic16f877a and 1602B LCD and some leds (2 leds with 1Komh resistors) it is enough to supply all of them
 
Dear salman,
Use 100 ohms resistors for LCD backlight[pin 15,16]. Connect directly 5V to microcontroller & LCD. If you are not using 100 ohms resistor the regulator will get heat & you need heat sink.

Regards
Udhay
 
Better use 78S05 its 2A, and 7805 is 1A, L7805 is 1,5A, that means you will need smaller heatsking size. Price difference almost dont exist. You can also look 78T05 3A.
 
salman. said:
I have a 16 x 2 LCD
Click to expand...

You should be able to easily use the L7805. A 16 X 2 LCD shouldn't be drawing more than 200-300mA with backlight on.

udhay_cit said:
Dear salman,
Use 100 ohms resistors for LCD backlight[pin 15,16]. Connect directly 5V to microcontroller & LCD. If you are not using 100 ohms resistor the regulator will get heat & you need heat sink.
Click to expand...

Some LCD modules have this resistor on the PCB. So adding an additional resistor may reduce brightness. If the resistor is on the PCB or not should thus be checked.
 
For pic + lcd, 2A regulator is too much... LM7805 sufficient...
adding series resistor will not change brightness so much & you can add a provision for the series resistor. if you want use it, otherwise replace with jumper...
 

If he use stronger voltage regulator needed heatsink will be smaller for that job, and in some cases not even need. Its obvious situation.

I use 78M05 (0,5A SMD) often for LCD with backlights projects without problems.

I practice whenever should be used 7805 I use 78S05.

General answer on thread question is that L7805 is enoguh to handle LCD with uC.
 

tpetar said:
If he use stronger voltage regulator needed heatsink will be smaller for that job, and in some cases not even need. Its obvious situation.
Click to expand...

Can you please explain that?

I'm not so sure of that.

Whether you use 7805 (datasheet: https://www.st.com/internet/com/TECHNICAL_RESOURCES/TECHNICAL_LITERATURE/DATASHEET/CD00000444.pdf) or 78S05 (datasheet: https://www.st.com/internet/com/TECHNICAL_RESOURCES/TECHNICAL_LITERATURE/DATASHEET/CD00000449.pdf), input voltage, output voltage and output current are the same. So, device power dissipation is the same for both 7805 and 78S05. P = (Vin- Vout) * Iout

According to the datasheets, 7805 and 78S05 have same thermal resistances (for TO-220 vs TO-220 and TO-3 vs TO-3). So, they should produce the same heat at the same input voltage, output voltage and output current. So, required heatsink should be similar.
 

Dissipation is the same I didnt say anything about that. I say that 78S05 is stronger he can handle up to 2A and will easier handle this job, maybe even without heatsink or with small heatsink (soldered on PCB).

Where you see that I say that about dissipation please explain ?
 

The power dissipation is the same. And according to the datasheets, thermal resistance is the same. So that means that heat produced will be the same and will require the same amount of cooling/heatsinking. So, size of heatsink for both should be the same. So, that means that if 7805 requires heatsink, 78S05 should also require heatsink.
 
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tpetar said:
Heatsink is not required in this design,
Click to expand...

That's okay.

and 78S05 will be cooler then 7805.
Click to expand...

This is what confuses me. Can you please explain how the 78S05 will be cooler? If power dissipation and thermal resistance are same, shouldn't the temperature rise also be the same? So, shouldn't they both be at similar, if not the same, temperatures? So, how will the 78S05 be cooler?
 

Yes you are right about temp, I dont know why I think that 78S05 have better situation about thermal resistance. 78S05 will give higher current if needed later for circuit, anyway earlier conslusion about L7805 stays, that is enough for this design.
 

I think 7805 should work fine,because the current been drawn from the circuit is not up to 1 amp
 

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