what are the possibilities to avoid heating of 7805 Regulator ?

dear all
I have a system with circuit that converts RS232 signal into fiber optic signal and then at the receiver end again same circuit converts fiber optics signal to RS 232 signal and then to computer.

Problem I am facing is that I have to operate it at 12VDC (as power supply),
in converter module I used 7805 regulator (complete circuit of converter designed in my LAB),
both things a fixed; supply voltage is 12VDC and module needs 5VDC.
The regulator heat up extremely great temperature.

what would be possible solution for it.

in converter module I used serial, counter and crystal ICs that needs 5VDC.

Posting a schematic is helpful.

Guessing... If your 5V section of the circuit draws really anywhere over 100mA or so, but maybe more like >200mA (you said "great temperature"), then it may be dissipating too much power. The regulator can provide 1 Amp, but under very, very specific circumstances I would think, it's really more ideal for light loads.

It is pointless offering advice based on conjecture, all the same, maybe a pass transistor around the 7805, and heat-sinking both together is necessary, and sometimes a fan is also a great help along with these additional things.

use a heatsink

circuit current is about 350mA at 5VDC.

heating is due to 12VDC to 5VDC difference (7 volts)

I saw some modules having input from =12VDC to +24VDC, while there are same ICs like serial, counter ICs.

how to design this type of power supply of the circuit.

If you can bear to do the regulator heat rise calculations, you'll see that the junction temperature is probably well over 125ºC, really the answer is the same as both replies: use a heat-sink, and/or a pass transistor. There are multiple heat-sink calculators online and in quite a lot of regulator datasheets. The TPS73801 datasheet explains this matter very well with an example that is easy to understand and adapt to your needs at the end, and Micrel cover this well in some of their LDO datasheets and app notes.

I found this one helpful indeed:
View attachment 1111 Clear and concise Thermal Considerations for Linear Regulators Sipex .ashx.pdf

Pass transistors function as a ratio if I've understood what happens on circuit correctly, and for a max of 350mA, you could easily set a ratio of 10:1, or 8:2, and so on. Meaning ~35mA through the regulator and the rest of the current through the pass transistor.

For 350mA it's almost less hassle than having a heat-sink, others might disagree of course.

Look at the ST datasheet for the 7805, page 28 (I think?), whatever page it's definitely figure 13 - the formula is there, if you have difficulty underdstanding it, say so and I or some-one else can explain it to you - I found it really hard to calculate at first as I was unfamiliar with the terms it uses, but that doesn't mean you will.

Power Loss (W) is always (Vi-Vout)*I and thermal resistance of heat sink determines Temp rise in deg/W

PC PSU's are so cheap, why bother re-inventing the wheel with HOT regulators.

Make or buy is constant decision you face with every part of every design.

if there is enough space you can use Step-Down Switching Regulator such as LM2575 [ it is a 5-pins IC & need 330 uH coil & schottky diod & two caps to work ] and heat will be less than your case

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if there is enough space you can use a step-down switching regulator such as LM2575 [ it is a 5-pins & need a coil 330uH & schottky diod & 2 caps]. you will get less temp.

Why does your circuit use such a high current? There will be much less heat if its current is much less.

The regulator will work with 7V input so drop 5V @ .35A in a resistor in series with its input. This saves 1.5w, use a 15 ohm resistor.
Frank

chuckey said:

The regulator will work with 7V input so drop 5V @ .35A in a resistor in series with its input. This saves 1.5w, use a 15 ohm resistor.
Frank

Click to expand...

FvM said:

use a heatsink

Click to expand...

I would combine these two excellent & basic suggestions, and use a series 10 ohm 2-watt resistor, plus add a heatsink. This would allow upto 500mA if required, yet keep the temperatures of everything under control.

the solution (#10) is good
but if you need much current you should use Step-Down Switching Regulator (#7)
watch this video "Switch mode power supply tutorial: DC-DC buck converters "
https://www.youtube.com/watch?v=CEhBN5_fO5o

Hi,

about post#10.

It surely will reduce the heating of the LM7805, but the overall heating in all components is the same.
It is just spreading the heat to different devices.

Klaus

What uses that high current? An old incandescent light bulb (heater)? Vacuum tubes?

Thanks everyone
I will try solution # 10
then I will give feedback.

If you haven't decide anything yet it might be a good idea to take a look at Roman Black's ingenious SMPS designs to replace 78xx/79xx linear regulators. Cheap and efficient. Link:https://www.romanblack.com/smps/smps.htm

Maybe the high current is used by vacuum tubes or old fashioned TTL 74xx logic ICs. Cmos ICs should be used instead because they use very low current.

shahbaz.ele said:

Thanks everyone
I will try solution # 10
then I will give feedback.

Click to expand...

The user would be wiser to learn the relaxation oscillator SMPS mode such as solution #15 or better or learn how manage wasted heat.

SunnySkyguy said:

The user would be wiser to learn the relaxation oscillator SMPS mode such as solution #15 or better or learn how manage wasted heat.

Click to expand...

I believe OP's issue was not about "wasted heat", but about the extremely high temp of his regulator. Imho an smps though it is power efficient where it is required, is also a more complex & hence failure prone solution.

It's an engineering decision.