Power Usage Question for LM628

[Burner1]

I am working with an LM628 Motion controller using it to read the position of an Encoder.

Let me preface by saying that I am a hobbyist and not formally educated in electronics. I understand a lot but sometimes miss the obvious.

I have set up the circuit with an Arduino and have it working with a larger power supply. I made the circuit around an pic 18f2520 and a 2.1a 5v regulator but the regulator is getting warm...and I don't think it should be getting warm meaning I need to go over my circuit.....but...I have a question I should be able to answer and I can not figure it out.

I am trying to reed the datasheet for the LM628 and I cannot figure out what tells me how much power I can expect it to use. This is my weakness in reading and understanding the datasheet.

Under "Absolute Maximum Ratings" I see maximum power dissipation of 605 mW. Is this the figure I need? Does this mean that at 5 volts I need a supply of 121 mA?

Thanks
Gary

 

[andre_luis]

Absolute Maximum Ratings is not indeed for operational purpose, but merely to know the condition up to what device can still operate, but will not burn yet.


+++

 

[FvM]

The part has a maximum Idd specification of 110 mA in electrical characteristics. It's a kind of legacy chip made in NMOS technology and has considerable static supply current consumption.

 

[Burner1]

Thank you, that is what I needed.

Gary

 

[Burner1]

Something does not seem right as far as the numbers. Maybe it does and I am just not getting it. The circuit is working but the current draw has to be more then 110 mA. I am using a 2.2a voltage regulator with a 12v power supply; number MC7805BTG. When the PIC is pulled and the LCD not connected and only the LM628 in the circuit it takes about :30 for the voltage regulator to start warming up. After a minute it is starting to get quite warm. With everything together the circuit works but I just would not think that the LM628 by itself, if it really is 110 mA would cause the regulator to get that warm:

**broken link removed**

 

[Audioguru]

Heating of the regulator (its power dissipation) is calculated by the current in it (why didn't you measure the current?) times the 7V across it.
The datasheet says that without a heatsink then 1.5W will make it VERY hot but not too hot.
If it doesn't have the input and output capacitors shown on the datasheet then it will oscillate which makes it hot.
If the input power supply is reduced to 9V then the temperature will be almost cut in half.

 

[Burner1]

Thank You for the answer. I am going to measure it (and I need to), just have not yet. I can lower the voltage but I was just trying to understand it. It just doesn't seem that 110 mA would cause it to do so which makes me think I am missing something. I have used this thing for other things which draw more then 110 mA but I never recall this heat.

I did change the Lm629 just to ensure I was not working with a damaged chip and it is functioning.