JAW regulator discussion

[shameem]

I was just thinking about this - because i have been playing with a JAW unit. It uses a 7805 for the uc and display which draw about 200ma (together). The 7805 sinks an extra 500-600ma as heat - thats a lot of power wasted and lots of heat. I am looking into low cost alternatives - i might get one of those switching ICs from maxim. The trouble is ATMEGA has a sample/hold ADC - i dont know how that will hold up to a switching power supply......

[Fred]

It uses a 7805 for the uc and display which draw about 200ma (together). The 7805 sinks an extra 500-600ma as heat - thats a lot of power wasted and lots of heat.

Either your sentence or your math is broken :-)

If the draw of the devices is 200mA then the current flowing through the reg is also 200mA (with some small internal use) and the power dissipated is the voltage difference between load and source times the current drawn by the load. They don't work like a Zener and resistor setup AFAIK anyway.

Fred.

[shameem]

Too much coffee ....

Anyways - if i put the ammeter in series after 7805 the current draw is about 180-200ma. If i put it on the 12v side the current draw is 800ma.... which makes sense because 7805 "shunts" or drops the extra voltage across a "resistor" to reach a stunning conversion efficiency of 14-35% ......

[Delta]

either something is broken, or you took your measurement wrong - there is no way an LM7805 should draw that much with that output, I have one on the desk here supplying 130mA and it draws 140mA.

[shameem]

Its the JAW unit from here - http://14point7.com/Widebands/JAW_1_03/JAW_1_03.htm

Apparently it's a well know problem - http://14point7.com/forum/index.php?topic=150.0

Since the designer doesnt provide support or the schematics - it's a bit harder to troubleshoot....

[Delta]

That persons post shows a fundamental lack of understanding of how a voltage regulator works, so much so that I almost fell of my seat laughing. I have a VERY simple circuit that performs far better than what his maths suggests. The actual internals of a regulator are far more complex and hence they draw less than my circuit, but its a pretty close approximation.

Don't believe me - look at the datasheet for an LM7805 - its current draw over what it supplies is supposed to be 5mA - I've got some that take 10mA more (they are quite old), but thats it. Something is definately wrong if its drawing that kind of current.

[shameem]

look at the datasheet for an LM7805 - its current draw over what it supplies is supposed to be 5mA -

I did look at the datasheet - it seems 5mA figure is the quiescent current. That does not seem to be the same as extra (operational) current draw.

http://www.national.com/ds/LM/LM78L05.pdf

[Delta]

LM7805 != LM78L05

Which one is it??

If its the LM78L05 - then its only rated to supply 100mA 140mA max - so its probably in latch up and out of the supply transistors SOA.
If its the LM7805 - then its rated to 1A and this should not be happening - there must be a problem. Quiescent current is the current required to run the device at what ever voltage temp etc over and above the operating current ie the current thats being output.

Seriously get a spice simulator - and put in a simple zener opamp controlled transistor setup and you will see that the input current is only a few mA larger than the output - this is WELL KNOWN. They would NOT design a regulator worse than this in integrated form. Seriously I don't understand why people would argue the point when its probably one of the most simple circuits in analog electronics and one of the most well known - go look it up. Its simply not possible for it to draw that much current unless something is wrong. Take a LM7805 and run it into a single resistor and measure the input and output currents - thats whats sitting on my desk - and its only 5-10mA difference.

[Fred]

Shameem,

If you have a voltage source of 20V and you have a fixed load called R1 at 1 Ohm and must have the potential at the input to R1 at 5V you must place a 3 Ohm resistor, R2, between it and the Voltage source. if suddenly you halve R1 to 0.5 Ohm, you need to do the same to R2 and make it 1.5 Ohm to keep the voltage current. Note there is no "third leg" in this diagram so ALL current travels through both resistors. This is effectively what these regulators do. The vary their own "resistance" in order to keep the voltage on the load consistent. At 200mA if the regulator is not heatsinked it will get bloody hot. That is normal. Attach a heatsink to it and be happy :-)

I also cracked up at darrylp's math. Quite amusing.

I think I'll split these posts off into their own thread.

Fred.

[shameem]

Thanks for the note Fred - i am actually planning to get a switched regulator and experiment with it. The fact remains that 7805 draws too much current and puts out too much heat (cant touch the heatsink after a few minutes of normal operation). I replaced it with a 2940 and it is getting hot too. The designer of the circuit is MIA and the rest of the community that uses this circuit has rigged up many different kinds of fans and are content with it. I havent even connected the sensor yet - and that will bring up another heat issue with another regulator LM317 - that one comes with a heatsink in the kit and so it must get even hotter once the sensor is connected. I just have to figure out which switched regulator works best and how to deal with sample/hold ADC.