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Old 02-02-2012, 12:48 AM #33
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Default Re: Super Low voltage drop Linear Reg

Quote:
Originally Posted by lasersbee View Post
We never design on the edge... it is called a reliability buffer
factor..
That may be silly to you... but our defective or returned
products count is virtually non existent...
No, I'm not saying design on the edge. There's a curve, it's in the datasheet, and it's meant to be used. If you're designing a driver for a particular point on a curve, there's nothing wrong (in fact, it's appropriate) to plan for the corresponding values that appear on that curve. I don't have the 1085 datasheet in front of me, but often times they'll actually include 3 different slopes. One for typical, and the other two for min/max. Or sometimes three for different operating temperatures.

If 1.8A implies a different dropout than the same IC would at 3A, it's completely appropriate, and not "designing on the edge" to plan around the supplied data you have for your particular current.

I'm completely with you on the notion of not engineering around a purely ideal-world scenario. But I don't think that's what you're doing if you intelligently factor in a dropout that corresponds to the current level you're designing for. Especially now that some of these LDO ICs are meant for currents as high as 7 or 8A, it just wouldn't make sense to design around the max-dropout, if you're not ever going to exceed 2A, etc.


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Last edited by rhd; 02-02-2012 at 12:49 AM.
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Old 02-03-2012, 05:41 PM #34
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Default Re: Super Low voltage drop Linear Reg

Since it will take a while to get our KA278RA05 LDO order from
Mouser I bought a few similar 4 pin LDOs from DigiKey yesterday
P/N PQ070XF01 which is only a 1Amp variety and it does not have
the internal R1 and R2 of the KA278.
But for this test it really should no matter...

We hooked up the circuit as shown in the OP. To be able to mimic
an LD that changes it's internal resistance due to internal heat
buildup we chose to use an LED and an a current limiting resistor
Rv. We did this test twice using a 1000 Ohm current limiting Rv and
once with a 100 Ohm resistor and noted the voltages and currents
below. We used a 5.6 Ohm resistor for R2 since that is what we
had on hand....


Rv .......... Input mA ... Out - Adj mA ... Out V ... Adj - Gnd mA

LED+100 ........ 81.7 mA ............... 55.1 mA ............. 7.89V ............ 55.1 mA

LED+1K .......... 33.6 mA ................ 6.5 mA .............. 8.20V ............. 6.5 mA

If the the circuit shown in the OP was a true Constant Current Regulated
circuit the Current across the LED+Rv shoukd have stayed the same and
not varied..

If the the circuit shown in the OP was a true Constant Current Regulated
circuit the output voltage would have varied much more than it did to
keep the Current Constant...

As it is this seems to be set up as a Constant Voltage regulator...

I'll do this test again when the actual parts arrive but I don't think the
results will change.


The Drawing on the left shows that the LD is in the Voltage adjusting
circuit (R1/R2) of the actual Voltage Regulated circuit on the right shown
in the Spec Sheet.





Jerry
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Last edited by lasersbee; 02-05-2012 at 12:38 PM. Reason: Spelling errors
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Old 02-05-2012, 07:02 PM #35
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Default Re: Super Low voltage drop Linear Reg

I'll do some more tests tonight if I get a chance. I've been helping my dad and grandpa put down our new wooden floorings, and my hands/back are killing me.

And yes, I will hook up my multimeter the right way just for you Jerry.
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Old 02-05-2012, 08:48 PM #36
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Default Re: Super Low voltage drop Linear Reg

Quote:
Originally Posted by Le Quack View Post
I'll do some more tests tonight if I get a chance. I've been helping my dad and grandpa put down our new wooden floorings, and my hands/back are killing me.

And yes, I will hook up my multimeter the right way just for you Jerry.

So... you admit it was the wrong way....

Take a break... and rest up.. Tomorrow is another day...


Jerry
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Old 02-06-2012, 12:48 AM #37
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Default Re: Super Low voltage drop Linear Reg

Well, I did a quick test when I went downstairs;

Hooked up a 9 volt battery with 5 ohms of current sensing resistance; Got out 250 mA, which is in accordance with the formula used to calculate output current.

I then took TWO 9 volt batteries, and hooked it up the same;

Still 250 mA. So, it does work.
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