LM350 linear driver output problem

[dethlore]

I haven't been able to find any solid answer as to why this isn't working quite right... so I thought I"d post it.

The dip-switch is purely just for me to select a different resistance. Switch 4 gives me 250mA and switch 3(.5 Ohms) gives 980mA. I get the same readouts if I don't use the switch, so that's not hindering anything...

I've tested multiple batteries (2x 18650s). Batteries are fully charged and I have tested them at giving about 3A, so I know they're not bad.

In the picture, you can see the power being switched on with Switch 1, thinking that could be a problem, I ran the (+) straight to the LM350, but I still only get 980mA.

I have tested this on 3 different LM350s with the same results.

The current should be much higher than this with only .5 Ohms between Adj and Out on the LM350.

I must be missing something... thanks for your help!

 

[Cyparagon]

This is not an appropriate use of DIP switches - they are typically rated for very low switching currents (<25mA).

Your main problem is you don't have enough input voltage to power a load with a forward voltage that high. Use three batteries, power a lower voltage load, or use a different driver.

 

[dethlore]

The dip switches were just for messing around, I realize this is not the proper way to use those.

The lack of voltage sort of makes sense.. what confuses me is that I thought the LM350 worked on the formula: Iout = Vref/R1 where R1 is the resistance from Adj and Out pins. With a Vref of 1.25, the 0.5 Ohms I have shown in the picture should give current output of 2.5A. What I couldn't find anywhere, but am assuming based on what you're saying, is that the IC can only give that much current output if there is enough Vin. So whether I'm using 15V or 20V, using .5 Ohms should give 2.5A out, but if I'm not using enough then it will just use as much voltage as it can, but yield less current. ie. 8Vin will only yield about 1 Amp out, 12Vin yields about 2 Amp out.

Am I understanding correctly?

 

[Cyparagon]

Your Vin should be Vf+1.25+ the minimum dropout of the regulator. The latter depends on operating temp, current, and still varies slightly with each piece. See the datasheet for a graph. It looks to be about 2.25V under these conditions. You may have read that diodes drop 0.7V, but this is only at low currents. At currents this high, the drop can be up to 0.9V or more. There is also the drop of the sense resistor to consider (2.5V).

Theoretically you need at least 0.9*4+2+2.25+1.25 ≈9V to stay in regulation. Anything below that and the regulator starts dropping current.

You also want to keep in mind that battery voltage sags under load, so even if this figure was 8V, it still would not be regulated properly. Use nominal cell voltage (3.6V) when calculating these things, not the full charge voltage.

 

[dethlore]

oh yeah... this does make sense. I've always been building these drivers for smaller loads... up to 1.3A maximum which is maintainable with the 2x 18650s. Kept overlooking the thought of more voltage drop.

Thanks for your help Cyparagon!