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Posted

I just built a 0-28 VDC power supply using a 5A transformer and a LM317T. The problem is that I can only get 135 mA of DC current at the output terminals. I'm using a 250 ohm resistor between the output and adj. with a 10K ohm potentiometer for the output adjustment. Any ideas??? Thanks, John


Posted

Go to our articles section and read the tutorial for the LM317. There is a link at the top of the page. It might give you some ideas of what is not connected right.

Did you get the right pin-out of the chip? V-out is the center pin and it is also the tab on the top. So if you connected the tab to a grounded heat sink or a ground pad on your circuit board, you will certainly have a problem.
Otherwise, if you can post a drawing, we can look at it and comment.

MP

Posted

I'm getting 27.6 volts at the output. The output current remains constant at .135 A reguardless of voltage setting (from 1.25 to 28 V). The voltage at the input is 28.9 V.

Posted

The voltage can be adjusted smoothly using the pot from 1.25 all the way up to 27.6 V. This thing works great, with the exception of the low current output. I, too, an starting to think the chip might be at fault.

Posted

Ante,

The regulator gets hot only when the circuit output is drawing maximium current (in this case, .135 mA). In this case, the regulator gets VERY hot.

My current (I) testing rig consists of a digital ammeter in series with a 20W 100k ohm pot. I start with the pot at the maximium value and slowly decrease resistance while watching the ammeter. Admittedly, not the most scientific rig, but it seems to work.

I'll try to post the circuit.
John

Posted

I found the problem! Apparently when I installed the PC board on the standoffs, is was distorting just enough to break one of the traces on the board. Let this be a lesson . . . never assume ANYTHING is infallable.

John

Posted

According to my DVM, the ammeter resistance is 1.2 ohms (it's an older Radio Shack digital meter, 0-10 A)and the pot goes all the way down to .1 ohm. I had to install a heat sink on the unit while performing this current test because the unit would overheat quickly and start to shut down (totally understandable). Your resistance calculation does bring up some interesting questions, though.

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