LT1083CP Question

[Ken1]

Hi, I want to use an LT1083CP for the regulator in a variable linear power supply. The transformer puts out 12V and can supply up to 3 amps. This will be used on my service bench for powering up devices for short periods of time for testing and troubleshooting. I would like to know what values I should use for the resistors for the adjustment terminal on this device.

[Robert Reed]

Output--120 ohms in series with 1K pot to ground. Junction of 120 ohm and pot to Adj. terminal for 0 to approx. 12 VDC. Actually zero can only be obtained by taking low end of pot to a small neg. voltage (-1.25V). This is only the skinny as there are many more parameters to consider and you need to give more info on your tranny and exactly what you want out of the regulator. A decent heat sink will be required especially with the TO-220 package in lieu of the TO-3 . Good info by Googling data sheet from the web.

[haklesup]

Just remember that you need to remove as much heat as voltage you drop across the device, this is true of any LDO application. For example if the source is 12V and you set this up for 5V regulation into a 3A load, you will need to remove (12-5)*3=21W of power from the package. In most cases this requires a heat sink and fan. If you go even lower in voltage or higher in current (say you use a variable LDO like the LM338), then the situation just gets hotter.

For this reason, an adjustable LDO is not the best choice for a universal variable power supply. you can do it but you need much more heat sinking for the worse case than you need most of the time. an LDO is best used when the source is only a couple volts greater than the output regardless of the current needed.

http://www.linear.com/product/LT1083#simulate

[Robert Reed]

Actually you can reduce heat sink size required. At lower voltages with the introduction of a few power resistors and a 2 or 3 position wafer switch. These would be placed in series with the input to the regulator. Have done this several times in the past , but I cannot tell you the optimum setup and values without exact details of your circuit and requirements.