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Old 7th July 2004, 08:19 PM   #1
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Exclamation LM3x7 Datasheet Error?

I'm designing a PSU using LM317T/LM337T regs. The National Semiconductor datasheet states that the output voltage is calculated by:

Vout = 1.25 * (1 + (R2 / R1)) + (Iadj * R2)

where Iadj = 1.25 / R1

However, using this formula on the Typical Applications circuits, I cannot get anything like what voltage they say I should be getting.

This is further complicated by the LM337L datasheet stating that the output voltage is calculated by:

Vout = 1.25 * (1 + (R2 / R1))

Which gives apparantely correct results.

Only this datasheet has the formula in this way, the LM317L, LM317T and LM337T all have the formula as in the first case. The L suffix is TO92 package and the T suffix is TO220 package.

Has anybody actually got any proven formula or values I can try? I built one using L suffix parts and according to the second formula and it worked fine, but I want to use T suffix parts now.
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Old 7th July 2004, 08:22 PM   #2
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I(adj) is the current on the "adjust" pin, not the total current pulled through the device -- it's measured in microamps -- Nat Semi has min/max/avg values on the datasheet. -- something like 50uA
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Old 7th July 2004, 08:41 PM   #3
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Hi Richi,

I_adj is indeed the current flowing out of the adjust pin. If you keep the recommended value for R2, 220 - 240 ohms, you can safely ignore I_adj and use the formula for the LM317L as well for the LM317T.

Cheers
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Old 7th July 2004, 08:55 PM   #4
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Here is a spreadsheet which I made for the purpose.

Use it
Attached Files
File Type: zip lm317_calc.zip (2.8 KB, 20 views)
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Old 7th July 2004, 09:24 PM   #5
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Thanks guys.

I was slightly misreading the datasheet and confusing Iadj with the whole current flowing in the programming resistors.

Click the image to open in full size.
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Old 7th July 2004, 09:32 PM   #6
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Good that you found out. In real life you can have R1 up to 1 kohms without problems. Iadj which cleary is stated is far less than worst case.
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