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Old 9th March 2007, 04:18 PM   #1
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Default LM317 max input voltage??

Can an LM317 take an input voltage of 36v if its well heatsinked. If not whats the best way to drop voltage a few volts, is there a formula for using a resistor to drop voltage.
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Old 9th March 2007, 04:32 PM   #2
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Check the datasheet for your device.

Heatsinking depends on the voltage dropped and current required.
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Old 9th March 2007, 04:41 PM   #3
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The adjustable voltage regulators have a max voltage difference from input to output. that's why you can use them with a pre-regulator to regulate hundreds of volts.

So yes, you can use 36 Volts in on a 317 if your output voltage is more than a couple of volts.

What you have to be careful of is the power dissipation within the device. Every amp produced at the output is an amp going through the output transistors in the device. So the total power dissipated in the device is output current times the voltage drop across the device. That can be substantial if you are sourcing a lot of current.

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Old 9th March 2007, 06:19 PM   #4
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Be careful of output shorts.

The 317 may well shut down safely if the output gets shorted, but the differential voltage may then be well exceeded!
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Old 9th March 2007, 08:56 PM   #5
cpemma is offline cpemma  United Kingdom
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Default Re: LM317 max input voltage??

Quote:
Originally posted by imperfectcircle
... is there a formula for using a resistor to drop voltage.
Ohms Law; R = V/I.

If you know the maximum load current you can calculate a resistor to bring the supply down to around 3V more than the required output. However, at a very light minimum load it may then not drop enough to keep the differential safe, so take possible load range into account.

And for the resistor wattage, W = I2R
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Old 9th March 2007, 09:17 PM   #6
Nordic is offline Nordic  South Africa
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Going down the resistor line...
the best way I can see is with a voltage divider....

I.e. at least two resistors in series...

Imagine you want 2.5 amps max and 2.5V output (as and example) your input is 10V.

current flowing through two resistors in series will be the same regardless of their values using ohms law.
R= E/I, i.e. the total resistance for the two resistors will be 10V/2.5A = 4R

so the divider will have to split 4R into whatever the ratio of the output to the input is... i.e 1/4 +3/4

so one of the resistors is 1/4 of 4 and the other is 3/4 of 4.

Power over the 2 resistors will be 2.5V @ 2.5A and 7.5 @ 2.5A respectively.
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Old 9th March 2007, 10:08 PM   #7
cpemma is offline cpemma  United Kingdom
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Quote:
Originally posted by Nordic
Power over the 2 resistors will be 2.5V @ 2.5A and 7.5 @ 2.5A respectively.
And when the load is in parallel with the first resistor?
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Old 9th March 2007, 10:19 PM   #8
cpemma is offline cpemma  United Kingdom
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Quote:
Originally posted by Nordic
Power over the 2 resistors will be 2.5V @ 2.5A and 7.5 @ 2.5A respectively.
And when the load is in parallel with the first resistor?
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Old 9th March 2007, 10:32 PM   #9
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You could drop some of the voltage using a transistor as a pre-regulator. No matter whether you use a transistor or a resistor, you are going to be dissipating the same power. The transistor will actually be easier to get rid of the power since it's designed to be mounted to a heat sink.

Sheldon
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Old 9th March 2007, 10:44 PM   #10
cpemma is offline cpemma  United Kingdom
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The transistor's the better solution, little load dependency. A crude emitter-follower pre-regulator should do the trick.
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