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Why use an FET on battery charger output?

Hi,

I'm working on a battery charger for a Palm Pilot. This fits inside the modem docking case, where space is limited. It basically takes 12v, regulates it down to 4.1v with a LM317L, and feeds it to the docked palm pda.

There's a circuit on the web that does this:

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The circuit has an FET as a switch on the output, to prevent the palm from discharging into the LM317 when 12v is not applied. Can this FET simply be replaced by a diode? Since the circuit has a variable regulator, it can be adjusted up to compensate for the voltage drop across the diode.

If this is right, then a 3 lead part gets replaced with a smaller 2 lead part. And the diode actually takes up zero space, as it could be used as the output wire with some spaghetti tubing or heat shrink.

Am I missing something here?

Reply to
nospam
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One day snipped-for-privacy@vegan.net got dressed and committed to text

Seems feasible to me, dont forget the diode will drop more volts than the FET.

-- Regards ..... Rheilly Phoull

Reply to
Rheilly Phoull

You could do a LM431 and a few PNP's to pull down the 12vdc to 4.1 . But the LM431 needs to draw thru a 300 ohm R to start a PNP or the 431 will turn on prematurely at 2.0 vdc. and there are hundreds of circuits .... I don't like the 317 .

It's easy to add a coil and a diode to make the LM431 circuit a "switcher" or "buck" regulator ....

Reply to
Paul Jones

A larger voltage drop is the only difference, right? The drop can be adjusted out on the LM317.

Is there any other benefit in using an FET?

Rheilly Phoull wrote:

Reply to
Lee

The mosfet works as a kind of variable resistor. At higher output currents the gate voltage drops and its resistance increases, limiting current. A series diode won't do that. As a current-limiting scheme, the mosfet is about as simple as you can hope for. Also, he says exact voltage is critical with lithium batteries; you can get the 317 output dialed in very precisely. With a diode on the output, you will have a varying voltage drop across the diode according to the current.

Reply to
kell

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