7809 regulator consumption?

[JimRayden]

I was wondering, how much current do the regulator suck... I got an idea of putting two batteries in a stompbox for 18V and regulate it to 9V. That would be more stable than a single bat, but what about the lifespan? How much lower will it be compared to the lifespan of two single unregulated batteries?

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Jimbo

[Paul Perry (Frostwave)]

If I understand you right, the life might be LONGER than that of a single battery, because the two 9v batteries (if equal) can drain right down to 6v each, and still give 9v via the reg.
Drain current (no load) on a bog standard LM7809 is between 3 and 5 ma, but there are a LOT of regs out there now, and some have virtually no stanfdby current at all. Linear Technology re strong on regs.
The most efficient thing to do, is to make a switchmode power supply... before you write off this idea, there may be some kits or 'simple' plans available from the DIY LED torch guys. (if you think WE are crazy..!)

[JimRayden]

If I understand you right, the life might be LONGER than that of a single battery, because the two 9v batteries (if equal) can drain right down to 6v each, and still give 9v via the reg.
Drain current (no load) on a bog standard LM7809 is between 3 and 5 ma, but there are a LOT of regs out there now, and some have virtually no stanfdby current at all. Linear Technology re strong on regs.
The most efficient thing to do, is to make a switchmode power supply... before you write off this idea, there may be some kits or 'simple' plans available from the DIY LED torch guys. (if you think WE are crazy..!)

You mean it will be longer than two regular battery lives? Because it's two batteries that'll go way below their normal "okay-I'm-dead-now"-voltage.

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Jimbo

[Hal]

ehhhhh

voltages of a battery doesn't really decrease...the cell should always produce a constant voltage.  Just the internal resistance increases, so with any normal load, the battery will apear to have a lower voltage.  I'm not totally sure that alkalines work this way, but i think they should. 

yes, yes, yes, though, everythign said was correct.  Most of the power lost in a regulator is lost to the voltage drop - 10ma, at 9v, from 18v drops 9v...p=90mw. 

[JimRayden]

So you're saying that I will have just one battery life of regulated 9 volts? That means connecting two batteries parallel and not regulating it will last way longer?

I have a high consumption circuit and I'm trying to find the best solution for not using the power supply.

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Jimbo

[RedHouse]

Data sheet says between 1.5mA and 2.5mA depending on temperature.

[JimRayden]

Hmm, how much current does a fetzer valve suck?

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Jimbo

[vanhansen]

Hmm, how much current does a fetzer valve suck?

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Jimbo

I think this will depend on which JFET you use.  Each have their own current properties so you'd have to do the math for each one.

[niftydog]

Just a heads-up - be careful with the power dissipation here.

Taking 18V down to 9V means you're dumping 9V in the regulator. Pulling just 100mA through the regulator will result in it dissipating almost 1W, which doesn't sound like much, but it's enough to make the regulator run pretty hot.

[JimRayden]

I think this will depend on which JFET you use.  Each have their own current properties so you'd have to do the math for each one.

Come on, you know I ment this question generally for a FET stage...

Just a heads-up - be careful with the power dissipation here.

Taking 18V down to 9V means you're dumping 9V in the regulator. Pulling just 100mA through the regulator will result in it dissipating almost 1W, which doesn't sound like much, but it's enough to make the regulator run pretty hot.

...but I don't think it'll be that high either.