Hi Mike,

I have searched data sheets from natsemi and TI, and cannot see how to
calculate power disapation.

Could you please provide a reference.

I disagree with your comment, to me a linear device working in a linear
region has to dissipate the power not consumed by the load ie

If I have 24 volts in, 5volt out to the load, load consumes say 1 amp.

Power into regulator is 24W, power out is 5W, so the regulator must consume
the missing 19W.

With your idea, the max power consumption would be 152mW ( vin - vout * Qc)
= (24 -5) *0.008A (from the TI data sheet maximum value)

Regards,

Kat.

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----- Original Message -----
From: "Mike Blakey" <mike.blakey@BAESYSTEMS.COM>
To: <PICLIST@MITVMA.MIT.EDU>
Sent: Monday, November 12, 2001 3:58 PM
Subject: Re: [EE]: PIC overheating a 7805 !


> The power used and dissipated by the voltage regulator is NOT calculated
that way.
> the power dissipated by the device is approx. = (Vin-Vout) * quiescent
current and
> NOT load current. Check the data sheets!
>
>
>
>
>
> haroldhallikainen@JUNO.COM on 09/11/2001 18:44:56
> Please respond to PICLIST@MITVMA.MIT.EDU@INTERNET@wtgw
>
> To:     PICLIST@MITVMA.MIT.EDU@INTERNET@wtgw
> cc:
> Paper Mail:
> Subject:        Re: [EE]: PIC overheating a 7805 !
>
>                    *** WARNING ***
>
> This mail has originated outside your organization,
> either from an external partner or the Global Internet.
>      Keep this in mind if you answer this message.
>
>
>         24V is a bit high! Calculate out the power dissipation
> (24V-5V)*LoadCurrent. Take this (in watts) and multiply by the rating of
> your heat sink in degrees C/ watt. This should give temperature rise from
> ambient.
>         Another thing to watch for is oscillating regulators, which get
hot.
> Check the output on a scope and make sure you've got DC without a bunch
> of RF.
>         Finally, consider a switching regulator. The inefficiency of using
a
> linear regulator to go from 24V down to 5V is pretty impressive.
>
> Harold
>
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