Power dissipation by an LM78XX regulator

[brett]

Ok, here's one for the techos.
I'm looking at using a regulated supply for a small class A amp.  It'll draw about 1 amp.
I've got a wall wart that gives 24VAC at 1 amp.  Rectified and filtered a bit (2200uF), that'll give me about 30V DC +/- 3 volts ripple.
I'm thinking about using an LM7824 regulator, but I'm not sure how much heat it will be disipating.  Is it 30V minus 24V times 1 amp = 6W ??  Or can I ignore the 2volt minimum drop across the regulator ie 4 volts x 1 amp = 4W.  (I don't think that the turn-on voltage for the regulator would be contributing to heat production, but I wouldn't be surprised if I was wrong, either :oops: )
If it's 6W, I need a fairly big heatsink (I might even use 2 regulators in parallel), while if it's 4W, a smallish heatsink will do the job.
Any suggestions?
thanks

[Basile]

it depends on the fact if teh regulator is enclosed in a box or not... perhaps a little fan xwill be useful if a heatsink doesn't do its job...

[R.G.]

The power dissipated in a device is always, always, always, always the voltage across it times the current through it.

If your amp will have a constant 1A of current, then the regulator will be dissipating 6W. Since the 78xx regulators self limit current at about 1A, it may or may not be in current limit part of the time if the amp uses more.

You may not ignore the minimum 2V drop across the regulator. It's still 2V times 1A.

One regulator is darned marginal in this application.  Consider two in parallel with load sharing resistors so one regulator does not take all the load until it sags and the other regulator  curs in.

[Gringo]

You can also try using a lm317 set for 24vdc output. IIRC it's rated at 1.5A...

[niftydog]

OR; use the "current bypass" configuration that is present in the better datasheets such as those available from Fairchild Semiconductor.

[brett]

Hi.  Thanks all.
So simple on reflection - V x A !! :oops:
I'll use either parallel devices or the PNP bypass method.
cheers

[loscha]

Power Dissipated by Regulator

PD = (VoltageRippleMax - VoltageOut) * CurrentOutMax

to work out what heatsink you need:

ThetaJunctionAmbient = (TemperatureJunctionMax - TemperatureAmbient) / PD

ThetaSinkAmbient = ThetaJunctionAmbient - ThetaJunctionCase - ThetaCaseSink

ThetaSinkAmbient is the value you look up in the catalog.

All those terms mean, as best as I can explain them....
Voltage Ripple Max = the maximum voltage, the top of the waveform in a filtered rectified signal
Voltage Out = the output of the regulator, a 7809 has an output of 9 volts
CurrentOutMax = Voltage output / load resistance
PD = Power dissipated in Watts

ThetaJunctionAmbient = the amount of heat that will be transferred from the actual chip inside to the case of the chip.
Temperature Junction Max = usually 150 degrees
Temperature Ambient = Maximum temperature the unit will be run under. If it's in a rack amongst other stuff, could be as much as 75-80 degrees. In Australia, for the open air, we use 40 degrees, cause it can get pretty damn hot up here!

ThetaSinkAmbient = the size of heatsink you need. A smaller number is better! In yr catalog of parts, a larger heatsink will have a lower value.
ThetaJunctionCase = 4 degrees per watt
ThetaCaseSink = thermal resistance of the mica washa .5 degrees / watt

sorry, it's a bit garbled, hope it helps someone!

[aron]

Good thread!!!!!

ARCHIVE

[R.G.]

I'd agree with most of that except

Power Dissipated by Regulator

PD = (VoltageRippleMax - VoltageOut) * CurrentOutMax

gives you an overly pessimistic figure. If the ripple is small, that's OK, but for larger ripple, the thermal mass of the chip/heatsink averages the heat over the ripple period, and the time that the ripple voltage is below the average level gives a lower dissipation matching the time the ripple is above it.

The correct expressions will be the time integral of V(t) * I(t) over a cycle.

But for low ripples, you're being conservative going with the max. Also with the average and min, for the same reasoning.

[George Giblet]

Voltage Ripple Max = the maximum voltage, the top of the waveform in a filtered rectified signal

The equation is slightly more accurate if you use the average DC voltage, as would be measured by your multimeter.  The current is largely independent of the input voltage so the average power is obtained by Vaverage*CurrentOutMax.

[R.G.]

Yep, that's what I said.

[GFR]

The power dissipated in a device is always, always, always, always the voltage across it times the current through it.

Well... not so emphatically always - not if the current and voltage are out of phase.

[R.G.]

Always. It's only if you're measuring current and voltage with AC reading meters that phase angle comes into it. Otherwise, it's always the integral of the absolute value of the instantaneous current and voltage.

The way phase angle gets into it is to make the high voltages appear only during low currents and vice versa. The time integral has no such confusion.

[Sir H C]

Remember the Tj for the part too.  If you are putting 6 watts through it, you need to know about heat sinking and what sort of self-heating you will get from the part or whatever else you are heating up.  Get things too hot and they will go pop.