LM7805 voltage regulator help. TIA
Rubicon
Happy New Year.
When using a LM7805 voltage regulator I see from the datasheet that it
requires a minimum of 7.5 volts to maintain line regulation.
To work out it's power dissipation with 7.5 volts Vin from a 500mA
unregulated power pack I understand it's:
(Vin - Vout) * Iload
so
(7.5V - 5V) * 0.44A = 1.1 Watts.
However under two different load conditions I measured the Vin as 7.2V
and 9.11V. Therefore I must take the highest value as worst case so:
(9.11V - 5V) * 0.44A = 1.81 Watts worst case.
The datasheet states that the Internal Power Dissipation is internally
limited. The note to it also states: "Thermal resistance of the TO-220
package (T) is typically 4degC/W junction to case and 50degC/W case to
ambient.
I have a snap-on TO-220 heatsink on it and it's Thermal Resistance
rating is 32degC/W.
How do you properly use these figures? The heatsink feels hot but not
painfully so at 1.81Watts. How do you work out it's actual temp to
make sure that it's not so hot that surrounding wire insulation, the
case etc aren't damaged?
In addition the single regulator is on a dual channel audio power
meter and I'm wondering if one regulator is good enough for all the
surging currents from both channels?
Also it seems I can't have a diode (1N004) for reverse polarity
protection due to the 0.6V drop from minimum Vin of 7.5V but if the
Vin rises to 9.11V due to not all the current being drawn from the
unregulated power pack can I in fact put one in safely?
Hold on here! The other measurement I took was Vin 7.2V. Damn!
Any help appreciated.
Many thanks.
Andrew.
John Popelish
>
> Hello,
>
> Happy New Year.
>
> When using a LM7805 voltage regulator I see from the datasheet that it
> requires a minimum of 7.5 volts to maintain line regulation.
> To work out it's power dissipation with 7.5 volts Vin from a 500mA
> unregulated power pack I understand it's:
> (Vin - Vout) * Iload
> so
> (7.5V - 5V) * 0.44A = 1.1 Watts.
At 500 ma output current, I think only about 2 volts differential is
needed to maintain regulation. The data sheet specifies 2 volts
(typical) drop out at 1 amp out.
> However under two different load conditions I measured the Vin as 7.2V
> and 9.11V. Therefore I must take the highest value as worst case so:
> (9.11V - 5V) * 0.44A = 1.81 Watts worst case.
>
> The datasheet states that the Internal Power Dissipation is internally
> limited. The note to it also states: "Thermal resistance of the TO-220
> package (T) is typically 4degC/W junction to case and 50degC/W case to
> ambient.
> I have a snap-on TO-220 heatsink on it and it's Thermal Resistance
> rating is 32degC/W.
>
> How do you properly use these figures? The heatsink feels hot but not
> painfully so at 1.81Watts. How do you work out it's actual temp to
> make sure that it's not so hot that surrounding wire insulation, the
> case etc aren't damaged?
You add the thermal resistance die to tab of 4 degrees per watt to the
thermal resistance of the heat sink of 32 degrees tab to ambient for a
total of 36 degrees die to ambient. At a 1.8 watt load, this implies
that the die will operate at 1.8*36=65 degrees C above ambient (say,
30 degrees) for a die temperature of 95 degrees C. It is rated to
operate at an absolute maximum of 125 degrees C. So you are within
spec at this ambient temperature (assuming you get an efficient
coupling between the tab and the heat sink). I would be more
comfortable if the heat sink were a bit larger, though.
By the way, this data sheet:
http://www.fairchildsemi.com/ds/KA/KA7805.pdf
Specifies the die to tab resistance as 5 degrees per watt, but this is
a minor increase at 1.8 watts.
> In addition the single regulator is on a dual channel audio power
> meter and I'm wondering if one regulator is good enough for all the
> surging currents from both channels?
> Also it seems I can't have a diode (1N004) for reverse polarity
> protection due to the 0.6V drop from minimum Vin of 7.5V but if the
> Vin rises to 9.11V due to not all the current being drawn from the
> unregulated power pack can I in fact put one in safely?
> Hold on here! The other measurement I took was Vin 7.2V. Damn!
>
> Any help appreciated.
>
> Many thanks.
>
> Andrew.
You could use a schottky diode or a 3 amp rectifier to lower the diode
drop a bit. But you have to worry about the bottom of the ripples,
not the measured (average) voltage. And it is a good idea to allow
for low line voltage. Could you use schottky diodes in the supply
rectifier to get a bit higher minimum voltage?
--
John Popelish
Rich Grise
You calculte the temperature rise pretty much the same way as
you would voltage drop. You've got 4 degC/watt junction to case,
and 50 degC/watt case to ambient; but with a heatsink, you need
to know its (NOTE NO APOSTROPHE) thermal resistance. And the 7805
has internal overtemperature protection, such that if your heatsink
can't dissipate 2 watts fast enough to keep the chip below its
(NOTE NO APOSTROPHE) thermal shutdown temperature, the chip will
just shut itself down and quit supplying voltage to the load.
(or current, or power, whatever - let's not get into that can
of worms!)
And yes, use the diodes for reverse polarity protection - see
the app notes - if you're competent to connect a 7805, I'm sure
you're competent to look up app notes. :-)
Happy Brew Beers!
Rich Grise (say 'Gryce'),
Self-Appointed Chief, Apostrophe Police.
"We have met the enemy, and he is us!"
- Pogo Possum, ca. 1950's
Bitbanger
If the voltage dips below that on the trough of the ripple, the
regulator will follow it down somewhat during that time. I call this
the "Kilroy" since the scope trace looks a little like the old WW II
Kilroy drawings. :) Sometimes the cure for this is to add capacitance
ahead of the regulator to bring up the trough voltage.
The 7805 will self-limit on temperature and current. In either case,
the regulation goes away and you don't get the output you want.
The input has to meet the minimum requirement and the heatsinking
has to provide dissipation for the maximum average input voltage at the
maximum load condition. The input voltage is an instantaneous
requirement, the dissipation is an average requirement.
There are other regulators that should work that have lower dropout
voltages.
-- BB
Rubicon
Thanks for all your help.
Rich Gryce you are completely correct regarding my punctuation. I
stand corrected. No apps notes on my datasheet. Will look for another
larger version of the datasheet.
With further thought and experimentation I have worked out what I
think is the correct voltage to use, heat dissipation of the regulator
etc.
Please correct me if I'm wrong.
My audio power meter has two modes, Dot and Bar.
When in Bar mode:
7.5V 500mA unregulated power pack setting.
Current drawn is .44A
7.19 V out of the power pack to the regulator
4.91 V out of the regulator
Dot Mode:
Current drawn is 0.08A
9.06 V out of the power pack to the regulator
5.04 V out of the regulator
Also when there is no signal input:
Current drawn is 0.04A
9.35 V out of the power pack to the regulator
5.04 V out of the regulator
And now:
When in Bar mode:
9V 500mA unregulated power pack setting.
Current drawn is .44A
8.7 V out of the power pack to the regulator
5.03 V out of the regulator
Dot Mode:
Current drawn is 0.08A
10.89 V out of the power pack to the regulator
5.04 V out of the regulator
Also when there is no signal input:
Current drawn is 0.04A
11.21 V out of the power pack to the regulator
5.04 V out of the regulator
Using a spreadsheet I've worked out the power dissipation of the
regulator for each mode using a 32degC/W heatsink and excluding the
reverse polarity diode drop as:
HEATSINK DISSIPATION
DIE-TAB Deg 5
HEATSINK Deg 32
AMBIENT Deg 30
DIODE DROP V 0
(Die-to-Tab deg) + (Heatsink deg) * (Regulator Wattage) +(Ambient Temp
deg)
Deg %125degC Max W
7.5V BAR 67.1024 53.68192 1.0032
7.5V DOT 45.2912 36.23296 0.3216
7.5V NIL INPUT 40.5168 32.41344 0.1724
Deg % W
9V BAR 86.6736 69.33888 1.6148
9V DOT 49.976 39.9808 0.468
9V NIL INPUT 42.8976 34.31808 0.2468
With 9V I can have a reverse polarity protection diode, ensure
regulation and keep the heat dissipation of the regulator within
limits.
A 28degC/W heatsink with the diode drop of 0.6V brings the dissipation
down to:
Deg %125degC Max W
9V BAR 72.8224 58.25792 1.3508
9V DOT 46.76 37.408 0.42
9V NIL INPUT 41.2384 32.99072 0.2228
One other thing.
The datasheet states that the operating temp of the LM78XXC series is
0 - 70degC which looks to be a problem with the 9V Bar heatsink deg of
76.23degC. Plenty of ventillation required I expect.
Any help appreciated with my workings.
Cheers,
Andrew.
On Mon, 31 Dec 2001 23:56:18 GMT, ZZru...@netaccess.co.nz (Rubicon)
wrote: