7809 replacement

M

Marcellus Pereira

Guest
Dear ALL,

I'm using a 7809 ( voltage regulator ) with 18~20V In and 40mA ( 1A,
50 ms peaks ).

But it is very hot... I can't lower the input voltage. Any ideas, like
a pin-compatible more robust device?

Thanks a lot,
Marcellus
 
Marcellus Pereira wrote:
Dear ALL,

I'm using a 7809 ( voltage regulator ) with 18~20V In and 40mA ( 1A,
50 ms peaks ).

But it is very hot... I can't lower the input voltage. Any ideas, like
a pin-compatible more robust device?

Thanks a lot,
Marcellus
Click to expand...

First, scope it to satisfy yourself that the regulator isn't
oscillating. You *are* asking that poor little device to dissipate a
heck of a lot of power. Help it out; introduce it to a good heatsink.
 
"Michael" <NoSpam@worldnet.att.net> wrote in message
news:3FB92413.63364E83@worldnet.att.net...
Marcellus Pereira wrote:

Dear ALL,

I'm using a 7809 ( voltage regulator ) with 18~20V In and 40mA ( 1A,
50 ms peaks ).

But it is very hot... I can't lower the input voltage. Any ideas, like
a pin-compatible more robust device?

Thanks a lot,
Marcellus


First, scope it to satisfy yourself that the regulator isn't
oscillating. You *are* asking that poor little device to dissipate a
heck of a lot of power. Help it out; introduce it to a good heatsink.
Click to expand...
I assume that you mean a 78M09, and that your average current is 40mA; a
TO-3 package should be able to dissipate 440mW without a heatsink.

In a linear regulator such as the 7809 the difference between the input and
output voltages is burned off as heat. So no matter what regulator you
choose you're going to dissipate the same amount of heat. If you use the
same package as your current 7809 it'll get just as hot. The only things
that you can do to control the situation are to control the amount of power
that you're dissipating with a pass resistor or transistor, or to control
the surface area that you're dissipating from with a heatsink.
 
"Marcellus Pereira" <mpereira.nospam@myrealbox.com> skrev i melding
news:ap6irvoogbqlt1snd5sb753n5652bg0119@4ax.com...
Dear ALL,

I'm using a 7809 ( voltage regulator ) with 18~20V In and 40mA ( 1A,
50 ms peaks ).

But it is very hot... I can't lower the input voltage. Any ideas, like
a pin-compatible more robust device?

Thanks a lot,
Marcellus
Click to expand...
What you can do is to couple an PNP power-transistor in parallell with your
7809 like this:

.--------------------------.
| |
| E |
| | T1 -> PNP
| |<
| .------------||
| | B |\
| | |
| | ____ K |
| ___ | | | |
o---o--|___|---o--|7809|-------o---------o
Vin R |____| Vout
|
|
|
|
|
===
GND
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

When the voltage over R exceeds ca 0.6V, T1 starts conducting, so the most
of the current goes through T1 instead of the 7809 ic.
 
On Mon, 17 Nov 2003 22:41:20 +0100, "Geir Klemetsen"
<regeir@N0ZPÁM.start.no> wrote:

"Marcellus Pereira" <mpereira.nospam@myrealbox.com> skrev i melding
news:ap6irvoogbqlt1snd5sb753n5652bg0119@4ax.com...
Dear ALL,

I'm using a 7809 ( voltage regulator ) with 18~20V In and 40mA ( 1A,
50 ms peaks ).

But it is very hot... I can't lower the input voltage. Any ideas, like
a pin-compatible more robust device?

Thanks a lot,
Marcellus

What you can do is to couple an PNP power-transistor in parallell with your
7809 like this:

.--------------------------.
| |
| E |
| | T1 -> PNP
| |
| .------------||
| | B |\
| | |
| | ____ K |
| ___ | | | |
o---o--|___|---o--|7809|-------o---------o
Vin R |____| Vout
|
|
|
|
|
===
GND
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

When the voltage over R exceeds ca 0.6V, T1 starts conducting, so the most
of the current goes through T1 instead of the 7809 ic.

Thank you all, guys!
Click to expand...

I'm trying a heatsink. If not works, I will use the transistor.

[]s
Marcellus
 
Thank you all, guys!

I'm trying a heatsink. If not works, I will use the transistor.

[]s
Marcellus
Click to expand...
Just one more thought, If the transistor is a very powerful one, it's gain
may be little (less than 20 in some cases), wich means that your 7809 may
still conduct more current than it can endure. What you have to do then is
to use a second transistor. There's two ways to that (as far as I know).
Here is how:

Darlington
Transistor

o E o E
| |
| |
| |
| |
|< Power .---------o
.----| transistor | |
| |\ | |
| | B |< |
B |< | o------o-| |
o------| | |\ |
|\ | Small | |/
A smaller | | PNP '-------| Powerful
transistor | | |> NPN
'---o--' |
| o
| |
| |
o K |
|
|
oK

This one use a combination
of an PNP and NPN
transistor. Much the same
as the Darlington.
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de
 
On Mon, 17 Nov 2003 17:04:19 -0200 Marcellus Pereira
<mpereira.nospam@myrealbox.com> wrote:

I'm using a 7809 ( voltage regulator ) with 18~20V In and 40mA ( 1A,
50 ms peaks ).

But it is very hot... I can't lower the input voltage. Any ideas, like
a pin-compatible more robust device?
Click to expand...
If you're sure about this being the max load current, then add a 120
Ohm, 1 Watt resistor in series with the input to the 7809. This will
allow about 1/2 of the power to be dissipated in the resistor and the
other half in the 7809.

-
-----------------------------------------------
Jim Adney jadney@vwtype3.org
Madison, WI 53711 USA
-----------------------------------------------
 
Jim Adney wrote:
On Mon, 17 Nov 2003 17:04:19 -0200 Marcellus Pereira
mpereira.nospam@myrealbox.com> wrote:

I'm using a 7809 ( voltage regulator ) with 18~20V In and 40mA ( 1A,
50 ms peaks ).

But it is very hot... I can't lower the input voltage. Any ideas, like
a pin-compatible more robust device?

If you're sure about this being the max load current, then add a 120
Ohm, 1 Watt resistor in series with the input to the 7809. This will
allow about 1/2 of the power to be dissipated in the resistor and the
other half in the 7809.
Click to expand...
The trouble with that is his 1 amp peaks. The 120 ohm resistor
wouldn't allow that. However, a 5 ohm 10 watt resistor in series
with the input would be a good choice, and might solve the
problem. It would move 5 watts of heat away from the 7809
during the 1 amp peaks. 440 mW in a TO-3 should not get
hot, so the heat is either coming from the 1 amp peaks,
or he's using the smaller package.

The best solution, if there is room, is a TO-3 7809
on a heat sink.


-
-----------------------------------------------
Jim Adney jadney@vwtype3.org
Madison, WI 53711 USA
-----------------------------------------------
Click to expand...
 
On Wed, 19 Nov 2003 06:34:52 GMT ehsjr@bellatlantic.net wrote:

Jim Adney wrote:

On Mon, 17 Nov 2003 17:04:19 -0200 Marcellus Pereira
mpereira.nospam@myrealbox.com> wrote:

I'm using a 7809 ( voltage regulator ) with 18~20V In and 40mA ( 1A,
50 ms peaks ).

But it is very hot... I can't lower the input voltage. Any ideas, like
a pin-compatible more robust device?

If you're sure about this being the max load current, then add a 120
Ohm, 1 Watt resistor in series with the input to the 7809. This will
allow about 1/2 of the power to be dissipated in the resistor and the
other half in the 7809.

The trouble with that is his 1 amp peaks.
Click to expand...
You're right, I overlooked the 1A peaks. In that case, just change the
resistor to 100 Ohms and add a 2000 uF capacitor to ground between the
resistor and the 7809. I'm not sure if the 2000 uF is big enough to
give you sufficient "ride-thru" but you can increase it if it's not
enough.

I admit that this is not the most elegant solution, but given what he
has installed there already I think it's the easiest and simplest.

-
-----------------------------------------------
Jim Adney jadney@vwtype3.org
Madison, WI 53711 USA
-----------------------------------------------
 
You must log in or register to reply here.

Welcome to EDABoard.com

Sponsor

Back
Top