[OZAPRS] Re: EMI Help (fixed)

[Andrew Rich]

thanks Chris !

----- Original Message -----
From: "Chris Hill" <chris.hill at crhtelnet.com.au>
To: <ozaprs at marconi.ics.mq.edu.au>
Sent: Sunday, August 15, 2004 3:39 PM
Subject: RE: [OZAPRS] Re: EMI Help (fixed)


Hi Andrew,

Assuming vehicle voltage (when operating) of 14V, then your poor old 7805
is
having to drop 9V, which with a load of 170mA, means it has to dissipate
1.53W.  Assuming a heatsink is fitted with a thermal coefficient of 27C/W,
then the 7805 will be running at 41 degrees above ambient;  perhaps 71 C
or
thereabouts.

If you aren't using a heatsink, then the 7805 (in TO-220 case) thermal
coefficient is somewhere around 54C/W, in which case the 7805 will be
running at 83 degrees above ambient;  perhaps 113 C or higher.  Some
7805's
include thermal protection, and will shutdown when the die temperature
hits
a threshold (usually 150C).


Here's another approach, which would drop the temperature of the 7805 by
15
degrees (assuming a heatsink of 27C/W is fitted):

1.  Determine the maximum current consumption of your system (you measured
170mA, so let's assume Max I = 200mA).

2.  Determine the dropout voltage of your regulator.  (Let's say it's 7V).

3.  Determine the minimum output voltage from your power source (Lets say
11.0V).

4.  Drop 4V prior to input to 7805, by (eg) fitting a 5W resistor is
series.
    Resistor value R = V_drop / I_max
    R = (11.0V - 7.0V) / 0.2A
    R = 4 / 0.2A
    R = 20 Ohms

5.  The 20 Ohm resistor in series with the input to the 7805 will have to
dissipate the following power:
    P_min = (V*V)/R
    P_min = (4*4)/20
    P_min = 16/20
    P_min = 0.8W

6.  Assuming vehicle voltage (when operating) of 14V, and with 20 Ohm
resistor in series:
    I = 170mA
    V_in = 14.0V
    V_drop_20R = 0.17A * 20R = 3.4V
    V_in_7805 = 14.0V - 3.4V = 10.6V
    P_7805 = (10.6V - 5.0V) * 0.17A = 0.952 Watts

7.  Assume heatsink thermal coefficient = 27C/W, and ambient = 30C
    T_case_7805 = ambient + 0.952W * 27C/W
    T_case_7805 = 30C + 25.704C
    T_case_7805 =~ 56C


Hope this helps.


Regards,



Chris

p.s  A useful side-effect is that you can observe the system current
consumption, just by measuring the voltage drop at any given time across
the
20R resistor;  I = V/20




-----Original Message-----
From: ozaprs-bounces at marconi.ics.mq.edu.au
[mailto:ozaprs-bounces at marconi.ics.mq.edu.au]On Behalf Of Hamish Moffatt
Sent: Sunday, 15 August 2004 12:01 PM
To: ozaprs at marconi.ics.mq.edu.au
Subject: Re: [OZAPRS] Re: EMI Help (fixed)


On Fri, Aug 13, 2004 at 02:50:08PM +1000, Andrew Rich wrote:
> What is the best way to get from car volts (14Volts) down to something
> acceptable for a 5 volt regulator ?
> Most of the problem from running off the car is heat dissapation.
Getting
> from 14Volts down to 7ish volts.
> Voltage divider ?

Voltager divider won't help; you'll be dissipating the same amount of
heat either way. You need the regulation too.

Guess you are using a 7805 in the TO-220 package. It's good for 1 amp in
that package, but at 7V drop times 1 amp, it's dissipating 7W. In short
it needs a heatsink!

Unless you can improve your efficiency, then all solutions mean using a
heatsink, and possibly a fan (probably not needed). There are other
devices available in other packages which are easier to bond to big
heatsinks, like a TO-3 package. There's such a thing as an 7805K (TO-3)
but you probably won't be able to find one. Instead consider an LM317K.
The 317 is an adjustable regulator (unlike the 78xx family) and needs
some resistors to set its output voltage.

The only way to improve efficiency would be a switchmode solution. There
are devices in a TO-220-like package that are switchmode, but I don't
know anything more about them.

Hamish
--
Hamish Moffatt VK3SB <hamish at debian.org> <hamish at cloud.net.au>
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