Question about 78L05 (To-92)

[Budiman]

78l05

I want to use 78L05 (To-92) with this condition :
Vin = 13.5V
Vout = 5.0V
IMax = 90mA

My question is :
1. Can I use the 78L05 without any heatsink ?
2. How to calculate the power dissipation ? ( 8.5V x 90mA or 5V x 90mA ?)

Thank you

 

[IanP]

reg circuit using 78l05

it is 8.5V * 90mA ..
78L05 will go into thermal overload protection mode - it's just to small package for this job ..
try to use 7805, and <1W (heat) can be tolerated by this one without heatsink ..

rgds,
ianp

 

[flyingsky]

78l05 power

maybe this circuit ji a good choose for you.
good luck!

Added after 2 minutes:

 

[Budiman]

to-92 heatsink

I cannot use heatsink, or 7805 because the space is very limited.
And flyingsky circuit cannot fit in there.



I search yahoo and found some information :

Thermal resistance = (TJ - TA) / PD
TJ = thermal junction
TA = thermal ambient
PD = Power dissipation

Thermal resistance for TO 92 is about 180 degree C ( I found in several dataheets).

So the PD for TO 92 is = (125 - 25)/180 = 0.56 Watt.

Is this formula/equation correct ?

I think to use a 5.1Volt 1W zenner to decrease the Vin.

If the equation is correct than adding the zener should solve my problem =
(13.5V - 5.1V - 5V) x 0.1A = 0.34A.



Thank you

 

[max0412]

to92 power dissipation

No your way off. The 78xx series need about 2V headroom to maintain regulation so your input voltage should be minimum 7V.

I don’t know what your doing with the zener?

Ianp gave you the correct info. You have these choices switch to TO220 use a resistor to drop Vin, or diodes to an acceptable level for the T0-92.

If you place a 68 ohm 1W or 2W would be better you drop .

68 ohm* 0.09A = 6.12V

The resistor will dissipate
6.12V * 0.09 = 0.551W

This would leave
13.5V – 6.12V = 7.38V

7.38V at the input of your 78l05

7.38V – 5V = 2.38V

Now your 78l05 only has to dissipate.

Pd = 2.38V * 0.09A = 0.2142W

Check if this acceptable for your maximum ambient temperature.You check this as shown in the data sheet.

"The maximum allowable power dissipation at any allowable ambient
temperature is PD = (TJ(max) − TA)/θJA. Operating at the absolute maximum TJ of 150°C can affect reliability"

θJA=140°C/W TO-92

Use 100 deg C for Tjmax then minus whatever your max ambient is this will give you the max allowable power dissipation.

You can vertically mount the resistor if space is that tight.

Next time you may want to check all this before you do a board up.

Read through the example here.


**broken link removed**

 

[yus]

7805 (to-220 package) or 78l05 (to-92)

...
You have these choices switch to TO220, use a resistor to drop Vin, or diodes to an acceptable level for the T0-92.
...

max0412's suggestion of using a resistor is, I believe, the cheapest solution..

a 56ohm (or 68ohm) 1Watt resistor should be enough.

with 56ohm resistor, there will be a 0.3W dissipation on 78L05 regulator at 90mA load.
with 68ohm, its only 0.2W. (with 2.38V drop across the regulator)