Question about L7805 Voltage regulator
Russell McMahon email (remove spam text)
> Add a 7912 regulator before the 7905. This will reduce the ripple and
> spread the heat. You can put it on the same heatsink or on another (beware
> the tab on 79xx is not ground).
It will spread the heat to another component (the 7912) BUT if they are on
an identical heatsink as before the heatsink temperature will be the same as
before. This may be acceptable but be aware of this.
If the input voltage is always about the same (24 volts) then a series
resistor is probably the simplest solution (but not the most elegant).
I mA load,
Vo output volts,
Vin supply input volts,
Vhr regulator "headroom" (minimum allowable drop across regulator) .
Vri regulator input volts ( = Vin-Vo-Vrh)
Leave the regulator a little more headroom than the databook says and ALWAYS
design for worst case load. AFAIR the 7905 needs about 2v headroom at full
For a linear (non switching regulator)
Power total = Vin * I
Power in load = Vo * I
Power lost in regulator)(s) = (Vin - Vo) * I
R = (Vin - Vo - Vrh) * 1000 / I
Power in R is (I/1000)^2 * R
For example for 24 volts in, 5 volts out, 3 volts regulator headroom, 500 mA
MAX load current.
R = (24 - 5 - 3) * 1000 / 500 = 32 ohms
Power in R = 0.5^2 * 32 = 8 watts
Resistor should be substantially higher powered than power dissipated.
Say in this case 2 x 10 watt resistors.
Note that at 500 mA total power = 24 x 0.500 = 12 watts so there is quite
some heat to dissipate.
A switching preregulator looks quite attractive unless you have power to
burn and don't mind the heat.
If you told us what the application was and why you had to use 24v it may
assist with a better solution.
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