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based on a suggested circuit from the datasheet, supplied current of LM7805 voltage regulator can be increased. the circuit

I can't source that BD536 pnp transistor or any equivalent listed on some websites. What parameters I should consider when searching for an transistor to use instead of the suggested BD536 ? (hfe, Ic, Vce-sat, Vceo ...)

I need to supply around 0.9A, which is just about the max current 7805 can supply. I'm trying to make the circuit above to avoid over-heating. can this transistor addition reduce the heat produced by the regulator when high currents are required ?

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    \$\begingroup\$ The 7805 is not an LDO! If is far from ‘low drop out’. Adding the transistor is not going to help with the heat. I’d suggest you find yourself a buck regulator. This will address your concerns with current and heat. \$\endgroup\$
    – Kartman
    Apr 29 at 21:43
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    \$\begingroup\$ For a linear regulator like this the amount of heat is going to be equal to the difference between the input and output voltages, times the supplied current. The heat will be in the regulator or the transistor, but it will be there. \$\endgroup\$ Apr 29 at 21:56
  • \$\begingroup\$ Yep, the 7805 is an HDO regulator, in anyone's book. You can make a step-down regulator circuit or you can buy a module for simplicity with expense. Take a look and there's some that are like bigger versions of the 7805 (TO-220) package. \$\endgroup\$
    – TonyM
    Apr 29 at 21:58
  • \$\begingroup\$ sorry, i meant to say linear voltage regulator. edited the question \$\endgroup\$ Apr 29 at 22:17
  • \$\begingroup\$ You need to spec. Vin tolerances at 0.9A. It can supply 1.5A with a 2V typ. Dropout and mention peak current if any more. It needs cooling air even with a bypass for the other part. Which could be a power R or a power PNP depending on your Vin tolerances \$\endgroup\$ Apr 29 at 22:18
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First, to answer your question, you want a PNP power transistor rated for at least twice the peak input voltage and twice the max output current. The transistor will see less than either of these max parameters, but that is design margin to increase reliability. Since you are trying to make 0.9 A, go for a 2A to 5A rated transistor. More is fine, but more expensive for no benefit. You don't say what the input voltage is. If it is 12 V, use a transistor rated for at least 25 V. etc. When the circuit is running, Q1 will see only 7 V, but again, design margin.

The value of R1 determines when Q1 starts to supply current to the output. You can select this resistor such that when the circuit is running at its max output current, approx. 50% of the regulator circuit heat is dissipated in Q1 and 50% in U1 (the 7805). If you wanted the circuit to supply more than U1 can handle on its own, such as 5 A, then you pick R1 such that the U1 current is about 75% of its rated max., and then Q1 would pass everything else and dissipate most of the heat.

Vbe is larger in power transistors than in small signal transistors, so let's assume that Q1 "comes on" at 0.7 V. The current through R1 is the same as the current through U1. At 0.45 A (1/2 the output current) through U1, you can solve for R1 with Ohm's Law. For currents above 0.45 A, the difference will go through Q1.

To drill down further, the heat dissipated in Q1 and U1 will not be exactly equal because some power is dissipated in R1. Also, the Q1 base current flows through U1, so it adds a little heat to both devices. Neither of these is a big deal, just things that will make the actual results slightly different from calculations.

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The 78XX series has been around for more than 50 years. If you look in more recent data sheets/app notes you will find more recent recommendations for Q1. Q1 connected as shown will certainly (obviously?) reduce the power dissipation in the 78XX. There are other ways of doing so if you must have a 78XX or some other linear regulator. Think of resistors in series/parallel. No linear regulation scheme will reduce the overall power dissipation.

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