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I want to understand the concept of current regulation in the circuit, In my research I came across the following link Boosting Regulator Current for IC 78xx by MJ2955 enter image description here

where in the circuit schematic in link above, the author claims he has obtained 12V,5A as output with 20V input voltage.

I tried to simulate the same circuit but with input of 12V, 0.1A and 7805 regulator as shown in this schematic: enter image description here

I see that when input current is 0.1A the output is 94mA approx. which agrees with ohms law V = I*R = 94.086mA * 53.33 ohm = 5.01V

But when I have my input current as 0.06A the output voltage is 3V (why not 5V? )and current is 57mA, as shown in this schematic enter image description here

This also agrees with ohms law V = I*R = 57.267 mA*53.33 ohm = 3.05V

The results are same as the circuit without the PNP transistor. schematic: enter image description here

So what is the impact of the PNP transistor in the circuit? How can I boost the current in my circuit? If I want to see 5W output across my load(53.33ohm in the schematic) what are the changes I should make in the circuit?

Note: I assume capacitors are only for smoothing the input and output and does not have much effect in simulation (correct me if wrong). I tried with 470uF in all three cases the results were same. (sorry was bit lazy to copy new screenshots and I didnt have enough reputation to embed schematics in my question)

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  • \$\begingroup\$ please copy links and paste to view schematics, sorry did not have enough reputation to embed links \$\endgroup\$ – sristisravan Jan 5 '16 at 14:57
  • \$\begingroup\$ The PNP allows the circuit to handle more current. It can't pull more current out of its supply than the supply is willing to provide. \$\endgroup\$ – brhans Jan 5 '16 at 15:13
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With any linear voltage regulator (current boosted or not current boosted) you are going to get this: -

Output current = Input current - a few mA to power the regulator

You are never (ever) going to see an output current greater than the input current. If you want that then you must consider a buck voltage regulator.

So what is the impact of the PNP transistor in the circuit? How can I boost the current in my circuit? If I want to see 5W output across my load(53.33ohm in the schematic) what are the changes I should make in the circuit?

If you want 53.33 ohms to dissipate 5 watts then the voltage needed is governed by the formula: -

Power = \$V^2\div R\$ or

Voltage = \$\sqrt{P\cdot R}\$ = 16.329 volts

I tried to simulate the same circuit but with input of 12V, 0.1A and 7805 regulator

The 7805 produces 5V on its output and not 16.329 volts.

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  • \$\begingroup\$ I have my input current as 0.06A the output voltage is 3V (why not 5V? ) \$\endgroup\$ – sristisravan Jan 5 '16 at 15:10
  • \$\begingroup\$ You can't cheat ohm's law (that easily). V=IR, so 0.06 x 53.33 = 3.2V. Now take into account that a little of that 0.06A is lost to the 7805 and doesn't flow through the load and it all works out... \$\endgroup\$ – brhans Jan 5 '16 at 15:12
  • \$\begingroup\$ So what does the article eleccircuit.com/… mean boosting current for 78XX? \$\endgroup\$ – sristisravan Jan 5 '16 at 15:14
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    \$\begingroup\$ A 78xx can only output a maximum of about 1A but if you need a regulated voltage, 3 amp supply you need to add components to it (or around it) so that the 3A is possible. \$\endgroup\$ – Andy aka Jan 5 '16 at 15:16
  • \$\begingroup\$ @SristiSravan, the usual way to supply a 7805 is with a fixed voltage supply. You are supplying it with a fixed current supply. The 12 V supply in your model has no effect because it's in series with the current supply. Your model is trying to predict the results of a gross abuse of the 7805. \$\endgroup\$ – The Photon Jan 5 '16 at 16:27

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