I need a power supply capable of supplying 7.5V at 6A using linear devices. To minimize power dissipation at this current level I have decided that using a LDO would allow me to lessen the power I need to dissipate by allowing me to have a lower input to output voltage.

I have selected to use the TL1963AKTTR. To acheieve the required current I think I should be able to use a PNP transistor such as the TIP2955 to boost the current similar to how it is shown in this circuit.

EDIT: Sorry, would this work as I expect


1 Answer 1


This is the circuit you are proposing: -

enter image description here

I've put a circle in red around the 1 ohm resistor because this is important about your concept. To activate the TIP2955 you'll need at least 0.7V across the 1R resistor and now your attempts to make a LDO circuit are reduced because.

With 0.1A flowing, the regulator drop out is added to 0.1V across the resistor With 0.5A flowing, the regulator drop out is added to 0.5V across the resistor

Have you taken this into account?

EDIT - following advice offered on an edit suggestion (which somehow disappeared) I'll add to the scenario. What I said above still stands but as the demand for more load current increases, so does the base emitter voltage on the TIP2955 (at 6A this may be in the order of 1.4V). And, for a 6A load, it is likely that the Hfe of the transistor will be about 30 (typical) implying the base current will need to be about 200mA. The 10R base resistor will need to pass this current and therefore, in adddition to the 1.4V required by the base there will be 2V across the 10R resistor - this equals 3.4V - all required to be generated across the 1R series resistor. This implies a 1R current of 3.4A and of course this is not viable through the regulator chip (1.5A max). In short this circuit will never be able to provide a 6A drive to the load.

If you got rid of the 10R resistor and replaced it with a short, you'll still need to produce 1.4V across the 1R and this just about means the circuit will work with 6A BUT, it's on the edge and there has to be a better way of doing this. The CE saturation voltage is not that good at 6A either so I'd abandon this idea and look for alternatives.

  • \$\begingroup\$ No I had not, so in this case I would need to have an input voltage of at least 1.05V at input? \$\endgroup\$ Commented Jun 20, 2013 at 9:54
  • \$\begingroup\$ The regulator drops 340mV and you need at least 0.7V across the resistor (which can drop to 0.68ohms because the regulator current can be higher). \$\endgroup\$
    – Andy aka
    Commented Jun 20, 2013 at 11:07
  • \$\begingroup\$ Would the circuit found as Circuit B found here be a better fit. I wish to implement it dropping the value of the 6 ohm resistor to 1.4ohms and setting Rsc=.11 ohms. Would you happen to know the purpose of the other two resistors? Thank you very much \$\endgroup\$ Commented Jun 20, 2013 at 18:20
  • \$\begingroup\$ @secretformula Circuit B is basically what I proposed in the final part of my answer minus the current limit transistors Q1 and Q3. The purpose of the two resistors: The one in Q1 collector is to protect Q3 should the output become shorted. The other one reduces the onset of the current limit by keeping Q3's base closer to its emitter when Q1 tries to conduct. Try something like 220R for both but, you've still got the basic problem in that the TIP2955 is a pig to get to conduct effectively. \$\endgroup\$
    – Andy aka
    Commented Jun 20, 2013 at 18:57
  • \$\begingroup\$ I was thinking about using MJE15028 due to its lower VCe of .5V. I really appreciate your help. \$\endgroup\$ Commented Jun 20, 2013 at 19:04

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