I've built a PCB with a 7805 based power supply on it. I tested the other parts of the PCB on a breadboard before I had the PCB made, but as the power supply circuit looked straightforward, I just took it out of a local electronics catalog without testing it.

The PSU part of my PCB

The problem I'm having is the power supply only outputs 2.8V until you put it under some load. Surely the TIP2955 should not affect the circuit in any way until the voltage drop across the 3 ohm resistor rises above 0.6V? What on earth is going on? I've even taken the components off the PCB and built them on a tag strip. Still only 2.8V until some current is drawn.

It's also worth pointing out that the 7805 is getting quite hot with the circuit at rest. But seems fine when a load (10 ohm 5W resistor, so about 1/2A) is applied.

I'm baffled...

  • \$\begingroup\$ It sounds like the 7805 might be oscillating. What does the output look like on a 'scope? \$\endgroup\$
    – Dave Tweed
    Jul 24, 2014 at 14:23
  • \$\begingroup\$ is this not one of those linear regulators that needs a 10mA load current (just a load resistor calculated from the desired output voltage, to give the current) for minimum 'no load' regulation? \$\endgroup\$
    – KyranF
    Jul 24, 2014 at 14:36
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    \$\begingroup\$ Output voltage is spec'd over a load range of 5 mA to 1 A. \$\endgroup\$ Jul 24, 2014 at 14:43
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    \$\begingroup\$ For lower current, the PNP is off, and the 7805 is fed by the 3 ohm resistor. Once the current reaches about .25 amps, the PNP starts turning on, and stabilizes the emitter base voltage at ~.7 volts. Essentially, the PNP acts to amplify changes in 7805 output current once the turn-on voltage has been reached. \$\endgroup\$ Jul 24, 2014 at 15:44
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    \$\begingroup\$ I wanted to do this in my capstone project, and my favorite professor talked me out of it. Adding an external transistor completely bypasses all the protection the regulator's designers built into it, ie. thermal shutdown, and short circuit protection. Don't do it, get a modern regulator capable of your current requirements. Linear Tech has a good selection of high current LDOs. \$\endgroup\$
    – Matt Young
    Jul 24, 2014 at 18:12

2 Answers 2


Very likely the 7805 is oscillating. 100 nF is a rather skimpy capacitor on its output. Also, make sure it is low ESR and physically close to the 7805 so that there is little inductance in the wires to and from the capacitor.

I'd make both caps at least 1 µF, maybe even 10 µF for the input cap. At low voltages, such values are cheap and readily available, and the same size as the 100 nF cap.


You should also decrease the resistor. Make it 1 Ω or 1.5 Ω. As it is now, the transistor is kicking in rather early. You want a decent current thru the 7805 when the transistor starts adding more to the output.

There should be no need for a quiescient load resistor. A properly connected 7805 is stable at 0 output current.

  • \$\begingroup\$ OK I deleted (add a resistor) comment \$\endgroup\$ Jul 24, 2014 at 18:30
  • \$\begingroup\$ Another solution relating to capacitance. It looks like this is how I'll attack things. Thank you for your suggestion. \$\endgroup\$
    – banzai
    Jul 24, 2014 at 23:26

You need to have a large input capacitor across the input, close to the emitter and pin 2 of the 7805. That's assumed to be present. Something like 1000uF electrolytic paralleled with 100nF ceramic. If you already have a big filter capacitor, you just need the ceramic capacitor.

I don't think you should alter the value of the 330nF capacitor too much, but you can increase the output capacitor to 1uF.

From a quick simulation, the 330nF will likely affect the frequency of oscillation but bypassing the input and increasing the output capacitance should stabilize it. 100-1000uF there too if you like.

The output of the 7805 has a resistive divider equivalent to about 15K ohms nominally, so that gives you some idea of the "load" on the regulator when you have connected no load.

This shows ringing behavior (in the input voltage to the 7805) with a step change in load, using your schematic with a plausible source impedance (but a lower base resistor). Frequency is about 50kHz.

enter image description here

Increasing the 0.33R resistor to 3 ohms makes the regulator have quite poor transient response. Here is the simulation of the regulator output (exactly your circuit) with a 2A load removed at t=10usec. With the 0.33 ohm resistor it only overshoots by 660mV. Adding 100uF across the output solves that.

Conclusion: add more capacitance (an electrolytic like 100uF/10V or 470uF/10V) on the 5V output and make sure there's 100nF across the input.

enter image description here

  • \$\begingroup\$ 1000uF ceramic? \$\endgroup\$ Jul 24, 2014 at 15:45
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    \$\begingroup\$ @WhatRoughBeast Oops, corrected. That would be an expensive capacitor. \$\endgroup\$ Jul 24, 2014 at 15:57
  • \$\begingroup\$ What happens in the second plot? Why the 16V spike when the load was removed? Also, why does the input oscillate with a switched load on the output? \$\endgroup\$
    – sherrellbc
    Jul 24, 2014 at 18:01
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    \$\begingroup\$ The transistor is "on" conducting current to the load, and when I suddenly flipped that switch ten microseconds in to remove the 2A load, it stays on for a bit, causing a nasty overshoot that could destroy a CMOS chip or whatever. A big(ish) output capacitor absorbs that and the voltage doesn't jump much. \$\endgroup\$ Jul 24, 2014 at 18:04
  • \$\begingroup\$ WOW, I never expected such a comprehensive answer. I'll get on to trying these suggestions over the week end. \$\endgroup\$
    – banzai
    Jul 24, 2014 at 23:24

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