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I am quite new to electronics and hoping to build a voltage regulator using the schematic below. I will be using a 7806 regulator and a 7.4 Lipo battery (yes, the schematic is probably a little overkill for 6V regulation but never mind that). The transistor required is a TIP2955 which is a PNP transistor.

A website I came across says this about PNP transistors:

Only when there is no current at the base will it turn on. And to ensure that no current enters the base, the base must be grounded (connected to ground). If any current goes into the base of a PNP transistor at all, the transistor will not conduct across from emitter to collector.

The schematic shows the PNP's base connected to positive, not ground. Will this schematic work and how? I'm rather confused as the preceding description seems to imply that it won't.

schematic
(source: zen.co.uk)

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  • \$\begingroup\$ Won't work at all, look at the drop out voltage in the 7806 datasheet. Further at 3A, this is not a good choice of topology. Let's pretend this will work for just a minute. ((7.4V-6)*3A)=4.2W, all wasted as heat in the regulator and pass transistor. Study buck convertors. \$\endgroup\$
    – Matt Young
    Jun 20, 2015 at 4:03
  • \$\begingroup\$ "the schematic is probably a little overkill for 6V regulation" No it is not overkill. Apart from the fact that there are more efficient architectures today, this is how it was done in the past. Matt is probably correct with the drop out voltage, but that doesn´t change the fact that the architecture in itself is pretty common. \$\endgroup\$
    – jippie
    Jun 20, 2015 at 7:09
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    \$\begingroup\$ That quote is truly amazingly wrong. You would be doing us a favour if you attributed it to the website where you found it, so that we could avoid it... \$\endgroup\$
    – user_1818839
    Jun 20, 2015 at 9:39
  • \$\begingroup\$ @BrianDrummond learningaboutelectronics.com/Articles/… \$\endgroup\$
    – jippie
    Jun 20, 2015 at 14:07
  • \$\begingroup\$ @Matt Young - If I were to swap out the 7806 for a buck converter, would the above schematic still allow me to draw higher currents? \$\endgroup\$
    – Mitchell Bennett
    Jun 20, 2015 at 23:43

1 Answer 1

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That quote is just wrong.

That circuit is correct, but not for your application as it will require about 8V minimum input to give 6V at the output.

It works by supplementing the output current of the 78xx (7806 for 6V) regulator if the regulator takes significant current (about 600mA) with additional current through the PNP transistor.

Your Lipo batteries will not have a constant output voltage, they will probably vary between 8.4V when fully charged down to about 6V or even less when discharged.

What do you need the 6V for? How much current do you need? What is the acceptable voltage range for the load?

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  • \$\begingroup\$ Thanks for your answer. Just out of interest, what makes you say that an 8V minimum input is required? Initially I will be using the 6V to power at least 3 servos - all of which are rated for 6V - hence I am looking to get as close to 6V as possible. I think the servos have a peak current draw of 1A each so 3A (max) will be drawn for now but I may add more servos later on which will increase this. \$\endgroup\$
    – Mitchell Bennett
    Jun 20, 2015 at 3:57
  • \$\begingroup\$ @MitchellBennett the drop out voltage for a 7806 is 2V. This means you need to be at least 2V higher than than the regulated output. Without the 2V increase, the regulator can't work properly. \$\endgroup\$
    – efox29
    Jun 20, 2015 at 4:07
  • \$\begingroup\$ Darn, I didn't see that. I guess I'll have to switch to a 11.1 Lipo then. \$\endgroup\$
    – Mitchell Bennett
    Jun 20, 2015 at 4:14
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    \$\begingroup\$ @mitchellbennett please don't go to an 11 V battery with a linear regulator! You'll be using only half of the energy on the battery, the other half wasted. Or, your battery will have to be twice as big to do the same job. And how will you get rid of the heat? (12 W is what a small soldering iron uses). Use a buck converter, they're small and cheap and over 90% efficient. Check at digikey for a Murata module, or you could even resort to eBay. \$\endgroup\$
    – tomnexus
    Jun 20, 2015 at 5:41

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