# LM Series Regulator With Transistor Question

Discussion in 'General Electronics Chat' started by PGB1, Jan 28, 2013.

1. ### PGB1 Thread Starter Member

Jan 15, 2013
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Hi All,
For background, I am very, very new at learning electronics & semi conductors.

Today's pesky question concerns using transistors to gain higher output from an LM series regulator. I understand how the LM78 & 79 regulators operate (basically). I also have a basic understanding of transistors. But, i can not seem to understand how the LM78xx controls the voltage level allowed through the transistor, especially in the second example below.

I have found two basic arrangements. A quick & rough sketch is attached to this post. For my examples, I'm using LM7805 & the source voltage is 18vdc.

One uses an NPN transistor with the base connected to the output of the LM7805 regulator. I think I understand this one-

Because the voltage on the output of the LM7805 is regulated to +5, the base of the transistor receives +5 volts. The input voltage of +18 at the emitter is allowed to flow out of the collector at only +5 volts due to the base receiving +5 (controlled by the LM7805's output terminal).
If the output were 12 volts from an LM7812, the base would receive +12 & the transistor would 'take in' +18 and send out +12.
Did I understand correctly or goof it all up?

The second method uses a PNP transistor. This is the one I can't figure out. (Assuming I did, indeed, understand the first example)

The voltage to the base is before the regulator.
So, if the input voltage was 18vdc & the LM7805 was installed as on the attached drawing, would not the base receive 18 volts, thus letting 18 through the transistor's emitter/collector to the project, rendering the LM7805 useless?

How does the circuit with the PNP keep the output voltage at 5 (if LM7805)?

I've searched a lot on this and, like usual, managed to confuse myself even more. If you all don't helping me understand the operation of this, I'll appreciate your words!
Thanks Very Much,
Paul

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2. ### BillB3857 Senior Member

Feb 28, 2009
2,402
348
Your basic understanding of the top circuit seems to be correct, to a point. The transistor configuration is known as an Emitter Follower. There will be a slight difference (~0.7V) due to the base-emitter voltage drop within the transistor. The second circuit makes no sense what so ever. A wire is shorting the emitter-base connection of the transistor and the collector isn't even connected. Where did you find that circuit?
The data sheets and application notes for the LM78XX regulators have many examples for their usage.

3. ### Audioguru New Member

Dec 20, 2007
9,411
896
Yes you goofed up.

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4. ### PGB1 Thread Starter Member

Jan 15, 2013
58
3
Thanks BillB3857 & Audioguru for taking the time to reply & correct my mistake.

I sure did copy the second one wrong. I left the resistor & capacitor out by error. It is a simplified version of one I saw at http://electronics-diy.com/12v-power-supply.php That's the only place I found one like this (so far).

The drawing I copied shows 1 Ohm 1/2w & two capacitors.

But, putting them back in, as Audioguru did for me, I'm still brain-stuck.
When the current through the regulator is high enough, the voltage across the resistor turns on the transistor. I understand this part. But...

When the transistor turns on, how is it controlled to only allow 5 volts to pass to the output? Is that based on the relationship between chosen resistance & the regulator's output voltage?

Paul

5. ### #12 Expert

Nov 30, 2010
16,684
7,322
I see you're having difficulty with the phrase, "turns on". I use the phrase, "whatever's necessary".

The definition of a 3 pin regulator like the 7805 regulator is that it passes whatever amount of current is necessary to get it's output pin to be 5 volts more positive than it's "ground" pin. The definition of "voltage regulator" is that the output voltage stays constant, regardless of the current needed by the load...within the limits of the chip's ability to pass current and not smoke.

In this circuit with a "helper" transistor, the regulator chip starts out doing its job, passing whatever amount of current is necessary to get the output voltage to the label on the chip, and we're assuming 5 volts today. If the load needs more than about .6 amps, the voltage developed across the 1 ohm resistor rises enough to start turning on the helper transistor. The helper transistor doesn't just slam on as hard as it can go. It is modulated by the regulator chip. If, perhaps, the load is 2 amps, the regulator chip will be supplying about .7 amps and the .7 volts developed across the 1 ohm resistor will have the helper transistor supplying the other 1.3 amps.

You will need this, "whatever's necessary" idea when you get to operational amplifiers. They do whatever's necessary to get their two input pins to have equal voltage and you can get them to do some mighty good tricks while they are trying to do whatever's necessary.

Jan 15, 2013
58
3