# How this ps unit works??

Discussion in 'General Electronics Chat' started by samy555, Aug 5, 2012.

1. ### samy555 Thread Starter Active Member

May 24, 2010
116
3
The circuit in the following figure (from: http://www.hobby-circuits.com/circuits/power-supply/dc-power-supply/891/variable-dc-power-supply) represents a variable power supply Transistor Q1 is npn type

This was corrected in the following figure

I have thought carefully about this circuit, but I could not get to the

mechanism of how it work!
I think that when the transistor Q1 operates, it must turn off Q2.
I try to simulate it using multisim10, but it didn't work, that is when I

change the value of any of P1 or P2 the output is fixed at near the

retectified voltage.
Any help in clarifying the way this circuit work would be welcome, thanks

2. ### #12 Expert

Nov 30, 2010
16,705
7,358
Yes, Q1 in an npn.

R2 provides drive current to Q2 and that current is multiplied by Q4, then it goes to the output.

If the voltage at the wiper of P2 is as much as the turn on voltage of Q1 (about .6 volts) then Q1 dumps the drive current from R2 thus stopping the votage from rising any more.

If the voltage at the wiper of P1 is equal to about .6 volts lower than the emitter of Q3 then Q3 sends current to Q1 which dumps the current from R2 and that stops the increase of current at the output.

3. ### samy555 Thread Starter Active Member

May 24, 2010
116
3
I did not understand the sentence in red
I want to know exactly what is happening at that moment
Thanks

4. ### #12 Expert

Nov 30, 2010
16,705
7,358
Q1, being an npn transistor, requires some positive voltage (about .5 to .6 volts) to drive current through its base to its emitter. When base to emitter current happens, then collector to emitter current can flow. (They are proportional to each other with the colector to emitter current being quite a bit more than the base to emitter current.) When the current through Q4 creates a voltage across P2, it is in the correct polarity to turn Q1 on. As you adjust P2 you adjust the voltage to the base of Q1. When that voltage is high enough, Q1 allows the current from R2 to go to ground instead of Q2. Thus, Q1 "dumps" the current that would have gone into Q2. Lacking that current, Q2 stops increasing the current to Q4 which stops increasing the current to P2. Thus, the output voltage is controlled.

5. ### samy555 Thread Starter Active Member

May 24, 2010
116
3
thanks alot
but please, the word "dumps" means take all current to ground?

6. ### #12 Expert

Nov 30, 2010
16,705
7,358
"dumps" means (in this case) a proportional amount of current, depending on the exact amount of current allowed into the base of Q2 by P2 and R3.

I understand your difficulty. A complete dump would be ALL the current, but that is not true in this case.

samy555 likes this.
7. ### samy555 Thread Starter Active Member

May 24, 2010
116
3
How Q3 controls output current???

Nov 30, 2010
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9. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
Are you committed to this design, or are you looking for ways to make a power supply? We have several good designs here and there.