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Integrated Regulated power supply
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xavier60:

--- Quote from: techguru on June 14, 2018, 06:44:52 am --- hi to all..

1)why we have Q1(pnp) and Q2(npn)?

                 According to me it is Q2 which is going to act as external series pass element.

2)But why do we have Q1 there and Internal series pass element collecter connected to base of PNP transistor what is the logic present there?
               
                 According to me collector of the series pass transistor should be connected to maximum positive voltage(not exceding 40V). what will happen now?

3)why emitter of internal series pass transistor connected to 220 ohm ant that is connected 68 ohm. A voltage divider is formed and it is connected to output?
           
                Emitter of internal pass transistor should be connected to external series pass transistor but why it is being connected to voltage divider and to output?

still i am not clear in this schematic? Kindly help me to understand?

--- End quote ---
R8 and R7 have nothing to do with each other.  R7 is there only to bleed off charge from the Base of  Q2 to help it turn off more quickly when needed to lessen voltage overshoot at the output when load is suddenly removed. R4 is for the same reason.
At up to about 3mA of output load, the internal transistor supplies all of the output current via R8. The transistor's Collector draws this current from the unregulated rail via R4.
Above 3mA, the voltage drop across R4 exceeds 0.6V causing some of the current to flow from Emitter to the Base of  Q1 causing it to partly turn on causing current to flow from its Emitter to Collector. This current is also supplied to the output via R7.
 At some higher load current, the voltage drop across R7 will exceed 0.6V causing Q2 to partly turn on and to supply current to the output.
The way that Q1 drives Q2 results in very high current gain, very approximately 1000. For example with 5A of load current, the internal transistor only need pass about 10mA.
R8 limits the amount current that the internal transistor can supply, likely to protect it from overload in case something goes wrong. 
techguru:
Dear all,
                   Thanks to cellularmitosis and xavier60 for their kind help.

 we have professor who does'nt spend time with us. all the good hearts in this Forum are thanked for this Service.

again thanks a lot..

xavier60 Sir,

                 I will be happy if u share the information how do you acquired this knowledge?

Because i didnt thought of 3mA concept.but the relationship reveals Vbe=Vt ln(Ic/Is),if we substitute that we will get 660mV.Which will help to turn on the transistor.. Thanks a lot Sir.

By
Student..
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