Working Of 7805 Linear Regulator IC

Discussion in 'General Electronics Chat' started by shankey_1919, Feb 17, 2010.

  1. shankey_1919

    Thread Starter New Member

    Feb 9, 2010
    Hello everyone,

    I was trying to understand the working of the IC7805 linear regulator. I tried to break up the complex circuit into many simpler ones in order to understand its working. I could break them up into several current mirrors, darlington amplifiers,voltage regulators etc but there were certain components which I had no idea about. (Q5,Q6,Q10,Q13,Q11 and Q14 in the attached image)

    The first attachment shows the ckt of 7805 and the 2nd one shows my attempt to break up the ckt.

    If you could point me in the right direction, it would be of much help.

    Last edited: Feb 17, 2010
  2. rjenkins

    AAC Fanatic!

    Nov 6, 2005
    This is an educated guess;

    The left section is the reference, OK.

    The middle part you have as the left darlington block and the undefined parts above that is the error amplifier; note the two base connections in to that from the ref and the resistor ladder at the output.

    That feeds out through the current mirror to the base of the output darlington - it's the only active connection to the output that turns it on.

    R11 and Q15 appear to be the main current limit, clamping off the output if the voltage across R11 gets too high.

    The zener D2 can also turn on Q15 to shut the output down if the inout-to-output voltage gets too high.

    Q14 looks to be another safety shutdown of some sort, either thermal or in case of a fault in the reference??

    Q4, Q3 & Q11 are yet another shutdown or limiting circuit, that appears to be output overvoltage protection or possibly the thermal limiter.
    shankey_1919 likes this.
  3. shankey_1919

    Thread Starter New Member

    Feb 9, 2010
    (This is my understanding of the 7805 circuit. If my understanding of the ckt is wrong, please point me in the right direction)

    I think,,,,,

    The resistor R4 limits current through the zener diode D1 and maintains the voltage across it constant thereby biasing the transistor Q12 and thereby causing the voltage across the resistors R7, R6 to be constant as long as the current through the zener diode is higher than the current at the breakdown voltage.

    So transistor Q12 is ON unless the i/p voltage decreases causing the zener diode to operate below its breakdown voltage.

    By considering the attachment "standard converter.jpg" and "7805_4.jpg", the transistors Q16 and Q17 which form a darlington pair (marked in red) controls the amount of current flowing through the load so as to maintain the o/p voltage constant at 5V.

    Now when there is a short circuit in the o/p, the voltage will be zero at the o/p thereby causing Q15 to be ON thus reducing the current to the base of the Q16 of the darlington pair since it is connected to a const current source formed by Q8 and Q9(current-mirror). This will cause the pass transistors( darlington pair) to be off thus effectively shutting down the o/p.

    Now whenever there is a decrease in the input voltage, the zener diode falls out of the operating breakdown voltage causing the current i/p to transistor Q12 to decrease to a great extent causing the voltage drop across R7 to be greatly reduced thereby shutting down Q14. And since this is connected to the same current mirror as mentioned before (Q8 and Q9), current to the base of the transistor Q16 increases causing higher amount of current to flow through the darlington pair(Q16 and Q17) causing the o/p voltage to increase across the load.

    The error amplifier formed by Q5, Q6, Q10 and Q13 obtain i/p from the voltage reference set up by (R6+R7) through Q13 and the output voltage Vo through Q10. The transistor Q6 functions only when there is sufficient voltage across R20 which inturn determines the functioning of Q5.

    Let us consider 3 cases....

    1: If there is a short circuit across the o/p, Vo is zero hence voltage across R20 is zero effectively shutting down Q6 which inturn shuts down Q5. This causes a reduction of reference current through the current mirror formed by Q1,Q2 and Q7 which gets reflected proportionally onto Q2(due to R3). This causes reduction of base current to the darlington pair (formed by Q4 and Q3) which causes reduction of current drawn from the current mirror(Q8 and Q9) at Q9 thereby increasing the current to the base of the darlington pair(Q16 and Q17) but Q15 which operates as a short circuit protector forms a conduction path for the increased current thus causing no harm... Moreover if o/p is short circuited Vo = 0, Q5 and Q6 are OFF. Thus current flowing through Iref of current mirror(Q8 and Q9) is lowered and causes less current to be reflected across Q9 causing less bias current to flow to the base of the Darlington pair(Q16 and Q17) thereby reducing short circuit current further.

    2: If o/p voltage is below normal but o/p is not short circuited, current at emitter of Q5 is low(when compared to the ideal case(Vo is as expected) only) but not zero as in the above case, causing decreased current to flow to the base of the darlington pair(Q4 and Q3) thereby decreasing the current drawn from current mirror(Q8 and Q9) at Q9. This effect inturn increases the current flowing to the base of darlington pair(Q16 and Q17) at Q16 causing increased curretn to flow through it thus increasing the voltage across the o/p.

    3: If o/p voltage is higher than normal, increased current flows through Q5 => increased current to base of darlington pair (Q4 and Q3) ar Q4 => increased current drawn from current mirror(Q8 and Q9) at Q9 => decreased base current to darlington pair (Q16 and Q17) => reducing current flowing to the load thus reducing the o/p voltage.

    I think the transistor Q11 is for thermal protection as the transistor is connected in such a way mentioned in the block diagram(between base of series pass element, gnd and error amp)... I am not sure about the functioning of Q11 but i think it bypasses current to the Gnd from current mirror(Q8 and Q9) thus reducing Vo thus reducing o/p voltage causing reduction of temperatue.

    I really have no idea why the capacitor C1 is inbetween current mirror(Q1,Q2,Q7) and darlington pair (Q4 and Q3).

    Thanks "rjenkins" for your reply... I had no idea for the usage of diode D2 and thanks for your clarification of it...

    (And sorry for my ugly red drawing of the 7805. Had little time to draw it)
  4. Jony130

    AAC Fanatic!

    Feb 17, 2009
    Well its not that easy analysis this type of circuits.
    And I think that Q10, Q5, Q6, Q4 and Q3 work as a reference voltage.
    and Q3, Q4 and Q11 is a error amplifier.
    Q12 is a starting circuit.
    And I'm not sure but maybe Q13 work as a thermal protection.
    Although it seems more logical that Q14 is a thermal protection.
    But this is only my assumption
  5. rjenkins

    AAC Fanatic!

    Nov 6, 2005
    Q12 is purely an emitter follower buffer for the zener D1. It's emitter is the output of the reference voltage section.

    Q14 is some sort of protection or limiter; the voltage on the resistor divider feeding it's base should be absolutely constant (as a fixed fraction of the reference).

    If it ever turned on during normal operation, the output voltage would drop and destroy the regulation.