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KCC's Quizzes: Linear Power Supply

See below for a classic linear power supply using bipolar transistors.  

Questions: 

  1. Is it an LDO? 
  2. Copy and paste the table below and enter in the suitable values for voltages on points C, F, and B.

Vin 

(volt) 

Vz

(volt) 

R1 (kΩ) 

R2 (kΩ) 

R3 (kΩ) 

R4 (kΩ) 

R5 (kΩ) 

Rz (kΩ) 

VC 

(volt)

VF 

(volt)

VB 

(volt)

24 

6.2 

2.2 

33 

47 

2.2 

3.3 

2.2 

 



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[edited by: emassa at 2:35 PM (GMT -5) on 9 Jan 2023]
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  • Should be piece of cake for our engineers!

  • 1. No, it is not an LDO, as the Darlington stage drops at least 1.4V. 
    2. VB=6.2V*(R3+R2)/R3=15V
    VF=6.2V-0.7V=5V
    VC=VB+1.4V=16.4V

  • 1. No. VIN>VOUT+1.2

    2. 

    Vin 

    (volt) 

    Vz

    (volt) 

    R1 (kΩ) 

    R2 (kΩ) 

    R3 (kΩ) 

    R4 (kΩ) 

    R5 (kΩ) 

    Rz (kΩ) 

    VC 

    (volt)

    VF 

    (volt)

    VB 

    (volt)

    24 

    6.2 

    2.2 

    33 

    47 

    2.2 

    3.3 

    2.2 

    VB+1.2 5.6 VZ*(R2+R3)/R3
  • 1. Yes, it's a linear regulator (Q3, Q4 = error amplifier + Q1, Q2 = power stage).
         But no, it's not "Low Drop Out" because of its NPN darlington power stage which
           needs a base voltage (Vc) which is 1.2V higher than the output voltage resulting in a minimum drop of 1.2V.
    2. Assuming Base-Emitter voltage (Vbe) = 0.6V:
       Vd = Vz = 6.2V
       Vb = (R2+R3) * (Vd/R3) = 10.55V
       Vc = Vb + (2*Vbe) = 11.75V
       Vf = Vz - Vbe = 5.6V

  • It looks like a linear regulator, with a differential pair formed by Q4/Q3, and the Darlington Q2Q1 being the pass element.

    I don’t understand why the Rz is supplied from Vout instead of Vin, but if we consider a perfect match between Q4 & Q3 (hence the same current following to the current sink R4), VB (Vout) would be extracted from the voltage divider of the feedback VD = Vz = 6.2V, then VB = 10.55V, with an assumption of Vbe3 = 0.6V, VF = 5.4V, and with an assumption of the Darlington pair Vbe12 = 1.2V, VC = VB + 1.2V => VC = 11.75V (With respect to the GND).

  • Good catch concerning Rz!
    As a first thought, you would connect Rz to the input to provide power to the zener. Doing so, the current through the zener would vary with input voltage variations (thus also the zener voltage) resulting in a bad line regulation.
    Connecting Rz to the (regulated) output keeps the zener current independent of the input voltage, thus improving line regulation.

  • Simulating it in LTspice reveals a subtlety about the circuit that isn't obvious in a static analysis.

  • 1. Is it an LDO? 

    Yes, as the drop from vin to vout is Vce of darlington pair.

    2. Vc  =  11.95
        V f =   5.5

        Vb =  10.55

    But, will the  regulator start if Rz is connected to output instead of input ?

  • Hmmmm... you make me think...
    R1 and R5 are quite small in relation to R4. This might cause node F to rise and thus bring the circuit out of regulation.
    Without starting LTspice, I think using 10 times higher resistors for R1 and R5 should make the circuit work as expected.

  • Yes, the circuit will start: R5 turns the darlington pair on without further help so the zener is supplied before regulation comes in.