Discrete voltage regulator (help needed)

Discussion in 'Homework Help' started by stefan.54, May 23, 2016.

  1. stefan.54

    Thread Starter New Member

    Dec 26, 2015
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    Please help me sort out this problem.

    I have the following circuit but my values are messed up (done tones of calculations and nothing) so please *ignore* the values on the schematic.

    I need to output 8V given that input is between 10V-15V, P_load is 3 W and tolerance is 5%.
    I do not need to use those transistors (though I have a handful), nor that specific Zener.
     
  2. crutschow

    Expert

    Mar 14, 2008
    13,056
    3,245
    Change R4 to zero ohms ( a short) and change the RV1 pot to 10kΩ.
    That will allow the output voltage to be adjusted from about 6.3V to 10.8V with the pot (using the 5.6V zener specified).
     
  3. stefan.54

    Thread Starter New Member

    Dec 26, 2015
    28
    4
    @crutschow
    Input and output voltage is " nan " after doing so. What next?
     
    Last edited: May 23, 2016
  4. AnalogKid

    Distinguished Member

    Aug 1, 2013
    4,546
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    And by the way, the current limit circuit will not work. Is it part of the assignment?

    ak
     
  5. stefan.54

    Thread Starter New Member

    Dec 26, 2015
    28
    4
    @AnalogKid

    Yes, I have to build and write the documentation of a voltage regulator using discrete components (IC would've made life so much easy, right?).
    The specifications it has to meet are :
    Vin=10-15V
    Vout=8V
    P_load=3W
    Tol= 5%
    Bonus points for protection circuit with indicating LED and diy PCB.

    Also, I do not have to use any of the components in the above schematic. It was merely a sketch I fetched from a class from first semester.
     
    Last edited: May 23, 2016
  6. crutschow

    Expert

    Mar 14, 2008
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    :confused: Seems like it will to me.
    When the current through R3 exceeds the base-emitter drop of Q2 it will start to turn on and starve Q3 of its base current.
     
  7. crutschow

    Expert

    Mar 14, 2008
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    Do a point-to-point check of the circuit to look for wiring errors.
    If the circuit is wired properly it should work.
     
  8. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    Transistor Vbe = 0.6
    LED Vf = 2.0
    Even if Q2 is saturated there still will be a voltage greater than Vbe across b-e. Don't know for sure, but seems to me...

    ak
     
  9. stefan.54

    Thread Starter New Member

    Dec 26, 2015
    28
    4
    Does any of you guys have time for some in depth analysis? It's due Friday and I still have lots to do :(
     
  10. crutschow

    Expert

    Mar 14, 2008
    13,056
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    You're right.
    I missed the obvious effect of the LED forward drop. :oops:
     
  11. crutschow

    Expert

    Mar 14, 2008
    13,056
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    Since this is homework help, not homework do, then I suggest you get busy. :rolleyes:
    We are happy to help with specific questions, but we don't do analyses for you.
     
  12. stefan.54

    Thread Starter New Member

    Dec 26, 2015
    28
    4
    @crutschow
    Never asked you guys to do it for me, just to guide me through :D.

    So, I'll write here how I thought I should deal with this problem.

    *First, I determined the P= (Vin_max-Vout)*Iout to choose a suitable "power transistor", given that Iout=P_load/Vout

    *Next, I_b1=Iout/hfe of power transistor
    Knowing this value, I can determine Rb (R2 in this case) which is R2=(Vin_min-Vout+Vbe)/I_b1. From I_b1 we can choose Q2 and Q3.

    *I randomly picked a Zener (with low dynamic resistance and, low temp coef. and a current similar to Q2)

    *Rd (R1 in this case) should be Rd=(Vin-Vz)/I_zener

    *I_b2 (for Q2) should be I_b2=Ic/hfe
    Knowing this value, I imply an I_r current 10 times bigger than I_b2 which goes through the R4-R5-Rv and figure the whole value of this branch ( 8V/I_r= x ohms)
    This is where I get stuck.
    In the algorithm I saw, it was implied that Vbb=Vref+Vbe ~ 1v and using the voltage divider they had a Vout_min and a Vout_max (but this algorithm was used in a power supply with variable output 1 to 5 V so it was obvious which one was the min and which the max).

    *If I manage to figure out the values of R4, R5 and RV (hopefully standard values), all that is left to do is to calculate R3 (protection resistor) and voila, that should do it.

    Any ideas?
     
  13. MrAl

    Well-Known Member

    Jun 17, 2014
    2,440
    492
    Hello,

    Might be some problems.

    First, with 10v in and 8v out, R2 only has about 1v across it so we only get about 1ma drive to Q3. With maybe 0.5 amp output, that means the transistor has to have a min beta of 500. Do you think it has that high of a gain?
    That means R2 has to be lowered, probably at least to 100 ohms just to get going, maybe lower depending on the gain of the transistor.

    Second, the values of the three resistances depend on the voltage of the zener. Figure that the base of Q1 has to be 0.7v higher than the zener voltage to get into regulation. That means that for any voltage output the arm of the pot must be at that voltage. The voltage divider formula will tell you what the upper pot node voltage should be and the lower node voltage should be, given zero current into the base. That can be modified to include a small base current later depending on what final value of R2 is made to be, but zero current into the base will get you thinking in the right way.

    Third, you probably have to take a closer look at the current limit circuit for critical times like with 10v in and 8v out. With only a small voltage differential it may not be able to steal enough base drive from Q3. Also, a base resistor is a good idea on Q2 to prevent a high base emitter current spike if the load is suddenly applied, especially a short circuit which then draws all it's current through that base emitter at least for some time until it can cut back. 100 ohms is a typical value. Also, calculate the current through the LED to make sure it works properly without blowing out or being too dim. Pay close attention to the voltage at the base of Q3 for all expected operating points.

    Fourth, some capacitance helps with transients. One across the input, one across the zener, one across the output, just to start with.
     
    Last edited: May 24, 2016
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