Voltage Regulator Using 2 Diodes

Discussion in 'Homework Help' started by jegues, Mar 13, 2011.

  1. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
    735
    43
    I'm not sure if I'm going about this question properly. Can someone quickly check if I'm on the right track?

    If not, can someone get me started in the right direction please?

    Thanks again!
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    Well I get 68.41978355514435mA for I2 current
     
  3. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
    735
    43
    Did you calculate it differently than how I did it?

    Also, I think I made a mistake in this question. It says that it should provide a reference voltage of 1.5V with a 150ohm load.

    I did my analysis with no load to provide a reference of 1.5V.

    Which is correct?

    Can you further explain how you attacked this problem?

    Thanks again!
     
  4. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    If we want 1.5V across the the two series-connected diodes we need Id = 68.4197mA of current. (Your equation for I2 looks ok)
    Load current is equal 1.5V/150 = 10mA
    So the resistor current is equal ID+IL
    R = (5V - 1.5V) / ( Id + Iload) = 3.5V/78.41mA = 44.63 = 47Ω.
     
  5. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
    735
    43
    If my equation for i2 is okay why are we calculating different values?

    I keep checking it and I am still obtaining 73.891mA, how are you obtaining your value for Id?

    EDIT: On a second attempt with Id=i2=73.891mA and iL = 10mA, I through R is found to be 83.891mA giving me an R = 41Ω.

    See figure attached for my 2nd attempt.

    Also, how do we deal with the last part where RL changes to a range a given values?

    As we vary RL, we vary IL as well, correct? So what exactly is the rest of the circuit going to do when RL is varied? Will the source simply supply more current? Or will that current remain the same and it is the current across the diodes that changes?

    Thanks again!
     
    Last edited: Mar 13, 2011
  6. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    You don't don't state anywhere what your value of VT is. Most people use the approximate VT=0.026 at room temp. I think this is the value Jony130 used.
     
  7. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Using VT=.026 I obtain the same values as Jony130.

    If the load is disconnected I have the diode current increasing to ~78.27mA which gives a new regulated voltage of 1.507V, using the 'exact' value of R=44.63Ω.
     
  8. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
    735
    43
    Ah that's what it was, we have been using VT = 0.025V. This must be the reason for the difference.
     
  9. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
    735
    43
    I've got a question with regards of how to preform the analysis for a range of different RL's.

    What's the correct procedure to do this analysis for multiple RL's?

    My professor mentioned that we can assume the current coming from Vin is the same, and also assume that Vo is still 1.5V for the new value of RL. From here we can solve IL and by KCL get the new current i2 (current across diodes)for the given RL. He then mentioned that with this you can calculate delta Vo. (He was mentioning how this was a quick way of doing it)

    I'm not sure if I fully understood what he was trying to explain, so don't be surprised if that sounds a little goofy or incorrect.

    Anyways, is the correct procedure to carry out the analysis for multiple RL's? We just want to see how it's going to affect the output, so what should we be considering? Are there any valid assumptions we can make?

    Thanks again!
     
  10. Adjuster

    Well-Known Member

    Dec 26, 2010
    2,147
    300
    Your professor's approximate method most likely assumes that RL is only allowed to vary between very large values (or open-circuit) and a minimum such that the diode current changes by not more than about an order of magnitude. This will constrain delta Vo within a range of perhaps 120mV, so that the feed current from Vin changes by no more than a few percent.

    Assuming the feed current to be constant, KCL tells you that the sum of the diode current and the load current must also be constant. Accordingly, any change in load current would result in an opposite change in the diode current. You can thus easily get a rough estimate of the diode current with a range of different load resistors, but the results will not be exact.

    This approximation gets more inaccurate with lower load resistances, as the remaining current in the diodes becomes comparable with the increasing change in feed current. If you calculated the diode current to be reduced to 10% of its un-loaded value in this way, the actual diode current would be more like 13.5% of unloaded, if I have my sums right.
     
  11. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Your prof's method works OK. It produces a result very close to the 'exact' method which involves the numerical solution of two equations - one of them being the Shockley diode equation.

    For instance, the 'exact' solution I got for the 50Ω load case was Vout=1.4828V.

    And the Prof's method would be something like ...

    Taking the 1.5V as the first pass value for Vout would give Itot=78.423mA with R=44.63Ω.

    The load would take 1.5/50 or 30mA leaving 48.423mA in the diodes.

    This would give Vd=0.741V with VT=.026 & Is=2.02981e-14

    Hence, Vout=2*Vd=1.482V

    Which is very close to the 'exact' value of 1.4828V.

    In the end this is all very academic given the temperature sensitivity of VT.
     
  12. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
    735
    43
    Here's my crack at it. (see figure attached)

    I tried to follow another example in the textbook.

    I used VT = 25mV and Is = 10^-15A. (hence why i2 is different than your Id, it corresponds with the work on my 2nd attempt in the pdf)

    Does this look alright?
     
  13. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Picking up on Adjuster's comments, I checked what happens when the load is reduced to 15Ω (10% of 150Ω), for instance.

    Vout falls to around 1.25V (a decrease of 250mV) and the diode current falls to ~573uA. Itot is ~84mA.

    So as Adjuster correctly points out, the assumption of stability over a wider range of loads is not valid at the increasing load current end of range. One needs to apply an 'exact' or more suitable method of solution.
     
  14. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
    735
    43
    But as an approximation, is my method of solution demonstrated in my previous post correct for RL's of 100, 75, and 50 ohms?

    It's good to know that what I am actually calculating is indeed an approximation, and with that the limitations of my results.
     
  15. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Your solution accounts for a single diode voltage change. Presumably the effective output voltage change would be twice that of a single diode.

    I have no concerns with your using the notional dynamic resistance, provided your recognize the limitations of the approach - which you have explicitly stated anyway.

    As to the validity of the approximation one can only make a comparison with a more 'exact' approach.
     
  16. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Also keep in mind the assumed change in diode current is the new load current minus the load current at 150Ω.

    So the load current at 150Ω would be 10mA.

    With a 50Ω load the new load current is notionally 30mA, which equates to a 20mA change in current.

    So at RL=50Ω the Vd change would be rd*20mA = 0.338*20e-3 = 6.76 mV.

    This would give a total Vo change of 2*6.76mV=13.52mV.

    An exact solution yields 15.2mV change based on your VT=0.025 and R=41.72Ω values.
     
  17. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
    735
    43
    Ahhh now I've seen what I've done wrong, this makes much more sense!

    Thank you for clearing things up!
     
Loading...