any Professional could write equations for IRF510 MOSFET for Id=30mA;Vo=5;Vdd=9?

Discussion in 'General Electronics Chat' started by vcoasm, Jan 12, 2012.

  1. vcoasm

    Thread Starter New Member

    Jan 12, 2012
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    any Professional could write equations for IRF510 MOSFET for Id=30mA;Vo=5;Vdd=9?
    guys i had spend more times on how to determine the drain current exactly using
    the prefer equations , but I failed i think the problem comes from extracting the correct
    Kp (constant of mosfet)
    any one can put here in details how to calculating + writing the procedures&equations
    step by step; please help me find Kp to determine the current i would like
    please please ..............etc
    extremely; DATASHEET did not including Kp;Ids(on); only they mentioned Rd(on)
    WHY;;;;;;;;;;;;;;;;;;;;;;;;;;????????????????????? ???????

    ok here my tries :=
    IRF510 Datasheet including this INFO:
    gfs = 1.3 @ 3.4 A and sometimes in other datasheet 1.3@4 A. & VgsOFF=4V
    i do not know what if that gfs is same as Kp or gm ;
    i assume it is gm ; thus:
    as Id = Kn (Vgs - VgsOFF)^2 .............................. (1)
    also
    gm = 2 root (Id * Kn) then ---> Kn = (gm)^2 / (4 * Id)
    apply this equation yiald: Kn = (1.3)^2 / (4* 3.4) = 124.5 mA/V^2

    the required ( given ) parameters are : design MOSFET with Id=10mA with output = 5V
    Vdd=9 V ; with using the voltage-divider ... here i want not to includes Rs ( resistor at
    source ) .
    as Vdd= Vds + Id*Rd
    thus Rd= (Vdd - Vds)/ Id = (9-5)/10mA = 400 ohms
    at this point all things correct right? theoretically YES yes yes
    then as Vgg = Vgs
    Vgg = Vdd * (R2/R1+R2) = Vgs
    multiply both denominator and divisor by R1/R1 gives :

    Vgs = Vdd * Rbb/ R1----------------------- (2)

    for MOSFET :
    for Id-= 10mA ; assume R1= 10K then
    using Equation (1) yiald: Vgs = root(Id/Kn)+VgsOFF = Root(10/124)+4V=4.28 V
    Assume R1 =10K
    using Equations (2) and solve for Rbb yealds: Rbb = (4.28/9V) * 10K = 4759.9 ohms

    and as 1/Rbb = 1/R1 + 1/R2 -----> thus R2 = 1 / ( 1/Rbb - 1/R1) = 1/(1/4759.9 - 1/10K)
    thus R2 = 9083 =~ 9K ohms

    THE RESULTs :
    Id = 10mA
    Vgs = 4.28 > VgsOFF (i,e MOSFET is ON )
    Id = 10mA @ V0 =5v for Vdd = 9 V

    The DISASTERs :
    when apply the design on LT-Spice it is not success WHY WHY WHY ....

    PLEASE ANYONE GIVE ME AN ANSWERS

    YOU HAVE TO LOOK THAT I FORETHOUGHT ( DELIBERATE) NOT TO INCLUDE RS IN MY DESIGN FOR ANALYSIS PURPOSES
    SO PLEASE CALCULATE WITHOUT RS
     
  2. Adjuster

    Well-Known Member

    Dec 26, 2010
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    IRF510: Do you mean this, rated at 5.6A Id??? http://www.ee.nmt.edu/~wedeward/EE443L/FA99/IRF510.pdf

    If you want to bias something like that consistently at just 30mA Id, you would really need some kind of feedback.

    Different samples of the same device are likely to behave radically differently at the sort of small gate voltage required, so that some may be barely on at all, others will conduct too much. The VGS(TH) is specified as being between 2.0V and 4.0V!

    Even with a source resistor dropping several volts, the drain current accuracy for a fixed gate voltage at such a low level would be poor. The gate voltage would need individual adjustment for accurate drain current setting.
     
  3. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Try use this equation:

    Id = K * ( Vgs - Vt )^2

    then

    K = Id / ( Vgs - Vt )^2

    So from the data sheet
    http://www.micropik.com/PDF/irfp140.pdf

    Vgs(th) = Vt = 3V

    And from figure 5

    Vgs = 5V; Id =10A

    K = 10A / ( 5V - 3V )^2 = 2.5

    But these calculations are not very accurate because MOSFET show great process spreader. For example on Vgs(th) = Vt = 2V...4V.

    So only real measurement can give you exact K value.
     
  4. vcoasm

    Thread Starter New Member

    Jan 12, 2012
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    first i'm really so happy on your reply
    i had tried upto 2 forum to answer but i think this forum will be my Favorited Forum

    what if i used the tested conditions of RDS(on) =0.055@Id=19A;Vgs(ON)=10 V

    thus ;
    Kn = (19)/(10-4)^2 = 527.77 mA/v2

    i see someone used this does this another right method?
     
  5. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    You cannot do that.
    To find K the MOSFET must be in saturation mode.
    Look here:
    I build this circuit in LTspice to find K factor.
    From the LTspice IRF510 model I know Vt = 3.8V and from DC bias analysis.
    Id = 0.894A for Vgs = 5V and Vds = 5V

    So

    K = Id / ( Vgs - Vt )^2 = 0.894/(5 -3.8)^2 = 0.62

    So we need Vgs ≈ 4.01V to get Id = 30mA
     
    Last edited: Jan 12, 2012
  6. vcoasm

    Thread Starter New Member

    Jan 12, 2012
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    amazing !!!!!!:eek:
    but i see you depend on using lt-spice to find k

    what if i want to find it with more accurating in my lab?

    how to use gfs to find Kn from this equation :
    gfs is same as Gm
    so ---> gfs = 2 root(Kn 8 Id)
    thus kn= gfs^2 / Id

    please correct this info to me:D
     
  7. vcoasm

    Thread Starter New Member

    Jan 12, 2012
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    think you gentlemen for feedback on my problem
    many guys maybe look at my problem not interesting
    but the challenge here is about how to biasing the IRF510 gate at the
    accurate value that achieve your requirements

    those two guys supply me with the big info that i had not got it from the american friends .

    now for Adjuster+jony130 can you give us an ensure on finding the Kn of MOSFET practically on LAB?
    i do not like to depending on LT-Spice for all my circuit design
    we really need for practical method to find the more accurate Kn specially
    when we want to design Oscillators (My interesting to finish my book ) and MIXERS for Frequency doubler

    best regard;
    Mr.\ Mamdouh .O. Khadidos
    Saudi Arabia
     
  8. Audioguru

    New Member

    Dec 20, 2007
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    The datasheet for every Mosfet shows a wide range of gate-source voltages to turn it on. Your old Mosfet is guaranteed to turn on very well with a gate-source voltage of 10V.
    The SIM program shows only a "typical" Mosfet but you cannot buy a "typical" one, you might get one that is much less sensitive.
    But maybe your teacher does not know DAT.
     
  9. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    To find Kn factor for the MOSFET in the LAB you need this simply circuit

    [​IMG]

    This symbol on the top represents the constant current source.
    For this circuit Vds = Vgs(off)
    Now we need to find Vgs for given Id current.
    So the only thing we need to change in this circuit is to increase the value of a constant current source.

    And of course you could use this circuit to plot the transfer characteristics.

    Also if you want to use MOSFET in small-signal circuit use small-signal MOSFET eg. BS170; 2n7002
     
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    Last edited: Apr 28, 2012
  10. vcoasm

    Thread Starter New Member

    Jan 12, 2012
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    Dear Jony130 (Senior Member);
    I would like to think you very much and i hope you help me next time to continue my
    practicals circuits to finalizing my book . Next time i'm go to calculating my circuit as i wrote in my book to design MIXERS+ DOUBLERS ; so i wish if you can help me .I am a teacher of math and i am interesting in design many types of oscillators theoretically then practically . At this times i'm on my final stages on my last chapter of designing mixers ;so please please try to help me when i'll facing difficult practices .

    about Kn , you are right i am now very sure that using the curves in datasheet and choosing a points are very good way to calculate Kn
    but what about gfs? hwo can i take its advantages ?
    for what gfs used?

    another thing... i see you are very smart...are you in MIT ? huhu i hope you will be someday if you are not

    good wishes; Mr.Mamdouh Khadidios
     
  11. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    In the lab it is much easier to use the method that I described to find K factor..
    How you are going to measure gm in the lab?
    Also don't forget that this equations Id = K*(Vgs - Vt)^2 is just a first approximation. This formula don't describe the behavior of real device.
    Also Vgs will change with temperature, and Vds voltage also has influence on Id current. Additionally don't forget that MOSFET parameters (such as Vgs(off) ) have a wide range of manufacturing spread among device of the same type.


    No, I'm not from MIT. Electronics is just my hobby.
     
  12. Adjuster

    Well-Known Member

    Dec 26, 2010
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    I once had to bias some quite large FETs at a tiny fraction of their rated drain current. This was done as follows:

    1. The FETs we started with were expensive ones, with a fairly narrow VGS(TH) range.
    2. A substantial source resistor was used, to drop a couple of volts.
    3. A select-on-test resistor was chosen to set the gate voltage as required for each circuit individually.
    I was aware at the time that what I was doing was pretty kludgey, but it was carefully done, and the horrid little circuits did what they were required to do. They worked for many years without failing.

    I do not think that this would have stood any chance of working properly without the big source resistors.
     
  13. vcoasm

    Thread Starter New Member

    Jan 12, 2012
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    Dear adjuster;
    I want to think you first for your care about my complaint of MOSFET Biasing

    I'm so happy to prove that each MOSFET with same number not idintical as Mr.Jony130 said. yes i had test this concept on my lab
    i fairly find out that i must first draw the curve of every mosfet I buy ;suddenly.

    BUT :
    i tried to find Vth voltage threshold @ 1 mA then use this value as (Vgsth)
    then substitute it in the general equation Kn=Id/(Vgs-Vgsth) after change the voltage divider at the gate
    finally i 'm so glad to successfully find any type i've in my junk box
    so you said that it is must be to use the source resistor.in my view let me say to you; no not recommanded
    I have build my colpitts oscillator with this results after find the characteristics of two type of MOSFET IRF510 & IRF470 and it is work
    but as you said source resistor must be there only for stable biasing not at all


    but I wonder if this method of testing MOSFET is a very good method to apply for any MOSFET as a practical for any hobby?

    I need for your feedback and also for you Mr.jony150


    BY:
    Mamdouh.O.Khadidsos
     
  14. Audioguru

    New Member

    Dec 20, 2007
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    Isn't the IRL510 one of the oldest and poorest Mosfets ever made?
     
  15. vcoasm

    Thread Starter New Member

    Jan 12, 2012
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    emmmmmm I see that it was older but the component had been enhanced with a diode connected between the drain and the source to protected from the static electricals.. Really i do not know the history of this part , sorry
     
  16. Audioguru

    New Member

    Dec 20, 2007
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    Old Mosfets like the IRF510 needed 10V from gate to source to source to fully turn on. Then most logic ICs cannot turn them on with only 3.5V or 5V.

    ALL Mosfets have an avalanche diode (zener diode) from drain to source due to the physical way they are made. But the gate-source max allowed voltage is only 20V for most Mosfets and if exceeded then the Mosfet will be blown up. Some "logic-level" Mosfets have a max allowed gate-source voltage of only 10V.

    If you don't know how to read datasheets then maybe you should do gardening instead.
     
  17. Adjuster

    Well-Known Member

    Dec 26, 2010
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    I would like to help, but I cannot quite understand what you have written.

    Biasing a high-current device to work at a low current consistently is difficult. If a large resistance in the source is not acceptable then other methods of stabilisation may be required.

    Possibly an oscillator may be made to stabilise by deriving bias from its own output amplitude, but the initial bias must be appropriate for it to start in the first place.

    In an extreme case, a separate amplifier may be used to control the bias. This is how many current controllers work. Typically, a small current sensing resistor is placed in series with the source (or sometimes the drain). The relatively small voltage drop across the resistor provides a feedback signal.
     
  18. vcoasm

    Thread Starter New Member

    Jan 12, 2012
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    For adjuster:
    I know i got a weak english learning in my home but that not mean no one could not understand what i have written .So if you had read this thread from the first time thus you will understand.Why did not asked your self why Mr.Jony can read and understand what i am writing?
    Hence no need to hints on my English sir

    To Audioguru:
    I do not like to speak with you my son and be polite and respecting the people is the first step to deal with them

    BYE
     
  19. Adjuster

    Well-Known Member

    Dec 26, 2010
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    Jony130 speaks at least two languages. He may be more skilled than I am at understanding broken English. I however have real difficulty following much of what you have written, whatever you think I should be able to understand: I am an electrical engineer, not a teacher of English as a second language.

    It is reasonable for us to respect each other here, and I am sorry if one of our members has upset you. This applies equally to yourself however: if you are not understood you must expect to be told this.

    Remember that the people who are trying to advise you here are not paid to do so.
     
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