modifying a circuit to use a mosfet

Discussion in 'The Projects Forum' started by Fenris, Oct 23, 2007.

  1. Fenris

    Thread Starter Active Member

    Oct 21, 2007
    288
    2
    hi all

    i need a little help if possible with a circuit i am modifying.
    the original circuit is a sound to light, activated by an
    electret mic, drives 4 LEDS in series and has a VR to
    adjust sensitivity. i have breadboarded it and it works.
    what i am attempting to do, and i have used a circuit
    simulator in which it works, is remove the LEDS and
    put a mosfet in to drive 2 X 20W filiment bulbs @ 12v.
    on the pic the transistor on the top left is a BC558
    original to the circuit the resistor on its collector is
    as was (almost) the only addition are the mosfet, bulbs
    and the 47K resistor. the theoretical FET is an IRF630.
    i have adjusted the value of the resistors to get the
    gate voltage right. however i know there is more to
    how they function. is this way of doing it correct?
    sadly although i can construct from schematics and
    have a basic understanding of electronics i lack the
    ability to correlate the datasheets information or :confused:
    understand the maths of it, so im stuck. help would be
    much appreciated

    Regards

    Fenris
     
  2. nanovate

    Distinguished Member

    May 7, 2007
    665
    1
    The IRF630 has a high on resistance. Since you are using 12V you can select a lower voltage FET (lower Rds_on). The IRLZ44N is one that pops up. The gate voltage you are supplying in your circuit is too low you should give it at least 10V to 15V for the IRF630 and 8V to 12V for the IRLZ44N. So you could eliminate the 63k resistor.

    You should also calculate how much current will go through the FET. Once you have this number then calculate how much power will be dissipated in the FET:

    P = I^2 * Rds(on)

    If it is less than a watt then you will be safe without heatsink using the TO-220 package (this leave margin since the 2W is the MAX)

    A small gate resistor will not hurt either.

    There are other MOSFETs besides the IRLZ44N theat would work also. Look for low Rds(on) types.

    There may be other tips-n-tricks I am forgetting which others may point out.
     
  3. Fenris

    Thread Starter Active Member

    Oct 21, 2007
    288
    2
    hi there
    thanks for that. i have looked up the formuler in the 'e-book volumes'
    i think i understand.

    power= (IxI)xRds(0n)?

    so 2 x 20 watt bulbs @12V in my simulator give 3.61A >

    P=(3.61x3.61)=13.03A

    the IRLZ244N has an Rds(on) of 0.044ohms >

    P=13.03x0.044=0.6W =600mW? so thats less than the 1W you mentioned.
    have i understood this?
    my local maplins has only the IRF540A in stock the datasheet gives this an
    Rds(on) of 0.052ohms so >

    P=13.03Ax0.052=0.7W or 700mW?

    i dare say i have got it wrong, as i mentioned, algebra still gives me the
    cold sweats thanks to a pointless teacher who all but killed any determination
    in me to get to grips with Maths

    Regards

    Fenris
     
  4. nanovate

    Distinguished Member

    May 7, 2007
    665
    1
    I believe you need to give the IRF540 10Vgs to get Rds(on) of 0.052 Ohms.
     
  5. Fenris

    Thread Starter Active Member

    Oct 21, 2007
    288
    2
    thanks again

    i have got the IRF540A and it works with 2 X 20W halogen bulbs so im
    quite happy. could you just tell me if my workings out were correct
    please. also im still at a loss with the business of the gate voltage
    needing to be ie. 2 - 4V greater then the supplied voltage.

    so 12V needs min 14V, max of 16V to the gate
    or is this just the maximum you can drive the gate over the supply
    voltage without damaging the Mosfet? having said that i see what
    you say for the IRF540A it only needs 10V would i be right in thinking
    that the Vgs(th) means 2V-4V above supply but can go 2V-4V below
    supply? sorry to be a pain but i understand plian english better than
    technical data sheets

    Regards

    Fenris
     
  6. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    657
    Lamp Power=Voltage *Current
    40W=12V*I
    I=3.33A
    MOSFET Power=(Lamp Current)^2*Rds(on)
    I think you can take it from there.
     
  7. nanovate

    Distinguished Member

    May 7, 2007
    665
    1
    The threshold voltage is where the MOSFET starts to turn on and conduct. This would be referenced to the source terminal.

    There should be a section called "Absolute Maximum Ratings" where you find the maximum values. For the IRF540 the gate to source voltage should always be less than 20V otherwise you risk damaging the device. In the section that specifies the Rds(on) they also specify the Vgs and drain current that the Rds(on) was measured at -- 10V and 11A in my datasheet.
     
  8. Fenris

    Thread Starter Active Member

    Oct 21, 2007
    288
    2
    hi guys

    thanks again thats cleared up a bit more.
    so with my current setup;
    (12V*3.33A)*0.052 =2.07W a little on the hot side.
    so if i went for the IRLZ44N then i would have;
    (12V*3.33A)*0.022 =879mW

    looking at the data sheets of say the IRF540A, specifically the
    absolutes. im working at 12V so do i take it im working at 12%
    of the absolutes and apply that accordingly? sorry if im being a pain
    i am reading the E-books and various other sources to try and get to
    grips with it. pic of part of the absolutes of the data sheet attached.

    Regards

    Fenris

    PS i have now, since this post, changed the 47K resistor for a 33K this has dropped
    the the current draw slightly of the bulbs but has reduced the W's to 1.8 on the
    Mosfet. incidently it has also increased the working range of a VR in the circuit
    that adjusts the sensitivity.
     
  9. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    657
    Your MOSFET power calculation is wrong.
    P=I^2*R . That's I*I*R , not V*I*R as you assumed. The supply voltage has no role in calculating the dissipation of the MOSFET when it is on.
    Power=3.33*3.33*0.052=0.578W
    Alternately, you could calculate the voltage across the MOSFET when it is on: V=I*R=3.33*.052=0.173V
    Then P=V*I=0.173*3.33=0.578W.
    You need to study up on Ohms Law and power calculation.

    Percentages are not relevant.

    I think Fairchild has a funny font in their datasheet. The International Rectifier IRF540N datasheet says Vgs(max) is +/- 20V.
     
  10. Fenris

    Thread Starter Active Member

    Oct 21, 2007
    288
    2
    hi there thanks for that
    it looks like my first workings
    out were right. I*I*Rds i got
    confused i guess trying to cram
    all the info in to fast.
    so i am actually running at less than 500mW
    thanks again everyone

    Regards

    Fenris
     
  11. Fenris

    Thread Starter Active Member

    Oct 21, 2007
    288
    2
    hello all

    well it all went swimmingly until i made
    the circuit on veroboard. from that point
    the lights came on without audio input,
    flickering and the mosfet got bloody hot.
    i have checked everything and up to the
    BC558 transistor that triggered the mosfet
    gate it all still works. now when it was on
    the breadboard it worked for about 40 mins
    before the bulbs grew dim, i thought my 12V/12Ahr
    battery was stuffed but it turns out its fine! so whats
    gone wrong ( i think the mos is dead, both of them)
    it was running (using the math) at 428mW so well under
    the IRF540A 2W. help and advice would be appreciated
    so i dont make the same mistake again.

    Regards

    Fenris
     
  12. JoeJester

    AAC Fanatic!

    Apr 26, 2005
    3,373
    1,159
    I would suspect the 540A demise was caused by attempting to pass 40 W [3.333A].

    Depending on the current draw for the rest of the circuit, it could explain why the bulbs dimmed quickly.
     
  13. Fenris

    Thread Starter Active Member

    Oct 21, 2007
    288
    2
    ok. from the maths though the 540 was running at well under 1W (576mW), the max is 2W also the Ld is rated @ 28A i was running @ 3.33A. ive been trawling the net trying to find if i missed anything out but i cant find any answers (probably wouldnt see them if they bit me) so am i on a loser doing it this way? should i stick an NPN power transistor in instead? a TIP type? all help gratefully recieved im drowning in datasheets and electronic sites here.

    Regards

    Fenris
     
  14. JoeJester

    AAC Fanatic!

    Apr 26, 2005
    3,373
    1,159
    Can I assume you installed a heat sink on the IRF 540?
     
  15. Fenris

    Thread Starter Active Member

    Oct 21, 2007
    288
    2
    oh yes indeed. perhaps i needed a pullup resistor on the mos?
    because of the way the circuit works the moss would be
    'ramping'? rather than solid on / solid off.

    Rgards

    Fenris
     
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