# 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
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
understand the maths of it, so im stuck. help would be
much appreciated

Regards

Fenris

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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

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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
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

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12. ### JoeJester AAC Fanatic!

Apr 26, 2005
3,403
1,230
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
3,403
1,230
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