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Help curing light dimmer "lamp sing"

You could just cut the capacitor by a factor of 10 or so and get the frequency high enough that the filament doesn't bounce as much. If you are going to add an inductor, raising the frequency will make it more effective. But adding a filter inductor, turns this into a switching regulator with potentially quite a bit higher efficiency. You put the inductor between the lamp and the drain. You also need a diode (should be a Schottky for its speed and low voltage drop) between the drain and the positive 12 volts, to carry the inductor current when the mosfet switches off. The cathode (banded end goes to

+12). This will lower the current in the mosfet to about the average lamp current, instead of the peak current it is carrying, now.

The inductor needs to be a ferrite core type rated for lamp current, and the diode needs to be rated for almost lamp current, also.

Lets say the lamp draws up to 5 amperes and you raise the frequency to

30,000Hz (by making the cap .001uF). When the mosfet is on only half the time, the average voltage out will be about 6 volts, so the lamp current may fall to 4 amperes (lamps do not follow ohm's law when they change temperature). You want an inductor large enough to keep the current from changing drastically (lets say by not more than half) during the time the switch is off. So in half of 1/30,000 seconds the inductor with 6 volts across it must have its current fall from 5 to 3 amperes (the average being 4).

The formula that relates voltage, current, inductance and time is: V=L*(di/dt) or voltage across the inductor in volts equals inductance in henries times the rate of change of current in amperes per second.

so 6V=L*(5-3)A/(1/(30000*2)) so L=50 micro henries, for this set of assumptions.

If you had a frequency of 3000Hz, L would need to be 500uHy and if you stay at 300Hz, it would have to be 5mHy to keep the current ripple the same. A 5mHy 5 ampere inductor is a lot bigger than a 50uHy one. but of course, by the time you have raised the frequency to an inaudible

30,000Hz, you noise problem has gone away (but your mosfet may be quite a bit warmer from the 100 times more switching loss).

Whatever you do, do not add any inductor without also adding the diode to carry its current or the inductor will make large voltage spikes every time the mosfet turns off, and will probably destroy the mosfet.

Reply to
John Popelish
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Try changing the timing capacitor (the one connected to pin 2) from

100n to 1500pF.

Best regards, Spehro Pefhany

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"it\'s the network..."                          "The Journey is the reward"
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Reply to
Spehro Pefhany

The big problem with adding the cap is that it connects a near short circuit across the 12 volt battery each time the mosfet switches on and the cap has only a small part of the battery voltage across it. In effect, you are trying to charge the cap to 12 volts with a time constant of the on resistance and battery resistance times the capacitance. This can make the battery bounce down severely, heating it, and making the mosfet much hotter. It also risks destroying the capacitor with a ripple current higher than it is rated for. I don't like it.

The capacitor from 555 output to ground moves this cap charging problem over to the output stage of the 555 and should make it very hot. It also replaces mosfet on off switching with slowly varying resistance that makes the mosfet into a series resistor that wasts all the extra voltage, and dumping almost as much heat into the mosfet as it puts into the lamp. Check how hot these two parts are. I don't like that, either.

Reply to
John Popelish

Hello:

I'm building a lamp dimmer for my 12vdc. video light using the circuit on this web site:

My problem is "lamp sing". There is a pretty clear buzzing at the 300hz. switching rate coming from the lamp when at less than the full output. The mic will surely pick up this noise.

What can I do? Can I use an inductor in-line with the lamp to suck up the noise and if so, how do I compute the inductance given the 300hz. known frequency?

Thanks for the help and any suggestions.

Reply to
G.I.O.

I don't know the answer for certain but the first thing I would try is a resistor in parallel with the FET that will keep about 3/4s of the full on current constantly thru the lamp. The Fet then will be switching less current and the shock to the filament will less. Be sure to calculate the power in the resistor to properly size its rating.

Dan

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Dan Hollands

1120 S Creek Dr Webster NY 14580 585-872-2606 snipped-for-privacy@USSailing.net
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Reply to
Dan Hollands

John:

Thank you for the detailed reply. Your solution of reducing the timing cap to ..001uf will work and there is no audible noise but I worry about the much higher switching frequency getting into something else... like my radio mics. I would think the lower the frequency the less likely it will affect other systems.

While waiting to get some response to my posting I tried to solve this issue by adding a 47uf cap across the lamp (this cured the buzzing when fully on) and a 20uf cap from output pin #3 to ground (to remove the buzzing when dimmed off).

The lamp now has a very low glow when fully dimmed down, which is just fine with me. What have I done, here, and can I get away with it? Do I need that diode with this scheme?

Thanks, George

Reply to
G.I.O.

the lamp.

source

Yes. A smaller wattage lamp will pull less current through the mosfet. If you adjust both lamps to the same light output, the difference will be even greater, because the lower wattage one running hotter will be more efficient at turning current into light.

Reply to
John Popelish

Yes, everything you say is correct. The MOSFET is way too hot.

So I'm back to square one. I will need to see if the 30k switching noise will be a problem for the radio mics. If not, then this is the simplest solution.

Reply to
G.I.O.

Using the .001 timing cap I get a pulse output of around 22k. No noise from the lamp.

Question: If I use a 60 amp MOSFET can I expect lower heat with a 20 watt lamp source than using one rated for, say 30 watts?

Reply to
G.I.O.

keep the heat

The power dissipation of the MOSFET is due to losses in the transistor due to on resistance and switching. The power dissipation *rating* of the MOSFET is more related to the packaging and heatsink (and maximum die temperature), and does not directly related to the losses. A big high-voltage MOSFET such as an IRFPE40 (800V/5.4A) in a TO-247 package can dissipate 150W according to the data sheet (if you could magically keep the case at 25°C), but with 2 ohms Rds(on) it would get quite hot. You don't want that.

What you want is very low Rds(on). A 20W lamp will draw less than 2A from your 12V supply. A part such as an IRF3707Z ($1.40 one-off)

30V/42A has Rds(on) of less than 0.01 ohm at 25°C. Conduction loss will only be 40mW at 2A (which translates into "barely warm" with no heatsink), even though it can "only" dissipate 28W at Tc=100°C.

Best regards, Spehro Pefhany

--
"it\'s the network..."                          "The Journey is the reward"
speff@interlog.com             Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog  Info for designers:  http://www.speff.com
Reply to
Spehro Pefhany

Actually, I meant this the other way around.

That is, given the same wattage lamp, say 20 watts. Can I assume using a MOSFET that's rated for 200 watts will run cooler than one rated for 100 watts?

If so, is there any harm in using the larger wattage device in an effort to keep the heat down within the casing?

Thanks!

Reply to
G.I.O.

Hello Spehro:

Thank you for the detailed reply. Very much appreciated.

George

Reply to
G.I.O.

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