Ripple filter

I would like to use a brushless 12v (0.07a) computer fan to vent a small area in my RV. I tried this, but the battery charger cooked the fan after a while due to ripple- it could not handle the peak voltage I guess. I replaced the fan but placed a 2 ohm (5w) resistor in series with it, and a

0.1mfd cap and 15v (3w) zener in parallel with it to filter and hopefully protect the fan- but now the 15v zener gets too hot as it "clamps off the ripple" (I only want it to conduct for transient protection). Fan still going ok though.

A transistor regulator would not be good as- correct me if wrong- it would show a 0.7 Vce drop. This would cut the fan performance and wasting current for regulation is not desirable when on battery power. Would a bigger capacitor provide enough filtering to prevent the zener from conducting (0.1mfd was a guess)? The resistor needs to be as smaller (ohms) if possible for faster fan. Can anyone improve on this? I am pushing my design abilities already. Thanks for any ideas!

Don't assume the butler (ripple) did it. This is 60 cycles from line power we're talking about, isn't it? If your battery charger is connected to the RV's battery, then the battery should absorb all the 60 Hz ripple and whatever runs off the battery will be running off smooth DC, and I'm not kidding you about that. If ripple is getting into stuff there's something wrong with the way you are using the battery charger, or your battery is shot, or something... by the way 0.1 uF is WAAAYYY too small to do anything about ripple, if indeed you have a ripple problem. Think 1000 uF. At 120 Hz (full wave rectification), a load of .07 amps would need a smoothing cap of 560 uF to get ripple to 1 volt.

You may just have too high a voltage. If your zener is dissipating more than 3 watts then you have at least .2 amps flowing through it. With that plus the fan current, say a total of .3 amps, going through the resistor, the voltage across the resistor would be .6 volts. Meaning 15.6 volts DC or more coming in. Considering that the zener voltage will rise because of the high current going through it, this is a conservative number.

Have you put a voltmeter on the battery while you are using the charger?

There's no way to give you a definitive answer based on very limited information in your post.

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John,

If the 15v zener is getting hot, then your battery charger is outputting something greater than 15V so a 12V regulator would still have at least a 3v drop to work with. Your zener is having to dissipate 15V times the current output of the battery charger minus the fan current so it is no wonder that it is getting hot. A voltage regulator would only need to dissipate whatever the drop across it actually turns out to be times the fan current. Likely a simple LM7812 would work fine.

Hope this helps.

--
James T. White

--- For your application, here's what I'd do: (View in Courier)

Vin Vout / / CHARGER+>---[1N4001]--+--[7812]---+ +| | +| [ ] | [FAN] | | | GND>------------------+----+------+

Assuming that your charger output is unsmoothed and outputs 16V peaks, then the cap will charge up to about 15.3V.

Since the 7812 will output 12VDC for your fan, and the cap will be charged up to 16V, the differential voltage across the regulator will be:

dV = Vin - Vout = 16V - 12V = 4V

with a fan current of 0.7A, the power dissipated by the regulator will be:

P = dVI = 4V * 0.07A = 0.28 watts

With a dropout voltage of 2.5V for the 7812, a 70mA load, and a full-wave rectified, unsmoothed output from the charger at 120Hz, the value of the capacitor will be: I dt C = ------ dV

where C is the capacitance of the capacitor in farads I is the load current in amperes dt is the period of the ripple dV is the allowable ripple in volts.

Now, since we have 16V across the cap and the output of the regulator is at 12V, that gives us 4V of headroom. Since the regulator requires that its input never fall more than 2.5 above its output, that means that we must make sure that the voltage across the cap never go below:

Vin(min) = Vout + Vdo = 12V + 2.5V = 14.5V

Since we'll have 16V across the cap when it charges, somewhere on the rising edge of the charger's output peaks, we have to make sure that it never falls below 14.5V when the cap is dicharging into the load and the charger's output isn't charging the cap and driving the load. Since the difference between 16V and 14.5V is 1.5V, that means that the allowable ripple voltage out of the cap is 1.5V

If we lower the stress on the cap and lower the ripple to 1V we can solve for the capacitance like this:

I dt 0.07A * 8.3ms C = ------ = --------------- = 581µF dV 1V

Most common small aluminum electrolytics have a capacitance tolerance of +/- 20%, so to make sure you get at least 581µF you need to increase its value to 697µF.

The closest commonly available value is 820µF, which would be fine for full-wave rectified 60Hz.

However, on the chance that the output of your charger is half-wave rectified 60Hz, the capacitance would need to be doubled for the same ripple, bringing the value to 697µF * 2 = 1394µF.

A 1500µF/25V Panasonic EEU-FC1E152 would work, as would anything you might have lying around with at least that capacitance and a working voltage somewhat greater than the peak output voltage from your charger.

There's not much reason to worry about transients since the cap will soak them up, but if you want to you could put a Zener in parallel with the cap with a voltage rating equal to greater than the peak output of the charger but less than the working voltage of the cap.

-- John Fields Professional Circuit Designer

I have an RV and its charger output is almost raw unfiltered pulsating DC, unless I hook the battery to it, in which case the 13.6 or whatever it is is as smooth as silk. I can see the difference on my 12V TV. :-)

Lead-acid batteries love a float charge. :-)

Cheers! Rich

--
Aaarghhh...

The foregoing has some haste-makes-waste errors in it.  Primarily,
with 16V out of the charger, the input to the regulator will be
15.3V which, with a dropout voltage of 2.5V for the regulator, means
that the ripple out of the cap can\'t exceed 0.8V at 120Hz.  That, in
turn, means the cap has to have a capacitance of at least 813
(820)µF with 120Hz ripple or 1626 (1800µF) with 60 Hz ripple.

After all is said and done, I\'d put a 2000µF/25V aluminum
electrolytic in there and be done with it.

do you have the battery connected to the charger also or just the fan?

connect the battery also, chargers do not put out DC...

if you want to run the fan from AC, get a small DC power supply, not a battery charger...

Mark

THANKS for all replies! John I agree that that would be the easiest thing to try so will look for a large electrolytic capacitor and let you all know how it works. Thanks for doing the math- I was way off with my .1mfd. If that don't work will look look into the regulators mentioned. Do they still make germanium transistors (low voltage drop)? I looked for them once -not too hard- and all I could find was antiques.

I wasn't too clear on the ripple source. The charger for the RV -called a converter- has 2 outputs, one for battery, the other powers the fuse bus for the RV accessories. I think they do this so charge rate won't be affected by accessory load. So I guess the battery might filter it's output but the fuse bus is not- wish I had a scope.

Just use John's design, as he designed it. If you use a filter cap and nothing else, the fan will still be exposed to too much voltage.

Ed

Do they still make germanium

--
If you were blowing up fans before, and you connect up that 2000µF
cap without using a regulator, you\'ll blow them up even quicker.

You _must_ use a regulator or a series resistor if you expect your
fan(s) to survive.