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!

 

A saturated transistor drops only 0.3v on load, Germaniums are even
lower. A transistor regulator is the way to go.

 

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.

 

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.

 

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

 

---
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.

 

---
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

 

##snip ##

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.
---

---
You don't need a germanium transistor in there, you need a
regulator.
---

---
Do you have a multimeter?

If you do, why don't you measure the DC voltage and the AC voltage
on the fuse bus and post back with what you find?

If you don't, why not buy a cheap one? They're handy...